I have to design a class D power amplifier, able to deliver 200 W on 50 ohm load at 100 kHz. Power supply should be 14 or 24V, I don't know at the moment so I need an output trasformer. I did few simulations on LTSpice with a resistive load and works well, anyway when I add a traformer something is not working. The inductor current increase without limit and the voltage output decreases. So initially the circuit delivers 200W with good efficiency byt after few milliseconds power decreases and also efficiency, because the inductor current raise without limit. I don't understand why. The LC filter is tuned at 120 kHz assumed a load of 0.5 ohm on primary side and a Q = 0.7. When I'm wrong?

enter image description here

enter image description here

enter image description here

  • 2
    \$\begingroup\$ 200 watts into a 50 ohm load requires power supplies that are a lot more than 24 volts. \$\endgroup\$
    – Andy aka
    Commented Sep 28, 2023 at 9:52
  • 4
    \$\begingroup\$ The output contains DC, you must have a coupling capacitor. \$\endgroup\$ Commented Sep 28, 2023 at 10:22
  • \$\begingroup\$ You need a diode between the +14V and "center point" Vs ... What kind of driver do you use? \$\endgroup\$
    – Antonio51
    Commented Sep 28, 2023 at 14:37
  • \$\begingroup\$ @Andyaka Not with the matching transformer, which is the whole point of the question.. \$\endgroup\$
    – tobalt
    Commented Sep 30, 2023 at 8:50

1 Answer 1


The series L1+L2 is connected to ground, and you are using a unipolar power supply for M1 and M2. This puts an average DC on the inductors, and so the current ramps to infinity.

One option is a large capacitor in series with L2, to create a phantom mid point. Often this sort of design will use two capacitors, one to each rail.

Another option is to drive M1 and M2 with a balanced bi-polar power supply.

Yet another option, especially if you are struggling with a low voltage power supply, is to bridge drive the load. Duplicate M1 and M2, drive them in antiphase to the first pair, and connect them to the bottom end of L2.

Only the first, capacitor, option will automatically balance. While the second and third options reduce the DC offset by an order of magnitude or two, they still won't eliminate it, and current may still grow, just much more slowly. You may need to monitor the DC current and correct for it by tweaking the drive duty cycle.

  • \$\begingroup\$ Acutally, with a H bridge configuration, there is still a DC component on the primary side of the transformer, and it is the same order of magnitude, so the current drift away. I don't understand if it is a weak design or I can tweak something to make things working. \$\endgroup\$
    – mtx4
    Commented Oct 18, 2023 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.