Gain of a Parallel Input

[populartsl]

Thinking of building a standalone power amp using the parallel input 12au7 as shown on AX84 standalone power amp https://www.ax84.com/corepoweramps.html

My question is how much gain the parallel input has. With my previous amp i used a half of a 12AX7 for recovery with a gain of 60, I reduced the signal with a 60 to 1 potential divider in the preamp. So i wondering how much i need to reduce the signal with the 12AU7 parallel input.

[pdf64]

There are 6 power amps on that page, which one are you thinking of?

[populartsl]

Here's the direct link.

https://www.ax84.com/static/corepoweram ... ematic.pdf

I'm guessing the gain of the parallel 12au7 input would be around 10

Cheers

[pompeiisneaks]

I thought parallel tubes gave you twice the gain, so if your 12au7's have a gain of 20, that woudl be 40, not 10 no?

~Phil

[didit]

Hello -

Voltage gain, which is usually what is of interest, remains roughly equal whether it's one or a pair of triode elements in parallel--roughly doubling the available current, and therefore lowering output impedance roughly by half.

Roughly.

Best .. Ian

[Malcolm Irving]

If you think of the plate characteristic curves, when you parallel the triodes everything on the curves stays the same, except that all the currents (on the vertical axis) are doubled.
If you keep the plate load and cathode resistor the same, the voltage gain will be doubled. This is because twice the current swing through the plate load will give twice the output voltage swing.

[pompeiisneaks]

Oh ok cool, thanks, so the voltage gain remains at 20, cool!

~Phil

[tubeswell]

...
If you keep the plate load and cathode resistor the same, the voltage gain will be doubled. ...

... but the bias point will change, which will affect the way the tube responds to input signal

[Malcolm Irving]

Yes, you are correct. You would need to halve the value of the cathode resistor to keep the bias voltage the same.

[SoulFetish]

So this is true. You don't achieve any greater voltage gain out of the amp, assuming you adjust your plate load and cathode resistances to maintain DC operating point. However, if you consider the AC loading of the following stage, you may observe less signal loss at the input (dependent on the load impedance) because you've lowered the driving impedance by half.

[pdf64]

I seem to remember reading that with 2 devices in parallel (and the operating point maintained), another benefit is that the stage noise is lowered a bit.

[roberto]

Well, resistors used to polarize the tube need to be half in resistance and double in power, so the Johnson–Nyquist noise is of course lower.

[SoulFetish]

Right, but one thing to consider is that in order to lower the Johnson noise, the tube(s) must share the halved value resistors or the sum of individual parallel resistors results in the same Johnson noise as the one resistor of twice the value. Of course, if they share a resistor which has half the resistance value, then twice the current flow through the resistor, resulting in more shot noise and contact noise.
The real benefit of paralleling two triodes in a single gain stage is due to the fact that that the noise generated by the two individual triodes are uncorrelated and add as " the square root of the sum of the squares of the individual noise sources, while the signal is correlated and adds directly, giving a 3dB improvement in signal-to-noise ratio."
- taken from "Resistor Types - Does It Matter?" - http://aikenamps.com/index.php/white-papers

[Malcolm Irving]

Two 100k resistors in parallel produce the same Johnson noise as a single 50k resistor (assuming all these resistors are at the same temperature).

[SoulFetish]

Then my explanation above is incorrect. Thank you for correcting that, Malcolm