Where's my voltage gain?!

[jimboyogi]

I put this thread here because this question is part theory and part practice.

I'm designing/building a pre-amp, and the 1st gain stage is using the triode of a ecf82/6u8-a, as used in V1 of the normal channel of the Vox AC10.

Valve specs are:
u = 40, ra = ~ 5.3K ohms

Circuit specs:
B+ = ~250V, input grid leak = 1M, Ra = 100K, following stage Rg = 1M, bias is a blue LED giving a bias voltage of 2.8V.

I did load lines and my sums, and in the setup I am using was expecting a voltage gain of ~38.

I have tried 4 different valves, and I'm only measuring a gain of 17 to 24 with a 1V pp input. The 17 was with a NOS one, the 24's were all used ones from my stash.

I tried using a 220K value for Ra, and got a gain of.............. 24!

What am I doing wrong?

Help!

[Merlinb]

Firstly it is a very non-linear triode. It's mu is only 40 for quite small bias voltage, between 0 and -2V really. Below that it drops quickly to about 30.
Secondly, with a 100k load you are operating in the extreme low-current area of the graph, where mu also falls!

If you want more gain you'll need a hotter bias and possibly a slightly lower anode resistor (47k to 22k maybe).

[mooreamps]

What he fails to mention... a red LED will give you a bias of 1.7 volts ; and two 1n4007 in series will give you a bias of 1 volt. To calculate the gain, substitude 1.7K for 1.7 volts ; and 1K for 1 volt. Also, if you should find any of this published in his books later on, you will now know where it came from.

-g

[Merlinb]

To calculate the gain, substitude 1.7K for 1.7 volts ; and 1K for 1 volt.

Not sure how that is supposed to help you calculate gain??

Jim, I think mooreamps invented LED bias, so you'd better remove it if you don't want to get sued.

[Loogie]

Actually I invented the LED bias and you can read about it in my first book "Why I am not Bertrand Russell". In the much anticipated sequel, "Mr. Galileo's Profession" I follow the well-known heretic's ultimately failed attempt at the I III V blues.

[jimboyogi]

Ummmmm...............

Keen to stay out of any disagreements, but a big thankyou for the info. I saw on the grid curves that it wasn't very linear, but I didn't realise the big effect that this might have on mu.

Anyway, tonight I'll try hotter bias and lower Ra, and report back.
Cheers

[mooreamps]

Not sure how that is supposed to help you calculate gain??

Oh boy !!!! You kidding me ????
To calculate the unbypassed common emitter gain is vc/vb or -Rc / rE + RE in a transistor amplifier. For an unbypassed common cathode gain stage, we substitude -Rc for Ra and rE + RE for Rk.

Jim, I think mooreamps invented LED bias, so you'd better remove it if you don't want to get sued.

I don't sue people. I post engineering comments for educational purposes, "for free".

-g

[mooreamps]

Try a couple of 1n4007 in series for the cathode, and 100K 1/2 watt for the plate. Also, since there is no cathode bypass cap, use 4n7 for the plate coupling cap or you will get too much mud in the bass.
Good luck.

-g

[mooreamps]

Actually I invented the LED bias and you can read about it in my first book "Why I am not Bertrand Russell". In the much anticipated sequel, "Mr. Galileo's Profession" I follow the well-known heretic's ultimately failed attempt at the I III V blues.

Actually ; diode amplifiers were first used in US Navy satellite communications Parametric receive amplifiers dating back as early as the late 1960's.
It's just that I choose to use parametric amplifier technology inside my preamps.

-g

[tubeswell]

I'm designing/building a pre-amp, and the 1st gain stage is using the triode of a ecf82/6u8-a, as used in V1 of the normal channel of the Vox AC10.

Valve specs are:
u = 40, ra = ~ 5.3K ohms

Circuit specs:
B+ = ~250V, input grid leak = 1M, Ra = 100K, following stage Rg = 1M, bias is a blue LED giving a bias voltage of 2.8V.

I did load lines and my sums, and in the setup I am using was expecting a voltage gain of ~38.

I have tried 4 different valves, and I'm only measuring a gain of 17 to 24 with a 1V pp input. The 17 was with a NOS one, the 24's were all used ones from my stash.

I tried using a 220K value for Ra, and got a gain of.............. 24!

What am I doing wrong?

Help!

Ummm... (just trying a guess about why you might have different measured results) besides what the others have said, did you figure out the AC load? (because AC is what you will be measuring when you measure the gain). The AC load (is different from the DC load and) is primarily affected by the plate resistance, plate resistor and the following stage's grid leak resistor (as well as by whether the stage's cathode is bypassed or not).

[Merlinb]

Oh boy !!!! You kidding me ????
To calculate the unbypassed common emitter gain is vc/vb or -Rc / rE + RE in a transistor amplifier. For an unbypassed common cathode gain stage, we substitude -Rc for Ra and rE + RE for Rk.

That is an approximation which only works well for transistors (and to some extent pentodes) because of their very high current gain, and their extremely high "anode resistance" which makes them act like a constant-current source. That formula cannot be applied to triodes -even the 12AX7- because the open-loop gain (mu), and their internal anode resistance, are both quite small .

For example, with Ra = 100k and Rk = 1k, your formula would predict a gain of 100 no matter what triode is used, which is obviously impossible.

If you use diode bias then you effectively have no Rk, so subbing 1k for 1V would be even more incorrect.

[jimboyogi]

tubeswell, I used the formula for Amplification using the AC load, accounting for the following stage grid leak. I suspect that the variables that gave me the wrong figure are the varying mu, and perhaps the wrong figure for ra for my operating conditions (although the influence of ra on the gain is much less than the influence of mu).

[jimboyogi]

Ummmmm...............

Keen to stay out of any disagreements, but a big thankyou for the info. I saw on the grid curves that it wasn't very linear, but I didn't realise the big effect that this might have on mu.

Anyway, tonight I'll try hotter bias and lower Ra, and report back.
Cheers

OK, I've tried a few of all of your suggestions, and here is what I found -

Vk = 2.8V(blue LED), Ra = 100k, Rg = 1M, Gain = 24

Vk = 1.9V(red LED), Ra = 100k, Rg = 1M, Gain = 28

Vk = 1.9V(red LED), Ra = 47k, Rg = 1M, Gain = 26

Vk = 1.1V(2*1n4007), Ra = 47k, Rg = 1M, Gain = 26 (with the beginnings of grid current limiting @ 1Vpp input)

So I got the most gain, and the most appropriate input sensitivity with this setup-
Vk = 1.9V(red LED), Ra = 100k, Rg = 1M, Gain = 28

Still a long way short of the ~ mid 30's gain expected. I'd prefer a bit more gain, but whaddayou do?

The behaviour of the 1.1V bias makes me wonder if this triode exhibits forward grid current early, like 12AT7?

Thanks again for all of the advice so far, and if there are any other ideas I'd love to hear them.

Cheers

[mooreamps]

For example, with Ra = 100k and Rk = 1k, your formula would predict a gain of 100 no matter what triode is used, which is obviously impossible.

If you use diode bias then you effectively have no Rk, so subbing 1k for 1V would be even more incorrect.

Which is a voltage gain of 100 and a power gain of 20.
Get A Grip ....


-g

[Merlinb]

Which is a voltage gain of 100 and a power gain of 20.

You're saying gain of 100 is a gain of 100? How enlightening.