Why some amps use two triode in parallel?

[Alan0354]

Matchless use both half of a 12AX7 in parallel for the first stage. But still use 1.5K cathode resistor and bypass cap. Use 100K plate resistor. What is the theory behind it?

I tried, seems like it sounded a little more aggressive like you bias the power tube a little hotter.

[Dave H]

It gives a tad more gain than a single triode, sounds a little ‘thicker’ and on the Watkin’s Dominator / Marshall 18W Normal channel it makes a handy mixer for the two inputs without the interaction and low impedance you get with the usual 68k resistors.

[soundmasterg]

Using a parallel triode doesn't give as much gain as those same two stages cascaded...because when you cascade then the gains multiply whereas when they are in parallel they add. The parallel stage will have a thicker sound than a single stage, but some of the crisp highs are lost as a result too. If you have a single triode with a shared 100k plate resistor and a shared 1k5 cathode resistor, to get the same gain out of a parallel stage would use a 1k5 shared cathode resistor, but would use a 50k shared plate resistor. The grids can be tied together, or can be separate also. You can also use a cathode resistor for each triode instead of a shared one, and voice it so one has higher gain than the other...say one gets a cathode cap and the other doesn't....then as they amplify, they kind of interact with each other....its a more harmonically rich sound. I've used it on the input in one amp I built for myself and I like it, but I wouldn't use it everywhere...its a nice option to have though. Merlin has some really good stuff on it in his preamp books.

Greg

[Gingertube]

My parallel input stage. The other thing it gives you is noise reduced by root 2.

Cheers,
Ian

[nashvillebill]

How about as a efx loop send (or preamp out)? Would the parallel stages allow driving lower impedance loads, as an alternative to using a cathode follower?

[Alan0354]

I don't think two in parallel gives lower output impedance. If you use 100K plate resistor, it is still mostly govern by that.

I am not sure about the lowering of the noise. I have to dig up some material. It is not that straight forward in calculating noise.

[Gingertube]

Alan,
No it's not.
Zout (if the cathode bias resistor is bypassed) is rp parallel RL.
For single 12AX7 triode say 65K || 100K = 39K
For parallel 12AX7 triodes say 65K || 65K || 100K = 24K5
or if using RL=47K 65K || 65K || 47K = 19K2

If the cathode bias resistor(s) is unbypassed then you are correct and it is largely RL which will set Zout.

That is still way higher than the 600 to 700 Ohms Zout of a 12AX7 triode stage cathode follower.

I used the parallel input stages to add some harmonics emphasis. One triode is biased at typical operating point using typical cathode resistor and bypass cap. The added parallel triode is running at low current with some high frequency emphasis to give a bit of emphais to the harmonic content coming from the guitar (and generating a bit more again in this triode stage).

Cheers,
Ian

[nashvillebill]

Makes sense, thanks.

[Alan0354]

Alan,
No it's not.
Zout (if the cathode bias resistor is bypassed) is rp parallel RL.
For single 12AX7 triode say 65K || 100K = 39K
For parallel 12AX7 triodes say 65K || 65K || 100K = 24K5
or if using RL=47K 65K || 65K || 47K = 19K2

If the cathode bias resistor(s) is unbypassed then you are correct and it is largely RL which will set Zout.

That is still way higher than the 600 to 700 Ohms Zout of a 12AX7 triode stage cathode follower.

I used the parallel input stages to add some harmonics emphasis. One triode is biased at typical operating point using typical cathode resistor and bypass cap. The added parallel triode is running at low current with some high frequency emphasis to give a bit of emphais to the harmonic content coming from the guitar (and generating a bit more again in this triode stage).

Cheers,
Ian

I always a little confuse how the rp comes into play in parallel triodes. So what you are saying is with RL=100K, Zout is 24.5K for parallel vs 39K for single. In parallel with cathode bypass, gm doubles so you have ac current doubled. So even if Zout is lower, you get more voltage out from parallel triodes? ie:

For single ip=gm X Vin. Vout=ip X Zout= gm X Vin X (rp//RL)=gm X Vin X 39K.

For parallel, ip=2 X gm X Vin. Vout = 2 X gm X Vin X(rp//rp//RL)=2 X gm X Vin X 24.5K.

[Gingertube]

Alan,
You are "over thinking" it.
Just think about "looking" into the anode.

With the cathode bias resistor bypassed you will see just rp "down" to 0V. AC wise this in parallel with the 100K RL (because the B+ rail is also at 0V AC, tied there by the filter capacitors).

With the cathode bias resistor (RK) unbypassed then looking into the anode you will see rp in series with RK times mu +1 ie you see rp + (u+1)Rk

Cheers,
Ian

[Alan0354]

Alan,
You are "over thinking" it.
Just think about "looking" into the anode.

With the cathode bias resistor bypassed you will see just rp "down" to 0V. AC wise this in parallel with the 100K RL (because the B+ rail is also at 0V AC, tied there by the filter capacitors).

With the cathode bias resistor (RK) unbypassed then looking into the anode you will see rp in series with RK times mu +1 ie you see rp + (u+1)Rk

Cheers,
Ian

I follow what you are talking, I want to know why the gain is higher with two tubes in parallel even though the Zout is lower.

[Gingertube]

See this old thread on parallel triodes
Gain calcs I did are at post #16
http://music-electronics-forum.com/t29302/
Cheers,
Ian

[Alan0354]

See this old thread on parallel triodes
Gain calcs I did are at post #16
http://music-electronics-forum.com/t29302/
Cheers,
Ian

I have to read this later on and do the gain calculation. One thing, how do you justify the noise is divided by root two? I don't think it's that simple. Noise have different sources.

1) The input resistance from external that general the thermal noise. That does not change whether you have one or two in parallel.

2) Then you have voltage noise referred back to the input from the tube. For voltage noise from the tube and plate resistor referring to input, you divide by the voltage gain. From you calculation, you have 78 for parallel, 54 for single, one case you have noise of two triodes which is 1.414 times the single triode=1.414Vn. For two in parallel, noise refer to input = 1.414Vn/78=0.01812Vn. For single triode, you have Vn/54=0.0185Vn. The advantage is very tiny. I don't thing there is an advantage. Now look at RL, it is the same in both cases, so when refer to input, the one with two triodes has slight advantage as it is divided by 78 instead of 54. So this you have slight advantage.

3) Then you have current noise generated by grid current. You have two triodes in parallel, you have to find out the totally grid current vs a single triode. I suspect it is going to be higher than a single triode. If both draw the same current as individual triode, again, it is multiply by root two!!!

You see, #1 is same for both. #2 show slight advantage for two triodes. #3 all likely be a little higher for two triodes. I don't see the justification of noise reduced by root two. Noise calculation is very complicated, it is nothing clean cut. You have to take into account every single noise source in the system and refer back to the input. I used Excel spread sheet to do noise calculations.

Do you have articles that I can look at?

[J M Fahey]

I worked with a lot of low noise designs, I don't know of reducing noise by parallel transistors.

I'm quite surprised that you don't , doubly so considering the Electronics expertise you claim, because it's a tried and true and widely studied technique, which applies to *any* active gain stage, be it tube, bipolar, Fet or anything they invent in the future.

The physical basis behind it is that audio signal processed by 2 parallel gain elements will be in phase and add up, while noise is generated by each of them individually, will NOT be in phase (precisely because being statistically random is inherent to *noise*), or, to be more precise, phase will be changing ar random all the time, so sometimes it will add, sometimes it will substract, the average is that you *lose* sqrt 2 or 3dB for 2 elements, 6dB with 4 and so on.

3dB less noise for each transistor/tube count doubling.

So much so, that "super low noise transistors" are manufactured etching dozens of parallel transistors in the same chip.

As of a classic paper explaining this, here you have one by Marshall Leach.
http://users.ece.gatech.edu/mleach/papers/Parallel.pdf

Not too practical with tubes (beyond 2 that is) but for very specific uses (LNA amplifiers for Satellite dishes or Space communications) composites have been built with *hundreds* of parallel transistors.

And everybody knows that the Military guys get to play with the most expensive "toys".

EDIT: here I found a Commercial application, the Jensen JE990 discrete low noise Op Amp , presented to the AES in 1980 (so it's not exactly "new unknown" Technology)
http://www.hairballaudio.com/blog/di...nal-amplifier/

It was built around the LM394 IC, which is nothing more than a lot of low noise transistors in parallel:

The design started with the LM394 dual super matched transistor as the input device. The LM394 is actually a monolithic device consisting of two pairs of 50 transistors connected in parallel

[jazbo8]

Can't find the link now (not sure he posted), but if you have Merlin's book, there are pages dedicated to the subject on tube noise. Gingertube provided a good summary of the mechanisms at work.