OUTPUT CAPACITOR CALCULATION

In simulation Output capacitor value is 330uFX8 8 numbers
how to arrive 2640uF @ 10A current

f

Hi,

As a general rule of thumb you can use this formula. I normally use it and works fine for me, but take it at your own risk.

Co = (Io * 10) / (Voripple * fmin)

10 is the number of internal clock cycles needed by the control loop to reduce the duty cycle from máximum to mínimum. This is a general condition with no direct dependency of the driver used.

If you use that formula, you will get Co = 2,53mF for a ripple of 2,5%.

So, that is why the PIXI design is suggesting you to use 330uF*8 which is obviously 2640uF.

Anyway, take also into account parameters like ESR and Iripple for better performance and derating too. And check all your warnings in your design.

Regards,

From your answer i cant get clarity

12VX2.5%=0.3V

From 2.5mF how to arrived at 330uF
please explain in brief

From your formula

Co=(10*10)/(0.3*132000)
=2.53mF

From 2.53mF how to arrived at 330uF & 2640uF
Please explain in brief

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Hello,

That's it. You got the value for Co I was talking about in my first email.
The fact is that the formula of Co is the minimum value for the output capacitance.

So, if you put 8 capacitors of 330uf in parallel you will get 2640uf which fits your needed output capacitor value, according to the formula.

Hope you understand me now and of course check all your design again.

Regards,

Thanks for your answer

How to decide internal clock cycles needed by the control loop to reduce the duty cycle from máximum to mínimum

in this above mentioned calculation your are assumed 10
any thumb rule to decide internal clock cycle or mentioned in Datasheet

In general how to decide internal clock cycle?

Hello,

The internal clock cycles has been set by design as an empyrical parameter and I have seen this assumption described in such a way in many datasheets and design guidelines.

In fact, the assumption is to be between 10 and 20, but 10 works fine and it's enough.
If you try to find this formula for a flyback converter you will see exactly this assumption.

Anyway, I have always used it and never had problems.

Regards

thanks for reply