**Part Number:**LM5143

**Other Parts Discussed in Thread:**CSD18533Q5A, CSD18563Q5A, LM5145

Hi,

Just downloaded the Synch Buck calculator from ti.com, and for one of their synch buck chips,

i got it to calculate the switching loss. It said top FET switching loss was 2.7W....

I only get this from the "Laszlo Balogh" method if i use external R(source)=6.8R and external R(sink) = 0.5R.

I wonder what total R(source) and R(sink) the ti.com loss calculator uses?

Laszlo Balogh" method (pages 4 - 12)

http://www.radio-sensors.se/download/gate-driver2.pdf

Our LM5143 based synch buck uses...

NFET CSD18533Q5A was bottom FET

NFET CSD18563Q5A was top FET

Buck was 12.5A out, 13v5 out, 27V5 Vin , 450kHz.

ti.com Synch Buck Switching Loss calculator....

www.ti.com/.../SYNC-BUCK-FET-LOSS-CALC

They give the equation for the switching loss, which indeed is the correct one, but that equation does not give 2.7W for switching loss.

It gives less than 0.5 Watt, when only the internal gate drive res is used.

I also calculated the switching loss in the manner described in the document by Laszlo Balogh...

www.radio-sensors.se/.../gate-driver2.pdf

..and again, the figure comes out nowhere near 2.7W.

As you know the principle for finding FET_ON switching loss is simple...

..You find the power for two time periods..

1...Time to charge Ciss from Vgs(th) to Vgs(i_pedestal) [i(pedestal) is the "valley" of the inductor current]

2...Time to discharge Cds from Vin to Zero. (Qgd ends up getting used here)

...(1)'s calculations is ultimately from dt = C.dv/i(drive)

where...

i(drive) is the current sourced by the driver, to find which, you need to know the drive output voltage

and the drive series resistance. (And also the "average" gate voltage over the time interval concerned).

-And then its Ohms Law.

Anyway, with the FET_OFF switching loss calculated in the same vain, it comes out as less than 0.5W..

So how did ti.com come up with 2.7W?

Must admit the driver source and sink resistances are low, and i believe on the real product we will have to add series

gate resistance, otherwise the FET will hard-switch too fast, and resulting noise could ruin the operation.