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Old 27th August 2010, 01:14 AM   #1
bobluis is offline bobluis  Argentina
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Default voltage divider in WE91 driver

Hello, I'd like to know the reason Western Electric used a voltage divider (VD)in the driver tube (310) in G2 instead of a more standard dropping resistor. The image shows the SP 91 version with divider in R8/R7 (arrow) and standard setup in generic pentode (right).

Although the voltage divider is easy to understand there is an aditional current to ground. How do you manage to calculate Ig2 current regarding the VD. Just trying to understand the circuit well.

Best regards
Roberto luis, Argentina
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File Type: jpg WE91 and pentode setup.jpg (90.4 KB, 80 views)

Last edited by bobluis; 27th August 2010 at 01:33 AM.
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Old 27th August 2010, 01:54 AM   #2
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The voltage divider is more stable, voltage level wise, than just a pullup resistor (versus varying screen current). To measure screen current, just subtract the pulldown resistor current from the pullup resistor current. (measure voltages across the resistors, calc the currents from the resistances, subtract to get the difference which departed thru the g2)
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Last edited by smoking-amp; 27th August 2010 at 01:57 AM.
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Old 27th August 2010, 02:37 AM   #3
bobluis is offline bobluis  Argentina
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Thanks Smoking, quite clear. Apparently it is a try and true method to calculate screen current, modifying VD resistors order of magnitude while maintaining its relationship.

You cannot use the data sheet tables in advance as they use a more standard topology regarding screen grid.

Best regards, Roberto
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