Non c'est un article qui montre que le paramétre le plus important pour le gain, c'est la résistance de "plaque" du tube, pas le µ, et qui montre que les différences de gain entre deux tubes sont bien inférieures à ce que le µ laisse supposer au premier abord.
Quand on calcule, une 12AT7 a la plupart du temps un gain supérieur à une 12AX7, et oué.
Citation:
Old Tele man Old Tele man is offline
Techus Maximus
Join Date: Nov 2004
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PREAMPS: its RP, not MU, that's important!
It’s the RP-value, NOT the MU-value that’s Important!
To many people, the most important characteristic of a vacuum tube is its Amplification Factor (mu). Unfortunately, this is not true.
Of the vacuum tubes’ three operating characteristics – Dynamic Plate Resistance (rp), Transconductance (gm) and Amplification Factor (mu) - two are explicit values and one is an implicit value. The gm and rp values are explicit because they come directly from the tubes’ actual operating voltages and currents. The mu value, however, is an implicit value that is mathematically-derived from the product of the rp and gm values:
mu = gm * rp
Gm and rp are almost (but not exactly) inversely related. That is, as gm increases with increasing plate current, rp proportionately decreases. This inverse relationship causes mu to be virtually a constant that typically varies less than 10-15%.
The vacuum tube operating characteristics gm, rp and mu are “open-circuit” values, that is, they are calculated with NO external loads applied. In the real world, however, tubes are operated in circuits having both input and output loads. It is the effects of these “loads” (especially upon the plate) that make rp more important than mu. Here’s why...
When resistances are connected in parallel (symbol “||”), their resulting resistance is always LESS than the smallest resistance value. For example, consider two resistances, A = 10K ohm and B = 100K ohm, connected in parallel:
R = (A||B) = (A*B)/(A+B)
R = (10K*100K)/(10K + 100K) = 9.09K ~ 9.1K ohms
The same thing happens when a tube is operated with a plate load resistor (RP), its rp value is “loaded” in the same fashion as resistance-B “loaded” resistance-A in the example above. For instance, operating a 12AX7 (rp = 62.5K ohm) with a 100K ohm plate load resistor produces a circuit gain of only 61.5, not 100!
Why? Because, the ‘effective’ plate resistance (rp’) is no longer 62.5K ohms – it’s 38% LESS – only 38.5K ohms:
rp’ = rp||RP
rp’ = (62.5K*100K)/(62.5K + 100K) = 38.46K ~ 38.5K ohms
...where:
rp = Tube dynamic plate resistance, ohms (12AX7: 62.5K ohm)
RP = Circuit plate load resistance, ohms (typically 100K ohm)
Now, when tube gm is multiplied times rp’ a much LOWER amplification factor (mu’) value is obtained because of the LOWER ‘effective’ rp’ value:
mu’ = gm * rp’
...for 12AX7 gm = 0.0016 A/V, then:
mu’ = (0.0016 * 38.5K) = 61.5
And, when the output loading of the NEXT circuit (tube, tone stack, etc.) is included, the effective plate load becomes even LOWER. For example, assume the 12AX7 plate load resistor RP = 100K ohm feeds its signal into a tone stack having a resistance value of Ro = 500K ohms. Now, rp’ becomes three loads in parallel:
rp’ = (rp||RP||Ro)
rp’ = 1/( 1/rp + 1/RP + 1/Ro )
rp’ = 1/( 1/62.5K + 1/100K + 1/500K ) = 35714.3 ~ 35.7K
mu’ = (0.0016 * 35.7K) = 57.1
Thus, while the “open-circuit” gain of a 12AX7 is 100 with no load, its gain in a “real world” circuit (tube with loads) will be nearer to 60 (ie: 57-62), depending upon RP and Ro.
SUMMARY: “Mu is what the tube (alone) is capable of...Mu’ is what you actually get (in a circuit).”
Additionally, just as circuit AMPLIFICATION is dependent upon rp, so also is circuit FREQUENCY response or BANDPASS (BP), ie:
BP = 1/(2*PI*R*C)...in Hz
...where: R = rp||RP||Ro
So, whenever we “swap” tubes, it’s the rp value of the new tube interacting with the RP, Ro, R and C values of the existing circuit that causes the sometimes “subtle” changes in amplification (gain) and tone (frequency response) that we hear.
