Oups!! I've been biasing EL34's for 10 years with a bias-probe that has no pin #1

[Iplayloud]

Back then, knowing even less about tube amps I built a home-made bias probe... using a 6L6 socket, that doeasn't have pin #1 which is the suppressor grid. I have used that contraption to bias a BUNCH of amps since.. Now I'm biting my nails and wondering what is the impact of leaving out that pin on EL34s for just one of a pair or quad of EL34s? Could it have affected the current draw or done damage to something?

[überfuzz]

Back then, knowing even less about tube amps I built a home-made bias probe... using a 6L6 socket, that doeasn't have pin #1 which is the suppressor grid. I have used that contraption to bias a BUNCH of amps since.. Now I'm biting my nails and wondering what is the impact of leaving out that pin on EL34s for just one of a pair or quad of EL34s? Could it have affected the current draw or done damage to something?

I wouldn't worry to much about that. I often... eh always, bias by ear. When I dialled something that sounds all right I do some control measurements.

If you have any of those amps close by you can do like this. Do this for both sides of the push pull power stage.

Resistance R = pin 3 to centre tap.
Measure voltage drop U = centre tap - pin 3 (pin 3 same points as above )
Calculate current I = U/R
Finally W = I*U

(This might be somewhat crude if it's more than a pair of tubes in the power stage, but it gives you a fair idea of what's happening in the power stage.)

[Iplayloud]

Wouldn't worry about bias or the disconnected suppressor grid? I don't feel that I could do it by ear, for example I installed a tube/ss rectifier switch on my princeton clone and on the ss setting the 6v6s go from drawing 20 mA to 29mA which is way too high and just by ear I couldn't have heard it. Your formula is interesting though..

[Dave H]

Resistance R = pin 3 to centre tap.
Measure voltage drop U = centre tap - pin 3 (pin 3 same points as above )
Calculate current I = U/R
Finally W = I*U

Why calculate W? It looks like it's the bias current power in one half of the primary (probably a few 100mW).

[MandyWeidemann]

Why calculate W? It looks like it's the bias current power in one half of the primary (probably a few 100mW).

+1

[Iplayloud]

Cool, now why don't we talk about the original question

[J M Fahey]

Cool, now why don't we talk about the original question

Slim chance; Mandy's here just to promote his blog

[mhuss]

Do an experiment: Set the bias with the suppressor disconnected, then connect it and measure the bias again. This will give you a better feel for how it will behave than any theoretical discussion. I would guess it would act like a tetrode with the suppressor floating, but I've never actually tried that.

[Dave H]

I don't know of any effect on the bias current but I did once forget to connect pin 1 to pin 8 on a 2 x EL34 cathode biased amp. It was a long time ago but I seem to remember that it was unstable at high volume (funny noises) and didn't make full output power.

[nsubulysses]

Just for reference I once had an Ampeg V4 with EL34s installed but without pin 1 tied to 8. It put out about 75W at clipping and this is what the screen grid resistors looked like.



After connecting pin 1 to pin 8 and putting in some 1.5K screen grid resistors it put out ~105W