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# TPS566238PEVM: Input capacitor selection

Part Number: TPS566238PEVM

Hi,

I'm using TPS566238PEVM for my design which has the next specifications:

Vin=3.3V or 12V

Vo=0.8V

Io=1.14A

Datasheet explain how to select input capacitor for my IC, you must use the equation I show you below where fsw=600khz according datasheet, Vin=3.3V (worst case) and I assume a Vinripple=0.1V. So the result is to implement a Cin(min)=4.6uF

However, I have simulated my circuit in webench and it recommends me to implement two capacitors: one of 0.1uF (which I agree because it will response faster to current demands from IC) and another one of 47uF (which I disagree, in my opinion this value is too big and footprint will be affected. According to datasheet,10 uF capacitor would be enough). Webench is assuming a Vinp-p=23.66mV that it is more optimistic than my assumption so I don't understand why it set 47uF for input capacitor.

I show you webench solution, I have modified inductor and output capacitor that it recommended to me according to datasheet recommendation:

• Hello Yolanda,

Equation 7 in the datasheet calculates the minimum required capacitor. TI recommends using high-quality X5R or X7R input decoupling capacitors of 30 µF on the input voltage pin VIN. The voltage rating on the input capacitor must be greater than the maximum input voltage (for your case 12V is expected at the input, I would recommend at least 25V cap rating). The capacitor must also have a ripple current rating greater than the maximum input current ripple of the application.

Thanks!

Tahar

• Okey then, so I'll use two 22uF in parallel to reach the value. I have another question about this, just to be sure, if I would want to make sure input voltage supply location is not a problem in my design I have to add 100 uF capacitor apart of these two 22uF or I would only have to add 3x22uF capacitors to reach these 100uF?

• Hi Yolanda,

That larger bulk cap of 100uF is required mainly when you are testing in the lab with a power supply for example and is to bypass the long leads. Otherwise, if the supply is co-located near on the PCB, you will not require this bulk cap and can just use capacitance required to meet your input ripple requirement.

Hope this helps.

Thanks,

Amod