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# LM60430: Output capacitor calculation: I don't find the same value as the one given in the datasheet

Part Number: LM60430

Hi,

I have to design a dc-dc based on the LM60430.

I would like to calculate the output capacitor and I am used to do it with the cut-off frequency of the loop. Here I do not have this information so I must use the formula given in the datasheet.

I use the "Equation 5" to check Cout. First thing I have done is to check that I type the formula correctly by using the data of the example.

In the example, we have:

• Vin = 12V
• Vout = 5V
• IoutMax = 4A
• deltaIout = 4A
• fsw = 400kHz
• Expected deltaVout in transient = 0.3V
• K for inductor calculation = 0.3 -> real K=0.27 after inductor selection @ 6.8uH

By doing that, I find that Cout > 93uF and ESR <66mohm while datasheet gives Cout > 86uF and ESR <22mohm. (using Cout of 86uF in the formula gives 324mV of deltaVout!)

I checked the same formula with the example of the LM63635 and there it is spot on. So I guess I type my formula correctly.

Would it be possible for someone to double-check the example in the datasheet ?

Thanks!

Sébastien

• HI

You result is right.

The load transient, both Cout, ESR will contribute to the deltavout, smaller Esr or larger Cout will be reduce the deltaVout. so if we choosing cermaic capacitor, normally its ESR is small which will helpful to the transient, in the same time, it means the output capacitance can be reduced.

the formula is for choosing a preliminary value of output capacitor which can meet transient performance ,but the final result is sitll depend on the ciruict test , it is related to the current slew rate, output capacitance derating and PCB design.

Thanks

• Hi,

Thank you! I know it will depend on many parameters but I am happy you find the same results ;).

It means there is a mistake in the datasheet. It is not a big deal but it seems better to update it.

Regards,

Sébastien

• Hi

Thank you for choosing TI device.