Question about Bias setting in Carvin TS100 when using EL34 vs 6L6GC

[dinkyguitar]

Hi,

I have a Carvin tube stereo amp, 50watts per channel that I am using as a power amp for my Fractal Axe FX III.

I replaced my GT EL34R power tubes with JJ 6L6GC.

As per Carvins instructions for bias using 2 meters.

Inside the amplifier, on the component side of the power tube printed circuit card, are switches to select the tube type for each channel. Above those
switches are markings that indicate the different tube types. Bias potentiometers are also near those switches for fine-tuning. Each side of the amplifier must
be biased with ALL four power tubes installed.

a) Remove from the printed circuit board the red wires connecting from the output transformers to QC2 and QC16.

b) Two milliamp meters will be needed. Insert a milliamp meter in series with each of the red wires to the printed circuit board. Current can now be
measured through both of the output transformers while no input signal is present. Make sure the leads don’t touch anything such as the chassis, each
other, or you- VERY HIGH VOLTAGES ARE PRESENT!

Power up the amplifier – switch the stand-by switch on. Adjust the bias pots accordingly to obtain a 50mA reading on each of the milliamp meters

My question is, since the 6L6GC is a 30 watt tube, as opposed to the EL34 25watt tube, should it be biased higher?

My B+ as measured at pin 3 of the power tube is 468vdc.

Maybe I'm getting confused or not understanding what bias really is.

Plus, 50mA per side is cold...that's 25mA per tube, something like 55% bias point.

But on the reverse side, it should be a transparent amp which should not color the sound....maybe that's why it's a cold bias.

dinky

 

[Jon Snell]

Valve biasing is an interesting yet simple subject.
Consider the EL34. All time favourite power pentode. Need about -30 - - -40 volts on the control grid, dependant upon anode and screen grid voltages.
The 6L6 is a kinkless beam power pentode and requires a different and higher negative voltage on the control grid.
What we are doing is adjusting the knee point or the point in which the valve, rather like a MosFet transistor, has a turn on poinr beyond that conducts too much and over heats or can damage the equipment it is fitted to.

Here is how to work out the anode dissipation of a valve;

Some like them hot some cold. Without getting too heavy with maths and algebra, a tech from a market leader amp maker (no names mentioned) who tends to bias on the hot side, told me this.

The maximum output from an EL34 is about 25 watts. So measure the anode voltage of the amp, for this exercise we will assume that to be 445 volts. Divide the 25 by the 445 then multiply by 0.7 the answer in ma. is the 70% plate load bias setting required for that amplifier.

The 0.7 means 70% Hot anode / plate dissipation. Therefore you can use 0.5 (50%) as a Cool AB setting, 0.6 (60%) as a Warmer AB setting the one I would prefer. Or go for the 0.7 or even 0.75 (70 - 75%) if you like it hot.

Example: 25 / 445 X 0.7 = 39.3 ma. Bias

The bias figure will depend on the anode voltage which will vary according to the current drawn by the valve. More current less volts, les current more volts. So you may need to do the calculation several times if large adjustments are made.

Example 2. 25 / 480 X 0.7 = 36.4 ma. Bias

Example 3. 25 / 410 X 0.7 = 42.6 ma. Bias

If looking for the Class A requirement, that's 100% plate dissipation. The figure to multiply by would be 1 but check the valve is operating within recommended plate voltages. Here is an EL34 example with 300 volts on the plate / anode.

25 watts max o/p of the valve divide by plate volts 300 multiply by 1 = bias current requirement for class A operation. Example. 25 / 300 X 1 = 0.083 = 83ma. or 25 / 350 X 1 = 0.071 = 71ma.

No rocket science - dead easy elementary stuff.

If you have a cathode biased amp, you can measure the bias current drawn by dividing the voltage drop across the cathode resistor, by its resistance. Example Cathode resistor 470 ohms voltage measured across it 25 volts.

25 / 470 = 0.053 = 53ma

53 ma flowing through the cathode resistor is the sum of the anode and the screen grid currents together. Deduct about 5% screen current 2.65 subtracted = 50.35ma anode current.

I will talk more about cathode bias later, but as a general rule a bigger value resistor the valve will draw less current, but the plate voltage will rise. Lower value resistor the valve will draw more current, the plate voltage will drop.
​

Basically, for a Marshall sound, bias cold - cool, for a Fender sound bias warm - hot.