# Can V²/R be used to find energy of a DC transformer pulse?

A person I am working with is trying to determine the efficiency of a pulse transformer and I am confused by the procedure.

We know the input energy because a capacitor is discharged across the primary.

The way my colleague measured output energy was look at an oscilloscope measurement of the output voltage pulse across a non inductive resistor, and find V2/R integrated across the pulse duration.

This formula assumes that I=V/R, but is that true for a step up transformer? If we doubled turn ratio and doubled pulse voltage we would calculate bigger energy with this formula, even though the current is reduced so energy is the same.

Can someone explain if this measurement approach is right or wrong? And why or why not?

This formula assumes that I=V/R, but is that true for a step up transformer?

Yes it is but....

If we doubled turn ratio and doubled pulse voltage we would calculate bigger energy with this formula

If you doubled the turns ratio you won't necessarily get twice the pulse voltage (except in the case of an open circuit). You have a fixed resistive secondary load and, it presents itself as a much lower value resistance on the primary when you have a step-up transformer. If you increase the turns ratio then the resistor looks even lower when seen from the primary.

All of this means that when you discharge the capacitor into the primary, the secondary voltage waveform is constrained more when the step-up turns ratio is higher (because of the new lower value load resistor presented at the primary).

So, you may get a slightly higher voltage pulse but, the width of the pulse will be smaller (constrained) and, when you do the energy calculation/integration, the energy liberated into the load will be about the same.

Can someone explain if this measurement approach is right or wrong?

The measurement technique appears good to me.

Integrated over what? ;)

Remember that, for a higher turns ratio, the primary-referred (reflected) load resistance is lower, so the RC time constant is shorter.

• Integrated over time Jul 8, 2022 at 0:29

If you want to maximize power output then match the source impedance $$\Z_o=\sqrt {L/C}\$$ to the load impedance by choice of n turns ratio along with Q reactive gain. The result will be a critically damped pulse with a smooth triangular power shape.

Using $$\R/4\$$ to reduce load resistance seen from the input for 2x the step-up turns ratio with square power law may be computed.

Although the pulse width changes with peak pulse power, the energy transferred ought to be the same. You can calculate the input energy $$\E_c=½CV^2\$$ minus losses from ESR, DCR, and switch, for a reasonable range of load R values or n chosen. Coupling factor =<1 loss ought to be included in impedance transformation.

• In this case my source is a capacitor that is charged up by a flyback converter and then discharged across the primary coil. Will the source impedance in this case simply be the capacitance of that capacitor or will it include anything else like the inductance of the pulse transformer's primary coil or the inductance of the flyback converter's secondary coil? Jul 8, 2022 at 0:28
• It’s both C & L as I said Jul 8, 2022 at 1:45