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Old 26th May 2009, 05:33 PM   #1
flysig is offline flysig  United States
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Default Why does lowering screen voltage reduce headroom?

Lowering the screen voltage on a pentode will move all the curves down, resulting in less gain. Why/how is the headroom reduced? Is it simply that the plate current is reduced, resulting in the cathode current being less and thus the drop across the cathode resistor is less, causing the grid bias to reduce a little bit?
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Old 26th May 2009, 07:58 PM   #2
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Yes, Flysig,

But I would argue the other way round. Sure, the maximum grid swing is reduced. But with the 'anode family' moving down and the same load, the anode swing (downwards) is quite reduced - that is, if the anode load line was initially optimal, i.e. with the bottoming anode excursion right at the bend in the Va-Ia graph at Vg1=0. Another way, the load line will cut the Vg1=0 line at a higher point (Va) than before. You now have reduced anode current to work with, thus less amplitude before it 'bottoms'. Drawing a load line on a set of graphs will pictorially illustrate.

(By increasing the load until the anode can again bottom at a low voltage will improve matters, but your available Ia swing would still be lower.)
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Old 27th May 2009, 05:20 AM   #3
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There are several ways you can look at it. I tend to think of a pentode screen as the equivalent of the plate in a triode: the lower its voltage, the lower the cathode current and so the lower the negative bias has to be. This reducess the maximum amplitude of signal that can be input to the control grid, hence you get reduced headroom.
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