transconductance

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bradzatitagain

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Joined
Oct 7, 2004
Messages
295
How can I express transconductance in microhmos if I know the current per volt? For instance, Euro tube mnuals, like Phillips and Telefunken list the tranconductance of the PF86 as 2.0mA/V. American manuals express it in microhmos; I'd like to compare some pentodes that show up only in European or NA manuals, both not both.
 
What about Larry and Curly?

A mho is an ohm walking backwards. Because he's been huffin. Which would also make him a cool ohm, right? :?
 
> Euro tube mnuals.. list the tranconductance.. as 2.0mA/V.

You have to keep your millis and micros straight.

2 milliAmps is 2,000 microAmps. So 2mA/V is 2,000 microMho.

US data goes back to days when a few microAmps was a lot of change for a volt of grid swing. So uMho was "logical". And if you express your total load in KK (later call Meg) Ohms, such as 0.050Meg for 50K, then you just multiply-through to get your gain. 2,000uMho * 0.050Meg = 100.

I think the US got stuck in a rut, and Europe moved to the better idea of listing newer hotter tubes in mMho instead of uMho.

Oh, just to confuse you further, some older US lit spells "microMho" as "mMho".

Then instead of turning Mr Ohm on his head, someone was inspired to honor Siemen. I know several generations of Siemens played MAJOR roles in the electric and electronic industries, but I don't know that any of them invented anything about transconductance. Ah, well, it beats having the coax cable (a Siemens invention; though rarely used for the purpose it was invented for) named after you.
 
Is there an easy way to explain (understand?) the relation between amplification factor and microMhos? I'm stumbling through Kimmel's Mu follower thing, a pentode CF sitting on top of a triode, the gain is the transconductance of the triode at (near) constant current.

So, a tube like 6DJ8 has 12.5K mA/V but an amplifiction factor of 33; 12AX7 has a trnsconductance of 1.6mA/v with an amplification factor of 100. Listed amp factors seem to have nothing to do with output in Kimmel's mu follower; and I'm a little lost on that detail. Is it because the output is coming from the voltage swing on the resistor between the pentodes cathode and the triodes plate; the signal flows from the cathode on top, not the plate below? The triode is amplifying voltage, the pentode is amplifying current? And "amp factor" listed in a tube manual is what, current amplification in a typical plate follower triode circuit?

Sorry we're a little ways away from my original question...
 
There is the parameter called plate resistance. Rp is operating point dependent and is based on how much the plate current changes for a small change in plate voltage. Rp = delta Vp / delta Ip.

When you put a load resistor in series to B+, you now have that resistance effectively in parallel with the Rp; also any loading by the next stage is in parallel as well. Call this the net plate circuit resistance.

Next you consider the transconductance, the change in plate current for a small change in grid voltage.

The voltage gain from grid to plate is the (negative sign) product of transconductance and net plate circuit resistance.

In the limit as the plate load resistor and the loading resistance of the next stage get very large, the maximum gain is the transconductance times Rp. This is what is called mu.
 
[quote author="bradzatitagain"]Is there an easy way to explain (understand?) the relation between amplification factor and microMhos? I'm stumbling through Kimmel's Mu follower thing, a pentode CF sitting on top of a triode, the gain is the transconductance of the triode at (near) constant current.

So, a tube like 6DJ8 has 12.5K mA/V but an amplifiction factor of 33; 12AX7 has a trnsconductance of 1.6mA/v with an amplification factor of 100. Listed amp factors seem to have nothing to do with output in Kimmel's mu follower; and I'm a little lost on that detail. Is it because the output is coming from the voltage swing on the resistor between the pentodes cathode and the triodes plate; the signal flows from the cathode on top, not the plate below? The triode is amplifying voltage, the pentode is amplifying current? And "amp factor" listed in a tube manual is what, current amplification in a typical plate follower triode circuit?

Sorry we're a little ways away from my original question...[/quote]

To further put some numbers to this: your spec for the AX7 of 1.6mA/V and amplification factor ("mu") of 100 means that the plate resistance Rp is 100/1.6*10^-3, or 62.5k ohms. With a perfect current source load, i.e. infinite resistance, for the plate load resistor, the voltage gain would be (-) 100.

For a plate load R of 100k, the gain would be 1.6mA/V * (100k || 62.5k). That equivalent R is about 38.5k, hence the gain would be (-) 61.5.
 
[quote author="PRR"]Ah, well, it beats having the coax cable (a Siemens invention; though rarely used for the purpose it was invented for) named after you.[/quote]
Don't just tell part of the story :grin:

What was coax cable invented for?

Best regards,

Mikkel C. Simonsen
 
> an easy way to explain (understand?) the relation between amplification factor and microMhos?

Amplification Factor (Mu) is a geometric quantity; you can measure it with a ruler, and actual value will be very nearly constant.

The cathode is a perfect infinite amplifier plus a dead resistance of 1/Gm.

The plate is a perfect source of the same current modulated in the cathode, shunted by a resistance of Mu+(1/Gm), the plate resistance Rp.

Both Gm and Rp vary significantly with current, are at best semi-constants.

> I'm stumbling through Kimmel's Mu follower thing

In My Humble Opinion, most such circuits are a waste of tubes. Kimmel's plan at least is no worse than a simple gain stage plus a cathode follower (except for wanting a higher supply voltage for the same output swing). It does not work like simplified analysis suggests: the top tube is NOT a cathode follower even though we take output at the cathode.

> gain is the transconductance of the triode at (near) constant current.

Incomplete. Voltage-gain in terms of Gm is meaningless without a resistance. What resistance is the triode working into????? That is NOT easy to answer correctly (I won't attempt it).

The cannonical Mu-follower with identical tubes/FETs will give a voltage gain of Mu/2. I don't know how much this mixed-tube scheme messes-up that simple result. With two different triodes it might be (Mu1+Mu2)/4. With a triode-strapped pentode, the pentode Mu will be around 20. With a self-bootstrapped-screen pentode, the screen load can easily eat all the advantage promised by pentode operation.

> "amp factor" listed in a tube manual is what, current amplification in a typical plate follower triode circuit?

No, it is the Voltage amplification with fixed plate current. On the plate-curves, draw a horizontal line at a reasonable current, mark off 1V or 10V of grid swing, see how much the plate voltage must be changed to keep the current on the horizontal line. If you try this all over the volt/amp quadrant, you will get nearly the same number everywhere, with a slight drop of Mu at lower currents or the highest plate voltages. If you bust the tube open and measure the grid and plate dimensions, you can calculate a Mu which will be nearly the same value.
 
[quote author="PRR"]
In My Humble Opinion, most such circuits are a waste of tubes. Kimmel's plan at least is no worse than a simple gain stage plus a cathode follower (except for wanting a higher supply voltage for the same output swing). It does not work like simplified analysis suggests: the top tube is NOT a cathode follower even though we take output at the cathode.
.[/quote]

I am so happy you said that.
 
> What was coax cable invented for?

Underground power distribution.

To make money in the electric racket, you have to deliver large energy to customers (mostly industrial at first) without killing anybody along the way.

Overhead distribution is obvious, but ugly and still dangerous. The Great Blizzard of 1888(?) in NYC, some people were snowed-in for a week, and a large part of the problem was the maze of telegraph and telephone wires in the streets after the poles collapsed from ice. Edison's early attempts to distribute electric used underground vaults, like small subway tubes with wires strung on the walls, very expensive.

Siemens was watching insulation technology (before good insulation, there was little alternative to bare wires on glass knobs). He could make an insulated cable, but proximity to ground would suck a lot of power. And there is the backhoe effect (even before backhoes were common): if you lay a wire underground, a backhoe WILL find and cut it. So high-power electric underground was considered lossy and dangerous.

So he made a (single-phase) cable with the hot lead in the center and the grounded lead as a shield around the outside. Ground-loss was small. He rigged a run from his generator to a big factory, showed it powering huge motors. Then, in public demonstration, with part of the cable left above ground, he took his coat off, picked up an axe, and said he would cut the cable with full power flowing. After a moment of thought, some people fainted. Nevertheless, he swung and cut the cable. He was not harmed. The axe blade was wedged in firm contact with the grounded shield before it hit the center hot conductor. The generators shut-down as, of course, they have to do when bad things happen.

That may not be the first use of coax. But I like the drama even more than Westinghouse's streetcars.

If you want an exact citation I'll have to go into the bowels of the library.
 
Brilliant!

-Imagine if PRR was your uncle or something... You'd have the Coolest uncle of any of your friends: -Even the one with the "Concorde-pilot uncle" and the one with the "World-Trade-Center-architect uncle... -about the only kid who cold still go toe-to-toe with you might be Tom Dowd's nephew!

:wink:

Thanks for the great bit of history as well as correcting the assumption that coax was invented for its present use! :thumb:

Keith
 
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