fixed bias

[ed]

I've been using the fixed bias method shown in the attachment for some time now. It's been bugging me that I might be able to do better by losing the reverse cap to ground. I'm not sure what effect it has on the signal vis the load on stage one and also effect on the filter made of coupling cap and Rg.

If the divider network on the bias supply(i.e parallel 20k + 10k) were high enough resistance wouldn't it be possible to inject the bias straight into the grid without unduly affecting Rg, and by so doing omit the cap to ground.

I'm all at sea here because Jones doesn't mention methods of grid bias and Langford Smith favours injecting fixed bias onto the cathode via a divider on the B+.

Is there a better method?

[Mike H]

Interesting, I wouldn't bother with the cap at all. If Rg is large enough, and the preset will be a fraction of the total resistance, what's the point...

[ed]

well that's the thing, as I see it anyway.....

Rg = 100k

so if I inject the bias above Rg the first stage will see a load in the region of 10K (100k in parallel with possible 10k)......

which was why I thought if I increase the resistance in the divider network to maintain my 100-60v bias the parallel resistances might not alter Rg too much.......

Am I looking at this the wrong way?

[Mike H]

All the ones I've seen (I'm pretty sure) have Rg between the grid and the bias preset (and you put the signal onto the grid end), precisely because Rg is likely to be the largest value so the remainder can be largely ignored. (?)

You would want sufficient supply smoothing for the bias supply however so hum doesn't get through. But as the current is so low this should be easy with basic RC filters.

HTH

[ed]

do you have an example?

It sounds to me from your description that you are grounding the bias.

The bias supply is heavy clc filtered...220uf 20H 220uf.

[Mike H]

Well just as you've drawn it but without the cap

[ed]

Well just as you've drawn it but without the cap

but surely that would mean dropping ground potential to -100v. Even my power supply is not that strong, and if it were then nobody's electrical equipment would work on planet earth...

I'm misunderstanding aren't I ?

[ed]

what about this??

If I reduce Rg to 95K and introduce Rg1 at 10k then the parallel network of Rg1 and the bias pot network would vary the bottom resistance from 5k to 7k5 depending on the bias pot setting.

Changing the load by this much would not make much difference to the signal in itself but it would mean that the left and right channels would have a different load on the first stage....depending on the setting of the bias pot.

[Paul Barker]

Changing the load by this much would not make much difference to the signal in itself but it would mean that the left and right channels would have a different load on the first stage....depending on the setting of the bias pot.

I wouldn't consider it because the valves should be close enough matched to make the position on the two pots close enough.

[ed]

Hi Paul...

Are you saying that the circuit with Rg1 instead of reverse cap is likely to work?

Any idea what the current reverse cap on attachment to first post is doing to my coupling cap/Rg filter?

I have a box full of so called 'matched valves'.........My current cooking set require -102 on one side and -97 on the other to maintain 50ma.....so much for matched valves.

[Paul Barker]

Hi Paul...

Are you saying that the circuit with Rg1 instead of reverse cap is likely to work?

Yes

Any idea what the current reverse cap on attachment to first post is doing to my coupling cap/Rg filter?

On the original schematic the driver sees Rg as a load impedance, on the most recent schematic the driver sees the Thevenin resistance of the whole network depending where in the pot you end up.

All the purpose of this cap to a bias supply is to filter the bias supply, if your bias supply is already clean dc leave it out.

I have a box full of so called 'matched valves'.........My current cooking set require -102 on one side and -97 on the other to maintain 50ma.....so much for matched valves.

Not actually matched then. But the difference in impedance shown to the driver and the effect of that impedance on gain is minimal.

~If it really worries you you could make a combination of fixed and auto bias, use the same fixed supply for both valves, and attach the wiper of a linear pot to ground the two poles to each valve cathode. just a thought. I would stick with what you have now drawn.

[Paul Barker]

If you give me the driver valve it's anode load resistor it's cathode resistor and whether it is bypassed I will work out for you the effect on gain of the possible extremes you may encounter.

[ed]

thanks for looking Paul........

I'm really struggling here because your take on the cap seems to be the same as Mike's and I don't follow this at all....

All the purpose of this cap to a bias supply is to filter the bias supply, if your bias supply is already clean dc leave it out.

I had it in because it is reversed and it stops the bias voltage grounding and thus preserving the bias voltage on the output valve grid. If I leave the cap out surely the bias will be 0v.

Have I got this wrong? I may be having a senior moment, in which case I apologise.

The driver in this case is a 9pin aikido 12ax7/12au7. I have 100k Rg with 0.33uf coupling cap, which I make gives 5hz cutoff. It was the effect of this reverse cap that I was querying in regard to what it does to my cut off.

[Nick]

If I leave the cap out surely the bias will be 0v.

Only if you replace it with a wire.

To my eyes, teh cap is doing two things.

1. Yep, its further smoothing the bias supply, but it should be smooth anyway

2. (and this is the important one to me), its providing the AC path to ground of the driving signal, without it, the signal will "see" the network between the end of the grid resistor and eventually finding its way to ground, the cap bypasses all that.

[Paul Barker]

I can't work it out from that I need resistances. As I don't work with the Aikido I can't say what those might me. The grid resistor of the driver isn't in the equation. The additional information of the driver: anode load resistance and any unbypassed cathode resistance are all that is required to calculate stage gain. The rest we know.

Your bottom resistor is grounding the bias voltage so has to be power rated accordingly. It will pull some current from the bias supply but it will not change the bias voltage of the output valve except by pulling it down, which is adjusted away as you twiddle your bias pot to arrive at what you want.

Regarding Nick 2nd use of the cap. Not required imho. You want the valve to take the AC you don't really want any shunt caps stealing some of it away.