# Design Buck Converter

Discussion in 'The Projects Forum' started by gusmas, Aug 2, 2011.

1. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Ok so this is my dilemma

Is there a general way to start designing the buck converter because I have used a lot of formulas and when I simulate my outputs are not even close to what I desire. for now i just want simulations to work although in about a months time i want to build a buck converter but first I need to get my design procedure correct:

So this is what I have so far:

Input voltage(Vi): 38V
Output Voltage(Vo): 10V
Output Current(Io): 10A
Fs = 100Khz

Calculating Inductor Value:

D = Vo/Vi
R = Vo/Io
Lmin = (1-D)/(2*Fs)*R
L must be at least 10times the value of Lmin.

Calculating Capacitor Value:

assuming voltage ripple is 1%:

1% of 10V = 100mV so therefore:

100mV = (1-D)/(8*L*C*Fs^2)

then making C the subject of the formula i get my Capacitor value.

Am I going in the right Direction?

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Do it have to operate only in the continuous conduction mode or it doesn't matter?

What is the minimum output current?

3. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Continues can't allow the inductor current to go to zero

4. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Ok.

What is the minimum output current?

5. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
I would say 9.5 amps I can't affor to much current variations since the load requires a 10amp, 10 to 14V signal.

6. ### mik3 Senior Member

Feb 4, 2008
4,846
63
What kind of load is it?

7. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Its a controller for hidrogen fuel cell

8. ### mik3 Senior Member

Feb 4, 2008
4,846
63
L=$\frac{Vin*(1-D)*D}{2*Ir*f}$

Assume Ir to be 30% of the maximum output current or less and caclulate Lmin.

9. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
D = Vo/Vin
= 10/38
= 0.26315

Calculating Lmin using your formula:

Lmin = 12.28.

10. ### mik3 Senior Member

Feb 4, 2008
4,846
63
OK.

For Cout:

Cout=$\frac{Ir*D}{8*f*Vr}$

Vr is the output peak-peak voltage ripple

11. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Sorry Lmin = 12.2807 * 10^ -6

Ok so for Cout:
Assuming the voltage ripple is 100mV
Cout = 9.8642uF

12. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Verify it with the simulation.

13. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Simulation indicates my Vo is 20V and my Current is 20 Amps.... My circuit is attached

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14. ### mik3 Senior Member

Feb 4, 2008
4,846
63
What is the duty cycle?

15. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
The duty cycle is 50%

16. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Then you should get 19V.

Set it to 23.6%

17. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
Ah cool I see.I set the duty cycle to 28%, Althoug there is fluctuations in the voltage and current between 10.2 and 10.8. Is there a a way to make the ripple less? Although I don't think it will effect the fuel cell controller to much?

18. ### mik3 Senior Member

Feb 4, 2008
4,846
63
It is possible to increase the value of the inductor or the capacitor. Increase that of the capacitor.

Is it pure capacitance or does it model the ESR and ESL of the capacitor?

19. ### gusmas Thread Starter Active Member

Sep 27, 2008
239
0
no unfortunatly not. But ya like I said that should not effect the controller correct? Thanks a lot for the help you saved me a lot of time

20. ### mik3 Senior Member

Feb 4, 2008
4,846
63
Increase the output capacitor to 100uF. If you build a real circuit you should check for the ESR of the capacitor.

The ESR affects the output voltage ripple by Vr=ESR*Ir.

Usually, if the ESR meets the ripple requirements then the capacitance is enough as not not increase the ripple too much. You might need a large capacitor with 3A of ripple current. You can increase the value of L to reduce Ir and thus decrease C.