This opus will be about 3 heroes. Why heroes?))) Since ancient times, heroes are the defenders of the Motherland, people who “stole”, that is, saved, and not, as now, “stole”, wealth.. Our drives are pulse converters, 3 types (step-down, step-up, inverter ). Moreover, all three are on one MC34063 chip and on one type of DO5022 coil with an inductance of 150 μH. They are used as part of a microwave signal switch using pin diodes, the circuit and board of which are given at the end of this article.
Calculation of a DC-DC step-down converter (step-down, buck) on the MC34063 chip
The calculation is carried out using the standard “AN920/D” method from ON Semiconductor. The electrical circuit diagram of the converter is shown in Figure 1. The numbers of the circuit elements correspond to the latest version of the circuit (from the file “Driver of MC34063 3in1 – ver 08.SCH”).
Fig. 1 Electrical circuit diagram of a step-down driver.
Chip pins:
Conclusion 1 - SWC(switch collector) - output transistor collector
Conclusion 2 - S.W.E.(switch emitter) - emitter of the output transistor
Conclusion 3 - TS(timing capacitor) - input for connecting a timing capacitor
Conclusion 4 - GND– ground (connects to the common wire of the step-down DC-DC)
Conclusion 5 - CII(FB) (comparator inverting input) - inverting input of the comparator
Conclusion 6 - VCC- nutrition
Conclusion 7 - Ipk— input of the maximum current limiting circuit
Conclusion 8 - DRC(driver collector) - collector of the output transistor driver (a bipolar transistor connected according to a Darlington circuit, located inside the microcircuit, is also used as an output transistor driver).
Elements:
L 3- throttle. It is better to use an open type inductor (not completely closed with ferrite) - DO5022T series from Oilkraft or RLB from Bourns, since such an inductor enters saturation at a higher current than the common closed type CDRH Sumida inductors. It is better to use chokes with higher inductance than the calculated value obtained.
From 11- timing capacitor, it determines the conversion frequency. The maximum conversion frequency for 34063 chips is about 100 kHz.
R 24, R 21— voltage divider for the comparator circuit. The non-inverting input of the comparator is supplied with a voltage of 1.25V from the internal regulator, and the inverting input is supplied from a voltage divider. When the voltage from the divider becomes equal to the voltage from the internal regulator, the comparator switches the output transistor.
C 2, C 5, C 8 and C 17, C 18— output and input filters, respectively. The output filter capacitance determines the amount of output voltage ripple. If during the calculations it turns out that a very large capacitance is required for a given ripple value, you can do the calculation for large ripples, and then use an additional LC filter. The input capacitance is usually taken 100 ... 470 μF (TI recommendation is at least 470 μF), the output capacitance is also taken 100 ... 470 μF (taken 220 μF).
R 11-12-13 (Rsc)- current-sensing resistor. It is needed for the current limiting circuit. Maximum output transistor current for MC34063 = 1.5A, for AP34063 = 1.6A. If the peak switching current exceeds these values, the microcircuit may burn out. If it is known for sure that the peak current does not even come close to the maximum values, then this resistor can not be installed. The calculation is carried out specifically for the peak current (of the internal transistor). When using an external transistor, the peak current flows through it, while a smaller (control) current flows through the internal transistor.
VT 4 – an external bipolar transistor is placed in the circuit when the calculated peak current exceeds 1.5A (at a large output current). Otherwise, overheating of the microcircuit can lead to its failure. Operating mode (transistor base current) R 26 , R 28 .
VD 2 – Schottky diode or ultrafast diode for voltage (forward and reverse) of at least 2U output
Calculation procedure:
- Select the nominal input and output voltages: V in, V out and maximum
output current I out.
In our scheme V in =24V, V out =5V, I out =500mA(maximum 750 mA)
- Select the minimum input voltage V in(min) and minimum operating frequency fmin with selected V in And I out.
In our scheme V in(min) =20V (according to technical specifications), choose f min =50 kHz
3) Calculate the value (t on +t off) max according to the formula (t on +t off) max =1/f min, t on(max)- the maximum time when the output transistor is open, toff(max)- the maximum time when the output transistor is closed.
(t on +t off) max =1/f min =1/50kHz=0.02 MS=20 μS
Calculate ratio t on/t off according to the formula t on /t off \u003d (V out + V F) / (V in (min) - V sat - V out), Where V F- voltage drop across the diode (forward - forward voltage drop), V sat- the voltage drop across the output transistor when it is in a fully open state (saturation - saturation voltage) at a given current. V sat determined from the graphs or tables given in the documentation. From the formula it is clear that the more V in, V out and the more they differ from each other, the less influence they have on the final result V F And V sat.
(t on /t off) max =(V out +V F)/(V in(min) -V sat -V out)=(5+0.8)/(20-0.8-5)=5.8/14.2=0.408
4) Knowing t on/t off And (t on +t off) max solve the system of equations and find t on(max).
t off = (t on +t off) max / ((t on /t off) max +1) =20μS/(0.408+1)=14.2 μS
t on (max) =20- t off=20-14.2 µS=5.8 µS
5) Find the capacitance of the timing capacitor From 11 (Ct) according to the formula:
C 11 = 4.5*10 -5 *t on(max).
C 11 = 4.5*10 -5 * t on (max) =4.5*10 - 5*5.8 µS=261pF(this is the min value), take 680pF
The smaller the capacitance, the higher the frequency. Capacitance 680pF corresponds to frequency 14KHz
6) Find the peak current through the output transistor: I PK(switch) =2*I out. If it turns out to be greater than the maximum current of the output transistor (1.5 ... 1.6 A), then a converter with such parameters is impossible. It is necessary to either recalculate the circuit for a lower output current ( I out), or use a circuit with an external transistor.
I PK(switch) =2*I out =2*0.5=1A(for maximum output current 750mA I PK(switch) = 1.4A)
7) Calculate R sc according to the formula: R sc =0.3/I PK(switch).
R sc =0.3/I PK(switch) =0.3/1=0.3 Ohm, We connect 3 resistors in parallel ( R 11-12-13) 1 ohm
8) Calculate the minimum capacitance of the output filter capacitor: C 17 =I PK(switch) *(t on +t off) max /8V ripple(p-p), Where V ripple(p-p)— maximum value of output voltage ripple. The maximum capacity is taken from the standard values closest to the calculated one.
From 17 =I PK (switch) *(t on+ t off) max/8 V ripple (p — p) =1*14.2 µS/8*50 mV=50 µF, take 220 µF
9) Calculate the minimum inductance of the inductor:
L 1(min) = t on (max) *(V in (min) — V sat— V out)/ I PK (switch) . If C 17 and L 1 are too large, you can try to increase the conversion frequency and repeat the calculation. The higher the conversion frequency, the lower the minimum capacitance of the output capacitor and the minimum inductance of the inductor.
L 1(min) =t on(max) *(V in(min) -V sat -V out)/I PK(switch) =5.8μS *(20-0.8-5)/1=82.3 µH
This is the minimum inductance. For the MC34063 microcircuit, the inductor should be selected with a deliberately larger inductance value than the calculated value. We choose L=150μH from CoilKraft DO5022.
10) Divider resistances are calculated from the ratio V out =1.25*(1+R 24 /R 21). These resistors must be at least 30 ohms.
For V out \u003d 5V, we take R 24 \u003d 3.6K, thenR 21 =1.2K
Online calculation http://uiut.org/master/mc34063/ shows the correctness of the calculated values (except Ct=C11):
There is also another online calculation http://radiohlam.ru/teory/stepdown34063.htm, which also shows the correctness of the calculated values.
12) According to the calculation conditions in paragraph 7, the peak current of 1A (Max 1.4A) is near the maximum current of the transistor (1.5 ... 1.6 A). It is advisable to install an external transistor already at a peak current of 1A, in order to avoid overheating of the microcircuit. This is done. We select transistor VT4 MJD45 (PNP type) with a current transfer coefficient of 40 (it is advisable to take h21e as high as possible, since the transistor operates in saturation mode and the voltage drops across it is about = 0.8V). Some transistor manufacturers indicate in the datasheet title that the saturation voltage Usat is low, about 1V, which is what you should be guided by.
Let's calculate the resistance of resistors R26 and R28 in the circuits of the selected transistor VT4.
Base current of transistor VT4: I b= I PK (switch) / h 21 uh . I b=1/40=25mA
Resistor in the BE circuit: R 26 =10*h21e/ I PK (switch) . R 26 =10*40/1=400 Ohm (take R 26 =160 Ohm)
Current through resistor R 26: I RBE =V BE /R 26 =0.8/160=5mA
Resistor in the base circuit: R 28 =(Vin(min)-Vsat(driver)-V RSC -V BEQ 1)/(I B +I RBE)
R 28 =(20-0.8-0.1-0.8)/(25+5)=610 Ohm, you can take less than 160 Ohm (same as R 26, since the built-in Darlington transistor can provide more current for a smaller resistor.
13) Calculate the snubber elements R 32, C 16. (see the calculation of the boost circuit and the diagram below).
14) Let's calculate the elements of the output filter L 5 , R 37, C 24 (G. Ott “Methods for suppressing noise and interference in electronic systems” p. 120-121).
I chose - coil L5 = 150 µH (same type choke with active resistive resistance Rdross = 0.25 ohm) and C24 = 47 µF (the circuit indicates a larger value of 100 µF)
Let's calculate the filter attenuation decrement xi =((R+Rdross)/2)* root(C/L)
R=R37 is set when the attenuation decrement is less than 0.6, in order to remove the overshoot of the relative frequency response of the filter (filter resonance). Otherwise, the filter at this cutoff frequency will amplify the oscillations rather than attenuate them.
Without R37: Ksi=0.25/2*(root 47/150)=0.07 - the frequency response will rise to +20dB, which is bad, so we set R=R37=2.2 Ohm, then:
C R37: Xi = (1+2.2)/2*(root 47/150) = 0.646 - with Xi 0.5 or more, the frequency response decreases (there is no resonance).
The resonant frequency of the filter (cutoff frequency) Fср=1/(2*pi*L*C) must lie below the conversion frequencies of the microcircuit (thus filtering these high frequencies 10-100 kHz). For the indicated values of L and C, we obtain Faver = 1896 Hz, which is less than the operating frequency of the converter 10-100 kHz. Resistance R37 cannot be increased by more than a few Ohms, as the voltage across it will drop (with a load current of 500mA and R37=2.2 Ohms, the voltage drop will be Ur37=I*R=0.5*2.2=1.1V).
All circuit elements are selected for surface mounting
Oscillograms of operation at various points in the buck converter circuit:
15) a) Oscillograms without load ( Uin=24V, Uout=+5V):
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Voltage +5V at the output of the converter (on capacitor C18) without load |
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The signal at the collector of transistor VT4 has a frequency of 30-40Hz, since without load, the circuit consumes about 4 mA without load |
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Control signals on pin 1 of the microcircuit (lower) and based on transistor VT4 (upper) without load |
b) Oscillograms under load(Uin=24V, Uout=+5V), with frequency-setting capacitance c11=680pF. We change the load by decreasing the resistance of the resistor (3 oscillograms below). The output current of the stabilizer increases, as does the input.
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Load - 3 68 ohm resistors in parallel ( 221 mA) Input current – 70mA |
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Yellow beam - transistor-based signal (control) Blue beam - signal at the collector of the transistor (output) Load - 5 68 ohm resistors in parallel ( 367 mA) Input current – 110mA |
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Yellow beam - transistor-based signal (control) Blue beam - signal at the collector of the transistor (output) Load - 1 resistor 10 ohm ( 500 mA) Input current – 150mA |
Conclusion: depending on the load, the pulse repetition frequency changes, with a higher load the frequency increases, then the pauses (+5V) between the accumulation and release phases disappear, only rectangular pulses remain - the stabilizer works “at the limit” of its capabilities. This can also be seen in the oscillogram below, when the “saw” voltage has surges - the stabilizer enters current limiting mode.
c) Voltage at the frequency-setting capacitance c11=680pF at a maximum load of 500mA
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Yellow beam - capacitance signal (control saw) Blue beam - signal at the collector of the transistor (output) Load - 1 resistor 10 ohm ( 500 mA) Input current – 150mA |
d) Voltage ripple at the output of the stabilizer (c18) at a maximum load of 500 mA
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Yellow beam - pulsation signal at the output (s18) Load - 1 resistor 10 ohm ( 500 mA) |
Voltage ripple at the output of the LC(R) filter (c24) at a maximum load of 500 mA
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Yellow beam - ripple signal at the output of the LC(R) filter (c24) Load - 1 resistor 10 ohm ( 500 mA) |
Conclusion: the peak-to-peak ripple voltage range decreased from 300mV to 150mV.
e) Oscillogram of damped oscillations without a snubber:
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Blue beam - on a diode without a snubber (insertion of a pulse over time is visible not equal to the period, since this is not PWM, but PFM) |
Oscillogram of damped oscillations without snubber (enlarged):
Calculation of a step-up, boost DC-DC converter on the MC34063 chip
http://uiut.org/master/mc34063/. For the boost driver, it is basically the same as the buck driver calculation, so it can be trusted. During online calculation, the scheme automatically changes to the standard scheme from “AN920/D”. Input data, calculation results and the standard scheme itself are presented below.
— field-effect N-channel transistor VT7 IRFR220N. Increases the load capacity of the microcircuit and allows for quick switching. Selected by: The electrical circuit of the boost converter is shown in Figure 2. The numbers of circuit elements correspond to the latest version of the circuit (from the file “Driver of MC34063 3in1 – ver 08.SCH”). The scheme contains elements that are not included in the standard online calculation scheme. These are the following elements:
- Maximum drain-source voltage V DSS =200V, because the output voltage is high +94V
- Low channel voltage drop RDS(on)max=0.6Om. The lower the channel resistance, the lower the heating losses and the higher the efficiency.
- Small capacitance (input), which determines the gate charge Qg (Total Gate Charge) and low gate input current. For a given transistor I=Qg*fsw=15nC*50 kHz=750uA.
- Maximum drain current I d=5A, since pulse current Ipk=812 mA at output current 100 mA
- voltage divider elements R30, R31 and R33 (reduces the voltage for the VT7 gate, which should be no more than V GS = 20V)
- discharge elements of the input capacitance VT7 - R34, VD3, VT6 when switching the transistor VT7 to the closed state. Reduces the decay time of the VT7 gate from 400nS (not shown) to 50nS (waveform with a decay time of 50nS). Log 0 on pin 2 of the microcircuit opens the PNP transistor VT6 and the input gate capacitance is discharged through the CE junction VT6 (faster than simply through resistor R33, R34).
— the coil L turns out to be very large when calculating, a lower nominal value L = L4 (Fig. 2) = 150 μH is selected
- snubber elements C21, R36.
Snubber calculation:
Hence L=1/(4*3.14^2*(1.2*10^6)^2*26*10^-12)=6.772*10^4 Rsn=√(6.772*10^4 /26*10^- 12)=5.1kΩ
The size of the snubber capacitance is usually a compromise solution, since, on the one hand, the larger the capacitance, the better the smoothing (less number of oscillations), on the other hand, each cycle the capacitance is recharged and dissipates part of the useful energy through the resistor, which affects the efficiency (usually A normally designed snubber reduces efficiency very slightly, within a couple of percent).
By installing a variable resistor, we determined the resistance more accurately R=1 K
Fig.2 Electrical circuit diagram of a step-up, boost driver.
Oscillograms of operation at various points in the boost converter circuit:
a) Voltage at various points without load:
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Output voltage - 94V without load |
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Gate voltage without load |
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Drain voltage without load |
b) voltage at the gate (yellow beam) and at the drain (blue beam) of transistor VT7:
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on the gate and drain under load the frequency changes from 11 kHz (90 µs) to 20 kHz (50 µs) - this is not PWM, but PFM |
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on the gate and drain under load without a snubber (stretched - 1 oscillation period) |
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on gate and drain under load with snubber |
c) leading and trailing edge voltage pin 2 (yellow beam) and on the gate (blue beam) VT7, saw pin 3:
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blue - 450 ns rise time on VT7 gate |
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Yellow - rise time 50 ns per pin 2 chips blue - 50 ns rise time on VT7 gate |
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saw on Ct (pin 3 of IC) with control release F=11k |
Calculation of DC-DC inverter (step-up/step-down, inverter) on the MC34063 chip
The calculation is also carried out using the standard “AN920/D” method from ON Semiconductor.
The calculation can be done immediately “online” http://uiut.org/master/mc34063/. For an inverting driver, it is basically the same as the calculation for a buck driver, so it can be trusted. During online calculation, the scheme automatically changes to the standard scheme from “AN920/D”. Input data, calculation results and the standard scheme itself are presented below.
— bipolar PNP transistor VT7 (increases load capacity) The electrical circuit of the inverting converter is shown in Figure 3. The numbers of circuit elements correspond to the latest version of the circuit (from the file “Driver of MC34063 3in1 – ver 08.SCH”). The scheme contains elements that are not included in the standard online calculation scheme. These are the following elements:
— voltage divider elements R27, R29 (sets the base current and operating mode of VT7),
— snubber elements C15, R35 (suppresses unwanted vibrations from the throttle)
Some components differ from those calculated:
- coil L is taken less than the calculated value L = L2 (Fig. 3) = 150 μH (all coils are of the same type)
- output capacitance is taken less than the calculated one C0=C19=220μF
- The frequency-setting capacitor is taken C13=680pF, corresponding to a frequency of 14KHz
- divider resistors R2=R22=3.6K, R1=R25=1.2K (taken first for output voltage -5V) and final resistors R2=R22=5.1K, R1=R25=1.2K (output voltage -6.5V)
The current limiting resistor is taken Rsc - 3 resistors in parallel, 1 Ohm each (resulting resistance 0.3 Ohm)
Fig.3 Electrical circuit diagram of the inverter (step-up/step-down, inverter).
Oscillograms of operation at various points of the inverter circuit:
a) with input voltage +24V without load:
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at the output -6.5V without load |
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on the collector – accumulation and release of energy without load |
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on pin 1 and the base of the transistor without load |
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on the base and collector of the transistor without load |
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output ripple without load |
When the developer of any device is faced with the question “How to get the required voltage?”, the answer is usually simple - a linear stabilizer. Their undoubted advantage is their low cost and minimal wiring. But besides these advantages, they have a drawback - strong heating. Linear stabilizers convert a lot of precious energy into heat. Therefore, the use of such stabilizers in battery-powered devices is not advisable. Are more economical DC-DC converters. About them that will be discussed.
Back view:
Everything has already been said about the operating principles before me, so I won’t dwell on it. Let me just say that such converters come in Step-UP (step-up) and Step-Down (step-down) converters. Of course, I was interested in the latter. You can see what happened in the picture above. The converter circuits were carefully redrawn by me from the datasheet :-) Let's start with the Step-Down converter:
As you can see, nothing tricky. Resistors R3 and R2 form a divider from which the voltage is removed and supplied to the feedback leg of the microcircuit MC34063. Accordingly, by changing the values of these resistors, you can change the voltage at the output of the converter. Resistor R1 serves to protect the microcircuit from failure in the event of a short circuit. If you solder a jumper instead, the protection will be disabled and the circuit may emit a magic smoke on which all electronics operate. :-) The greater the resistance of this resistor, the less current the converter can deliver. With its resistance of 0.3 ohms, the current will not exceed half an ampere. By the way, all these resistors can be calculated by mine. I took the choke ready-made, but no one forbids me to wind it myself. The main thing is that it has the required current. The diode is also any Schottky and also for the required current. As a last resort, you can parallel two low-power diodes. The capacitor voltages are not indicated on the diagram; they must be selected based on the input and output voltage. It's better to take it with double reserve.
The Step-UP converter has minor differences in its circuit:
Requirements for parts are the same as for Step-Down. As for the quality of the resulting voltage at the output, it is quite stable and the ripple, as they say, is small. (I can’t say about ripples myself since I don’t have an oscilloscope yet). Questions, suggestions in the comments.
Now there are many microcircuit LED current stabilizers, but all of them, as a rule, are quite expensive. And since the need for such stabilizers in connection with the spread of high-power LEDs is large, we have to look for options for them, stabilizers, and cheaper ones.
Here we offer another version of the stabilizer on a common and cheap chip of the key stabilizer MS34063. The proposed version differs from the already known stabilizer circuits on this microcircuit by a slightly non-standard inclusion, which made it possible to increase the operating frequency and ensure stability even at low values of the inductance of the inductor and the capacitance of the output capacitor.
Features of the microcircuit - PWM or PWM?
The peculiarity of the microcircuit is that it is both PWM and relay! Moreover, you can choose for yourself what it will be.
The document AN920-D, which describes this microcircuit in more detail, says something like this (see the functional diagram of the microcircuit in Fig. 2).
During the charging of the time-setting capacitor, a logical unit is set at one input of the AND logic element that controls the trigger. If the output voltage of the stabilizer is lower than the nominal one (at the input with a threshold voltage of 1.25V), then the logical unit is also set at the second input of the same element. In this case, a logical unit is also set at the output of the element and at the input “S” of the trigger, it is set (the active level at the input “S” is log. 1) and a logical one appears at its output “Q”, opening the key transistors.
When the voltage on the frequency-setting capacitor reaches the upper threshold, it starts to discharge, and a logical zero appears at the first input of the AND logic element. The same level is also supplied to the reset input of the trigger (the active level at the “R” input is logic 0) and resets it. A logical zero appears at the output “Q” of the trigger and the key transistors close.
Then the cycle repeats.
It can be seen from the functional diagram that this description applies only to the current comparator, functionally connected to the master oscillator (controlled by input 7 of the microcircuit). And the output of the voltage comparator (controlled by input 5) does not have such "privileges".
It turns out that in each cycle the current comparator can both open the key transistors and close them, unless, of course, the voltage comparator allows. But the voltage comparator itself can only give permission or prohibition to open, which can only be worked out in the next cycle.
It follows that if you short-circuit the input of the current comparator (pins 6 and 7) and control only the voltage comparator (pin 5), then the key transistors are opened by it and remain open until the end of the capacitor charging cycle, even if the voltage at the input of the comparator exceeds the threshold. And only when the capacitor begins to discharge will the generator close the transistors. In this mode, the power supplied to the load can only be dosed by the frequency of the master oscillator, since the key transistors, although closed forcibly, are only for a time of the order of 0.3-0.5 μs at any frequency value. And this mode is more similar to PFM - pulse frequency modulation, which belongs to the relay type of regulation.
If, on the contrary, you short-circuit the input of the voltage comparator to the housing, eliminating it from operation, and control only the input of the current comparator (pin 7), then the key transistors will be opened by the master oscillator and closed at the command of the current comparator in each cycle! That is, in the absence of load, when the current comparator does not work, the transistors open for a long time and close for a short period of time. When overloaded, on the contrary, they open and immediately close for a long time at the command of the current comparator. At some average load current values, the keys are opened by the generator, and after some time, after the current comparator is triggered, they are closed. Thus, in this mode, the power in the load is regulated by the duration of the open state of the transistors - that is, full PWM.
It can be argued that this is not PWM, since in this mode the frequency does not remain constant, but changes - it decreases with increasing operating voltage. But with a constant supply voltage, the frequency remains unchanged, and the load current is stabilized only by changing the pulse duration. Therefore, we can assume that this is a full-fledged PWM. And the change in operating frequency when the supply voltage changes is explained by the direct connection of the current comparator with the master oscillator.
When both comparators are used simultaneously (in the classical circuit), everything works exactly the same, and the key mode or PWM is switched on depending on which comparator is triggered at the moment: when there is an overvoltage - the key one (PWM), and when there is an overload on the current - PWM
You can completely eliminate the voltage comparator from operation by shorting the 5th pin of the microcircuit to the housing, and also stabilize the voltage using PWM by installing an additional transistor. This option is shown in Fig. 1.
Fig.1
Voltage stabilization in this circuit is carried out by changing the voltage at the input of the current comparator. The reference voltage is the gate threshold voltage of field-effect transistor VT1. The output voltage of the stabilizer is proportional to the product of the threshold voltage of the transistor and the division coefficient of the resistive divider Rd1, Rd2 and is calculated by the formula:
Uout=Up(1+Rd2/Rd1), where
Up – Threshold voltage VT1 (1.7…2V).
Current stabilization still depends on the resistance of resistor R2.
The operating principle of the current stabilizer.
The MC34063 chip has two inputs that can be used to stabilize the current.
One input has a threshold voltage of 1.25V (5th pin ms), which is not beneficial for fairly powerful LEDs due to power losses. For example, at a current of 700mA (for a 3W LED), we have losses on the current sensor resistor of 1.25*0.7A=0.875W. For this reason alone, the theoretical efficiency of the converter cannot be higher than 3W/(3W+0.875W)=77%. The real one is 60%...70%, which is comparable to linear stabilizers or simply current limiting resistors.
The second input of the microcircuit has a threshold voltage of 0.3V (7th pin ms), and is designed to protect the built-in transistor from overcurrent.
Typically, this is how this microcircuit is used: an input with a threshold of 1.25V - to stabilize voltage or current, and an input with a threshold of 0.3V - to protect the microcircuit from overload.
Sometimes an additional op-amp is installed to amplify the voltage from the current sensor, but we will not consider this option due to the loss of the attractive simplicity of the circuit and the increase in the cost of the stabilizer. It will be easier to take another microcircuit...
In this option, it is proposed to use an input with a threshold voltage of 0.3V to stabilize the current, and simply turn off the other one, with a voltage of 1.25V.
The scheme turns out to be very simple. For ease of perception, the functional units of the microcircuit itself are shown (Fig. 2). 
Fig.2
Purpose and selection of circuit elements.
Diode D with choke L— the elements of any pulse stabilizer are calculated for the required load current and the continuous mode of the inductor current, respectively.
Capacitors Ci and Co– blocking at the entrance and exit. The output capacitor Co is not fundamentally necessary due to small ripples of the load current, especially at large values of the inductor inductance; therefore, it is drawn as a dotted line and may not be present in the real circuit.
Capacitor CT– frequency-setting. It is also not a fundamentally necessary element, so it is shown with a dotted line.
The datasheets for the microcircuit indicate a maximum operating frequency of 100 KHz, the table parameters show an average value of 33 KHz, and the graphs showing the dependence of the duration of the open and closed states of the switch on the capacitance of the frequency-setting capacitor show the minimum values of 2 μs and 0.3 μs, respectively (with a capacitance of 10 pF).
It turns out that if we take the last values, then the period is 2μs+0.3μs=2.3μs, and this is a frequency of 435KHz.
If we take into account the operating principle of the microcircuit - a trigger set by a master oscillator pulse and reset by a current comparator, it turns out that this ms is logical, and the logic has an operating frequency of at least several MHz. It turns out that the performance will be limited only by the speed characteristics of the key transistor. And if it did not operate at a frequency of 400 KHz, then the fronts with pulse decays would be delayed and the efficiency would be very low due to dynamic losses. However, practice has shown that microcircuits from different manufacturers start up well and operate without a frequency-setting capacitor at all. And this made it possible to increase the operating frequency as much as possible - up to 200 KHz - 400 KHz, depending on the type of microcircuit and its manufacturer. The key transistors of the microcircuit maintain such frequencies well, since the pulse rises do not exceed 0.1 μs, and the fall times do not exceed 0.12 μs at an operating frequency of 380 KHz. Therefore, even at such elevated frequencies, dynamic losses in transistors are quite small, and the main losses and heating are determined by the increased saturation voltage of the key transistor (0.5...1V).
Resistor Rb limits the base current of the built-in key transistor. The inclusion of this resistor shown in the diagram allows you to reduce the power dissipated on it and increase the efficiency of the stabilizer. The voltage drop across the resistor Rb is equal to the difference between the supply voltage, the load voltage and the voltage drop across the microcircuit (0.9-2V).
For example, with a series chain of 3 LEDs with a total voltage drop of 9...10V and powered by a battery (12-14V), the voltage drop across the resistor Rb does not exceed 4V.
As a result, the losses on the resistor Rb are several times smaller compared to a typical connection, when the resistor is connected between the 8th pin ms and the supply voltage.
It should be borne in mind that either an additional resistor Rb is already installed inside the microcircuit, or the resistance of the key structure itself is increased, or the key structure is designed as a current source. This follows from the graph of the dependence of the saturation voltage of the structure (between pins 8 and 2) on the supply voltage at various resistances of the limiting resistor Rb (Fig. 3).
Fig.3
As a result, in some cases (when the difference between the supply and load voltages is small or losses can be transferred from resistor Rb to the microcircuit), resistor Rb can be omitted, directly connecting pin 8 of the microcircuit either to the output or to the supply voltage.
And when the overall efficiency of the stabilizer is not particularly important, you can connect pins 8 and 1 of the microcircuit to each other. In this case, the efficiency may decrease by 3-10% depending on the load current.
When choosing the value of the resistor Rb, you have to make a compromise. The lower the resistance, the lower the initial supply voltage the load current stabilization mode begins, but at the same time the losses on this resistor increase over a large range of supply voltage changes. As a result, the efficiency of the stabilizer decreases with increasing supply voltage.
The following graph (Fig. 4), as an example, shows the dependence of the load current on the supply voltage at two different values of the resistor Rb - 24 Ohm and 200 Ohm. It can be clearly seen that with a 200 Ohm resistor, stabilization disappears at supply voltages below 14V (due to insufficient base current of the key transistor). With a 24 Ohm resistor, stabilization disappears at a voltage of 11.5 V.
Fig.4
Therefore, it is necessary to carefully calculate the resistance of the resistor Rb to obtain stabilization in the required range of supply voltages. Especially with battery power, when this range is small and only a few volts.
Resistor Rsc is a load current sensor. The calculation of this resistor has no special features. You should only take into account that the reference voltage of the current input of the microcircuit differs from different manufacturers. The table below shows the actual measured reference voltage values of some microcircuits.
| Chip |
Producer |
U reference (V) |
| MC34063ACD | STMicroelectronics | |
| MC34063EBD | STMicroelectronics | |
| GS34063S | Globaltech Semiconductor | |
| SP34063A | Sipex Corporation | |
| MC34063A | Motorola | |
| AP34063N8 | Analog Technology | |
| AP34063A | Anachip | |
| MC34063A | Fairchild |
Statistics on the value of the reference voltage are small, so the given values should not be considered as a standard. You just need to keep in mind that the actual value of the reference voltage may differ greatly from the value indicated in the datasheet.
Such a large spread in the reference voltage is apparently caused by the purpose of the current input - not load current stabilization, but overload protection. Despite this, the accuracy of maintaining the load current in the above version is quite good.
About sustainability.
The MC34063 chip does not have the ability to introduce correction into the OS circuit. Initially, stability is achieved by increased values of the inductor inductance L and, especially, the capacitance of the output capacitor Co. In this case, a certain paradox arises - when working at higher frequencies, the required pulsations of voltage and load current can be obtained with small inductance and capacitance of the filter elements, but at the same time the circuit can be excited, so it is necessary to install a large inductance and (or) a large capacitance. As a result, the dimensions of the stabilizer are overestimated.
An additional paradox is that for step-down switching stabilizers, the output capacitor is not a fundamentally necessary element. The required level of current (voltage) ripple can be obtained with one choke.
You can obtain good stability of the stabilizer at the required or reduced values of inductance and, especially, output filter capacitance by installing an additional RC correction circuit Rf and Cf, as shown in Figure 2.
Practice has shown that the optimal value of the time constant of this chain should be no less than 1KOhm*uF. Values of chain parameters such as a 10KΩ resistor and a 0.1μF capacitor can be considered quite convenient.
With such a correction circuit, the stabilizer operates stably over the entire supply voltage range, with low values of inductance (units of μH) and capacitance (units and fractions of μF) of the output filter or without an output capacitor at all.
The PWM mode plays an important role in stability when used to stabilize the current input of the microcircuit.
The correction allowed some microcircuits that previously did not want to work normally at all to operate at higher frequencies.
For example, the following graph shows the dependence of the operating frequency on the supply voltage for the MC34063ACD microcircuit from STMicroelectronics with a frequency-setting capacitor capacity of 100 pF.
Fig.5
As can be seen from the graph, without correction this microcircuit did not want to operate at higher frequencies even with a small capacity of the frequency-setting capacitor. Changing the capacitance from zero to several hundred pF did not fundamentally affect the frequency, and its maximum value barely reaches 100 KHz.
After the introduction of the RfCf correction chain, this same microcircuit (like others similar to it) began to operate at frequencies up to almost 300 KHz.
The above dependence can perhaps be considered typical for most microcircuits, although microcircuits from some companies operate at higher frequencies without correction, and the introduction of correction made it possible to obtain for them an operating frequency of 400 KHz at a supply voltage of 12...14V.
The following graph shows the operation of the stabilizer without correction (Fig. 6).
Fig.6
The graph shows the dependences of the consumed current (Ip), load current (In) and output short-circuit current (Isc) on the supply voltage for two values of output capacitor capacitance (Co) - 10 µF and 220 µF.
It is clearly seen that increasing the capacitance of the output capacitor increases the stability of the stabilizer - the broken curves at a capacitance of 10 μF are caused by self-excitation. At supply voltages up to 16V there is no excitation; it appears at 16-18V. Then some kind of mode change occurs and at a voltage of 24V a second kink appears. At the same time, the operating frequency changes, which is also visible in the previous graph (Fig. 5) of the dependence of the operating frequency on the supply voltage (both graphs were obtained simultaneously when examining one instance of the stabilizer).
Increasing the output capacitor capacity to 220 µF or more increases stability, especially at low supply voltages. But it doesn't eliminate the excitement. More or less stable operation of the stabilizer can be achieved with an output capacitor capacity of at least 1000 µF.
In this case, the inductance of the inductor has very little effect on the overall picture, although it is obvious that increasing the inductance increases stability.
Changes in operating frequency affect the stability of the load current, which is also visible in the graph. The overall stability of the output current when the supply voltage changes is also not satisfactory. The current can be considered relatively stable in a fairly narrow range of supply voltages. For example, when running on battery power.
The introduction of the RfCf correction chain radically changes the operation of the stabilizer.
The following graph shows the operation of the same stabilizer but with the RfCf correction chain.
Fig.7
It is clearly visible that the stabilizer began to work as it should be for a current stabilizer - the load and short circuit currents are almost equal and constant over the entire range of supply voltages. In this case, the output capacitor generally ceased to influence the operation of the stabilizer. Now the capacitance of the output capacitor only affects the level of ripple current and voltage of the load, and in many cases the capacitor can not be installed at all.
Below, as an example, are the values of the load current ripple for different capacitances of the output capacitor Co. LEDs are connected 3 in series in 10 parallel groups (30 pcs.). Supply voltage - 12V. Choke 47 µH.
Without capacitor: load current 226mA +-65mA or 22.6mA +-6.5mA per LED.
With a 0.33uF capacitor: 226mA +-25mA or 22.6mA +-2.5mA per LED.
With a 1.5uF capacitor: 226mA +-5mA or 22.6mA +-0.5mA per LED.
With a 10uF capacitor: 226mA +-2.5mA or 22.6mA +-0.25mA per LED.
That is, without a capacitor, with a total load current of 226 mA, the load current ripple was 65 mA, which, in terms of one LED, gives an average current of 22.6 mA and a ripple of 6.5 mA.
It can be seen how even a small capacitance of 0.33 μF sharply reduces current ripple. At the same time, increasing the capacitance from 1 µF to 10 µF already has little effect on the ripple level.
All capacitors were ceramic, since conventional electrolytes or tantalum do not provide even close ripple levels.
It turns out that a 1 µF capacitor at the output is quite sufficient for all occasions. Increasing the capacitance to 10 µF with a load current of 0.2-0.3 A hardly makes sense, since the ripple no longer decreases significantly compared to 1 µF.
If you take the inductor with a higher inductance, then you can do without a capacitor even at high load currents and (or) high supply voltages.
The ripple of the input voltage with a 12V supply and the capacity of the input capacitor Ci 10 μF does not exceed 100 mV.
Power capabilities of the microcircuit.
The MC34063 microcircuit operates normally at a supply voltage from 3V to 40V according to datasheets (MS from STM - up to 50V) and up to 45V in reality, providing a load current of up to 1A for a DIP-8 package and up to 0.75A for an SO-8 package. By combining serial and parallel connection of LEDs, you can build a lamp with an output power from 3V*20mA=60mW to 40V*0.75...1A=30...40W.
Taking into account the saturation voltage of the key transistor (0.5...0.8V) and the permissible power of 1.2W dissipated by the microcircuit case, the load current can be increased up to 1.2W/0.8V=1.5A for the DIP-8 package and up to 1A for the SO-8 package.
However, in this case, a good heat sink is required, otherwise the overheating protection built into the chip will not allow operation at such a current.
Standard DIP soldering of the microcircuit body into the board does not provide the required cooling at maximum currents. It is necessary to mold the DIP housing pins for the SMD version, removing the thin ends of the pins. The remaining wide part of the pins is bent flush with the base of the case and only then soldered onto the board. It is useful to position the printed circuit board so that there is a wide area under the microcircuit body, and before installing the microcircuit you need to apply a little thermal conductive paste to its base.
Due to the short and wide pins, as well as due to the tight fit of the housing to the copper polygon of the printed circuit board, the thermal resistance of the microcircuit body is reduced and it will be able to dissipate slightly more power.
For the SO-8 case, installing an additional radiator in the form of a plate or other profile directly on the top of the case helps.
On the one hand, such attempts to increase power look strange. After all, you can simply switch to another, more powerful microcircuit or install an external transistor. And with load currents of more than 1.5A, this will be the only correct solution. However, when a load current of 1.3A is required, you can simply improve the heat dissipation and try using a cheaper and simpler option on the MC34063 chip.
The maximum efficiency obtained in this version of the stabilizer does not exceed 90%. Further increase in efficiency is prevented by the increased saturation voltage of the key transistor - at least 0.4...0.5V at currents up to 0.5A and 0.8...1V at currents 1...1.5A. Therefore, the main heating element of the stabilizer is always the microcircuit. True, noticeable heating occurs only at the maximum power for a particular case. For example, a microcircuit in an SO-8 package heats up to 100 degrees at a load current of 1A and, without an additional heat sink, is cyclically turned off by the built-in overheating protection. At currents up to 0.5A...0.7A the microcircuit is slightly warm, and at currents 0.3...0.4A it does not heat up at all.
At higher load currents, the operating frequency can be reduced. In this case, the dynamic losses of the key transistor are significantly reduced. The overall power loss and case heating are reduced.
External elements that affect the efficiency of the stabilizer are diode D, inductor L and resistors Rsc and Rb. Therefore, the diode should be selected with a low forward voltage (Schottky diode), and the inductor should be selected with the winding resistance as low as possible.
You can reduce the losses on the resistor Rsc by reducing the threshold voltage by choosing a microcircuit from the appropriate manufacturer. This has already been discussed earlier (see the table at the beginning).
Another option for reducing losses on the resistor Rsc is to introduce an additional constant current bias for the resistor Rf (this will be shown in more detail below using a specific example of a stabilizer).
Resistor Rb should be carefully calculated, trying to take it with as much resistance as possible. When the supply voltage changes within large limits, it is better to replace the resistor Rb with a current source. In this case, the increase in losses with increasing supply voltage will not be so sharp.
When all of the above measures are taken, the share of losses of these elements is 1.5-2 times less than the losses on the microcircuit.
Since a constant voltage is supplied to the current input of the microcircuit, proportional only to the load current, and not, as usual, a pulse voltage proportional to the current of the key transistor (the sum of the load currents and the output capacitor), the inductance of the inductor no longer affects the stability of operation, since it ceases to be an element correction chain (its role is played by the RfCf chain). Only the amplitude of the key transistor current and the ripple of the load current depend on the inductance value. And since the operating frequencies are relatively high, even with low inductance values the load current ripple is small.
However, due to the relatively low-power key transistor built into the microcircuit, the inductor inductance should not be greatly reduced, since this increases the peak current of the transistor while its average value remains the same and the saturation voltage increases. As a result, the losses on the transistor increase and the overall efficiency decreases.
True, not dramatically - by a few percent. For example, replacing the inductor from 12 µH to 100 µH made it possible to increase the efficiency of one of the stabilizers from 86% to 90%.
On the other hand, this allows, even at low load currents, to choose a choke with low inductance, making sure that the current amplitude of the key transistor does not exceed the maximum value allowed for the microcircuit, 1.5A.
For example, with a load current of 0.2A with a voltage of 9...10V, a supply voltage of 12...15V and an operating frequency of 300KHz, a choke with an inductance of 53µH is required. In this case, the pulse current of the key transistor of the microcircuit does not exceed 0.3A. If we reduce the inductance of the inductor to 4 μH, then at the same average current, the pulse current of the key transistor will increase to the limit value (1.5A). True, the efficiency of the stabilizer will decrease due to increased dynamic losses. But perhaps in some cases it will be acceptable to sacrifice efficiency, but use a small-sized inductor with small inductance.
Increasing the inductance of the inductor also allows you to increase the maximum load current up to the maximum current value of the key transistor of the microcircuit (1.5A).
As the inductor inductance increases, the current shape of the switching transistor changes from completely triangular to completely rectangular. And since the area of the rectangle is 2 times larger than the area of the triangle (with the same height and base), the average value of the transistor current (and load) can be increased by 2 times with a constant amplitude of the current pulses.
That is, with a triangular pulse shape with an amplitude of 1.5A, the average current of the transistor and load is:
where k is the maximum pulse duty cycle equal to 0.9 for a given microcircuit.
As a result, the maximum load current does not exceed:
In=1.5A/2*0.9=0.675A.
And any increase in load current above this value entails exceeding the maximum current of the key transistor of the microcircuit.
Therefore, all datasheets for this microcircuit indicate a maximum load current of 0.75A.
By increasing the inductance of the inductor so that the transistor current becomes rectangular, we can remove the two from the maximum current formula and get:
In=1.5A*k=1.5A*0.9=1.35A.
It should be taken into account that with a significant increase in the inductance of the inductor, its dimensions also increase slightly. However, sometimes it turns out to be easier and cheaper to increase the load current by increasing the size of the inductor than installing an additional powerful transistor.
Naturally, with the required load currents of more than 1.5A, there is no way around installing an additional transistor (or another controller microcircuit), and if you are faced with a choice: a load current of 1.4A or another microcircuit, then you should first try to solve the problem by increasing the inductance by going to increasing the throttle size.
The datasheets for the chip indicate that the maximum duty cycle does not exceed 6/7 = 0.857. In reality, values of almost 0.9 are obtained even at high operating frequencies of 300-400 KHz. At lower frequencies (100-200KHz) the duty cycle can reach 0.95.
Therefore, the stabilizer works normally with a small input-output voltage difference.
The stabilizer works interestingly when the load currents are lower than the rated ones, caused by a decrease in the supply voltage below the specified one - the efficiency is at least 95%...
Since PWM is implemented not in the classical way (full control of the master oscillator), but in a “relay” way, using a trigger (start by the generator, reset by the comparator), then at a current below the rated one, a situation is possible when the key transistor stops closing. The difference between the supply and load voltages is reduced to the saturation voltage of the switching transistor, which usually does not exceed 1V at currents up to 1A and no more than 0.2-0.3V at currents up to 0.2-0.3A. Despite the presence of static losses, there are no dynamic ones and the transistor works almost like a jumper.
Even when the transistor remains controlled and operates in PWM mode, the efficiency remains high due to the reduction in current. For example, with a difference of 1.5V between the supply voltage (10V) and the voltage across the LEDs (8.5V), the circuit continued to operate (though at a frequency reduced by half) with an efficiency of 95%.
The current and voltage parameters for this case will be indicated below when considering practical stabilizer circuits.
Practical stabilizer options.
There will not be many options, since the simplest ones, repeating the classic options in circuit design, do not allow either increasing the operating frequency or current, or increasing efficiency, or obtaining good stability. Therefore, the most optimal option is one, the block diagram of which was shown in Fig. 2. Only the component ratings can change depending on the required characteristics of the stabilizer.
Figure 8 shows a diagram of the classic version.
Fig.8
One of the features is that after removing the current of the output capacitor (C3) from the OS circuit, it became possible to reduce the inductance of the inductor. For the test, an old domestic choke on a DM-3 rod with 12 μH was taken. As you can see, the characteristics of the circuit turned out to be quite good.
The desire to increase efficiency led to the circuit shown in Fig. 9
Fig.9
Unlike the previous circuit, resistor R1 is connected not to the power source, but to the output of the stabilizer. As a result, the voltage across resistor R1 became less by the amount of voltage across the load. With the same current through it, the power released on it decreased from 0.5 W to 0.15 W.
At the same time, the inductance of the inductor was increased, which also increases the efficiency of the stabilizer. As a result, efficiency increased by several percent. Specific numbers are shown in the diagram.
Another characteristic feature of the last two schemes. The circuit in Fig. 8 has very good stability of the load current when the supply voltage changes, but the efficiency is rather low. The circuit in Fig. 9, on the contrary, has a fairly high efficiency, but the current stability is poor - when the supply voltage changes from 12V to 15V, the load current increases from 0.27A to 0.3A.
This is caused by the wrong choice of resistor R1, as mentioned earlier (see Fig. 4). Since the increased resistance R1, reducing the stability of the load current, increases the efficiency, in some cases this can be used. For example, with battery power, when the limits of voltage change are small, and high efficiency is more relevant.
A certain pattern should be noted.
Quite a lot of stabilizers were manufactured (almost all of them were used to replace incandescent lamps with LED lamps in the car interior), and while stabilizers were required from time to time, microcircuits were taken from faulty boards of network “Hubs” and “Switches”. Despite the difference in manufacturers, almost all microcircuits made it possible to obtain decent stabilizer characteristics even in simple circuits.
The only chip I came across was the GS34063S from Globaltech Semiconductor, which in no way wanted to operate at high frequencies.
Then several microcircuits MC34063ACD and MC34063EBD from STMicroelectronics were purchased, which showed even worse results - they did not work at higher frequencies, poor stability, high voltage of the current comparator support (0.45-0.5V), poor stabilization of the load current with good efficiency or poor efficiency with good stabilization...
Perhaps the poor performance of the listed microcircuits is explained by their cheapness - the cheapest ones that were available were purchased, since the MC34063A (DIP-8) microcircuit from the same company, removed from a faulty Switch, worked normally. True, at a relatively low frequency - no more than 160 KHz.
The following microcircuits, taken from broken equipment, worked well:
Sipex Corporation (SP34063A),
Motorola (MC34063A),
Analog Technology (AP34063N8),
Anachip (AP34063 and AP34063A).
Fairchild (MC34063A) - I'm not sure I identified the company correctly.
ON Semiconductor, Unisonic Technologies (UTC) and Texas Instruments - I don’t remember, since I began to pay attention to the company only after I was faced with the unwillingness to work with ms of some companies, and they didn’t buy chips from these companies on purpose.
In order not to throw away the purchased, poorly performing, MC34063ACD and MC34063EBD microcircuits from STMicroelectronics, several experiments were carried out, which led to the circuit shown at the very beginning in Fig. 2.
The following Fig. 10 shows a practical circuit of a regulator with an RfCf correction circuit (in this circuit R3C2). The difference in the operation of the stabilizer without and with a correction chain was already discussed earlier in the section “On stability” and graphs were presented (Fig. 5, Fig. 6, Fig. 7).
Fig.10
From the graph in Fig. 7 it can be seen that current stabilization is excellent over the entire range of supply voltages of the microcircuit. The stability is very good - as if PWM is working. The frequency is quite high, which makes it possible to use small-sized chokes with low inductance and completely eliminate the output capacitor. Although installing a small capacitor can completely eliminate load current ripple. The dependence of the load current ripple amplitude on the capacitor capacity was discussed earlier in the section “On stability”.
As already mentioned, the MC34063ACD and MC34063EBD microcircuits from STMicroelectronics that I received turned out to have an overestimated reference voltage of the current comparator - 0.45V-0.5V, respectively, despite the value indicated in the datasheet of 0.25V-0.35V. Because of this, at high load currents, large losses occur on the current sensor resistor. To reduce losses, a current source was added to the circuit using transistor VT1 and resistor R2. (Fig. 11).
Fig.11
Thanks to this current source, an additional bias current of 33 μA flows through resistor R3, so the voltage across resistor R3, even without load current, is 33 μA * 10 KΩ = 330 mV. Since the threshold voltage of the current input of the microcircuit is 450 mV, then for the current comparator to operate, the current sensor resistor R1 must have a voltage of 450 mV-330 mV = 120 mV. With a load current of 1A, resistor R1 should be at 0.12V/1A=0.12Ohm. We set the available value to 0.1 Ohm.
Without a current stabilizer on VT1, resistor R1 would have to be selected at the rate of 0.45V/1A=0.45Ohm, and the power would be dissipated on it at 0.45W. Now, at the same current, the loss on R1 is only 0.1 W
This option is powered by a battery, load current up to 1A, power 8-10W. Output short circuit current 1.1A. In this case, the current consumption decreases to 64mA at a supply voltage of 14.85V, respectively, the power consumption drops to 0.95W. The microcircuit in this mode does not even heat up and can be in the short circuit mode for as long as you like.
The remaining characteristics are shown in the diagram.
The microcircuit is taken in an SO-8 package and the load current for it is 1A. It gets very hot (the temperature of the pins is 100 degrees!), so it's better to put the microcircuit in a DIP-8 package, converted for SMD mounting, make large polygons and (or) come up with a radiator.
The saturation voltage of the microcircuit key is quite large - almost 1V at a current of 1A, which is why the heating is like that. Although, judging by the datasheet for the microcircuit, the saturation voltage of the key transistor at a current of 1A should not exceed 0.4V.
Service functions.
Despite the absence of any service capabilities in the microcircuit, they can be implemented independently. Typically, an LED current stabilizer requires turning off and adjusting the load current.
Turn on-off
Turning off the stabilizer on the MC34063 chip is implemented by applying voltage to the 3rd output. An example is shown in Fig.12. 
Fig.12
It was experimentally determined that when voltage is applied to the 3rd output of the microcircuit, its master oscillator stops, and the key transistor closes. In this state, the consumed current of the microcircuit depends on its manufacturer and does not exceed the no-load current indicated in the datasheet (1.5-4mA).
The rest of the options for turning off the stabilizer (for example, by applying a voltage of more than 1.25V to the 5th output) turn out to be worse, since the master oscillator does not stop and the microcircuit consumes more current compared to the board on the 3rd output.
The essence of such management is as follows.
At the 3rd output of the microcircuit, a sawtooth voltage of charge and discharge of the frequency-setting capacitor acts. When the voltage reaches the threshold value of 1.25V, the capacitor begins to discharge, and the output transistor of the microcircuit closes. This means that to turn off the stabilizer, you need to apply a voltage of at least 1.25V to the 3rd input of the microcircuit.
According to the datasheets for the microcircuit, the timing capacitor is discharged with a maximum current of 0.26 mA. This means that when an external voltage is applied to the 3rd pin through a resistor, to obtain a switching voltage of at least 1.25V, the current through the resistor must be at least 0.26mA. As a result, we have two main figures for calculating the external resistor.
For example, if the stabilizer supply voltage is 12...15V, the stabilizer must be reliably turned off at the minimum value - at 12V.
As a result, the resistance of the additional resistor is found from the expression:
R=(Up-Uvd1-1.25V)/0.26mA=(12V-0.7V-1.25V)/0.26mA=39KOhm.
To reliably turn off the microcircuit, select the resistor resistance less than the calculated value. In the fragment of the circuit Fig. 12, the resistor resistance is 27KOhm. With this resistance, the turn-off voltage is about 9V. This means that if the stabilizer supply voltage is 12V, you can hope to reliably turn off the stabilizer using this circuit.
When controlling the stabilizer from a microcontroller, resistor R must be recalculated for a voltage of 5V.
The input resistance at the 3rd input of the microcircuit is quite large and any connection of external elements can affect the formation of a sawtooth voltage. To decouple the control circuits from the microcircuit and thereby maintain the same noise immunity, diode VD1 is used.
The stabilizer can be controlled either by applying a constant voltage to the left terminal of resistor R (Fig. 12), or by short-circuiting the connection point between resistor R and diode VD1 to the body (with constant voltage present at the left terminal of resistor R).
Zener diode VD2 is designed to protect the input of the microcircuit from high voltage. At low supply voltages it is not needed.
Load current regulation
Since the reference voltage of the microcircuit current comparator is equal to the sum of the voltages on resistors R1 and R3, by changing the bias current of resistor R3, the load current can be adjusted (Fig. 11).
Two adjustment options are possible - variable resistor and constant voltage.
Figure 13 shows a fragment of the diagram in Figure 11 with the necessary changes and design relationships that allow you to calculate all the elements of the control circuit.
Fig.13
To regulate the load current with a variable resistor, you need to replace the constant resistor R2 with an assembly of resistors R2’. In this case, when the resistance of the variable resistor changes, the total resistance of resistor R2’ will change within 27...37KOhm, and the drain current of transistor VT1 (and resistor R3) will change within 1.3V/27...37KOhm=0.048...0.035mA. In this case, the bias voltage across resistor R3 will vary within 0.048...0.035mA*10KOhm=0.48...0.35V. To trigger the current comparator of the microcircuit, a voltage of 0.45-0.48 ... 0.35V \u003d 0 ... 0.1V must drop on the resistor-current sensor R1 (Fig. 11). With resistance R1=0.1Ohm, this voltage will drop across it when the load current flows through it within 0...0.1V/0.1Ohm=0...1A.
That is, by changing the resistance of the variable resistor R2 'within 27 ... 37 KΩ, we can regulate the load current within 0 ... 1A.
To adjust the load current with a constant voltage, you need to put a voltage divider Rd1Rd2 in the gate of the transistor VT1. With the help of this divider, you can match any control voltage with the one required for VT1.
Figure 13 shows all the formulas needed for the calculation.
For example, it is required to adjust the load current within 0 ... 1A using a constant voltage that varies within 0 ... 5V.
To use the current stabilizer circuit in Fig. 11, we put the voltage divider Rd1Rd2 in the gate circuit of the transistor VT1 and calculate the resistor values.
Initially, the circuit is designed for a load current of 1A, which is set by the current of the resistor R2 and the threshold voltage of the field-effect transistor VT1. To reduce the load current to zero, as follows from the previous example, you need to increase the current of the resistor R2 from 0.034mA to 0.045mA. With a constant resistance of the resistor R2 (39KΩ), the voltage across it should vary within 0.045 ... 0.034mA * 39KΩ = 1.755 ... 1.3V. At zero voltage at the gate and the threshold voltage of the transistor VT2 is 1.3V, a voltage of 1.3V is set on the resistor R2. To increase the voltage on R2 to 1.755V, you need to apply a constant voltage of 1.755V-1.3V = 0.455V to the VT1 gate. According to the condition of the problem, such a voltage at the gate should be at a control voltage of + 5V. Having set the resistance of the resistor Rd2 to 100KΩ (to minimize the control current), we find the resistance of the resistor Rd1 from the ratio Uу=Ug*(1+Rd2/Rd1):
Rd1= Rd2/(Uу/Ug-1)=100KOhm/(5V/0.455V-1)=10KOhm.
That is, when the control voltage changes from zero to + 5V, the load current will decrease from 1A to zero.
A complete circuit diagram of a 1A current stabilizer with on-off and current control functions is shown in Fig. 14. The numbering of new elements continues what was started according to the scheme in Fig. 11.
Fig.14
The circuit was not tested as part of Fig. 14. But the circuit according to Fig. 11, on the basis of which it was created, was fully tested.
The on/off method shown in the diagram has been tested by prototyping. Current control methods have so far been tested only by simulation. But since the adjustment methods are created on the basis of a really proven current stabilizer, during assembly you only have to recalculate the resistor values to match the parameters of the applied field-effect transistor VT1.
In the above circuit, both options for adjusting the load current are used - with a variable resistor Rp and a constant voltage of 0...5V. The adjustment with a variable resistor was chosen slightly differently compared to Fig. 12, which made it possible to apply both options simultaneously.
Both adjustments are dependent - the current set in one way is the maximum for the other. If the variable resistor Rp is used to set the load current to 0.5A, then by adjusting the voltage the current can be changed from zero to 0.5A. And vice versa - a current of 0.5A, set by a constant voltage, with a variable resistor will also change from zero to 0.5A.
The dependence of the load current adjustment by a variable resistor is exponential, therefore, to obtain linear adjustment, it is advisable to select a variable resistor with a logarithmic dependence of the resistance on the angle of rotation.
As resistance Rp increases, the load current also increases.
The dependence of load current regulation by constant voltage is linear.
Switch SB1 turns the stabilizer on or off. When the contacts are open, the stabilizer is turned off, when contacts are closed, it is on.
With fully electronic control, turning off the stabilizer can be achieved either by applying a constant voltage directly to the 3rd pin of the microcircuit, or by means of an additional transistor. Depending on the required control logic.
Capacitor C4 ensures a soft start of the stabilizer. When power is applied, until the capacitor is charged, the current of field-effect transistor VT1 (and resistor R3) is not limited by resistor R2, but is equal to the maximum for the field-effect transistor turned on in current source mode (units - tens of mA). The voltage across resistor R3 exceeds the threshold for the current input of the microcircuit, so the key transistor of the microcircuit is closed. The current through R3 will gradually decrease until it reaches the value set by resistor R2. As this value approaches, the voltage on resistor R3 decreases, the voltage at the current protection input increasingly depends on the voltage on the current sensor resistor R1 and, accordingly, on the load current. As a result, the load current begins to increase from zero to a predetermined value (by a variable resistor or a constant control voltage).
Printed circuit board.
Below are options for the stabilizer printed circuit board (according to the block diagram of Fig. 2 or Fig. 10 - a practical version) for different chip packages (DIP-8 or SO-8) and different chokes (standard, factory-made or homemade on a sprayed iron ring ). The board was drawn in the Sprint-Layout program version 5:
All options are designed for installation of SMD elements of standard sizes from 0603 to 1206, depending on the calculated power of the elements. The board has seats for all elements of the circuit. When desoldering the board, some elements may not be installed (this has already been discussed above). For example, I have already completely abandoned the installation of frequency-setting C T and output Co capacitors (Fig. 2). Without a frequency-setting capacitor, the stabilizer operates at a higher frequency, and the need for an output capacitor is only at high load currents (up to 1A) and (or) small inductances of the inductor. Sometimes it makes sense to install a frequency-setting capacitor, reducing the operating frequency and, accordingly, dynamic power losses at high load currents.
Printed circuit boards do not have any special features and can be made on both single-sided and double-sided foil PCB. When using double-sided PCB, the second side is not etched and serves as an additional heat sink and (or) a common wire.
When using metallization on the back side of the board as a heat sink, you need to drill a through hole near the 8th pin of the microcircuit and solder both sides together with a short jumper made of thick copper wire. If you use a microcircuit in a DIP package, then the hole must be drilled against the 8th pin and when soldering, use this pin as a jumper, soldering the pin on both sides of the board.
Instead of a jumper, good results are achieved by installing a rivet made of copper wire with a diameter of 1.8 mm (a cable core with a cross-section of 2.5 mm2). The rivet is placed immediately after etching the board - you need to drill a hole with a diameter equal to the diameter of the rivet wire, insert a piece of wire tightly and shorten it so that it protrudes from the hole no more than 1 mm, and rivet it thoroughly on both sides on the anvil with a small hammer. On the installation side, the rivet should be flush with the board so that the protruding head of the rivet does not interfere with the unsoldering of the parts.
It may seem strange advice to make a heat sink specifically from the 8th pin of the microcircuit, but a crash test of the case of a faulty microcircuit showed that its entire power part is located on a wide copper plate with a solid outlet to the 8th pin of the case. Pins 1 and 2 of the microcircuit, although made in the form of strips, are too thin to be used as a heat sink. All other terminals of the case are connected to the microcircuit crystal with thin wire jumpers. Interestingly, not all microcircuits are designed this way. Several more cases tested showed that the crystal is located in the center, and the strip pins of the microcircuit are all the same. Wiring - with wire jumpers. Therefore, to check it, you need to “disassemble” several more microcircuit housings...
The heat sink can also be made from a copper (steel, aluminum) rectangular plate 0.5-1 mm thick with dimensions that do not extend beyond the board. When using a DIP package, the plate area is limited only by the height of the inductor. You should put a little thermal paste between the plate and the chip body. With an SO-8 package, some mounting parts (capacitors and diode) can sometimes prevent a tight fit of the plate. In this case, instead of thermal paste, it is better to use a Nomakon rubber gasket of suitable thickness. It is advisable to solder the 8th pin of the microcircuit to this plate with a jumper wire.
If the cooling plate is large and blocks direct access to the 8th pin of the microcircuit, then you need to first drill a hole in the plate opposite the 8th pin, and first solder a piece of wire vertically to the pin itself. Then, thread the wire through the hole in the plate and press it against the chip body, solder them together.
A good flux for soldering aluminum is now available, so it is better to make a heat sink from it. In this case, the heat sink can be bent along the profile with the largest surface area.
To obtain load currents of up to 1.5A, the heat sink should be made on both sides - in the form of a solid polygon on the back side of the board and in the form of a metal plate pressed against the chip body. In this case, it is necessary to solder the 8th pin of the microcircuit both to the polygon on the back side and to the plate pressed to the case. To increase the thermal inertia of the heat sink on the back side of the board, it is also better to make it in the form of a plate soldered to the polygon. In this case, it is convenient to place the heat-sinking plate on the rivet at the 8th pin of the microcircuit, which previously connected both sides of the board. Solder the rivet and plate, and secure it with soldering in several places around the perimeter of the board.
By the way, when using a plate on the back side of the board, the board itself can be made of one-sided foil PCB.
The inscriptions on the board for the positional designations of the elements are made in the usual way (as are the printed tracks), except for the inscriptions on the polygons. The latter are made on a white service layer “F”. In this case, these inscriptions are obtained by etching.
The power and LED wires are soldered at opposite ends of the board according to the inscriptions: “+” and “-” for power, “A” and “K” for LEDs.
When using the board in an uncased version (after checking and tuning), it is convenient to thread it into a piece of heat-shrink tubing of a suitable length and diameter and heat it with a hairdryer. The ends of the heat shrink that has not yet cooled down must be crimped with pliers closer to the terminals. The hot-pressed heat shrink glues together and forms an almost airtight and fairly durable housing. The crimped edges are glued so tightly that when you try to separate, the heat shrink simply breaks. At the same time, if repair or maintenance is necessary, the crimped areas unstick themselves when reheated with a hairdryer, without leaving even traces of crimping. With some skill, you can stretch the still hot heat shrink with tweezers and carefully remove the board from it. As a result, the heat shrink will be suitable for re-packaging the board.
If it is necessary to completely seal the board, after shrinking the thermal shrink, its ends can be filled with a thermocouple. To strengthen the “case”, you can put two layers of heat shrink on the board. Although one layer is quite durable.
Stabilizer calculation program
For accelerated calculation and evaluation of circuit elements, a table with formulas was drawn in the EXCEL program. For convenience, some calculations are supported by VBA code. The operation of the program was tested only in Windows XP:
When you run the file, a window may appear warning you about the presence of macros in the program. You should select the “Don’t disable macros” command. Otherwise, the program will start and even recalculate according to the formulas written in the cells of the tables, but some functions will be disabled (checking the correctness of the input, the possibility of optimization, etc.).
After starting the program, a window will appear asking: "Restore all input data to default?", In which you need to click the "Yes" or "No" button. If you select "Yes", all input data for the calculation will be set by default, as an example. All formulas for calculation will also be updated. If "No" is selected, the input data will use the values saved in the previous session.
Basically, you need to select the "No" button, but if you do not want to save the previous results of the calculation, then you can select "Yes". Sometimes, when you enter too many incorrect inputs, some kind of malfunction, or accidentally deleting the contents of a cell with a formula, it is easier to exit the program and run it again by answering the question "Yes". This is easier than finding and correcting mistakes and rewriting lost formulas.
The program is a regular Excel worksheet with three separate tables ( Input data , Output , Calculation results ) and stabilizer circuit.
The first two tables contain the name of the entered or calculated parameter, its short symbol (it is also used in formulas for clarity), the value of the parameter and the unit of measurement. In the third table, the names are omitted as unnecessary, since the purpose of the element can be seen right there in the diagram. The values of the calculated parameters are marked in yellow and cannot be changed independently, since formulas are written in these cells.
To the table " Input data » the initial data is entered. The purpose of some parameters is explained in the notes. All cells with input data must be filled in, since they all take part in the calculation. The exception is the cell with the parameter “Load current ripple (Inp)” - it may be empty. In this case, the inductance of the inductor is calculated based on the minimum value of the load current. If you set the value of the load ripple current in this cell, then the inductance of the inductor is calculated based on the specified ripple value.
Some parameters may differ among different chip manufacturers - for example, the value of the reference voltage or current consumption. To obtain more reliable calculation results, you need to provide more accurate data. To do this, you can use the second sheet of the file (“Chips”), which contains the main list of different parameters. Knowing the chip manufacturer, you can find more accurate data.
In the table " Output » intermediate calculation results of interest are found. The formulas used for calculations can be seen by selecting the cell with the calculated value. A cell with the “Maximum fill factor (dmax)” parameter can be highlighted in one of two colors – green and red. The cell is highlighted in green when the parameter value is acceptable, and in red when the maximum allowable value is exceeded. In the cell note you can read which input data needs to be changed to correct it.
The AN920-D document, which describes this chip in more detail, states that the maximum duty cycle value of the MC34063 chip cannot exceed 0.857, otherwise the control limits may not coincide with the specified ones. It is this value that is taken as the criterion for the correctness of the parameter obtained in the calculation. True, practice has shown that the real value of the fill factor can be greater than 0.9. Apparently, this discrepancy is explained by “non-standard” inclusion.
The result of the calculations is the values of the passive elements of the circuit, summarized in the third table " Calculation results" . The obtained values can be used when assembling the stabilizer circuit.
Sometimes it is useful to adjust the obtained values to suit yourself, for example, when the obtained value of the resistor resistance, capacitor capacitance or inductor inductance does not coincide with the standard one. It is also interesting to see how changing the values of some elements affects the overall characteristics of the circuit. This feature is implemented in the program.
To the right of the table " Calculation results" There is a square next to each parameter. When you click the left mouse button on the selected square, a “bird” appears in it, marking the parameter that requires selection. In this case, the yellow highlight is removed from the field with the value, which means that you can independently select the value of this parameter. And in the table " Input data" The parameters that change are highlighted in red. That is, a reverse recalculation is performed - the formula is written in a cell of the input data table, and the parameter for calculation is the table value " Calculation results" .
For example, by placing a “bird” opposite the inductance of the inductor in the table “ Calculation results" , you can see that the “Minimum load current” parameter of the table “ is highlighted in red Input data ».
When the inductance changes, some parameters of the table also change " Output ", for example, "Maximum inductor and switch current (I_Lmax)". In this way, you can select a choke with the minimum inductance from the standard range and dimensions, without exceeding the maximum current of the key transistor of the microcircuit, but “sacrificing” the value of the minimum load current. At the same time, you can see that the value of the output capacitor Co also increased to compensate for the increase in load current ripple.
Having selected the inductance and made sure that the other dependent parameters do not go beyond dangerous limits, remove the check mark next to the inductance parameter, thereby securing the result obtained before changing other parameters that affect the inductance of the inductor. Moreover, in the table “ Calculation results" formulas are restored, and in the table " Input data" , on the contrary, are removed.
In the same way, you can select other parameters of the table " Calculation results" . However, you should keep in mind that the parameters of almost all formulas overlap, so if you want to change all the parameters of this table at once, an error window may appear with a message about cross-references.
Download the article in pdf format.
Mains power supplies are often used to power portable electronic equipment at home. But this is not always convenient, since there is not always a free electrical outlet at the place of use. What if you need to have several different power sources?
One of the right solutions is to make a universal power source. And as an external power source, use, in particular, the USB port of a personal computer. It is no secret that the standard version provides power for external electronic devices with a voltage of 5V and a load current of no more than 500 mA.
But, unfortunately, most portable electronic equipment requires 9 or 12V for normal operation. A specialized microcircuit will help solve the problem. voltage converter on MC34063, which will greatly facilitate production with the required parameters.
Block diagram of the mc34063 converter:
MC34063 Operating Limits
Description of the converter circuit
Below is a schematic diagram of a power supply option that allows you to get 9V or 12V from a 5V USB port on a computer.
The circuit is based on a specialized microcircuit MC34063 (its Russian analogue K1156EU5). The MC34063 voltage converter is an electronic control circuit for a DC/DC converter.
It has a temperature-compensated voltage reference (CVS), a variable duty cycle oscillator, a comparator, a current limiting circuit, an output stage, and a high-current switch. This chip is specially manufactured for use in boost, buck and inverting electronic converters with the smallest number of elements.
The output voltage obtained as a result of operation is set by two resistors R2 and R3. The choice is made on the basis that the comparator input (pin 5) should have a voltage of 1.25 V. You can calculate the resistance of the resistors for the circuit using a simple formula:
Uout= 1.25(1+R3/R2)
Knowing the required output voltage and the resistance of resistor R3, you can quite easily determine the resistance of resistor R2.
Since the output voltage is determined by , the circuit can be greatly improved by including a switch in the circuit that allows it to obtain various values as needed. Below is a version of the MC34063 converter for two output voltages (9 and 12 V) 
The microcircuit is a universal pulse converter, which can be used to implement step-down, step-up and inverting converters with a maximum internal current of up to 1.5A.
Below is a diagram of a step-down converter with an output voltage of 5V and a current of 500mA.
MC34063A converter circuit
Set of parts
Chip: MC34063AElectrolytic capacitors: C2 = 1000mF/10V; C3 = 100mF/25V
Metal film capacitors: C1 = 431pF; C4 =0.1mF
Resistors: R1 = 0.3 ohm; R2 = 1k; R3 = 3k
Diode: D1 = 1N5819
Choke: L1=220uH
C1 – capacitance of the frequency-setting capacitor of the converter.
R1 is a resistor that will turn off the microcircuit if the current is exceeded.
C2 – filter capacitor. The larger it is, the less ripple, it should be LOW ESR type.
R1, R2 – voltage divider that sets the output voltage.
D1 – the diode must be ultrafast or Schottky diode with a permissible reverse voltage of at least 2 times the output.
The supply voltage of the microcircuit is 9 - 15 volts, and the input current should not exceed 1.5A
MC34063A PCB
Two PCB options
Here you can download a universal calculator
