[R-390] OT(just a bit): From whence cometh the grid current?

[Al Parker]

and the screen grid does draw current, more significantly than a control 
grid might.
73,
Al, W8UT
----- Original Message ----- 
From: "David Wise" <David_Wise at Phoenix.com>
To: <r-390 at mailman.qth.net>
Sent: Monday, June 08, 2009 3:58 PM
Subject: Re: [R-390] OT(just a bit): From whence cometh the grid current?


> Voltage dividers present to their load, an impedance
> equal to the top and bottom in parallel.  See
>
> http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem .
>
> Dave Wise
>
> -----Original Message-----
> From: r-390-bounces at mailman.qth.net [mailto:r-390-bounces at mailman.qth.net] 
> On Behalf Of Barry
> Sent: Monday, June 08, 2009 11:06 AM
> To: r-390 at mailman.qth.net
> Subject: [R-390] OT(just a bit): From whence cometh the grid current?
>
> I have an HP-606A that I'm checking over and have found the voltage on the
> screen grid of one of the 6AW8A amplifiers in the power supply is off 
> (low)
> a bit.  The voltage is supplied to the screen grid via a voltage divider
> from B+ that consists of a 1.2M and a 100K resistor from B+ (500V) to
> ground.  With a 500V supply and nothing else connected, this theoritically
> gives 38V at the junction of these two resistors.
>
> If I unplug the tube and measure the voltage at the screen grid pin, I get
> close to 38V; however, when I plug in the tube, it drops to about 32V.
>
> My question is where is the grid current coming from?  Is it coming from
> ground up through the 100K or from B+ down through the 1.2M?  I assume it
> draws it from ground because as little as 1mA would drop a significant
> amount of voltage through that 1.2M resistor.
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