No load protection for Class D amp

[Gaz]

Thanks for the answers everyone, and scholarly discussion as usual!

[Helmholtz]

The leakage inductance+ speaker inductance is in series with resistive part of the load, limiting the rate at which current can change. There is no such protection for the transformer primary when the load is removed.

Can you explain?

I don't see how series resistance could mitigate current changes, especially it won't prevent current disruption. But it will damp the associated resonant ringing.
Parallel resistance, on the contrary, would provide a snubber effect.

[Mike Sulzer]

Can you explain?

I don't see how series resistance could mitigate current changes, especially it won't prevent current disruption. But it will damp the associated resonant ringing.
Parallel resistance, on the contrary, would provide a snubber effect.

When disrupting a current I in a time interval t, the derivative of the current should be proportional to I and inversely proportional to t. The bigger I, the bigger the derivative. A series resistor would be expected to limit I, and therefore the derivative. (The current I takes some time to reach a limiting value through an inductor, and therefore overdriving with a low frequency could be more dangerous to the amp than a high frequency when there is no load).

[Helmholtz]

When disrupting a current I in a time interval t, the derivative of the current should be proportional to I and inversely proportional to t. The bigger I, the bigger the derivative. A series resistor would be expected to limit I, and therefore the derivative. (The current I takes some time to reach a limiting value through an inductor, and therefore overdriving with a low frequency could be more dangerous to the amp than a high frequency when there is no load).

When driven from something like a current source, the series resistor won't change much. Of course the energy released from an inductor increases proportional to IČ, where I is the current just before the disruption. So a higher current will charge the parasitic capacitance to a higher ("flyback") voltage.

But actually I am not clear which 2 scenarios you are comparing.

[Mike Sulzer]

When driven from something like a current source, the series resistor won't change much. Of course the energy released from an inductor increases proportional to IČ, where I is the current just before the disruption. So a higher current will charge the parasitic capacitance to a higher ("flyback") voltage.

But actually I am not clear which 2 scenarios you are comparing.

Start with an inductor in the plate with no current through it. Hit the grid hard, driving it positive. Nearly all the power supply voltage appears across the inductor; the tube is not operating as a current source in this case, initially. Current through the inductor rises with time. Eventually the tube limits the current. A large enough resistor in series would limit the current first. I do not know the details, and I think it would take a good simulation to show what really happens. The soft screen supply in a guitar amp is a factor as well.

[Helmholtz]

Start with an inductor in the plate with no current through it. Hit the grid hard, driving it positive. Nearly all the power supply voltage appears across the inductor; the tube is not operating as a current source in this case, initially. Current through the inductor rises with time. Eventually the tube limits the current. A large enough resistor in series would limit the current first. I do not know the details, and I think it would take a good simulation to show what really happens. The soft screen supply in a guitar amp is a factor as well.

I think that the amplitude of periodic currents is mainly limited by the power tubes' internal plate resistance which is typically greater than 20k, i.e. considerably higher than the reflected series resistance. There will be some influence of the series resistance, though.
But what matters for the flyback peak is only the momentary inductor current at the instant of disruption.

[Helmholtz]

I just occurred to me that the series resistance makes a major difference, indeed. While I don't expect a strong influence on current amplitude, the series resistance part of the load impedance causes the phase delay between voltage and current to stay below 45°. Without the resistance the phase shift will be close to 90°, meaning that the current is maximum when voltage is zero. A current disruption closer to its peak will produce higher energy flyback spikes.