Shared input resistor

[dynaman1]

I'm eliminating the four individual inputs on a non-master Marshall and simply using a single input jack to feed both grids of a 12AX7. Would there be any probs by using a single shared input resistor as opposed to feeding each grid with its own separate resistor??

Thanks.

[sluckey]

No problem. Just tie the grids together and use a single resistor. Kinda like this Plexi style preamp...

http://home.comcast.net/~seluckey/amps/november/november.pdf

[jjasilli]

I'm thinking that a 1M grid leak resistor serving 2 input stages, is like a 500K resistor serving each of 2 stages.  If so, it would reduce gain & hi's.  This seems to fit into the circuit topology of the schematic shown, because:  Gain-wise neither input stage is followed by a voltage divider (except for the vol pot which when dimed = a 1M grid leak resistor to then next stage); and Tone-wise:  the dark channel is, well dark; and the bright channel with the small cathode bypass cap might benefit from hi's being attenutaed on the way in.  Any thoughts or comments?

[PRR]

> it would reduce gain & hi's.

How?

IMHO, the only real issue is not exceeding the "Max Grid Resistance" rating of the tube. And 12AX7 has no max resistance rating.

[FYL]

IMHO, the only real issue is not exceeding the "Max Grid Resistance" rating of the tube. And 12AX7 has no max resistance rating.

I've seen 10M in a few docs. Cf. for instance the (real) Svetlana 12AX7 datasheet:

"

Maximum ratings:
Anode voltage, DC 330 volts
Anode dissipation, per triode 1.2 watts
Cathode current, continuous, per triode 10 mA
Positive DC grid voltage 0 volts
Negative DC grid voltage -50 volts
Maximum grid-circuit resistance 10.0 megohms

"

http://www.drtube.com/datasheets/12ax7-sed2001.pdf

Anyway, this doesn't matter in real world applications.

[jjasilli]

> it would reduce gain & hi's.
How? . . .

My theory:  Because the signal is being split into 2, its strength should decline.  There would be no such decline if the 2 channels were switched so that only 1 channel is fed at a time.  But, if simulataneoulsy serving more than one channel, the grid leak resistor could be a larger value, per FYL's post, to compensate for signal loss due to the split. 

If my theory is wrong, then one input could support an infinite number of 1st preamp stages (but for ambient resistance in the connecting wires).  Is this the case?

[sluckey]

Because the signal is being split into 2, its strength should decline.

But the signal is not being split. Splitting the signal implies a voltage divider and current flow. There is no grid current so there is no voltage drop so all the voltage appears at the grid, or both grids in this case.

If my theory is wrong, then one input could support an infinite number of 1st preamp stages (but for ambient resistance in the connecting wires).  Is this the case?

Yes, that is the case, theoretically, kinda. But be practical. In the case of one input driving two preamps...

What's the resistance of the 12AX7 grid to ground? 100MΩ? 1GΩ? Infinity? Who knows. (My Fluke 87 says it's infinity) But just say it's the low value of 100MΩ. Put two of those in parallel and you still have 50MΩ. Now put a 1MΩ in parallel with that and you have 0.98MΩ. Certainly not 0.5Ω. Now try that math using 1GΩ instead of 100MΩ. The idea is that no grid current flows in a properly operating typical 12AX7 gain circuit. If no current flows, the grid resistance must be infinity. For all practical purposes, the input resistance is set by the value of the grid return resistor.

I probably made a mess of this, but you can prove it with a sig gen, scope and two channel amp.

[jjasilli]

OK, got it, thanks!  It's amazing how the "simple stuff" remains a plague on understanding.

[PRR]

> What's the resistance of the 12AX7 grid to ground? 100M?

The resistance ought to well over 100Megs.

Since guitar pickup impedance is 5K-100K, we'd get in real trouble around 1,000 grids on one pickup.

There's also 100pFd input capacitance. Two grids, 200pFd. This rolls-off the highs. The 200pFd against 100K alone is an 8KHz high-cut. Reality is more complicated. 10 feet of guitar cable is another 300pFd. IE, another 100pFd grid is like a 3 foot longer cable, and nobody thinks about that. The pickup is really an inductance so the L-C resonates at the top of the guitar band, and more C resonates lower. But nearly all guitars have volume controls which damp the resonance.

> signal is being split into 2

I'm having a small dog show. Need a table for the dog to stand on for judging. Normally one dog at 28 inches. However a table can hold 2, 3, maybe 10 small dogs and not sag below 27.9 inches.

I have 124V power from the street. I turn on a lamp and get 123.8V. Two lamps, 123.6V. I can "split the signal" a dozen ways (lamps, PC, TV, pump) and still have about 120V.

The loading of a triode grid is very small (thouh not zero). we can put several on a pickup and not have table-sag.

[dynaman1]

I'll hang a single resistor and be done with it. Thanks.