Dumb question regarding calculating the gain of a preamp stage

[theRagman]

So, you commonly see the formula for the gain of a preamp stage given as: mu*(Rload/(Rload+Ra)), where Ra is the plate resistance of the tube. That's all well and good, except that nowhere in this formula do we see plate voltage, which is counterintuitive to me.

Can somebody help me understand what is going on here? Is mu quietly varying with plate voltage, even though we call it a constant? Something else?

As a practical matter for engineering purposes, is it better not to use this formula at all and rely on the load line?

[pdf64]

u varies with anode current, so for a given typical 1k5 100k common cathode stage, the gain will be a bit higher with the supply voltage at 350V than at 250V.

[theRagman]

u varies with anode current, so for a given typical 1k5 100k common cathode stage, the gain will be a bit higher with the supply voltage at 350V than at 250V.

Thank you. That demystifies that!

[tubeswell]

Or to put it another way:


'mu' = amplification factor = change in plate voltage/change in grid voltage (for a fixed plate current).


whereas:


A = Voltage gain (i.e., a different thing to 'mu') = Vac out/Vac in (which will typically be less than the 'mu' of a tube), hence:



A(bypassed) = -mu x [Rp/(Rp + rp)]

A(unbypassed) = -mu x [Rp/(Rp + rp + Rk(mu + 1))

where:
A = voltage gain
mu = amplification factor (the minus sign is a mere technically indicating inverting amplification, but is important when considering the effect of feedback from an unbypassed cathode resistance)
Rp = the plate load resistor
rp = internal plate resistance
Rk = cathode resistor


'A' is directly related to load (because Ohms Law).

DC load resistance (no signal) is different to AC impedance (Rac). 'Rac' takes into account all current sinks (including loads on the other side of capacitors as well as current feedback e.g., from unbypassed cathode resistance etc), hence:

Rac = Rp || Rload = Rp.Rload/(Rp + Rload)

[HotBluePlates]

So, you commonly see the formula for the gain of a preamp stage given as: mu*(Rload/(Rload+Ra)), where Ra is the plate resistance of the tube. That's all well and good, except that nowhere in this formula do we see plate voltage ...

... Is mu quietly varying with plate voltage, even though we call it a constant? ...

As a practical matter for engineering purposes, is it better not to use this formula at all and rely on the load line?

You can have "The Right Answer."  Or you can have "an answer right now."

Experience shows that most audio does not require "high precision" especially when you need 10x the signal to sound "twice as loud."  So the Designer might use a number of simplifying assumptions when they are familiar with the the limiting cases where those assumptions no longer apply.

You picked a formula that provides a quick answer by assuming Mu is constant (pretty close to true), and that assumes the important factor is the ratio of Plate Load Resistor to Internal Plate Resistance.

Below is a graph from a 12AX7 data sheet that shows how Mu, Transconductance (Gm) and Internal Plate Resistance ("Rp" on the graph, "Ra" in your version of the formula, "rp" in Tubeswell's version of the formula).  Mu = Gm / Rp, and because the latter terms change by nearly the same proportion in opposite directions, Mu stays mostly the same.  However, we see Rp change a lot with plate current, and that has the great bearing on in-circuit Gain.




But take a typical 12AX7 gain stage with a 100kΩ plate load, and triple plate current from 0.5mA to 1.5mA.  How does Gain change (assuming no other limiting factors, as we might discover with a load-line)?

    Rp at 0.5mA --> 80kΩ
    Rp at 1.5mA --> 54kΩ

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  That a huge, very-obvious difference, and Designers chased adding gain stages rather than splitting hairs too finely with some of these other issues.

[theRagman]

So, you commonly see the formula for the gain of a preamp stage given as: mu*(Rload/(Rload+Ra)), where Ra is the plate resistance of the tube. That's all well and good, except that nowhere in this formula do we see plate voltage ...

... Is mu quietly varying with plate voltage, even though we call it a constant? ...

As a practical matter for engineering purposes, is it better not to use this formula at all and rely on the load line?

...But take a typical 12AX7 gain stage with a 100kΩ plate load, and triple plate current from 0.5mA to 1.5mA.  How does Gain change (assuming no other limiting factors, as we might discover with a load-line)?

    Rp at 0.5mA --> 80kΩ
    Rp at 1.5mA --> 54kΩ

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  That a huge, very-obvious difference, and Designers chased adding gain stages rather than splitting hairs too finely with some of these other issues.

Thank you, that's an incredibly helpful answer and example. I'm surprised that Rp is the value that's moving around, but the graph sure makes that clear.

One day, decibel math will become more intuitive for me, but right now my first impulse is, "OMG! It's 20% more gain! Aaaaugh, my headroom calculations!" even though, as you've shown, it's barely over the threshold of being audible. Ah, well. That's what practice is for.

[shooter]

One day, decibel math will become more intuitive for me

out in the backwoods intuiting


1.  It's log math so lets cut down a tree,
2. start walking up the side of the fresh-cut log, caution, it's pretty tough til ya cross over that there curve, then it's easy goin


 

[RoadShow]

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  That a huge, very-obvious difference, and Designers chased adding gain stages rather than splitting hairs too finely with some of these other issues.

I guess there's something I'm missing.
If I do 20 log (55x55) = 20 log (3025) I get 69.6 dB
If I do 20 log (55) = I get 34.8 dB

Am I looking at this wrong?

[tubeswell]

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  That a huge, very-obvious difference, and Designers chased adding gain stages rather than splitting hairs too finely with some of these other issues.

I guess there's something I'm missing.
If I do 20 log (55x55) = 20 log (3025) I get 69.6 dB
If I do 20 log (55) = I get 34.8 dB

Am I looking at this wrong?

34.8 '+35' = 69ish

[HotBluePlates]

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  ...

I guess there's something I'm missing.
If I do 20 log (55x55) = 20 log (3025) I get 69.6 dB
If I do 20 log (55) = I get 34.8 dB

Am I looking at this wrong?

A "logarithm" is a ratio, with a Value in relation to a Reference.

Let's consider the 66.2x Gain afforded by the 1.5mA gain stage with the 100kΩ plate load.
If we look to get Voltage Gain as a decibel logarithm for this stage alone, we notice the AC Plate Volts (our "Value") are 55.6x bigger than the AC Grid Volts (our "Reference").  So Voltage Gain = 20 log (Value / Reference) = 20 log (55.6 / 1) = 20 x 1.745 = ~34.9 decibels increase.

Now let's switch to the 55.6x Gain afforded by the 0.5mA gain stage with the 100kΩ plate load.
If we look to get Voltage Gain as a decibel logarithm for this stage alone, the AC Plate Volts are 66.2x bigger than the AC Grid Volts.
Voltage Gain = 20 log (Value / Reference) = 20 log (66.2 / 1) = 20 x 1.821 = ~36.42 decibels increase.

    If we have already calculated "decibels of Voltage Gain" then we can subtract to find the difference in decibels.
    "1.5mA Gain" (in decibels) - "0.5mA Gain" (in decibels) = 36.42 dB - 34.9 dB = 1.52 dB increase from 0.5mA to 1.5mA

But let's say we have not calculated individual "decibels of Voltage Gain" for each case in advance.  That's okay, because we can use one stage's amplification (55.6x, no units) as the "Reference" and the other stage's gain (66.2x) as the "Value."

    20 log (Value / Reference) = 20 log (66.2 / 55.6) = 20 log 1.19065 = 20 x 0.07578 = 1.5157 dB = ~1.52dB

    We see we can calculate a "difference of decibels" by calculating individual gains in dB and subtracting.
    We can also calculate a "difference of decibels" by using one as the "Reference" and thereby skip an extra step.

If I do 20 log (55x55) = 20 log (3025) I get 69.6 dB
If I do 20 log (55) = I get 34.8 dB

Yes, now notice as Tubeswell says:

   20 log (55) = 34.8 dB ----> Gain of 1 stage, referred to stage's grid.

   20 log (55x55) = 20 log (3025) = 69.6 dB ----> Gain of Stage 2, referred to 1st stage's grid.
   Compares Stage 2's Plate to Stage 1's Grid
   Or "Total Gain after 2 stages"

But you can also do "Log Math" and add decibels-gain afforded by each stage:
   Stage 1 Gain:  34.8 dB
   Stage 2 Gain:  34.8 dB

   Gain after 2 stages:  34.8 dB + 34.8 dB = 69.6 dB

[RoadShow]

    Gain at 0.5mA --> 100 x [100kΩ / (100kΩ + 80kΩ)] = 55.6 times
    Gain at 1.5mA --> 102 x [100kΩ / (100kΩ + 54kΩ)] = 66.2 times

    Additional Amplification: 20 log (66.2 / 55.6) = 20 log (1.19) = 20 x 0.07578 = 1.52 dB

+10dB is "double-loud" so while this ~1.5dB bump is audible, it's not "Wow!" (at least, not without a bunch of other circuit-changes/gains compounding the result).  Meanwhile, cascade 2 of the 0.5mA gain stages, and Gain goes from "~55x" to "~3000x" or about +35dB.  ...

I guess there's something I'm missing.
If I do 20 log (55x55) = 20 log (3025) I get 69.6 dB
If I do 20 log (55) = I get 34.8 dB

Am I looking at this wrong?

A "logarithm" is a ratio, with a Value in relation to a Reference.

Let's consider the 66.2x Gain afforded by the 1.5mA gain stage with the 100kΩ plate load.
If we look to get Voltage Gain as a decibel logarithm for this stage alone, we notice the AC Plate Volts (our "Value") are 55.6x bigger than the AC Grid Volts (our "Reference").  So Voltage Gain = 20 log (Value / Reference) = 20 log (55.6 / 1) = 20 x 1.745 = ~34.9 decibels increase.

Thanks for the great explanation!!