What's that second electrolytic doing in the Rangemaster?

[carboncomp]

OK, the first on the 9v is a filtering cap, but what's the second one attached from the emitter to ground doing bypassing the 3K9 resistor? and whats this configuration called, as I keep seeing it pop up in pedals like the Tone Bender?

[mac]

The 47uf cap is increasing voltage gain for all frequencies, because the 3db limit is given by f=2*pi/3k9/47uf=0.87Hz.
Voltage gain is A=10k/ri, where ri is the emiter internal resistance. I'm ignoring cap ESR.
IIRC the germ internal resistance is 0.026v/ie. Most RM work at 7v so ie=(9v-7v)/10k, and A is rouglhy 79.
The RM asymetrically distorts a bit when you hit the strings hard.

mac

[carboncomp]

Thank You Mac but I dint get any of that except "hit the strings hard" 

[samhay]

short answer -  it is a bypass capacitor, which gives it voltage gain without affecting the DC point and thus makes it a better amplifier while maintaing as much headroom as possible.
longer answer - http://www.youtube.com/watch?v=2UkuLxLu7AI

[duck_arse]

it is "the emitter bypass capacitor". and it BOOSTS. try yr rangemaster without it.

[carboncomp]

short answer -  it is a bypass capacitor, which gives it voltage gain without affecting the DC point and thus makes it a better amplifier while maintaining as much headroom as possible.
longer answer - http://www.youtube.com/watch?v=2UkuLxLu7AI

Thanks, got a little lost with that video so was wondering if you could explain in layman's terms how shunting the AC to ground would boost the signal, and why we don't see this configurations on all pedals (thinking the LPB1 or recovery stage of the BMP for example?)

[R.G.]

Thanks, got a little lost with that video so was wondering if you could explain in layman's terms how shunting the AC to ground would boost the signal, and why we don't see this configurations on all pedals (thinking the LPB1 or recovery stage of the BMP for example?)

The very common setup with a bipolar transistor of having a resistor from the power supply to the collector and another resistor to ground from the emitter is a form of gain reduction by negative feedback.

Let's pretend we have an NPN with a collector resistor Rc, an emitter resistor Re, and a base resistor Rb. Rb connects to a source of bias voltage which is magically correct somehow (that is, it's not important to this discussion).

The base is connected to the emitter of the transistor by a diode junction. This has the property that if the base-emitter voltage rises, the current through it rises *exponentially*.  Since the bipolar lets through the collector-emitter a current proportional to the base current, but larger by perhaps hundreds of times, a tiny rise in base current causes a big rise in emitter current. And a big rise in emitter current causes a big rise in emitter voltage, because the current in the emitter causes a voltage across the Re to ground.

So if base voltage goes up, base current goes up *exponentially* faster, emitter current goes up 100s of times more than that, emitter voltage rises, and this rise is in the direction to counteract the base voltage rise that created it. As base voltage goes up and down, the emitter voltage ... follows... it. This is where we get "emitter follower" by the way.

Essentially all of the base voltage is counteracted by the current amplification and the voltage caused across the emitter resistor.

What happens when we put a capacitor to ground across the emitter resistor Re? Capacitors don't do V=IR. They do I = C dv/dt, meaning that they let through ever more *AC* current as frequency (dv/dt) goes up. So the effective value of the emitter resistor for AC is dramatically reduced by putting a cap across it. It lets through much more AC current for the same change in base voltage because the emitter cannot follow the base as well - the capacitor has reduced the negative feedback that made the emitter follow.

The same current (minus the base current, which is usually very, very small) flows in the collector and emitter. With equal (mostly!) currents flowing in them, the voltages which appear across those two resistors have to be in the ratio of the resistors, right?  So if the collector resistor Rc is 10x the emitter resistor Re, then the voltage across Rc must be 10x as big, right?

That is true for AC and DC as long as there are no other complicating factors.

You can complicate the factors by placing a capacitor across the emitter resistor. Now the DC voltages across the resistors are the same as before, but if you amplify an AC signal, the capacitor across the emitter resistor "shorts out" the resistor for AC purposes as noted above. So the AC voltage on the emitter is a tiny fraction of what it was before you added the capacitor.

The AC current in the collector resistor Rc is now much bigger because the capacitor is preventing some of the negative feedback that limited the AC excursions of base current. So the AC gain goes up while the DC bias conditions remain the same.

So - the gain goes up because the cap to ground is removing some of the AC negative feedback that held the gain down.  And all pedals/circuits don't do this because it's not all that often that you just want all the gain you can get.

[PRR]

> the 3db limit is given by f=2*pi/3k9/47uf=0.87Hz

More like f=2*pi/27/47ufv = 125Hz

We must bypass the emitter impedance. Assuming 1mA, the naked emitter impedance is 27 ohms.

In parallel with the 3k9, but that's insignificant.(*)

In series with all the stuff at the base, divided by hFE. That can be a mess to calculate. And it can make a difference, since we like to make the base-stuff high impedance.

(*) In JFET and vacuum-tube work, the cathode internal impedance is usually "similar to" the cathode bias resistor. And we must assume that electrolytics decay, so we pick them large. So we can short-cut by figuring just the Rk, then rounding-up 3 or 4 times.

[mac]

> the 3db limit is given by f=2*pi/3k9/47uf=0.87Hz

More like f=2*pi/27/47ufv = 125Hz

It is F=1/(2*pi*R*C) 

mac

[PRR]

More like f=2*pi/27/47ufv = 125Hz

It is F=1/(2*pi*R*C) 

Well, yeah. I copy-pasted that part verbatim.

My point is that the node impedance is dominated by the Emitter Internal resistance, not the 3.9K resistor alone.

[R.G.]

It is, but the external emitter resistance is in series with the internal Shockley resistance, not parallel.

[PRR]

> the external emitter resistance is in series with the internal Shockley resistance, not parallel.

If the Base goes to a low impedance (it often does), they are effectively parallel.

Cowles says it well:

http://i.imgur.com/mbDZZvo.gif

[R.G.]

If the Base goes to a low impedance (it often does), they are effectively parallel.

Yep, thought about that. However, most times the base is held to a high impedance in amplifier practice. There are ways to get the base to a low impedance, but mostly the beginners won't be doing that.

[PRR]

> most times the base is held to a high impedance in amplifier practice.

But for the AC analysis, the base "looks into" some source impedance.

As a *general* design guide: low source impedance, high load impedance. (There's also matching and current-drive but less common today.)

But this thread is about a specific circuit.....

> in the Rangemaster

I *admit* that I overlooked this. A picture might have helped.



OC44 datasheet shows Stromverstärkung at 6V 1mA is 100 typ (45 to 225).

For *this* design, clearly the 0.005u at the input dominates. At max Beta it reflects to emitter as 0.005u*225= 1.125uFd. The emitter cap need only be much larger than 1.125uFd to not be the bottleneck. 5uFd would be groovy. 47uFd may have been the same price and already on the assembly bench (for battery bypass). And this is in line with Cowles' suggestion to over-size one of the two caps (base and emitter).

[Electric Warrior]

When Dallas returned to using OC44s after their OC71 period, they decreased the bypass cap value to 25µF. They kept using 50µF for the power filter for a while..

[mac]

I did a Spice model, including guitar circuitry, and 3 bypass caps, 5u, 22u and 47u.
Assuming SPICE is accurate, these are the freq distribution and schematic:





Let me know your suggestions to improve the model.

mac

[R.G.]

But this thread is about a specific circuit.....
For *this* design, clearly the 0.005u at the input dominates. At max Beta it reflects to emitter as 0.005u*225= 1.125uFd. The emitter cap need only be much larger than 1.125uFd to not be the bottleneck. 5uFd would be groovy. 47uFd may have been the same price and already on the assembly bench (for battery bypass). And this is in line with Cowles' suggestion to over-size one of the two caps (base and emitter).

Actually, I suspect that the idea was to put the gain falloff from the RC time constant of the emitter R/C well below audio - if any thought went into it at all. Once the cap at the emitter bypasses the emitter resistor sufficiently, then the capacitive divider effects are stable above that. In this case, 47uF gets you below 1Hz, which is overkill.

Cowles' suggestion is good; if you stagger too many rolloffs on top of one another, the asymptotic tails in the pass band give you a slushy, sloppy response at the breakpoint. Staggering is handy if you have room to do it.

[PRR]

> if any thought went into it at all.

It's interesting to think about that. It could be cut-paste from a stock transistor booster (fat emitter cap) with the input from a tube-amp (0.005u). Trying to recreate the "design" in my (well-worn) head is difficult, meaning that in 1970 it probably was not fully-planned in advance, because the computations are too tedious.

I started from 0.25mA (390K, 82K, small Vbe, and 3.9K), so 104 Ohms at the emitter. Nominal gain of 100, so 10K at the base. The bias-net modifies this to 8.7K. Neglecting source, this is a 3.6KHz bass-cut.

But 0.005uFd is 5,000pFd. We know that 300pFd cable capacitance against the pickup's inductance resonates well inside the audio band. So 5,000p (0.005u) will also ring in midrange. We have an R-L-C with more than one C and more than one R. With money, you turn to the network analyzer or the analog computer. And hope you keep track of scale-factors. And wonder what the output means. Or design with the parts-drawer and soldering-iron.

As Mac shows, now it isn't AS tedious, and the pictures are pretty.

> Assuming SPICE is accurate

Freq response, either it gets it utterly wrong (often an input mistake), or the answer is *exactly* the answer to the problem as-stated.

> suggestions to improve the model.

No objection. Myself I'd grab 10H inductance, 300pFd, but I don't think that will make much change. Being super fretful I might put 200K across the coil for core-loss, but you include the Volume control so that's minor.

Note that this is response for an infinitely TINY input. With peak gain of 19dB (about 9), this amplifier WILL clip any good strum.

[Electric Warrior]

Let me know your suggestions to improve the model.

mac

What happens when you increase the amp's input impedance to 1M? I think that's what those old AC30s and Marshall's had. Probably cuts a bit more low end..

[PRR]

> input impedance to 1M? ...Probably cuts a bit more low end..

No, "more" bass.

10nFd against 470 is about 35Hz bass-cut, 10nFd against 1Meg is 17Hz.

The 35Hz bass-cut is why Mac's plot trends down more in the left inch. 17Hz would be straighter further. Both are far below the guitar band, so no-difference. Especially since the lead guitar's strong 100hz-600Hz output, with HUGE top-boost, will overwhelm whatever happens in the bottom notes.