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Old 18th June 2009, 11:18 AM   #1
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Default Help with CATHODE resistor value on 6BM8 PP Amp

Guys... a cry for help!! Techno help that is.

I have recently got a Pioneer SX-410 receiver that I must service for a lady friend.

A friend of mine suggested that I must replace the SINGLE cathode resistor of the output section with four independent components for improved reliability. Apparently this single resistor blows up easily.

The tubes are 6BM8 / ECL82 in push-pull mode (I think).

All four the cathode connections of the tetrode part of the output tubes go to a single wire-wound resistor.

Should I use resistors with the same value or should the value be divided or multiplied by four?

See picture attached. The Cathode resistor is the black horizontal device on the bottom of the picture.
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Old 18th June 2009, 11:41 AM   #2
chrish is offline chrish  Australia
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I will give you a hint - Mr Ohm's Law...

The resistor is used to raise the voltage of the cathode with respect to the grid (biasing it). The current going through all of those four tubes will give a certain voltage drop across that resistor. If you divide that current by four (individual cathode resistors, that is four in stead of one to conduct the same current), to maintain the same voltage drop you must do what to the resistance?

Ohm's Law


E=voltage, I=current, R=resistance.



(If current is reduced by a factor of four, then resistance must be raised by a factor of four to keep the same bias voltage. The individual cathode resistors will be four times the value of the single one. Your next question should then be 'what about the power dissipation of each of these resistors........)
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Old 18th June 2009, 12:14 PM   #3
chrish is offline chrish  Australia
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With 6BM8s you will probably end up with a cathode resistor of about 390 Ohms and passing around 35mA will dissipate around 1/2 Watt. I would use 2 Watt resistors (or two 780 Ohm 1 Watt resistors paralleled, or 2 x 200 1Watt in series or example)

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Old 18th June 2009, 12:25 PM   #4
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Thank you Chris.
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