About input impedance and transistors

[Steben]

I felt an urge to evoke this topic. Impedance and (bipolar!) transistor circuits are connected to each other in such large amount I feel like needing some satisfaction to my quest.

Most know, some "bipolar" low input impedance circuits like the wah, fuzz face and rangemaster (and all their equivalents) tend to load a guitar and loose this when placed behind a buffer. Sometimes this is negatively called "tone sucking" (especially when in off-position however, "please give me a buffer") sometimes it is positively referred to as "smooth sound" (as in fuzzes, "go away with your buffers"). Especially the recent piggy and sziklai stuff is very interesting on this part.

It is my honest opinion that you can design your effect to whatever your desired impedance is AND this with easily as much components as the stock thing (without buffers). It is, however, very important to understand the importance of electronics. The following is what I learned over some time, but it is only 90% scientific (read: not 100% checked by experts). I'm pointing my arrows especially towards RG, who in my opinion can easily give feedback.

First the loading. Loading means the input impedance of your effect is substantially below that of your guitar. If you know your pickups have an impedance of a couple of kohms (humbuckers easily 10 kohms) and your volume pot 250 to 500 kohms. SO at the input of your circuit you get between a 5 to 500 kohms. Loading not only means volume loss in this case. The combination of guitar and circuits act as a sort of treble-off filter. The coils in the pick-ups have inductance (opposite of capacitance). The lower the ratio (input impedance)/guitar, the less treble! This is where bleed caps come in on the volume controls, they compensate treble loss.

Now the transistors. Were not speaking of buffer circuits here. The input impedance of a normal BJT amplifier (common - emtter) is related to its current gain and internal resistance (Since the bias resistors at the base are usually much higher than a couple of tens of kohms, they don't interfere) Please correct me if wrong but as a thumb rule I use the formula ("gain" x "25") / (1000 * transistor current). For a simple preamp (1mA) medium silicon tranny this 200x25/1000x0.001 = 5 kohms. For a germanium with gain 100 only 2,5 kohms (even more treble-off, we all know germ fuzzes sound warmer). This in the best case as much as a guitar at full volume. Instant loss of the half amplitude and a lot of treble.

How to raise the input impedance and lessen the treble loss without a buffer then? Very easy: use a darlington (miss piggy and mister sziklai ae entering). It's very common to have a combined gain of tens of thousands (gain "square"). Two of those silicons give this time 5 kohms x 200 = 1 Megaohm !!! Even two germs still give 250 kohms. Of course this sounds the same as with a buffer, but with even more gain (but don't forget a buffer results in more gain too). But that's ok, since there are million ways to reduce gain. For instance: guitar volume (who cares with such a massive input impedance now) or bleed resistors from emitter to base or to ground or piggy...

What with a Wah circuit for example with a darlington for Q1, but with a lot of Q-compensation of the resistor to ground? Instant tone spitting, instead of sucking?

[R.G.]

you can design your effect to whatever your desired impedance is AND this with easily as much components as the stock thing (without buffers). It is, however, very important to understand the importance of electronics.

Sometimes.  Sometimes, perhaps usually, it takes added parts. It often takes fewer parts to use a buffer. But sometimes you can fix the impedance with what you already have.

First the loading...

Yes, that's correct. Loading reduces signal in the ratio of the output impedance of the guitar and the input impedance of the effect. The output impedance of the guitar is frequency sensitive because the pickups are inductive and their impedance rises with frequency at 20db/decade. This causes the very-noticeable treble loss.

Loading means the input impedance of your effect is substantially below that of your guitar.

Actually, it means that the input impedance of your effect is less than *ten times* the impedance of your guitar. You need to get to at least ten times the guitar's impedance at a given frequency to have a negligible effect on the guitar sound.

The input impedance of a normal BJT amplifier (common - emtter) is related to its current gain and internal resistance (Since the bias resistors at the base are usually much higher than a couple of tens of kohms, they don't interfere)

It depends.
The input impedance of a CE amplifier is indeed the current gain times the unbypassed resistance on the emitter. This resistance is the sum of the internal Shockley resistance and the external unbypassed emitter resistor. The internal Shockley resistance is 25mV/emitter current, and the external resistance is whatever it is. Notice that the internal resistance is only substantial at quite low currents.

At 1ma, the Shockley resistance is 25 ohms, so a gain of 100 transistor has an input resistance of 2.5K unless you add external emitter resistance. At 100uA, it's 25K. That's still a huge load on a guitar pickup. You need to get to 100K to be acceptable if not good for loading, so that means you are forced to either high gain transistors or quite low currents. Both of theses have advantages and disadvantages.

How to raise the input impedance and lessen the treble loss without a buffer then? Very easy: use a darlington (miss piggy and mister sziklai ae entering). It's very common to have a combined gain of tens of thousands (gain "square"). Two of those silicons give this time 5 kohms x 200 = 1 Megaohm !!!

Again, maybe. That's the right way to think. The thing that causes a problem is the fact that the gain of a small signal transistor is a function of the emitter current. This is how multipliers work - they modulate the emitter current of a differential pair of transistors. The gain goes down at low currents.

If you have two 2N3904s with Hfe of 200 at 1ma, the high-current transistor at 1ma has a gain of 200. The driving transistor's emitter current is 1/200th of that, so it's gain is quite low - not unity, but much lower than 200. This is the real reason that single-package darlingtons are available. The input transistor is designed to have good gains at low currents to get high current gains. So with two ordinary transistors, the gain of a darlington connection may be thousands instead of hundreds of thousands.

Notice that using a discrete darlington is also inserting another transistor. You are only a few resistors away from the same cost as a buffer.

The real reason a separate buffer transistor gives high input impedance is that it must have a large unbypassed emitter resistor if it's a CC/emitter follower. The emitter resistor is usually on the order of 10K, and that with the hfe of say 200 gives an input impedance at the base of 2megohm.

If you have an emitter resistor composed of two resistors, one bypassed and one not bypassed, only the unbypassed part contributes to the input impedance.

The gain of the CE circuit is effectively the collector resistor divided by the unbypassed emitter resistor, which includes the internal resistor. So you can get high gain with no emitter resistor (that is, only the internal one) but you also get low input impedance. You can get higher input impedance with an unbypassed emitter resistor, but it lowers your gain. The two objectives conflict. If you can't get what you want in both gain and input impedance, your best option is to use another stage (transistor). It's all tradeoffs.

Notice that piggybacking is for reducing transistor gain. It's a special purpose thing for circuits like the Fuzz Face, not too generally useful. That's probably the only reason I figured it out fifty years after other transistor connections were already thought up.   I suspect it was thought up and tried many times, but the "inventor" thought - "Yuck! lower gain! Who'd want that? I'm not going to tell anybody I came up with that silly idea."

[bioroids]

Very interesting read.

Also you can raise a stages input impedance by the means of "bootstraping". But I think that only helps on the respect of the biasing resistors, having no effect on the transistors part of the input impedance.

Luck

Miguel

[Joe]

A bit mathy, but covers a lot:
http://portal.jce.ac.il/courses/Electronics_Engineering/analog_electronics2/supplement/notes_floyd/ch6.pdf