Tube Amps / Music Electronics
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|6/2/1999 11:09 PM|
|Jim L.||12AX7 Plate Resistors|
Does anyone know how I can calculate the gain of a 12AX7 tube stage?
I have seen 100k plate resistors used often, is there a reason for this?
How does this affect the sound if you have one amp running 300 volts B+ and another running 400 B+ but with a 100k plate resistor? If I increased the 100k to 300k on a Ebb of 300 volts would I lower the amplification factor?
Also, for maximum gain, would you ground the Cathode? or have a network of a 15k, 100uf,...like used in the McTube. What is the purpose of the capactior, shunting AC to ground? This would provide more gain, correct?
but why not just send the cathode directly to ground?
I'd apprecaite any help, I'm trying to rework my preamp tubes a bit for a different style of overdrive.,...
|6/2/1999 11:33 PM|
If you increased the plate voltage from 300vdc to 400vdc with the 12AX7 and the 100K plate load resistors, you will also increase the current through the tube.
If you then adjusted the current back down through the tube by changing the bias voltage and (or) the cathode bias resistor, you'd change the voltage on the plate again from lack of current flow through the tube and plate load resistor and consequent voltage drop through that very same plate load resistor!
Cause and effect.
If you took that same tube and stuck a 300K plate load resistor in it, the plate voltage would drop considerably. Less current would be the result again.
Like a balance beam scale.
There is a method to this madness and a balance must be achieved in order to get the right amount of idle current, plate voltage and voltage gain.
I guess you could have max gain and a very flat frequency response with a grounded cathode 12AX7, if it had it's grids controlled by a variable negative -3v bias voltage supply.
But since that's a drag, the obvious solution is to self bias them with a cathode resistor and get most of the gain back by bypassing the cathode resistor and subsequent negative feedback with a big old capacitor who's reactance is less then 1/10 of the DC resistance of the bias resistor! Whew!
Your example 15K and 100uF cap = < 1Hz, but the 15K sets a low current idle point to swing the AC around.
All the AC freqs above a few Hertz are at ground anyhow with a big bypass cap.
To a point, you could VERY loosely figure the gain by dividing the plate load resistor value by cathode resistor and multiplying that by the efficiency of the tube.
Not terribly accurate but fun to mess with.
100k/1k5*(.50)= gain of about 33...
220K/2K7*(.50)= gain of about 41... etc.
Loose and dirty.
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|6/3/1999 12:02 AM|
I'm glad Bruce explain all that.There are posts about the difference in headroom with different voltages, and brown sound with less. But you can run into problems, if you try to change the cathode resistor to get more gain, it will work to a point,as with the plate resistor.As Bruce said.there has to be a balance.Like a tetter totter, you can get some terrible sounds from changing the cathode resistor too much. It will sound like a metallic blown speaker.There are some on here that have put in switchs to change the bypass cap and resistor used. I would think too that if its a lower plate voltage you can get by with a 1/2 watt resistor.But to go above 350 you should use a 1 watt resistor. Look at the archives on some of the Marshall mods for alot of changes and tries. Depending on the amp and circut it will have a big impact on how it sounds. [Richie]
|6/3/1999 12:10 AM|
I forgot to add, 400 Volts is a bit too much. These tubes are not made to handle this for very long. Some of the best sounding amps run the lower voltages.150 to 180 range. I think thats why some of the boogies just eat tubes. they run them so hot.And even the good ones will not last as long.I know you all are going to ask .. but what about the reverb tubes....they run very high for a 12AX7. [Richie]
|6/3/1999 5:15 AM|
With the grid referenced to ground the cathode resistor will set the DC bias of the tube so you can’t connect the cathode directly to ground without upsetting the bias. Therefore for maximum gain you have to AC couple the cathode to ground with a big Cap. A 15k cathode resistor is a little on the large side with 100k plate R. Rk is usually 1k5 and a bypass cap of 22u is big enough to fully bypass it at the lowest guitar frequency.
With the cathode resistor fully bypassed the voltage gain is given by -
Where u is the amplification factor (100 for a 12AX7). Ra is the tube internal resistance and RL is the effective plate load resistance. e.g. if the plate resistor is 100k and you are driving a 500k vol pot then RL is 100k//500k or 83k. Ra depends on plate current. It’s about 60k at a plate current of 1mA but it goes up as the plate current goes down so you won’t get a huge gain increase by changing the plate load from 100k to 300k.
|6/3/1999 8:52 AM|
Thanks to all who replied,
|6/3/1999 1:47 PM|
Dave: (No comment on your fine explanation, just some add'l circuit info for comparative gain calculations)
A 15k cathode resistor, with a 100k plate resistor, would be unusual. It wasn't stated in Jim's question, but the Real McTube booster pedal circuit has a plate resistor of 330k (connected to a B+ about 170v) along with the 15k Rk (bypassed w/100uf) on the first stage. It also has a 1meg Rg, and a series .01uf from the input jack to the grid. That stage is driving a 500k volume pot, wiper to the next input grid. The next stage uses a 680k Rp, 4.7k Rk w/100uf cap, output shunt is essentially 1megohm (the lower end of the 1meg has a 500k pot paralleled w/100k resistor, to keep the final signal output reasonable). The whole setup is rather unconventional, when compared to usual guitar preamp stages we're familiar with.
I haven't gotten around to trying this circuit, and hoped by now someone would have, and commented on its performance. It appears the actual voltage at the plate terminal is much lower than the conventional fender preamp circuits. Plate A is 108v, with a 1.5v bias. Plate B is 60v, with a 0.7v bias. (At least that's what's listed in the article.)
Quick calcs using the equation you provided result in a stage A gain of 80, and a stage B gain of about 77. Current is only about 0.1ma, so I guessed at Ra=70k.
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