Choosing an Output capacitor for buck converter

Thread Starter

anishkgt

Joined Mar 21, 2017
521
Hello all,

First time doing buck converter so please bear with me.
Based on the BQ24640, i am trying to choose an output capacitor. In 8.2.2.3 section on page 18 shows how to calculate and i get
Iout = 4.26/2*SQRT(3)*1.31
= 4.83 (is it uf or V )
How do i go forward from here ?
 

Papabravo

Joined Feb 24, 2006
13,924
Your numbers don't make any sense 4.26/(2*SQRT(3)*1.31) = 0.938...
The units of Iout would be Amperes. Not volts, not μF
What is the matter with you?

Equation (9) on p. 18 of the datasheet is talking about the ripple current rating of the capacitor. The units should be Amperes, unless otherwise specified. there is nothing in that paragraph that talks about the capacitance of the output capacitor. You shouldn't be fooling around with things you don't understand. You'll shoot yer eye out!
 
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Thread Starter

anishkgt

Joined Mar 21, 2017
521
Got the math in the excel wrong. So the ripple current is 0.94A. and the datasheet says
The bq24640 has internal loop compensator. To get good loop stability, the resonant frequency of the output
inductor and output capacitor must be designed from 12 kHz to 17 kHz. The preferred ceramic capacitor is 25 V
or higher rating, X7R or X5R.
 

Papabravo

Joined Feb 24, 2006
13,924
So 25V is the maximum voltage rating of the capacitor. This means don't EVER put more than 25 VDC across the capacitor.
X7R and X5R are tolerance designations that tell you how much the capacitance will vary as a function of temperature. Commercial grade pats are usually characterized from 0-70°C. MIL-SPEC parts are usually characterized over a wider range.
 

Papabravo

Joined Feb 24, 2006
13,924
So with the ripple current how is the capacitor selected ? 25v what uf ?
I believe there are different criteria for selecting the capacitance of the output capacitor. It might be the case that the value is not critical since the load will have a much larger capacitance. I don't have much experience with this kind of circuit for this application.
 

Papabravo

Joined Feb 24, 2006
13,924
I don't know the answer to your first question on the LC filter. R1 & R2 form a voltage divider which is fed back to the converter at pin VFB. The 22 pf provides a very low AC impedance across the 300 kΩ resistor, effectively bypassing it as far as any AC voltage is concerned. You want the voltage divider to look at the low frequency DC voltage on the output pin.
 
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