Electronic – critical inductance value when designing a buck converter

From your numbers it seems like the instantaneous draw when the audio output is at the maximum peak of the waveform is too much for converter 1. This would explain why it works at higher frequencies, since converter 1 then sees more the average rather than peak instantaneous current draw.

You say converter 1 puts out 12V at 1A, which is 12W. To put out 10W RMS audio power would mean the peaks of the audio waveforms draw twice that instantaneously. This would overload converter 1 by almost 2x according to your description. At 1kHz audio output, the overload happens for only about 250µs at a time. Since the average draw is OK, converter 1 doesn't fault.

This is just a guess, of course, but it is consistant with the information you provided.

You don't specify which relay you're using so I'm taking this one as an example.

Rated voltage = 110V AC
Current = 21mA @ 60Hz
Coil resistance = 932\$\Omega\$

Now the total impedance

\$ Z = \dfrac{110V}{21mA} = 5240\Omega \$

That's the resistive part (932\$\Omega\$) with the reactive part at 90°. Then the reactance

\$ X_L = \sqrt{Z^2 - R^2} = \sqrt{5240^2 - 932^2} = 5154\Omega \$

Then, since

\$ X_L = 2 \pi f L \$

\$ L = \dfrac{X_L}{2 \pi f} = \dfrac{5154\Omega}{ 2 \pi \mbox{ } 60Hz} = 13.7H \$

That's a pretty high value, but if you do the same calculation for 50Hz, where the current is 24.2mA, you get a comparable value: 14.1H.