Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: worth on August 27, 2011, 11:09:04 am
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If I'm biasing , for instance , a parallel 6v6 Champ , and I get a reading of 56ma., does this mean that 56ma. is for BOTH tubes , and I would then halve the reading for 28ma. to set the bias ?
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That depends. We need more info. How and where do you get a reading of 56ma?
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I'm using a Bias-Rite type tube socket checker... plugged into one of the two output tube sockets, ( with , of course a tube in the other socket ) gives me a reading of 50ma. using a 200 ohm cathode resistor. This is a 5F1 with a plate voltage of 355V...sounds good , and not redplating , which I assume WOULD happen if the bias was twice as hot as what it should be.
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I'm using a Bias-Rite type tube socket checker... plugged into one of the two output tube sockets, ( with , of course a tube in the other socket ) gives me a reading of 50ma
Then that reading is for that tube only. You'll need to plug it into the other socket to measure current for the other tube.
We still need the voltage dropped across that 200Ω cathode resistor to get a reasonably accurate dissipation for the tubes. But, even neglecting the cathode voltage, that one tube is running cool. 355 x .05 = 17.75 watts static dissipation. The real dissipation number will be even lower when you factor in cathode voltage. I'm just ignoring any screen current.
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I knew that the tubes were running a bit cool... the next lowest cathode resistor I have is 150 ohms , which put the plate dissipation over 100%.. so I have to settle with what I have for now.
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> that one tube is running cool. 355 x .05 = 17.75 watts static dissipation.
Wasn't it 6V6? 17W is toasty for 6V6.
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Wasn't it 6V6? 17W is toasty for 6V6.
Yes it was. Don't know why I was thinking 6L6. Sorry for the confusion.
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What I REALLY need to know is.... if the plate dissipation of my 6V6 is 17.75V , then since my amp has TWO 6V6 in parallel, do I halve the 17.75V to get around 9V per tube ? It doesn't seem possible that two 6V6s' are reading 50ma. per tube when the plate V is 355 and I'm using a 200 ohm cathode resistor. I'd have to use something around 450 ohms to stay under 100% plate dissipation. Am I making sense ?
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As stated before...
I'm using a Bias-Rite type tube socket checker... plugged into one of the two output tube sockets, ( with , of course a tube in the other socket ) gives me a reading of 50ma
Then that reading is for that tube only. You'll need to plug it into the other socket to measure current for the other tube.
We still need to know the voltage across the 200Ω cathode resistor. Once you measure that voltage it's easy to say how much current both tubes are pulling.
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What I REALLY need to know is.... if the plate dissipation of my 6V6 is 17.75V , then since my amp has TWO 6V6 in parallel, do I halve the 17.75V to get around 9V per tube ? It doesn't seem possible that two 6V6s' are reading 50ma. per tube when the plate V is 355 and I'm using a 200 ohm cathode resistor. I'd have to use something around 450 ohms to stay under 100% plate dissipation. Am I making sense ?
It depends how your output sockets are wired. Assuming Pin 8 of each socket connected to the same 200R Rk, then the current you are measuring on either Pin 8 with be the sum of the current in both envelopes. i.e.: if you (only) have a single shared Rk for both tubes, then the current in Rk will be the sum of the currents in both tubes.
200R Rk is low for 2 x 6V6s with that sort of plate voltage. It is more typical to use 250R or 270R, (or 1 x 470R-500R per 6V6).
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The voltage across the cathode resistor measures 13.7 volts
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It depends how your output sockets are wired. Assuming Pin 8 of each socket connected to the same 200R Rk, then the current you are measuring on either Pin 8 with be the sum of the current in both envelopes.
I dissagree. The biasrite plugs in between a tube socket and a tube. Any current measured by the biasrite on pin 8, whether it's a direct current reading (series with pin 8) or voltage reading across a 1Ω resistor, will pertain to that tube only.
i.e.: if you (only) have a single shared Rk for both tubes, then the current in Rk will be the sum of the currents in both tubes.
I totally agree with this part of your statement, but this is not the same as the reading you will get with the biasrite.
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13.7 volts across the cathode resistor.
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The voltage across the cathode resistor measures 13.7 volts
OK. Assuming your cathode resistor actually measures 200Ω, then the current thru that resistor is 13.7/200 = 68.5ma. This is the TOTAL current that flows thru BOTH tubes. Since you say you have 50ma flowing thru one tube, that only leaves 18.5 to flow thru the other tube. One is way hot, the other is way cold.
Try measuring with the biasrite again and for each tube. Post both current readings.
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You're right.. the tubes are way mismatched.. 50ma and 18ma. Thanks for the replies .
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It depends how your output sockets are wired. Assuming Pin 8 of each socket connected to the same 200R Rk, then the current you are measuring on either Pin 8 with be the sum of the current in both envelopes.
I dissagree. The biasrite plugs in between a tube socket and a tube. Any current measured by the biasrite on pin 8, whether it's a direct current reading (series with pin 8) or voltage reading across a 1Ω resistor, will pertain to that tube only.
True if you were using a bias rite. I wasn't talking about a bias-rite, I was just talking about determining the total tube current when measuring the voltage at the cathode(s) that have a shared Rk.
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You're right.. the tubes are way mismatched.. 50ma and 18ma. Thanks for the replies .
Now swap the tubes around and remeasure with the bias rite. Have you got the same readings on the same sockets?, or did the readings move with the tubes? (Thinking to isolate the cause of the current mismatch)