Boost Converter with mc34063

[Waleed]

Hi
I'm trying to get 19V / 3.5A supply From 12V battey
I made a boost converter circuit using mc34063 ic
The simulation of the circuit was perfect and I got 19V/3.7A as output
This is my circuit

But the problem was in real time ... I made it but i couldn't get more than 600mA
I think there is two possibilities for that failure

• Maybe R1 value .. Because I couldn't get a small resistance with value of 0.07ohm so I replaced with a wire
• Or the inductor L1 which i used have a considered resistance

Please help me .. What is the problem and how to solve it

 

[duke37]

1. A poor quality L1 as you say.
2. C1 not having a low enough ESR so not able to provide the peak current. Try shunting with another similar capacitor.
3. C3 not able to accept the peak current.

Does anything get warm?

 

[Waleed]

1. A poor quality L1 as you say.
2. C1 not having a low enough ESR so not able to provide the peak current. Try shunting with another similar capacitor.
3. C3 not able to accept the peak current.

Does anything get warm?

Thank you duke37 for your help
Yes the TIP31 get warm ... and when the load get smaller (Greater Load current) it doesnt cut-off ever
about c1 and c3 .. i dont think they are the reason of failure because we're working with DC
thanks again

 

[duke37]

But you are not working with DC, if it were pure DC you would not need any capacitors and the inductor would have no effect.

The circuit takes an amount of energy from C1 and transfers it to L1 when the transistor in turned on.
This energy 'lump' is transferred to C3 when the transistor turns off.
Thus the current you measure is the average of a very jerky supply. The capacitors must have a low internal resistance (ESR) to deal with the pulses.

 

[(*steve*)]

If the TIP31 never turns off then you have a huge problem because the current will be shunted to earth.

How are you determining that this never turns off?

My guess is that the inductor has too high a DC resistance (or is saturating) and the transistor is not capable of being switched fast and/or hard enough.

 

[Waleed]

But you are not working with DC, if it were pure DC you would not need any capacitors and the inductor would have no effect.

The circuit takes an amount of energy from C1 and transfers it to L1 when the transistor in turned on.
This energy 'lump' is transferred to C3 when the transistor turns off.
Thus the current you measure is the average of a very jerky supply. The capacitors must have a low internal resistance (ESR) to deal with the pulses.

I used for C1,C3 ceramic capacitors which have less than 0.1 ohm as esr
i think the problem is in L1 .. so i added a resistor by series to L1 as it's shown in the figure


then i noticed that L1 has a big chance for being the reason of failure, It might has a considered esr
I made L1 by turning a 1 mm / 24 turns / 0.8 m wire around EI ferrite core .. I don't think it works well
I would ask you please if you have any information about designing inductors
Thank you again duke37

 

[duke37]

R1 is shown as 0.07Ω and you have replaced this with a link. The Texas sheet says that this resistor should be above 0.2Ω but perhaps this is only necessary without Q1 to limit the IC current.

Q1 is switched off when the voltage across R1 reaches 330mV. With your link in place, the current will be very high and will heat the transistor. Saturation of the core may well occur.

R5 will do nothing except waste power.

I cannot help with inductor design but the first thing you need will be data on the core.

 

[BobK]

R1 looks like it is functioning as a current sense resistor. If it is wrong, the chip is going to cut off the current at the wrong time.

You need to get a power inductor rated at higher than the peak current (which is going to be higher than your output current). It I highly doubtful that the one you make meets this spec.

The datasheet should have info on sizing the inductor.

Bob

 

[KrisBlueNZ]

I agree with the others that the most likely problems are (a) the inductor is saturating - you need an inductor rated for at least 5A saturation current; (b) the input and output capacitors must be low-impedance, high-ripple-current types; (c) replacing a current sense resistor with a piece of wire is never a good idea. I have a few more concerns.

First, if you're building the circuit on a breadboard, don't! You need low-impedance, high-current connections and a breadboard simply will not cut it. You should really use a PCB that has been carefully designed, considering the current paths, but it's possible to prototype on stripboard or pad-board if you follow the recommendations.

Second, the MC34063 is an ancient device and there are much better options nowadays. A big improvement can be gained by using a MOSFET instead of a bipolar transistor, especially instead of a slow one like a TIP31. I think you should abandon that circuit.

You can get ICs with switching devices built in - see the LM2587 and the LT1170. If you want to use an external switch, I recommend the UC3845 driving a MOSFET. The example applications show a flyback converter but it's equally at home as a boost converter.

Even if you're not going to use it, the LT1170 has a very detailed data sheet written by experts and it's well worth reading for guidelines on component selection and PCB layout.

 

[Waleed]

I agree with the others that the most likely problems are (a) the inductor is saturating - you need an inductor rated for at least 5A saturation current; (b) the input and output capacitors must be low-impedance, high-ripple-current types; (c) replacing a current sense resistor with a piece of wire is never a good idea. I have a few more concerns.

First, if you're building the circuit on a breadboard, don't! You need low-impedance, high-current connections and a breadboard simply will not cut it. You should really use a PCB that has been carefully designed, considering the current paths, but it's possible to prototype on stripboard or pad-board if you follow the recommendations.

Second, the MC34063 is an ancient device and there are much better options nowadays. A big improvement can be gained by using a MOSFET instead of a bipolar transistor, especially instead of a slow one like a TIP31. I think you should abandon that circuit.

You can get ICs with switching devices built in - see the LM2587 and the LT1170. If you want to use an external switch, I recommend the UC3845 driving a MOSFET. The example applications show a flyback converter but it's equally at home as a boost converter.

Even if you're not going to use it, the LT1170 has a very detailed data sheet written by experts and it's well worth reading for guidelines on component selection and PCB layout.

Thank you KrisBlueNZ for your helpful reply, i gave up using this circuit. I think I'm going to go with your advice, so i'm going to use the uc3845 driving a MOSFET.
I use proteus software for circuits simulation, but it doesn't have a model for the UC3845. I would ask you please if you know any simulation software has the UC3845 model

 

[KrisBlueNZ]

No, sorry. Texas Instruments may provide a SPICE model but then you would have to buy SPICE.

 

[ruabk]

Hi, Waleed .
Psim softwave can help you. It have one example in example foder use Uc 3844.

 

[electronicsbeliever.com]

YES the inductor is saturating. you can focus on redesigning the inductor. You can use toroid cores if you want. Consider also replacing the BJT with NMOS to really operate at saturation and store all the energy in the inductor during Ton period.