RMS value

Thanks for your comments. However, could you please give comments in details and explanations. The calculation of power efficiency seems to be simple. But some use RMS and some use average. It is not clear even though in publications.

Thanks Dr. Khan 

You meant if we use RMS, we have to use it for both of input and output? (also same with average?)

For a conclusion, the power efficiency can be calculated by one of 2 methods: RMS or average:

1. With P = I*V (when either or both I, V is/are constant), the average will be used.

2. With P = Irms^2*R: the RMS value will be used.

In particular, in the converter topologies like buck, boost, flyback: the efficiency can be calculated as below (wihout the loss caused by driver and controller)

a) Pin = Vin*Iin(average)

b) Pout = Vout(average)*Iout(average) or Pout = Rload*Iout(rms)

Efficiency = Pout/Pin

In a deeper understanding:

Pin = Pout + Ploss in which Ploss = Ploss(Pswitch) + Ploss(Nswitch) +Ploss(inductor) + Ploss(capacitor_ESR)

1.1. Ploss(Pswitch) or Ploss(Nswitch) = Average(Vds*Ids). This loss contains both of conduction loss and switching loss. In order to optimized and lowest the losses on these power switches. the optimized point should be found out. ( I am working on this)

1.2: Ploss(inductor) = Irms(inductor)^2*Rinductor

1.3 Ploss(cap) = Irms(cap)^2*ESR

In general, Ploss = Pin - Pout

If digital controller is used in the control of DC-DC converter, How to calculate controller loss?. controller loss also to be taken into account???? in the calculation of efficiency of DC-DC converter.

Hi Viswanatha,

The only one way is we have to calculate the controller loss with modern power measurement equipment (of course under difference mode of operations). I usually use the measurement equipment from Keithley, Keysight. those equipment can be used to tracked the power losses well.

Brs,

Ha

thanks@van ha Nguyen

Hello,

I think it depends on the type of the load, Voltage stiff or current stiff.

If the load is resistance with or without DC--> Irms* Vrms.

If the load is resistance with or without DC, and inductance--> Iavg*Vavg.

Regards,

Ahmed

Thanks

I think Vrms and Irms are always valid. P=Vrms×Irms is true for both constant as well as for variable values. P= Vavg×Iavg is valid only for pure contineous dc both voltage and current. However in case of inductor for contineous currrent the average and rms is always same.