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

[David Wise]

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.