# Designing flyback converter - Switch voltage drop

Switch voltage drop is often used in turn ratio calculations, however it is rarely specified in datasheets for converter ICs with built-in FETs. The closest I found is "Rds at Isw". Should I multiply these and use everywhere? Or should I multiply Rds by estimated Iavg or Ipeak? When Vin is low this produces rather huge difference, like 1:5 vs 1:10.

UPDATE:

I think I found an answer on powerelectronics site:
MOSFET conduction loss PCONDmosfet = Imosfeton(avg)^2 x RDSon x D

if PCONDmosfet = Vsw * Imosfeton then Vsw = Iavg x Rds x D
Looks pretty much as I expected, except for that D multiplier.

• What do you need that for? Estimating the thermal dissipation of the MOSFET? Estimating the efficiency? Estimating the minimum viable input voltage for a given power output? The voltage drop in a MOSFET isn't some nearly constant value like in a diode or even BJT. The datasheet doesn't have a fixed number for that reason. There are various plots though, typically including one showing Vds at various gate voltages and drain currents. – jms May 29 '18 at 0:13
• It's just a formula for turn ratio estimate I found on the web: Nps = (Vin - Vsw)/(Vout + Vfw). I realize that drop in FET is not constant. I believe it depends on current, that is why I asked which current is best to use. – Maple May 29 '18 at 0:32