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Re: one output tranny hotter than the other-need advise

The Scott LK-48 b is a great little integrated amp, and I'm certain that the phono L and R input biasing sockets help alot for AC signal balancing, but I can't figure out how to bias the DC standing current for the amp either. I've never had an owner's manual, so let's try it this way. What you're really interested in finding out is the amount of DC voltage across the 10 ohm/ 2 watt resistors that are on the cathodes (pin 3) of both output tubes for each channel. With that information, you can calculate how much current is running through both output tubes and how much wattage is being dissipated by the output tubes. So let's say that you break out your Digital volt meter, set it to DC voltage, and measure 0.8VDC across one of those 10 ohm/ 2 watt resistors with no input AC signal into the amp and your speakers still connected. .8V divided by 10ohms equals .08amps by Ohm's Law. The .08amps will divide (maybe) equally across both output tubes. So each tube has .04amps or 40 miliamps running through it. You already know that the plate voltage is about 420V DC. So, .040amps times 420V equals 16.8watts. Each output tube would be dissipating 16.8 watts. That's too much power for a 6BQ5,EL84,7189 type tube to be dissipating and you should adjust the negative voltage applied to the grids of the output tubes to a value which is more negative until the voltage on the 10 ohm resistor is around 0.6VDC or less. When you adjust the current in that way, be sure to remeasure the DC voltage on the plates of the output tubes before you recalculate the wattage dissipation. Also, one other problem which might foul your calculations is the value of the 10 ohm resistor. With the power off, check the value of the 10ohm/2 watt resistors and make sure they are accurate. If they aren't, your measurements will be off. Oh, and the output transformer which is closer to the 5AR4 is always hotter than the output transformer which is on the other side of the chassis.


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