Single ended amplier

[pwdrive]

Ref, load lines I posted in an earlier post, I also did a couple of dynamic transfer curves, Ra = 12k and Ra = 50k, not done too accurate, just done to show the difference:

       

Lawrence.

[Terry]

... if the OP valve is biased in Class A, how can the anode current double?

Andy, the anode current only remains constant if there is no input signal. It is then determined by the (fixed) bias but, when you apply an input signal, the anode current must increase for positive going half cycles of the input waveform. However, as it must also decrease on the negative going half cycles, the peaks and troughs must average out to the same level.

Because of the smoothing components, as far as the PSU is concerned, the current demand doesn't change for Class A, as you've said.

[Murphyv310]

... if the OP valve is biased in Class A, how can the anode current double?

Andy, the anode current only remains constant if there is no input signal. It is then determined by the (fixed) bias but, when you apply an input signal, the anode current must increase for positive going half cycles of the input waveform. However, as it must also decrease on the negative going half cycles, the peaks and troughs must average out to the same level.

Because of the smoothing components, as far as the PSU is concerned, the current demand doesn't change for Class A, as you've said.

Absolutely Terry.
In class B the current increases with higher volume levels. In class A the smoothing must be adequate to cope with the swings on positive drive especially if the amp has good LF response.

[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.

[Kalee20]

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.

Not quite.

You bias the valve so that it's sitting in the middle of the available CURRENT range, so that when clipping occurs it's hopefully symmetrical (like you said). Then, you check that the bias you need is at least equal to the peak input swing, so that the grid never has to try to go positive (if it does, you get a new distortion mechanism).

If this isn't satisfied, you have to operate at a lower standing current, or raise the HT voltage (so you can then increase the bias to throttle back to your favoured current). And that's why valves need high voltages!

[Mark Hennessy]

I was trying to keep it simple, which is why I went for the resistive load rather than a transformer - in which case current and voltage are proportional. In my experience of teaching electronics, people always find voltage is easier to visualise (and put on a 'scope) than current.

But yes, when you put the transformer in circuit, then it's the current that matters - and my explanation falls down because there's only ~20V DC across the transformer primary

[Kalee20]

Actually, even with a resistor, if you want maximum output swing, the principle, 'bias so that the voltage across the load is approx. halfway down from HT' is appropriate (same as saying that current is half of available maximum). It's not the same as saying, 'bias at centre of available input swing.'

Although, having reread your previous post, in the light of your comment above, I realise I might have misunderstood which voltage range you meant, output rather than input!

[Mark Hennessy]

OK - I've just re-read it as well, and I think it's clear that I've talking about the anode voltage, but if there's any doubt, then I'm happy to have it clarified. The object is to help understanding, after all. I've added (in red) the word "input" to one paragraph

The one thing I'm not sure about - because of my lack of experience with valves - is just how "saturated" a valve can be. With bipolar transistors, transistors, it's pretty easy at a first and even a second look (though to do saturation properly gets surprisingly complex, but it's mercifully rare to need to go there). With MOS-FETs and JFETs it's even easier because the channel is basically resistive. But honestly, I'd be guessing when it comes to valves.

Obviously that, plus whatever is across the cathode resistor (assuming it's held still by a bypass cap) determines how low the anode can swing. So that's where "deliberate vagueness" and "carefully chosen words" came in. Any other apparently vagueness was for the sake of teaching, where knowing what to leave out and even occasional "white lies" are wholly appropriate at a given level

[Mike Watterson]

As the valve ages you can run out of emission on the peaks.
You can see the curves at lower anode voltages.
Pentodes (And Tetrodes) are rather different to Triodes due to Vg2, the Screen. A resistor vs low impedance vs transformer tap (ultra linear) makes a big difference for Screen performance.
You want to be above the "knee" or Tetrode Kink (some Pentodes have a slight kink).

[Mark Hennessy]

Thanks for that - it's clearly more complex than I'm used to with semiconductors!

I've been doing quite a bit of reading up over the last couple of days, so although I've not been not replying directly to all the valve theory posts, they are appreciated

(still prefer transistors though )