Single ended amplier

[Mark Hennessy]

" provide enough drive to the OP valve to swing it from double the standing Ia, " not I sure I understand you here, if the OP valve is biased in Class A, how can the anode current double? I know the anode voltage swings higher than HT when current in the primary is 0, does current double too?

To keep it simple, let's assume the anode load is a simple resistor.

You choose the quiescent anode current that puts a voltage across this resistor that is half of what it could be when the valve is conducting as much as it possibly can be. It'll be roughly half the HT.

That's the quiescent standing DC condition. Now we superimpose the AC signal upon that.

As the input signal goes positive, the valve conducts more, and the anode voltage falls. But the current in the anode resistor goes up when this happens. And if you set up the DC conditions correctly*, then this should be a doubling of current.

Naturally, the opposite happens when the incoming signal goes negative. As a result, the anode current falls accordingly.

* We assume that audio signals are symmetrical either side of ground (largely true, but there are some exceptions). Therefore, we bias in the centre of the available voltage range, which means that when clipping occurs, it'll hopefully be pretty symmetrical.

If that is the case that means the PSU has to be capable of at least twice Ia.

The current drawn from the PSU stays constant when you average it out. The instantaneous current obviously changes over a cycle, but as well as doubling, it also goes to zero.


EDIT: Posts crossed...
EDIT: Minor clarification added in red.