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I am designing a synchronous buck converter for variable input from 12 to 36 V and output fixed at 14.4 V. The power is 400 W and switching frequency is 50 kHz.

While designing I came across 2 terms. One is ripple current that I kept 35% of rated current. The second is ripple voltage and I do not know what value to select to find the capacitance.

So want to know what ripple voltage is and what value should I select to find the capacitance?

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  • \$\begingroup\$ It may help your understanding if your load is a fixed constant voltage load or fixed constant current sink. But I get the feeling you don't know how to select your capacitor in the first place, is that correct? Do you have limits for your ripple voltage or current for your converter? \$\endgroup\$
    – winny
    Dec 10 '18 at 11:38
  • \$\begingroup\$ actually I am right now collecting all the data required for designing a Synchronous Buck Converter. Till now I have only calculated value of inductor that is 18uH (don't know if it is right but still made an effort). So you are right i don't know how to select my capacitor. \$\endgroup\$ Dec 10 '18 at 12:58
  • \$\begingroup\$ How did you choose inductance value without knowing the capacitance? (I'm not saying you are wrong, I'm asking leading questions) \$\endgroup\$
    – winny
    Dec 10 '18 at 14:04
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    \$\begingroup\$ Can you succeed, if the ripple voltage is 5 volts? or do you prefer 0.5 volts? or even 0.05 volts? \$\endgroup\$ Dec 10 '18 at 14:16
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The voltage ripple on the capacitor is produced by the ripple current in the inductor and value of the voltage also depends on the ESR (if ESR is big then the output voltage increase). I recommend you this chapter from Power Electronics Course from Coursera. At page 10/10 you can find the variation of the output voltage waveform with respect to inductor current. https://d3c33hcgiwev3.cloudfront.net/_e43e45ee7cdee1ba37a4864fd5081f9f_Sect2-3.pdf?Expires=1544572800&Signature=diQkKJv7VAgBajTK6uddgb3kpUZECb9crYQbEZLUUBjHWxAg8WcauOzj1Z41D3N8M0600UgnKGYSenId4pf71zDlUttVw1MztGUzTGmu0KycqfZHK43OC2ljaJ7SmIGl2j-1JMf0Uo7AxNS2mPyQkvx7zTf3kNTVGUpRMMt7FMs_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A

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As you might already know, the current flowing through the coil of a buck converter is changing with switching frequency in a sawtooth kind of way. This is the ripple current I_ripple that directly flows through the coil into the output capacitor.

Because the output capacitor has a (small) serial resistance (called ESR), this current produces a ripple voltage proportional to the ripple current across the output capacitor. So ripple Voltage Vripple = I_ripple * ESR.

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When you have a synchronous buck converter, the input to the LC stage is basically a "hard" square wave and this is averaged by the LC like so: -

enter image description here

Then, to reasonably calculate the output ripple voltage, consider the LC as a low pass filter and set the natural resonant frequency to be low enough to attenuate the 50 kHz square wave by a decent amount. If you chose the resonant frequency to be (say) 500 Hz there would be approximately 80 dB of attenuation at 50 kHz: -

enter image description here

Interactive filter calculator.

I've chosen values for L and C and R that produce a resonant frequency of about 500 Hz and R would be about 0.5 ohms for the required power throughput quoted. If you use the controls you can find that the attenuation at 50 kHz is 80 dB.

That 80 db is the reduction in the AC signal content that is produced by the switching action. So, if the AC content was a 20 volts peak-to-peak square wave, this could be reasonably approximated by a sinewave of about 25 volts peak-to-peak. This is attenuated by 80 dB (10,000 times) to produce an output ripple voltage of 2.5 mV peak-to-peak.

So, decide what ripple voltage is acceptable to meet your design aims and choose the output capacitor to work with the inductor to give you enough attenuation.

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