Can't get bias hot enough on rob's 6v6 jcm 800

[jl2556]

When running EL34s (the trannies will handle it per Rob's build notes), I can't get the bias hot enough. 17-19 MV at the most. From doing some research, I know there is a "range resistor" that need to be changed to increase the range of the bias trim pot. Is it the 470 ohm straight off the 50V bias tap or the 7.5K that goes from the bias pot to the middle leg of the bias balance pot? Those are the 2 that make the most since. Also, what do I need the value to be?

Here's a link to the page with the layout or the schematic (6v6, not the 1 watt):
https://robrobinette.com/RR2104_Master_Volume_Micro.htm

 

[Ten Over]

Change the 470 ohm resistor to 1.5K/1W. Half watt if you have those on hand.

 

[jl2556]

Change the 470 ohm resistor to 1.5K/1W. Half watt if you have those on hand.

Thanks man. Out of curiosity, what's the formula a guy can use to figure that out? Just trying to learn something as I go

 

[Kevin Wolfe]

It would be hard to nail with a formula since changing the resistor not only changes the bias but the plate voltage as well.

 

[Ten Over]

Thanks man. Out of curiosity, what's the formula a guy can use to figure that out? Just trying to learn something as I go

It's a surprisingly complicated situation. I use my own handwritten tables compiled from a mountain of data.

 

[jl2556]

Cool....thanks guys.

 

[Kevin Wolfe]

It's a surprisingly complicated situation. I use my own handwritten tables compiled from a mountain of data.

That’s very true. So many variables.

 

[Tom Kamphuys]

I'm completely lost here. I live in a happy bubble that's about to explode. I thought bias circuits were simple. Here we go:

When running EL34s (the trannies will handle it per Rob's build notes), I can't get the bias hot enough. 17-19 MV at the most.

MV seems a bit much, but mV seems way too little. What have you measured here?

Change the 470 ohm resistor to 1.5K/1W. Half watt if you have those on hand.

How does that help? I thought the bias circuits drew very little current, so I would expect this change to do ... nothing?

It would be hard to nail with a formula since changing the resistor not only changes the bias but the plate voltage as well.

It seems to me it is a voltage divider. And how is the plate voltage changed? And why does that matter?

Better prepare for impact, I feel I'm starting to slide down the Dunning-Kruger curve. Hope it's in the right direction...

 

[Kevin Wolfe]

The plate voltage matters because of Ohms law. You’re using voltage, current and resistance to calculate the plate dissipation of you power tube/tubes. Change any one of those three values and the solution to the math equation changes.

 

[jl2556]

MV seems a bit much, but mV seems way too little. What have you measured here?

Uh.....17-19mv??

How does that help? I thought the bias circuits drew very little current, so I would expect this change to do ... nothing?

Read this:
https://el34world.com/charts/Biascircuits.htm

 

[clintj]

I'm completely lost here. I live in a happy bubble that's about to explode. I thought bias circuits were simple. Here we go:



MV seems a bit much, but mV seems way too little. What have you measured here?


How does that help? I thought the bias circuits drew very little current, so I would expect this change to do ... nothing?


It seems to me it is a voltage divider. And how is the plate voltage changed? And why does that matter?

Better prepare for impact, I feel I'm starting to slide down the Dunning-Kruger curve. Hope it's in the right direction...

The bias circuit is just that, a basic voltage divider with one end at max negative (just after the diode) and the other end at ground.

Plate voltage changes because of power supply sag on the B+ rail. If you adjust the bias voltage to make the tubes pull more current, the voltage coming from the PT through the rectifier, filter caps, and output transformer windings lowers in response to that. You can kind of guesstimate how much to alter the bias circuit after a few tries, but just swapping resistors one standard value at a time is a safe approach.

Sent from my Pixel 2 using Tapatalk

 

[Tom Kamphuys]

Can't you use this as your guesstimate:

If the bias is reduced to zero volts then the valve will run red hot (and perhaps be destroyed), while if it is made very negative then it will reach cut-off. This covers the full range of useful adjustment from cold class-B to total meltdown. So how much negative voltage do we need to cover the full range? That’s easy: just take the screen-to-cathode voltage and divide this by the triode mu of the valve, i.e. the mu quoted on the datasheet when operating in triode mode. This will be enough to bias the valve almost to cut-off, and we don’t really need more than that! The triode mu of some popular valves are:
EL34 = 10
EL84 = 20
KT66 = 7.5
KT77 = 11
KT88 = 7.5
6L6GC = 8
6V6GT = 10

For example, if you’re using a 6L6GC with a screen voltage of 400V, then aim for a raw negative voltage of:
400/8 = 50V (negative).
By contrast, an EL84 running at 300V would need only:
300/20 = 15V negative.
Remember, this is not the normal operating voltage, it’s just the maximum voltage worth aiming for so you can then adjust it down to whatever normal operating value you like.

http://www.valvewizard.co.uk/bias.html

and then use the potmeter for the quick adjustment. No need to adjust the components to varying plate voltages, only adjust the potmeter setting (?).


I would have thought that the bias was ~-13.5V. Is the 17-19mV the voltage over a 1 Ohm cathode resistor?

Why is the influence of the 470 to 1500 Ohm resistor that large? Both are very small w.r.t. the other resistances in series (5k + 12.5k + 7.5k + 50k). Or is it because it only conducts a small portion of the time and the others conduct continuously, making the voltage drop across it effectively much larger?

 

[Ten Over]

Can't you use this as your guesstimate:


http://www.valvewizard.co.uk/bias.html

and then use the potmeter for the quick adjustment. No need to adjust the components to varying plate voltages, only adjust the potmeter setting (?).


I would have thought that the bias was ~-13.5V. Is the 17-19mV the voltage over a 1 Ohm cathode resistor?

Why is the influence of the 470 to 1500 Ohm resistor that large? Both are very small w.r.t. the other resistances in series (5k + 12.5k + 7.5k + 50k). Or is it because it only conducts a small portion of the time and the others conduct continuously, making the voltage drop across it effectively much larger?

It's one thing to have a general ballpark figure for your bias voltage and another thing to design a circuit that will deliver that voltage.

I used to always call the resistor between the transformer and the diode the "current limiting resistor". Not very many people seemed to know what I was talking about, so I took to calling it "that resistor between the transformer and the diode".

The capacitor in the bias circuit only charges when the transformer voltage causes the diode to be forward biased. Once the capacitor reaches a steady charge, the portion of the ac wave that forward biases the diode becomes rather small. Current only flows when the diode is forward biased, so there are a series of short current pulses charging the capacitor. The capacitor discharges through the bias circuit in between the current pulses. The average voltage on the capacitor is a function of the charging characteristics and the discharging characteristics making it very difficult to calculate. Putting a resistor in the charging system has a potent effect on the average voltage on the capacitor because, amongst other things, it limits the magnitude of the current during the charging current pulses.

 

[Tom Kamphuys]

I've tried to calculate the voltage at the capacitor (Vc) as a function of the maximum AC voltage (Vp), the current limiting resistor and the load (sum of all resistors in serie). To simplify the math I assume an infinite capacitance, no drop over the diode and a triangular AC (i.s.o sinusoidal). I think the first assumption is allowed because you want a stable bias. The triangular AC will make that the result is a bit off, but it's too late to bring out the asin and integration.

It's based on the assumption/fact that the charge that leaves the capacitor in a complete cycle has to be compensated by a short burst of current during the time the AC voltage is larger than Vc. (I will not pay attention to the sign of Vc)

This results in a quadratic equation for Vc, which can be solved as usual.
a = 1
b = -2*Vp -8*Vp*Rcl/Rl
c = Vp*Vp

The result for Vp = 50V, Rcl = 470R, Rl=75k is -40V.
The result for Vp = 50V, Rcl = 1500R, Rl=75k is -34V.

I almost never do things right the first time though...