# The way to efficiently operate some power devices with less loss, and appropriate heat sharing

I'm using a buck converter(OKR-T/10-W12-C made by Murata Power Solutions). The converter supports 4.5 ~ 14.0V input.

The input is 9.5V voltage from the another buck device.

And between of them, the schottky diode SR54L is stationed. So, because of the diode, some voltage drop occurs(0.2 ~ 0.3V) but the input is 9.5V so the voltage drop will not come a big problem.

Then, have to I use a schottky diode? Or, can I remove the diode?

Which way is more efficient to make less power dissipation? I want to use less power because of the efficiency but I'm worried about the heat dissipation of the converter. Will dividing the heat with schottky diode just not good solution for the heat sharing? Will using the diode can affect to the performance of the converter?

(Using heatsink is impossible because the converter is too small to use a heatsink.)

• "Then, have to I use a schottky diode? Or, can I remove the diode?" - that depends on its purpose. Why is it there? – Bruce Abbott Feb 7 '19 at 7:47
• @BruceAbbott I installed the diode initally because the previous circuit's input was directly connected by the SMPS(installed for reverse-polarity protection). Now the converter gets input by the buck converter so I came up with the possibility of uselessness of the diode. (Because the previous input is reverse-polarity protected too) But if there can be the 'significant' heat sharing I will not modity the circuit. – Chanho Jeon Feb 7 '19 at 7:55
• get rid of the unneeded part. – dandavis Feb 7 '19 at 18:48
• and keep the needed parts... that's where the question is about... – Huisman Feb 7 '19 at 23:22

You could calculate the dissipation of your application and compare it to the Temperature Derating Curves.

The last derating curve example is 12V in, 1V 10A out. From the Efficiency vs. Line Voltage graphs, I estimate an efficiency of 78% at conditions 12V in, 1V 10A.

Defining Ploss the power dissipagtion that causes the converter to heat up:

Ploss = Pin - Pout

Efficiency = Pout/Pin

Ploss = Pout/Efficiency - Pout

Ploss = (1V*10A)/0.78 - (1V*10A) = 2.82W

From the Efficiency vs. Line Voltage graphs, you could estimate the efficiency using your output voltage and current. And next, calculate the power dissipation.

If it is below 2.82W you can use the converter up to about 38 degrees C using no fan (nor heatsink, where would it be placed??) and need to derate according the last graph of the Temperature Derating Curves.

I didn't address your original question yet if it would make sense using a shottky to lower the input voltage. I think it would make sense, if you exceed the 2.82W above. The Efficiency vs. Line Voltage graphs show that a lower input voltage gives a higher efficiency, and so, less dissipation. Note that a standard diode would even better in this case than a schottky diode.