A few minutes ago, I wrote:
>
>So you have a voltage divider with 10M in the top leg and two 100k
>resistors in parallel (for safety) at the bottom. So the no-load voltage
>at the divider tap has got to be 2380V x 50k / (10M + 50k) = 11.84V.
>
>
>IF all the resistors are of the correct values, and are also wired
>correctly, and you do have 2380V connected to the top end, then there
>WILL be 11.84V at the tap, plus or minus a few percent.
>
>That would be exactly as designed, to provide a "HV OK" signal to the
>Triode Board.
>
>We're also presuming that the B-minus rail is closely tied to chassis...
>yes?
Apparently no. Scott reports an open-circuit resistor from B-minus to
ground.
Having fixed that, he's now seeing 11.68V from the divider, which is
within 1.4% of the expected value.
Many thanks to whoever pointed Scott towards that resistor. Normal
one-to-one support for the Triode and Tetrode Boards continues on
another channel.
(Rick, I'll be right with you.)
--
73 from Ian GM3SEK
http://www.ifwtech.co.uk/g3sek/boards
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