I'm thinking that with the drive pot @ -0-, the next stage grid leak resistance is only 120K, the resistor under the drive pot. Anything under 220K into a 12a_7 tube is really low and will probably darken tone.
Remember that resistance
by itself has no frequency-dependent effect. A purely-resistive circuit creates the same voltage drops at any given point for all applied frequencies.
It's when you add C or L that the circuit becomes frequency-dependent. In this case, it's C added which ultimately has the effect of a small capacitance to ground. That forms a low-pass (or hi-cut) filter. Making any series-R large moves the -3dB point of the filter lower and lower, cutting more highs.
Compare to the 68kΩ input resistors at a 1st gain stage. 0Ω causes no high-frequency roll-off, 34kΩ causes some roll-off, 68kΩ moves the -3dB point lower (which is equivalent to more roll-off at the 34kΩ -3dB frequency), 1MΩ moves the -3dB point very much lower.
I think you're thinking of a series-C and shunt-R, which forms a high-pass (low-cut) filter, where smaller R moves the -3dB point higher thereby cutting more lows.
I'm thinking that with the drive pot @ -0-, the next stage grid leak resistance is only 120K, the resistor under the drive pot. Anything under 220K into a 12a_7 tube is really low and will probably darken tone.
Oh yeah... Instead of passing the 1st gain stage output through a volume control and then a voltage divider which cuts output by half, why not just remove the cathode bypass cap? That will also cut stage gain by about half with the cathode resistor you have shown.
That will leave gain roughly the same.
Right. I wanted to leave roughly the same net-effect as the circuit he's using, but with fewer parts. Arguably, it might result in marginally-less noise, because there's fewer resistances in the circuit. Really, I just thought it more effective if he wants a net-gain of half (and max volume) of what is at the V1's plate to simply amplify only half as much at V1.
But that nice little 2.2uF bypass cap is "boosting" highs at the 1st stage; so it's removal might darken the amp.
It's best to see the cathode bypass cap as reducing local negative feedback to increase gain. At some frequency, the cap becomes a less-effective bypass and gain begins to revert to the with-feedback level. The cap value
by itself can't tell you what that frequency will be; you have to plug it into the formula along with R. Where R and X
C are equal, there is -3dB in response.
"R" for the equation is really R
k in parallel with the resistance seen looking into the cathode [which is r
p/(mu+1)], but it's usually close enough to use R
k, and know the frequency will be a little higher than calculated. It's even better to check by ear after you have a guess of a suitable bypass cap value.
F
-3dB = 1/(2*pi*R*C) = 1/(2*pi*1.5kΩ*2.2uF) = ~48Hz
You can estimate there will be roughly no roll-off an octave higher than the -3dB point, which means almost no cut at a guitar's low E. So removing the cap should cut gain by about half while leaving tonal balance substantially the same.
In any event, it's a quick test to remove 3 components and see if the result is judged worthwhile. If no-good, it's easy to put them back.
Topic: Need help with 2nd gain stage calculation (Read 3210 times)