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It is common when wiring a pentode so that it behaves as a triode, to connect the screen (g2) to the plate through a small resistor. Others have used diodes shunted by capacitors. The aim is to ensure plate and screen follow each other.
My question is this - what are the considerations for the best 'operating point' of the screen - is there a 'best' d.c. voltage differential between plate and screen even though they are a.c. coupled ?
My question is this - what are the considerations for the best 'operating point' of the screen - is there a 'best' d.c. voltage differential between plate and screen even though they are a.c. coupled ?
My conceptual image of a pentode gets me a spanking every time I mention it, but I don't care. Think of it like this:
From the point of view of voltage, the triode is defined by cathode, grid, and screen. Hence the mu of a triode-connected pentode is, more or less, screen mu. Choose the screen voltage in the same way as you would the anode voltage of a real triode. Don't exceed datasheet limits.
The anode of the pentode is a current drain. This needs only to be at high enough to protect the screen from excess current during use. Too low and the screen won't last long, too high is wasteful and the anode will have a shorter life because power dissipation is proportional to voltage for a given current.
So, find your best operating point from the datasheet's triode-connected characteristic curves using a loadline as usual, and then see how far you can reduce anode voltage whilst maintaining screen current within its limit.
The danger area to watch out for is when anode voltage swings low. If anode voltage drops too far during operation, anode current plummets and screen current soars. That's where you don't want to go. The main job of the screen resistor is to protect the screen, so that screen current causes a voltage drop that in turn reduces current. The ensuing distortion is definitely worth avoiding.
All in all, considering you probably want to err on the side of fidelity rather than efficiency, you are likely to end up close to what the datasheet suggested in the first place.
Ian
From the point of view of voltage, the triode is defined by cathode, grid, and screen. Hence the mu of a triode-connected pentode is, more or less, screen mu. Choose the screen voltage in the same way as you would the anode voltage of a real triode. Don't exceed datasheet limits.
The anode of the pentode is a current drain. This needs only to be at high enough to protect the screen from excess current during use. Too low and the screen won't last long, too high is wasteful and the anode will have a shorter life because power dissipation is proportional to voltage for a given current.
So, find your best operating point from the datasheet's triode-connected characteristic curves using a loadline as usual, and then see how far you can reduce anode voltage whilst maintaining screen current within its limit.
The danger area to watch out for is when anode voltage swings low. If anode voltage drops too far during operation, anode current plummets and screen current soars. That's where you don't want to go. The main job of the screen resistor is to protect the screen, so that screen current causes a voltage drop that in turn reduces current. The ensuing distortion is definitely worth avoiding.
All in all, considering you probably want to err on the side of fidelity rather than efficiency, you are likely to end up close to what the datasheet suggested in the first place.
Ian
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Joined 2009
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No. The anode voltage can be considerably lower than screen without causing a problem. Depends on the circuit and the application and the particular valve.
The limit is imposed by screen current, which in turn depends on Vak as well as the difference between it and Vas.
Consider that electrons reaching the screen have considerable velocity so they can lose plenty of energy on the way to the anode and still get there. As long as they do get there, so screen current limit is not exceeded, a lower anode voltage results in less power dissipation for the same AC output, so is more efficient.
I see I forgot something important from my first post, which is that the increase in efficiency means that more audio power can be extracted before exceeding the anode dissipation limit. Not worth it in my view, because where power is the big issue, overdrive is likely, and the screens would fry.
Thing is, efficiency isn't a big issue for us, and the added circuit complexity and cost isn't generally worthwhile.
Ian
This actually makes sense, no justification for a beating
I come from usenet.
p.s. I lived in Leeds when I was younger, not far from your place. Tetley's, best pint in the world !
So did I, when they played good football. Not exactly a romantic city. I moved to Newcastle and defected to Fed. Bradford is the dead centre of industrial civilisation.
Ian
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