modified 7 years ago

Am I doing my math right

00:08:45

Ok, so I did some math regarding the efficiency of my circuit http://everycircuit.com/circuit/4709847533879296 and I calculated an efficiency of 99.95% assuming the MOSFET was a perfect switch, which it pretty much is considering that it has a voltage drop of tens of millivolts when passing tens of amps. I did the math by seeing how much voltage in the 1.4nF capacitor was shorted out each time the MOSFET was turned on, and then calculating the power lost in watts due to that, which I calculated as 10mW, which seems unrealistically low, considering that the output power is around 20-25W. The equation I used was (V^2*C)/2*f, where V was 5.5V, C was 1.4nF or 1.4*10^-9F, and f was 500kHz, or 500,000Hz. Is it true that the majority of power loss is in MOSFET resistance rather than the small residual 5.5V of power in the capacitor being shorted out 500,000 times a second despite the MOSFET resistance being on the order of milliohms?

Edit: I did some more measurements in the simulator and found that the MOSFET has a dissipation due to internal resistance similar to the dissipation due to the shorting out of the capacitor before it is fully discharged, meaning it has an efficiency of 99.9% which seems a bit extreme to me. Is my math correct?

published 7 years ago

kiani

7 years ago

In my opinion accuracy can be checked with sim. But efficiency, not good idea. Cause the efficiency depends even on the solder joints.

jason9

7 years ago

The question here is whether my math is correct. I already know that any number of things could have a parasitic internal resistance which will reduce efficiency. I’m assuming all wires and components are perfect.

kiani

7 years ago

Oh i see.

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