Paralleled triodes resulting characteristic curves

Just to be sure...
paralleling two twin triodes, in the curve drawings, I will have a new resulting triode, which has:
- the same grid bias curves,
- the same plate voltages values,
- double current values,
- half internal resistance values,
- double the Mu and Gm.

Am I wrong?

When //-ing, what do you do for the plate resistor ? Halved or the same ?

i was pondering this a while back. I think when i was doing it, i imagined double mA values on the graph, used double Mu and Gm, and half Internal Plate resistance. I'm not sure about Ra.

At the time I was satisfied when I played with the numbers, but i never built anything with it. I don't know if what i was doing was correct, though. Back then, I had plenty of time on my hands to design, but not much money to build. Now I'm going back to school and don't really have time OR the money......


wait! now that i think of it... If you double the current values on the graph first. And then draw a line and solve for a plate resistor value, the value is 1/2 of what it would be before you doubled the current. If you use this line with doubled current values to solve for Mu, Gm they should be double. If you use this line to solve for the internal resistance, it should be half as much.

All of the curves on the graph should hold true for each triode. If the triodes are the same the current divided between them should be equal. Since it's two triodes, the current halves.



Or... you could convert to 'double-triode values' after you figure out what you want to use for the single triode circuit:

Draw your load line.
Pick an Operating point and estimate Mu, Gm and internal resistance.
Do all the math you want to calculate resistor values etc.


Then to find out what they would be if you put two in parallel, sharing one Ra and one Rk:

divide Ra and Rk by 2.
divide internal plate resistance by 2.
multiply Gm and Mu by 2.

Using your new 'double-triode values' you can solve for output z, gain, cap values etc.


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i think this will work. but i'm pretty much talking out of my ass here. if i'm way off, somebody correct me. please! like i said it's been a while for me.



-joe

You double gm but you halve Rp, so mu doesn't change.

> mu doesn't change

My thought exactly.

Or--- Mu is a voltage thing, not a current thing. (If you figure Mu from Rp and Gm, the currents cancel out.)

Or--- Mu is a geometric parameter, based on channel length and grid effectiveness. Build a cylindrical triode, you can measue Mu with a ruler. Build the same cylindrical triode, twice as long, you get the same Mu.

Good explanation!

Now - next question is what happens if we parallel two tubes of different mu, like 12AU7 and 12AX7. Will we get a "skewed" transition curve, like on remote-cutoff tubes?

I'm thinking possible simple alternatives to hard-to-get tubes for vari-mu.

Jakob E.

[quote author="PRR"]Or--- Mu is a voltage thing, not a current thing. (If you figure Mu from Rp and Gm, the currents cancel out.)

Or--- Mu is a geometric parameter, based on channel length and grid effectiveness. Build a cylindrical triode, you can measue Mu with a ruler. Build the same cylindrical triode, twice as long, you get the same Mu.[/quote]

So, just to sum up, the resulting triode has:

1) Same plate voltage value,
2) Double current values,
3) Double Gm value, but
4) Same Mu values,
5) Half of internal plate resistance

and doing the math as pointed out by joe:

Then to find out what they would be if you put two in parallel, sharing one Ra and one Rk:

divide Ra and Rk by 2.
divide internal plate resistance by 2.
multiply Gm and Mu by 2.

Using your new 'double-triode values' you can solve for output z, gain, cap values etc.

Click to expand...

Right ??

[quote author="gyraf"]next question is what happens if we parallel two tubes of different mu, like 12AU7 and 12AX7. Will we get a "skewed" transition curve, like on remote-cutoff tubes?

I'm thinking possible simple alternatives to hard-to-get tubes for vari-mu.
[/quote]

interesting... :?

Did anyone investigate on this?

===================

Respect,

Val

thanks for the clarification guys. i knew it was only a matter of time before someone who really knew what they were talking about sorted that out for me.


So other than the Mu correction, does the rest of my post hold up??

I like trying to post to help out and everything, but I don't want to add to all of the mis-information on the internet either. Most of the time i just wait and watch the threads. I hate trying to search for info only to find a bunch of crappy forum posts or websites that contradict each other. :guinness: :guinness:

gyraf wrote:
next question is what happens if we parallel two tubes of different mu, like 12AU7 and 12AX7. Will we get a "skewed" transition curve, like on remote-cutoff tubes?

I'm thinking possible simple alternatives to hard-to-get tubes for vari-mu.

Click to expand...

have no idea....

not sure if they would average out well or if certain features would dominate. the chain is only as strong as it's weakest link bla bla bla.... or best of both worlds!

that type of situation comes to mind for me, but i don't know.

-joe

[quote author="funkydiplomat"]
Draw your load line.
Pick an Operating point and estimate Mu, Gm and internal resistance.
Do all the math you want to calculate resistor values etc.

Then to find out what they would be if you put two in parallel, sharing one Ra and one Rk:

divide Ra and Rk by 2.
divide internal plate resistance by 2.
multiply Gm and Mu by 2.

Using your new 'double-triode values' you can solve for output z, gain, cap values etc
-joe[/quote]

I couldn't find this info elsewhere for Parallel triodes.

Does anyone know if the method for plotting Maximum Plated Dissipation is the same ? or do I need to half everything on that?

> Does anyone know if the method for plotting Maximum Plated Dissipation is the same?

In general, two plates is twice the heat-throwing area and can throw twice the heat.

In small-signal stages, it rarely matters. Take the mature Fender 12AX7 stage: 100K plate resistor, 1.5K cathode resistor, running say 333V supply: it sits at 1mA and about 231V across plate to cathode, 0.231W dissipation, rating is 1.0W. We could run over 600V supply and be inside dissipation rating. But this would give over 400V on the tube, past the plate voltage rating. In fact you can't hardly cook a 12AX7 if you try.

If you double the tubes and double the fixed conductances (halve the resistor values, 50K and 750R), it is still the same. If one tube didn't cook, two tubes won't.

If you double the tubes and stick with 100K 1.5K resistors, you have double the plate area to throw-off roughly the same heat. Actually with two 12AX7 and 100K and 1.5K resistors, the current will increase and the tube voltage will decrease. There may be some case where tube heat would increase, but I doubt it, not on any happy amplifier stage.

So a known-good one-triode plan, doubled-up, with either same or half resistor values, will be safe for heat.

You should ponder current-sharing. At same electrode voltages, two units can be as much as 40% apart in curent. So if you design for 100% Pd, one unity could be 80% and the other could be 120%. However we almost never see this much difference, especially in twin-triodes. If you are really working near 80% Pd, separate cathode resistors will reduce the difference. As you get past 90% you should test for current-hogging (or just be less reckless).

Some of the bigger twin-triodes, e.g. 6CG7, do have limits like 3.5W/plate but 5W total both plates. You can run one plate at 3.5W if the other is much less; if you push both then they are 2.5W max each.

I plotted 6CG7 and the only way I could find Pd trouble is with 500V supply and 18K load. Run 2.5K bias, you land near 250V and 14mA or 3.47W. One of those is fine; two of them (9K and 1.25K) in one bottle exceeds 6CG7 rating. But 500V and 18K load is a VERY untraditional affair, moreso for a 9K load. And raising that to 13K puts us back inside the 5W/bottle rating. And if you really need 500V and ~~10K, there are other (heftier) choices, like the TV V-sweep tubes, which today are often cheaper than the hi-fi fad preamp bottles.

If you are doubling BIG tubes, simple: one 6L6 at 350V and 5K will make 10W, two 6L6 at 2.5K will make 20W. (However once you spend for two tubes, your output transformer costs can be much less if you go push-pull rather than simple parallel.)