I came up with the idea of ​​creating this converter after buying an Asus EeePC 701 2G netbook. Small, comfortable, much more mobile than huge laptops, in general, beauty, and nothing more. One problem - you have to constantly recharge. And since the only source of power that is always at hand is a car battery, the desire naturally arose to charge the netbook from it. During the experiments, it turned out that no matter how much you give a netbook, it still won’t take more than 2 amperes, that is, a current regulator, as in the case of charging conventional batteries, is not needed. Beauty, the netbook itself will destroy how much current to consume, therefore, you just need a powerful step-down converter from 12 to 9.5 volts, capable of
give the netbook the required 2 amps.

The well-known and widely available MC34063 chip was taken as the basis of the converter. Since during the experiments a typical circuit with an external bipolar transistor has proven itself, to put it mildly, not very well (heats up), it was decided to attach a p-channel field device (MOSFET) to this mikruha.

Scheme:

A 4..8 uH coil can be taken from an old motherboard. Have you seen that there are rings on which several turns are wound with thick wires? We are looking for one on which 8..9 turns with a single-core thick wire - just the thing.

All elements of the circuit are calculated according to , in the same way as for a converter without an external transistor, the only difference is that V sat must be calculated for the field-effect transistor used. It is very simple to do this: V sat \u003d R 0 * I, where R 0 is the resistance of the transistor in the open state, I is the current flowing through it. For IRF4905 R 0 =0.02 Ohm, which at a current of 2.5A gives Vsat=0.05V. What is called, feel the difference. For a bipolar transistor, this value is at least 1V. As a result, the power dissipation in the open state is 20 times less and the minimum input voltage of the circuit is 2 volts less!

As we remember, in order for the p-channel field switch to open, it is necessary to apply a negative voltage to the gate relative to the source (that is, apply voltage to the gate, less than the supply voltage, since the source is connected to the power supply). For this we need resistors R4, R5. When the transistor of the microcircuit opens, they form a voltage divider, which sets the voltage at the gate. For IRF4905, with a source-drain voltage of 10V, to fully open the transistor, it is enough to apply a voltage to the gate 4 volts less than the source (supply) voltage, U GS = -4V current). Well, besides, the resistances of these resistors determine the steepness of the opening and closing fronts of the field device (the lower the resistance of the resistors, the steeper the fronts), as well as the current flowing through the transistor of the microcircuit (it should be no more than 1.5A).

Ready device:

In general, the radiator could even be taken smaller - the converter heats up slightly. The efficiency of this device is about 90% at a current of 2A.

Connect the input to the cigarette lighter plug, the output to the netbook plug.

If it’s not scary, then you can simply put a jumper instead of the R sc resistor, as you can see, I personally did it, the main thing is not to short anything, otherwise it will boom 🙂

In addition, I would like to add that the typical methodology is not at all ideal in terms of calculations and does not explain anything, so if you really want to understand how it all works and how it is calculated correctly, I recommend reading.

The MC34063 is a fairly common type of microcontroller for building low-to-high and high-to-low voltage converters. The features of the microcircuit are its technical specifications and performance indicators. The device holds loads well with switching current up to 1.5 A, which indicates a wide scope of its use in various pulse converters with high practical characteristics.

Description of the microcircuit

Stabilization and voltage conversion- This is an important feature that is used in many devices. These are all kinds of regulated power supplies, converting circuits and high-quality built-in power supplies. Majority consumer electronics designed specifically on this MS, because it has high performance and easily switches a fairly large current.

The MC34063 has a built-in oscillator, so to operate the device and start converting voltage to different levels, it is enough to provide an initial bias by connecting a 470pF capacitor. This controller enjoys great popularity among a large number radio amateurs. The chip works well in many circuits. And having a simple topology and a simple technical device, you can easily understand the principle of its operation.

A typical switching circuit consists of the following components:

• 3 resistors;
• diode;
• 3 capacitors;
• inductance.

Considering the circuit for lowering the voltage or stabilizing it, you can see that it is equipped with deep feedback and a fairly powerful output transistor, which passes the voltage through itself in forward current.

Scheme for switching on voltage reduction and stabilization

It can be seen from the diagram that the current in the output transistor is limited by resistor R1, and the time-setting component for setting the required conversion frequency is capacitor C2. The inductance L1 accumulates energy in itself when the transistor is open, and when it is closed, it is discharged through the diode to the output capacitor. The conversion factor depends on the ratio of the resistances of the resistors R3 and R2.

The PWM stabilizer operates in a pulsed mode:

When the bipolar transistor is turned on, the inductance gains energy, which is then stored in the output capacitance. This cycle is repeated constantly, providing a stable output level. Provided that there is a voltage of 25V at the input of the microcircuit, at its output it will be 5 V with a maximum output current of up to 500mA.

Voltage can be increased by changing the type of resistance ratio in the feedback circuit connected to the input. It is also used as a discharge diode at the moment of action of the back EMF accumulated in the coil at the moment of its charge with the transistor open.

Applying such a scheme in practice, can produce highly efficient step down converter. At the same time, the microcircuit does not consume excess power, which is released when the voltage drops to 5 or 3.3 V. The diode is designed to provide a reverse discharge of the inductance to the output capacitor.

Pulse buck mode voltage can significantly save battery power when connecting devices with low consumption. For example, when using a conventional parametric stabilizer, it took at least 50% of the power to heat it during operation. And then what to say if you need an output voltage of 3.3 V? Such a step-down source with a load of 1 W will consume all 4 W, which is important when developing high-quality and reliable devices.

The MC34063 has shown that the average power loss is reduced to at least 13%, which has become a major incentive for its practical implementation to power all low-voltage consumers. And given the pulse-width principle of regulation, then the microcircuit will heat up slightly. Therefore, it does not require radiators to cool it. The average efficiency of such a conversion circuit is at least 87%.

Voltage regulation at the output of the microcircuit is carried out due to the resistive divider. If it exceeds the nominal value by 1.25V, the comporator switches the trigger and closes the transistor. In this description, a voltage-down circuit with an output level of 5V is considered. To change it, increase or decrease, it will be necessary to change the parameters of the input divider.

An input resistor is used to limit the current of the switching key. Calculated as the ratio of the input voltage to the resistance of the resistor R1. To organize an adjustable voltage regulator, the midpoint of a variable resistor is connected to the 5th pin of the microcircuit. One output to the common wire, and the second to the power supply. The conversion system works in the frequency band of 100 kHz; when the inductance changes, it can be changed. As the inductance decreases, the conversion frequency increases.

Other operating modes

In addition to the operating modes for lowering and stabilizing, boosting is also quite often used. differs in that the inductance is not at the output. A current flows through it to the load when the key is closed, which, when unlocked, supplies a negative voltage to the lower output of the inductance.

The diode, in turn, provides the discharge of the inductance to the load in one direction. Therefore, when the key is open, 12 V from the power source and the maximum current are formed on the load, and when it is closed on the output capacitor, it rises to 28V. The boost circuit efficiency is at least 83%. circuit feature when operating in this mode, the output transistor is smoothly turned on, which is ensured by limiting the base current through an additional resistor connected to the 8th output of the MS. The clock frequency of the converter is set by a small capacitor, mainly 470pF, while it is 100kHz.

The output voltage is determined by the following formula:

Uout=1.25*R3 *(R2+R3)

Using the above circuit for switching on the MC34063A chip, it is possible to make a USB-powered boost converter up to 9, 12 or more volts, depending on the parameters of the resistor R3. To carry out a detailed calculation of the characteristics of the device, you can use a special calculator. If R2 is 2.4K and R3 is 15K, then the circuit will convert 5V to 12V.

Schematic on MC34063A voltage boost with external transistor

In the presented circuit, a field effect transistor is used. But she made a mistake. On a bipolar transistor, you need to change in some places K-E. And below is a diagram from the description. The external transistor is selected based on the switching current and output power.

Quite often, this microcircuit is used to power LED light sources to build a step-down or boost converter. High efficiency, low consumption and high output voltage stability are the main advantages of circuit implementation. There are many LED driver circuits with different features.

As one of the many examples of practical application, consider the following diagram below.

The circuit works like this:

When a control signal is applied, the internal trigger of the MS is blocked, and the transistor is closed. And the charging current of the field-effect transistor flows through the diode. When the control pulse is removed, the trigger goes into the second state and opens the transistor, which leads to the discharge of the gate VT2. Such an inclusion of two transistors provides quick on and off VT1, which reduces the likelihood of heating due to the almost complete absence of a variable component. To calculate the current flowing through the LEDs, you can use: I \u003d 1.25V / R2.

Charger on MC34063

The MC34063 controller is universal. In addition to power sources, it can be used to design charger for phones with an output voltage of 5V. Below is a diagram of the implementation of the device. Her principle of operation explained as in the case of a normal downcast. The battery charge output current is up to 1A with a margin of 30%. To increase it, you must use an external transistor, for example, KT817 or any other.

This opus will be about 3 heroes. Why bogatyrs?))) From ancient times, bogatyrs are the defenders of the Motherland, people who “stole”, that is, saved up, and not, as it is now, “stole”, wealth .. Our drives are pulse converters, 3 types (step-down, step-up, inverter ). Moreover, all three are on the same MC34063 chip and on the same type of DO5022 coil with an inductance of 150 μH. They are used as part of a microwave signal switch on pin diodes, the circuit and board of which are given at the end of this article.

Calculation of the step-down converter (step-down, buck) DC-DC on the MC34063 chip

The calculation is carried out according to the standard method "AN920 / D" 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 - SWE(switch emitter) - emitter of the output transistor

Conclusion 3 - TS(timing capacitor) - input for connecting a timing capacitor

Conclusion 4 - GND- ground (connected to the common wire of the step-down DC-DC)

Conclusion 5 - CII(Facebook) (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 is also used as a driver of the output transistor, connected according to the Darlington circuit, standing inside the microcircuit).

Elements:

L 3- throttle. It is better to use an open-type choke (not completely covered with ferrite) - the DO5022T series from Coilkraft or RLB from Bourns, since such a choke saturates at a higher current than the common Sumida CDRH closed-type chokes. It is better to use chokes with a larger inductance than the calculated value.

From 11- a 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- respectively, the output and input filters. The capacitance of the output filter determines the magnitude of the output voltage ripple. If during the calculation it turns out that a very large capacitance is required for a given ripple value, you can calculate for large ripples, and then use an additional LC filter. The input capacitance is usually taken 100 ... 470 microfarads (TI recommendation is at least 470 microfarads), the output capacitance is also taken 100 ... 470 microfarads (220 microfarads taken).

R 11-12-13 (Rsc) is a current sense 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, then the chip 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 be omitted. The calculation is carried out precisely for the peak current (of the internal transistor). When using an external transistor, peak current flows through it, less (control) current flows through the internal transistor.

VT 4 – an external bipolar transistor is put into 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 (ultrafast) diode for voltage (forward and reverse) 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) \u003d 20V (according to TK), 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- voltage drop across the output transistor when it is in a fully open state (saturation - saturation voltage) at a given current. V sat determined by the graphs or tables given in the documentation. It can be seen from the formula 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 \u003d 4.5 * 10 -5 *t on (max).

C 11 = 4.5*10 -5 * t on (max) \u003d 4.5 * 10 - 5 * 5.8 μS \u003d 261pF(this is the min value), take 680pF

The smaller the capacitance, the higher the frequency. Capacitance 680pF corresponds to a frequency of 14KHz

6) Find the peak current through the output transistor: I PK(switch) =2*I out. If it turned out to be more than the maximum current of the output transistor (1.5 ... 1.6 A), then a converter with such parameters is impossible. You either need to 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 Rsc according to the formula: R sc =0.3/I PK(switch).

R sc \u003d 0.3 / I PK (switch) \u003d 0.3 / 1 \u003d 0.3 Ohm, connect 3 resistors in parallel R 11-12-13) by 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)- the maximum value of the output voltage ripple. The maximum capacity is taken from the closest to the calculated standard values.

From 17 =I PK (switch) *(t on+ t off) max/8 V ripple (p — p) \u003d 1 * 14.2 μS / 8 * 50 mV \u003d 50 μF, we 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) \u003d t on (max) * (V in (min) -V sat -V out) / I PK (switch) \u003d 5.8µs *(20-0.8-5)/1=82.3 µH

This is the minimum inductance. For the MC34063 chip, the inductor should be selected with a known large inductance value than the calculated value. We choose L = 150 μH from CoilKraft DO5022.

10) Divider resistances are calculated from the ratio V out \u003d 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 Сt=С11):

There is also another online calculation http://radiohlam.ru/theory/stepdown34063.htm, which also shows the correctness of the calculated values.

12) According to the calculation conditions of clause 7, the peak current 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 the VT4 MJD45 transistor (PNP-type) with a current transfer coefficient of 40 (it is advisable to take h21e as much as possible, since the transistor operates in saturation mode and a voltage of about = 0.8V drops on it). Some transistor manufacturers indicate in the title of the datasheet about a low value of the saturation voltage Usat of the order of 1V, which 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 \u003d 10 * 40 / 1 \u003d 400 Ohm (we take R 26 \u003d 160 Ohm)

Current through resistor R 26: I RBE \u003d V BE /R 26 \u003d 0.8 / 160 \u003d 5mA

Resistor in the base circuit: R 28 =(Vin(min)-Vsat(driver)-V RSC -V BEQ 1)/(I B +I RBE)

R 28 \u003d (20-0.8-0.1-0.8) / (25 + 5) \u003d 610 Ohm, you can take less than 160 Ohm (of the same type as R 26, since the built-in Darlington transistor can provide more current for a smaller resistor.

13) Calculate snubber elements R 32, C 16. (see boost circuit calculation and diagram below).

14) Calculate the elements of the output filter L 5 , R 37, C 24 (G. Ott “Methods for suppressing noise and interference in electronic systems” pages 120-121).

Chose - coil L5 = 150 μH (same type inductor with active resistive resistance Rdross = 0.25 ohm) and C24 = 47 μF (a larger value of 100 μF is indicated in the circuit)

Calculate the filter damping factor xi =((R+Rdross)/2)* root(C/L)

R=R37 is set when the damping factor is less than 0.6 to remove the peak in the relative frequency response of the filter (filter resonance). Otherwise, the filter at this cutoff frequency will amplify the vibrations, not attenuate them.

Without R37: Xi=0.25/2*(root 47/150)=0.07 - there will be a rise in frequency response up to +20db, which is bad, so we set R=R37=2.2 Ohm, then:

C R37: Ksi = (1 + 2.2) / 2 * (root 47/150) = 0.646 - with xi 0.5 or more, the frequency response decline (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 (those filter these high frequencies of 10-100kHz). For the indicated values ​​of L and C, we obtain Fcp=1896 Hz, which is less than the frequencies of the converter 10-100 kHz. The resistance R37 cannot be increased more than a few ohms, because the voltage will drop on it (at 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):

Voltage + 5V at the output of the converter (on capacitor C18) without load

The signal on the collector of the transistor VT4 has a frequency of 30-40Hz, maybe without load, the circuit consumes about 4 mA without load

Control signals to 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 reducing the resistance of the resistor (3 waveforms below). In this case, the output current of the stabilizer increases, as does the input.

Load - 3 68 ohm resistors in parallel ( 221 mA) Input current - 70mA

Yellow beam - transistor-based signal (control) Blue beam - signal on the collector of the transistor (output) Load - 5 68 ohm resistors in parallel ( 367 mA) Input current - 110mA

Yellow beam - transistor-based signal (control) Blue beam - signal on the collector of the transistor (output) Load - 1 resistor 10 ohm ( 500 mA) Input current - 150mA

Conclusion: depending on the load, the pulse repetition rate changes, with a higher load, the frequency increases, then the pauses (+ 5V) between the accumulation and recoil phases disappear, only rectangular pulses remain - the stabilizer works “at the limit” of its capabilities. This can also be seen from the waveform below, when the “saw” voltage has surges - the regulator enters the current limiting mode.

c) Voltage on the frequency-setting capacitance c11=680pF at maximum load 500mA

Yellow beam - capacity signal (control saw) Blue beam - signal on 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 500mA

Yellow beam - output ripple signal (c18) Load - 1 resistor 10 ohm ( 500 mA)

Voltage ripple at the output of the LC (R) filter (s24) at a maximum load of 500mA

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 has decreased from 300mV to 150mV.

e) Oscillogram of damped oscillations without snubber:

Blue beam - on a diode without a snubber (you can see the insertion of a pulse with time not equal to the period, since this is not PWM, but PWM)

Oscillogram of damped oscillations without snubber (enlarged):

Calculation of the boost converter (step-up, boost) DC-DC on the MC34063 chip

http://uiut.org/master/mc34063/. For a boost driver, it is basically the same as the buck driver calculation, so it can be trusted. The circuit during online calculation automatically changes to the typical circuit from “AN920/D” Input data, calculation results and the typical circuit itself are presented below.

- field N-channel transistor VT7 IRFR220N. Increases the load capacity of the chip, allows you to quickly switch. Selected by: The electrical circuit of the boost converter is shown in Figure 2. The numbers of the circuit elements correspond to the latest version of the circuit (from the file “Driver of MC34063 3in1 - ver 08.SCH”). There are elements in the diagram that are not on standard scheme online calculation. These are the following elements:

• Maximum drain-source voltage V DSS =200V, maybe high voltage at the output + 94V
• Small channel voltage drop RDS(on)max=0.6Om. The lower the channel resistance, the lower the heating loss and the higher the efficiency.
• Small capacitance (input) that determines the gate charge Qg (Total Gate Charge) and low input gate current. For this transistor I=Qg*fsw=15nC*50 kHz=750uA.
• Maximum drain current I d=5A, mk pulse current Ipk=812 mA at output current 100mA

- elements of the voltage divider R30, R31 and R33 (reduces the voltage for the VT7 gate, which should be no more than V GS \u003d 20V)

- elements of the discharge of the input capacitance VT7 - R34, VD3, VT6 when switching the transistor VT7 to the closed state. Reduces VT7 gate decay time from 400nS (not shown) to 50nS (50nS waveform). Log 0 on pin 2 of the microcircuit opens the VT6 PNP transistor and the input gate capacitance is discharged through the VT6 CE junction (faster than just through the resistor R33, R34).

- the coil L in the calculation turns out to be very large, a smaller value is chosen L = L4 (Fig. 2) = 150 μH

- 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 value of the snubber capacitance is usually a compromise solution, since, on the one hand, the larger the capacitance, the better the smoothing ( less number fluctuations), 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 calculated snubber reduces the efficiency very slightly, within a couple of percent).

By setting a variable resistor, the resistance was determined more accurately R=1 K

Fig. 2 Electrical circuit diagram of a step-up (step-up, boost) driver.

Oscillograms of work at various points in the boost converter circuit:

a) Voltage at various points without load:

Output voltage - 94V without load

Gate voltage without load

Drain voltage without load

b) the voltage at the gate (yellow beam) and at the drain (blue beam) of the transistor VT7:

on the gate and on the drain under load, the frequency changes from 11 kHz (90 μs) to 20 kHz (50 μs) - those are not PWM, but PFM

on gate and drain under load without snubber (stretched - 1 oscillation period)

gate and drain under load with snubber

c) leading and trailing edge voltage pin 2 (yellow beam) and at the gate (blue beam) VT7, saw pin 3:

blue - 450 ns rise time on VT7 gate

Yellow - rise time 50 ns per pin 2 microcircuits blue - 50 ns rise time on VT7 gate

saw on Ct (pin 3 IC) with control overshoot F = 11k

Calculation of DC-DC inverter (step-up / step-down, inverter) on the MC34063 chip

The calculation is also carried out according to the standard method “AN920/D” from ON Semiconductor.

The calculation can be carried out immediately “online” http://uiut.org/master/mc34063/. For an inverting driver, it is basically the same as the buck driver calculation, so it can be trusted. The circuit during online calculation automatically changes to the typical circuit from “AN920/D” Input data, calculation results and the typical circuit itself are presented below.

- bipolar PNP transistor VT7 (increases the load capacity) The electrical circuit of the inverting converter is shown in Figure 3. The numbers of the circuit elements correspond to the latest version of the circuit (from the file “Driver of MC34063 3in1 - ver 08.SCH”). The scheme has elements that are not on the typical online calculation scheme. These are the following elements:

- elements of the voltage divider R27, R29 (sets the base current and mode of operation VT7),

- snubber elements C15, R35 (suppresses unwanted fluctuations from the throttle)

Some components differ from the calculated ones:

• coil L is taken less than the calculated value L=L2 (Fig. 3)=150 μH (the same type of all coils)
• the output capacitance is taken less than the calculated C0 \u003d C19 \u003d 220 μF
• the frequency-setting capacitor is taken C13 = 680pF, corresponds 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)

current limiting resistor taken Rsc - 3 resistors in parallel 1 ohm each (resultant resistance 0.3 ohm)

Fig. 3 Electrical circuit diagram of the inverter (step-up / step-down, inverter).

Oscillograms of work at various points in the inverter circuit:

a) at +24V input voltage without load:

at the output -6.5V without load

on the collector - accumulation and release of energy without load

on pin 1 and the base of the transistor without load

on the base and collector of the transistor without load

output ripple without load

The parts in the circuit are rated for 5V with a current limit of 500mA, with a ripple of 43kHz and 3mV. The input voltage can be from 7 to 40 volts.

The resistor divider for R2 and R3 is responsible for the output voltage, if they are replaced with a tuning resistor somewhere around 10 kOhm, then it will be possible to set the required output voltage. Resistor R1 is responsible for limiting the current. Capacitor C1 and coil L1 are responsible for the ripple frequency, capacitor C3 is responsible for the ripple level. The diode can be replaced with 1N5818 or 1N5820. To calculate the parameters of the circuit, there is a special calculator - http://www.nomad.ee/micros/mc34063a/index.shtml , where you only need to set the required parameters, it can also calculate the circuits and parameters of converters of two types not considered.

2 printed circuit boards were made: on the left - with a voltage divider on a voltage divider, made on two resistors of size 0805, on the right - with a variable resistor 3329H-682 6.8 kOhm. MC34063 microcircuit in a DIP package, under it are two tantalum capacitors of size D. Capacitor C1 is size 0805, output diode, current limiting resistor R1 is half a watt, at low currents, less than 400 mA, you can put a lower power resistor. Inductance CW68 22uH, 960mA.

Ripple waveforms, Rlimit = 0.3 ohm

These waveforms show ripples: on the left - without load, on the right - with a cell phone load, a 0.3 ohm limiting resistor, below with the same load, but a 0.2 ohm limiting resistor.

Ripple waveform, R limit = 0.2 ohm

Taken characteristics (not all parameters measured), at an input voltage of 8.2 V.

This adapter was made to charge a cell phone and power digital circuits on the go.

The article showed a board with a variable resistor as a voltage divider, I will place the corresponding circuit to it, the difference from the first circuit is only in the divider.

33 Responses to "DC-DC Buck Converter on MC34063"

Very much!
It's a pity, I was looking for 3.3 Uout, and I need more help (1.5A-2A).
Can you improve?

The article provides a link to a calculator for the scheme. According to it, for 3.3V, you need to set R1 \u003d 11k R2 \u003d 18k.
If you need more currents, then you need to either add a transistor, or use a more powerful stabilizer, for example LM2576.

Thank you! Sent.

If you put an external transistor, will current protection remain? For example, set R1 to 0.05 Ω, the protection should work at 3 A, because mikruha itself will not withstand this current, then the EU must be strengthened with a field worker.

I think the limitation (this chip has a current limit, not protection) should remain. The datasheet has a bipolar circuit and calculations for increasing the current. For higher currents, I can advise LM2576, it is just up to 3A.

Hello! I also assembled this circuit for car charging a mobile phone. But when he is “hungry” (discharged) he eats a very considerable current (870mA). for this mikruha it's still normal, it just needs to get warm. I collected both on a breadboard and on the board, the result is the same - it works for 1 minute, then the current just drops and the mobile phone turns off the charge.
I don’t understand only one thing ... why the author of the article does not have the same denomination from the calculated ones, in practice, with the calculator that cited the link in the article. according to the author's parameters "... with a ripple of 43 kHz and 3 mV." and 5V at the output, and the calculator with these parameters gives out C1 - 470 peak, L1 - 66-68 μH,
C3 - 1000uF. The question is: WHERE IS THE TRUTH?

At the very beginning of the article it is written that the article has been sent for revision.
I made mistakes during the calculations, and because of them the circuit gets so hot, you need to choose the right capacitor C1 and inductance, but so far all hands have not reached this circuit.
The mobile phone turns off the charge when a certain voltage is exceeded, for most phones this voltage is more than 6V with something volts. It is better to charge the phone with a smaller current, the battery will live longer.

Thanks to Alex_EXE for the answer! I replaced all the components using a calculator, the circuit does not heat up at all, the output voltage is 5.7V, and when loaded (charging a mobile phone) it gives out 5V - this is the norm, and for a current of 450mA, I chose the details using a calculator, everything came together in fractions of a volt. I took the coil for 100 μH (the calculator gave out: at least 64 μH, which means it can be more :). I will write all the components later, as I test it, if anyone is interested.
There are not so many sites like your Alex_EXE (Russian-speaking) on ​​the Internet, develop it further if you can. Thank you!

Glad it helped 🙂
Write it down, someone might find it useful.

Ok, I write:
The tests were successful, the mobile phone is charging (the battery in my nokia is 1350mA)
-output voltage 5.69V (apparently 1mV lost somewhere :) - without load, and 4.98V with a "mobile" load.
- input onboard 12V (well, this is a car, it is clear that 12 is ideal, and so 11.4-14.4V).
Ratings for the scheme:
- R1 \u003d 0.33 Ohm / 1W (because it gets a little warm)
— R2=20K /0.125W
— R3=5.6K/0.125W
— C1=470p ceramics
- C2=1000uF/25v (low impedance)
— C3=100uF/50v
- L1 (as I already wrote above 100 μH, it is better if it is 68 μH)

That's all:)

And I have a question for you Alex_EXE:
I can’t find information on the Internet about “Ripple voltage on the load” and “Conversion frequency”
How to correctly set these parameters in the calculator, that is, to choose?
And what do they mean anyway?

Now I want to do battery charging on this mikruha, but you need to clearly understand these two parameters.

The less fluctuations, the better. I have 100uF and the ripple level is 2.5-5%, depending on the load, you have 1000uF - this is more than enough. Pulsation frequency is within normal limits.

I somehow understood about the ripples, this is how much the “voltage jumps”, well .... approximately:)
And here is the conversion frequency. What to do with her? seeks to reduce or increase? Google is silent about this as a partisan, or that's what I was looking for :)

Here I can’t tell you for sure, although the frequency from 5 to 100 kHz will be normal for most tasks. In any case, it depends on the task, analog and precision devices are most demanding on frequency, where fluctuations can be superimposed on the working signals, thereby causing their distortion.

Alexander writes 04/23/2013 at 10:50

Found what you need! Very handy. Thank you very much Alex_EXE.

Alex, please explain to the kettle, if a variable resistor is introduced into the circuit, within what limits will the voltage change?

is it possible using this circuit to make a 6.6 volt current source with adjustable voltage, Umax so that it does not exceed these same 6.6 volts. I want to make several groups of LEDs (slave U 3.3 volts and current 180 mA), in each group there are 2 LEDs, the last. connected. 12 volt power supply, but if necessary I can purchase another one. Thank you if you answer...))

Unfortunately, I did not like this design - it was painfully capricious. If in the future the need arises, I can return, but so far I have scored on it.
For LEDs, it is better to use specialized microcircuits.

The higher the conversion frequency, the better. the dimensions (inductance) of the inductor are reduced, but within reasonable limits - for the MC34063, 60-100 kHz is optimal. Resistor R1 will heat up, because. in fact, this is a current-measuring shunt, i.e. all the current consumed by both the circuit itself and the load flows through it (5V x 0.5A \u003d 2.5Watt)

The question is of course stupid, but is it possible to remove +5, ground and -5 volts from it? You don't need a lot of power, but you need stability, or do you have to install something extra like 7660?

Hello everyone. Guys who can help make the output 10 volts or better with adjustment. Ilya can you ask me to paint. Please tell me. Thank you.

From the mc34063 manufacturer's spec sheet:
maximum frequency F=100 kHz, typical F=33 kHz.
Vripple = 1 mV - typical value, Vripple = 5 mV - maximum.
—
10V output:
- for step-down DC, if the input is 12 V:
Vin=12V, Vout=10V, Iout=450mA, Vripple=1mV(pp), Fmin=34kHz.
Ct=1073 pF, Ipk=900 mA, Rsc=0.333 Ohm, Lmin=30 uH, Co=3309 uF,
R1=13k, R2=91k (10V).
- for step-up DC, if the input is 3 V:
Vin=3V, Vout=10V, Iout=450mA, Vripple=1mV(pp), Fmin=34kHz.
Ct=926 pF, Ipk=4230 mA, Rsc=0.071 Ohm, Lmin=11 uH, Co=93773 uF, R=180 Ohm, R1=13k R2=91k (10V)

Conclusion: for step-up DC with the given parameters, the microcircuit is not suitable, since Ipk = 4230 mA > 1500 mA is exceeded. Here is an option: http://www.youtube.com/watch?v=12X-BBJcY-w
Install a 10 V zener diode.

Judging by the waveforms, your inductor is saturated, you need a more powerful inductor. You can increase the conversion frequency, leaving the inductor of the same dimensions and inductance. By the way, the MC-shka quietly works up to 150 kHz, the main thing is internal. transistors should not be turned on with a “darlington”. As far as I understand, it can be connected in parallel to the power circuit?

AND main question: how to increase the power of the converter? I look, the conduits are small there - 47 microfarads at the input, 2.2 microfarads at the output ... Does the power depend on them? Solder there one by one, one and a half microfarads? 🙂

What to do, boss, what to do?!

It is very incorrect to use tantalum capacitors in power circuits! Tantalum does not like high currents and ripples very much!

> It is very incorrect to use tantalum capacitors in power circuits!

and where else to use them, if not in switching power supplies ?! 🙂

Excellent article. It was a pleasure to read. Everything is understandable plain language without showing off. Even after reading the comments, I was pleasantly surprised, the responsiveness and ease of communication are on top. Why did I get on this topic. Because I'm collecting odometer winding for Kamaz. I found a circuit, and there the author strongly recommends that the microcontroller be powered in this way, and not through the roll. Otherwise, the controller lights up. I don’t know for sure, but probably the roll does not hold such an input voltage, and therefore the palitsa. Since there is 24 V on such a machine. But what I didn’t understand was that in the diagram according to the drawing, it seems to be a zener diode. The author of the odometer winding was assembled on smd components. And this ss24 zener diode turns out to be a smd Schottky diode. HERE on the diagram is also drawn as a zener diode. But it seems to be well understood, there is a diode and not a zener diode. Although maybe I'm confusing their drawing? maybe this is how Schottky diodes are drawn and not zener diodes? It remains to clarify such a small thing. But thanks a lot for the article.

Published on 16.09.2011

I needed to get 5V from a higher voltage (and subsequently 3.3V). At the same time, it was necessary to ensure efficiency, since the battery was the power source and its charge is not infinite. There will also be no opportunity to organize a heat sink, the circuit will be sealed. Linear voltage regulators such as LM7805 and others like them will not help here. You need a pulse converter (DC-DC Converter), i.e. step-down voltage converter. The advantages of a pulse converter are obvious - high efficiency, does not require a heat sink (at least, if they get heated, then not as much as linear converters).
There are many specialized circuits, for example LM2574, LM2594, LM267x, LT1073, L4971, ST1S03, AS1333, ST1S03, ST1S06, ST1S09, ST1S10, ST1S12 (ST1Sxx- a very worthy series). They exist in different packages for different output voltages and currents. The cost of such chips is about 3 euros, but I need a reliable and inexpensive solution. Chip MC34063- this is what we need now. MC34063 very common, you can buy without problems. The cost is only from 0.2 euros! Works with voltage from 3 to 40 volts, maximum current 1.5A, conversion frequency 100KHz. By the way, on its basis it is possible to assemble a boost converter (see also “”), but now we will deal with a step-down converter.

The scheme is taken from the documentation. I did not have a 0.33 ohm limiting resistor (Rsc), I removed it at my own peril and risk. Schottky diode put the one that was. The values ​​of the input and output capacitors are also different. For the first test option, it will do, but it's better not to save on this. This is how the scarf came out:

The photo shows a pulse buck converter with an output voltage of 3.3 V. The resistor values ​​are R1=5.1KΩ, R2=10KΩ.
According to documentation MC34063 the maximum switching current is 1.5A. I didn’t have to load more than 0.2A, so I can’t tell you the “practical ceiling”.
But with such a load at an input voltage of 12V, all elements of the circuit remain cold.
Here you can use the form for calculating the circuit parameters: http://www.nomad.ee/micros/mc34063a/index.shtml

See also: