Can anyone calculate the gain of big muff stages?

[brett]

Hi
this has puzzled me for a while, and I need to solve it before I get to pulling my BMP apart for some mods.

The gain stages of the BMP have a series resistor into the transistor base (e.g. R7, usually 8.2k, in the schematic at Tonepad), and a feedback resistor (R9, usually 470k, but also 390k).  There is also a feedback cap, but for simplicity let's ignore it, at least for now (and the diodes, too).  Other than that, it's a common emitter stage, with a medium sized drain resistor (R10, 15k, sometimes 12k) and an emitter resistor (R11, 100 ohms, but as high as 390 ohms in the second stage of the "Russian Green" BMP).

As a regular common source stage, the gain would be about R10/R11 = 15k/100 = 150. But the signals through the input resistor and feedback resistor meet and partially cancel at the base coz they're out of phase (ok, that's dead obvious  ).

I'm wondering whether, maybe, because the same input impedance (to the BJT + "base grounding" resistor) is seen by the feedback signal (via the feedback resistor) and the input signal (via the input resistor), whether the gain stabilises at the ratio of the feedback to input resistances (ie 470k/8.2k = 57), so long as this is less than the upper limit set by Rd/Re (150).  (A bit like an inverting opamp). But this is way too high for the three stages in the BMP (50 x 50 x 50 = a ridiculous number).

But then...  I thought that the total output impedance between the output of the previous stage and the "mixing point" might be more relevant. That would be approximately the sum of the input resistor and the drain resistor, or 470k/(15k+8.2k) = 20.  Yes, three stages of 20 would give a total for three stages of 8000, which would be out of control without the sustain pot and filtering reduce the gain to a workable level.

Does this sound right? Can anyone help me out?
thanks 

[R.G.]

Your analysis is on the right path.

Yes, you need to account for the collector resistor in figuring the equivalent series input resistance of the stage. You also need to account for the effect of the next stage's input impedance loading on the gain of each stage.

My network equation is rusty, so I did the lazy thing - I dumped the circuit into the simulator. The stage shows in-circuit loaded gains of 27db, a bit over 20.

[brett]

Thanks heaps RG.
You're worth your weight in gold.
No wonder my BMP gets so wild with the gain turned up!

[mac]

As 470K >> 15k, can the 470k negative feedback fx be ignored, I mean the cancelling effect? What happens with the 27db gain if the 470k is varied from say 220k to 680k?
What's happens if hfe is lowered/increased?

This is a good exercise... correct if I'm wrong, I never did AC analysis.

• The input resistance of the BJT is hfe*emiter resistor, 500*100 ohm=50k, for a 2n5088, bc549C. But there is a 8.2k in series, 58.2k. If this is true the effect of the 0.1uf can be neglected since all the guitar freqs are above the corner freq. ¿?
• Also there are two resistors to gnd, a 100k, and a 470k+15k (+ P.S resistance which can be ignored). They are in parallel, so it's like a single 82k to gnd. ¿?
• If all above is true, then the signal leaving the prior BJT sees two similar paths, one to gnd 82k, the other 58.2k to the next BJT base. Almost half the signal is shunted to gnd. ¿?

mac

[R.G.]

As 470K >> 15k, can the 470k negative feedback fx be ignored, I mean the cancelling effect?

No, it can't. The 470K's effect is on the base, both in bias and signal feedback. The big difference in the two do let you ignore the 15K in comparison to the 470K for this, though.

What happens with the 27db gain if the 470k is varied from say 220k to 680k?

The closed loop gain changes a bit, maybe a few db. This has dramatic changes in the bias point, though.

What's happens if hfe is lowered/increased?

Very little. With hfe over about 100, almost nothing. One reason this kind of circuit is used is to make it insensitive to hfe.

• The input resistance of the BJT is hfe*emiter resistor, 500*100 ohm=50k, for a 2n5088, bc549C. But there is a 8.2k in series, 58.2k. If this is true the effect of the 0.1uf can be neglected since all the guitar freqs are above the corner freq. ¿?

Not exactly. The feedback from the collector makes the junction of the 470K, 8.2K and base be kind of like an opamp inverting input. This lowers the apparent input impedance from what the transistor itself looks like. If the transistor's gain was high enough, this point would appear to be grounded, so the input signal would only see the 8.2K and ground. The 8.2K, transistor base and 470K feedback resistor are similar to an inverting opamp stage with 8.2K input resistor, 470K feedback resistor, and the - input of the opamp. This is not strictly true because the transistor's voltage gain is not high enough, but it's a good way to look at the circuit.

• Also there are two resistors to gnd, a 100k, and a 470k+15k (+ P.S resistance which can be ignored). They are in parallel, so it's like a single 82k to gnd. ¿?

The 100K goes to ground. The 470K goes to a signal generator, the collector of the transistor. The impedance of the node sees them as in parallel, but the signal effect at the node sees the signal voltage on the collector coming through 485K.

• If all above is true, then the signal leaving the prior BJT sees two similar paths, one to gnd 82k, the other 58.2k to the next BJT base. Almost half the signal is shunted to gnd. ¿?

No. See the discussion above.

[mac]

Doh!
Thanks RG, it seems that I underestimated the effect of the 470k.
Any good basic AC network analysis on the web?

mac

[Gus]

Don't forget about re of the stage when the emitter is a lower value you need to calculate re.  25/Ic(ma) ohms for 15K/ (100 + re).

The 3th stage 8.2 k input is added too from the output resistance of the stage before it.  We know it is less than 15K because of feedback.  So 470k/ (8.2k+source resistance)