Hi
A newbie here, searched the forum, but did not find an answer. I have some basic electronic knowledge, but I don't want to open my tube-amp to measure bias current via the voltage drop method. Scared to touch high voltage parts, which I think is reasonable for a newbie.
So I thought using one of these "adaptors", installed between the tube socket and the tube, that basically uses a octal adaptor probe with a 1 Ohm resistor in line with pin 8 and measure the voltage drop. Basic equation lets you calculate the current - you all know that.
Here's what confuses me:
- in my amp -if I properly understand the schematic - there already is a 2.7 Ohm resistor, that connects Pin 8 to ground (that would be my measuring point, if I WOULD open the tube-amp...)
- if I install one of those adaptor probes, doesn't that ADD 1 Ohm to this setup and create a voltage divider?
- so, if I measure the voltage drop across the adaptor's 1 Ohm resistor, doesn't that give me a completely different reading, because I measure the voltage drop at the "upper part" of a voltage divider?
- if the latter is true, what equation should I use to determine the bias current?
To even complicate things more, the manual of the bias measuring device ("Biasmaster" - it offers 4 probes) states:
…amplifiers, that have each one cathode resistor per tube or pair of tubes (like McIntosh MC30): using two or more adapters at the same time might lead to errors in measurements. So they suggest to only install ONE probe for certain types of amps - is mine such an amp?
I'll attach the schematic of the amp - it's not the original one, but a reverse engineered one I found on the web, so the 8.2 Ohm resistor connected to Pin 8 is the 2.7 Ohm one in my amp...
Thanks for helping!