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Old 21st September 2008, 12:56 PM   #1
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Join Date: Mar 2004
Location: Albury NSW Australia
Default Pentodes, triode mode and grid screen dissipation 101

Hi Guys
I was hoping to have a point clarified for me.
As a pentode we are given anode disipation and Screen Grid dissipation. Lets take the C3g as an example. Maximum ratings are 3.5W for the Plate and .75W for the Screen Grid.
So if I tie the screen grid to the anode and have lets say 120 V of B+ on the plate at 7mA then is not the total power dissipation VxI=.84Watts? Does this mean that we have exceeded the screen grids ratings even though it is tied to the plate?
I would appreciate clarification of this as it seems to me that the screen grid, albiet tied to the plate (which is more able to handle Power) is still a small easily friable entity.
in advance
Nick Mega
"Better to say nothing and keep them guessing than to speak and remove all doubt."
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Old 21st September 2008, 02:52 PM   #2
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You may use the full power dissipation rating of the plate plus that of the screen (or maybe a bit less in the interests of tube longevity ). The tube doesn't know that the plate is tied to the screen, and it makes no difference anyway under DC (no signal) conditions.

Electrons from the cathode are attracted by g2 but most of them continue on to the plate. True, the screen can be in danger of over-dissipation if the plate voltage falls well below that of the screen for a significant length of time. However, if you think about it, that can't possibly happen if the screen is strapped to the plate, so triode mode is actually safer for the screen than pentode mode!
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Old 21st September 2008, 02:56 PM   #3
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Location: Stittsville, Ontario, Canada
Nick :
You have to consider plate current and screen current separately.
At the voltage you use as an example the screen draws less current than the plate. For example at -1.5 volt grid, and 150V on plate and screen, the plate draws 19mA while the screen draws 5. This ratio of about 4 to 1 will hold fairly constant for voltages above the "knee" of the curves or about 50 volts.

So for your example of 120 Volts and 7 mA, the plate will draw an estimated 5.6 mA and dissipate .67 Watts, and the screen will draw 1.4mA and dissipate .17 Watts, and is well within its ratings.

The screen current rises above plate current as the voltage drops below about 50 Volts. This can be a problem when plate voltage is constant, but with both plate and screen decreasing together below 50 volts the total current is so low, and the voltage is low, that total dissipation is also low and you dont need to worry about it. At least not for this tube and the curves I am looking at, although there may be some pentodes for which the screen current rises at a high enough voltage to make this an issue. Even with such a pentode as that, you dont bias the tube at that point, and the load line is unlikely to pass through such a point, so you dont need to worry about it.

edit : I see Ray put in an answer while I was typing in mine, and he is right, the "knee" in the curve always happens at a plate voltage less than the the screen voltage, so the situation I talk about in my last paragraph cant happen, so even one more reason not to worry about it !
Robert McLean
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Old 22nd September 2008, 01:48 AM   #4
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Location: Albury NSW Australia
Default Thanks to both of you

Ok thanks I got it. Its just that I have never seen anyone discuss that concept when a pentode is driven in triode mode.
Onwards and upwards
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