MK3's Germanium Diode

[Megatron]

Is the MK3 germanium diode there for temperature stabilization only? Does it serve another purpose?

[jblack547]

That is a bit odd base bias. No matter what the pot is set to the base voltage will always be the forward voltage??? What circuit is this?

[jblack547]

What is the designator on the diode? You've made a half wave rectifier out of the base of the transistor and the diode itself and then send it al to ground. Interesting.

[jblack547]

The pot on the base must be controlling an eq?

[jblack547]

Q1, Q2 frequency compensated darlington pair. This gives lots of gain and stability for low cost.

[Megatron]

That is a bit odd base bias. No matter what the pot is set to the base voltage will always be the forward voltage??? What circuit is this?

I am not sure where you see that regarding the forward voltage. The diode is reverse biased. This is a MK3 Tonebender.

What is the designator on the diode? You've made a half wave rectifier out of the base of the transistor and the diode itself and then send it al to ground. Interesting.

It is a germanium diode.

The pot on the base must be controlling an eq?

That would be the Fuzz pot. The tone pot pans between the 0.1uF and 2200pF caps.

Q1, Q2 frequency compensated darlington pair. This gives lots of gain and stability for low cost.

Yes - all of these types (MK3/4, Roto, Park, Buzzaround) use a darlington pair.

[psychedelicfish]

My guess would be that the diode's reverse leakage is biasing the transistor. Its forward voltage probably also clips one half of the signal coming into Q3

[Thecomedian]

at first I thought it was a widlar current mirror diode 

[mac]

Is the MK3 germanium diode there for temperature stabilization only? Does it serve another purpose?

The reverse diode adds a bit of temp stabilization because the diode leakage sucks some excess transistor leakage to ground sending the collector up, helping to bias at the same time.
The PNP base current is going into the base, instead of out. As there is nothing connecting the base to Vcc, the reverse base current has to come from the collector. How much current? (-1)*diode leakage*hFE.
The voltage across the 18k will be (transistor leakage - diode leakage*hFE)*18k
You can estimate the diode leakage by connecting a 1k resistor and the reverse diode in series from Vcc to Gnd, reading the voltage across the resistor, and dividing by 1k.

Think of a resistor from B to Gnd, like in the MKII first stage.

mac

[jblack547]

According to "Horowitz and Hill: The Art of Electronics" a diode configuration like this one is to "prevent base-emitter reverse voltage breakdown." This makes sense since there is a darlington pair driving this stage. The signal is already maxed out before it gets to the diode/transistor.

[sevenisthenumber]

Can I use another type diode? One thats easier to find?

[midwayfair]

Can I use another type diode? One thats easier to find?

It has to be germanium, but the actual part number doesn't matter.

[sevenisthenumber]

Can I use another type diode? One thats easier to find?

It has to be germanium, but the actual part number doesn't matter.

thanks, I would love to understand why? Any one care to explain?

[Germanium_Boy]

Really, really interesting... I love this Tone Bender, less fuzzy than his big brothers, but very good sounding...

[midwayfair]

Can I use another type diode? One thats easier to find?

It has to be germanium, but the actual part number doesn't matter.

thanks, I would love to understand why? Any one care to explain?

I thought this was explained above ... germanium semiconductors have leakage. Silicon doesn't.

[sevenisthenumber]

Can I use another type diode? One thats easier to find?

It has to be germanium, but the actual part number doesn't matter.

thanks, I would love to understand why? Any one care to explain?

I thought this was explained above ... germanium semiconductors have leakage. Silicon doesn't.

So the leakage is necessary here? Sorry, Im just trying to get a good understanding.

[Digital Larry]

According to "Horowitz and Hill: The Art of Electronics" a diode configuration like this one is to "prevent base-emitter reverse voltage breakdown." This makes sense since there is a darlington pair driving this stage. The signal is already maxed out before it gets to the diode/transistor.

Even though I'm not certain, I vote for this answer.  It helps keep the transistor from blowing up given the large signal coming into the base.  All the other business about temperature stabilization and, err ... leakage compensation seems a bit overblown for a fuzz circuit.   I could be wrong.  Just playing armchair engineer over here, nothing to see, move along....

[jonnyeye]

All diodes leak when reverse biased.  All of them.  This includes the collector-base diode (present in every bipolar transistor) in every common-emitter stage ever made. 

Germanium diodes leak significantly.  I have measured the leakage of the collector-base junction of a few transistors in my collection (at basement temperature, with 9V reverse bias) - late production Russian germanium PNP leak from 0.25 to 0.65uA. Best behaved Tesla NPN leaks about 2-3uA; badly behaved examples (like those we'd more likely choose for Q3 in a Mk III!) leak 5-12uA. 

Plain Ge diodes leak too; some generic 1N34A I checked leak 4-6uA.  Oh, and this leakage is temperature and reverse voltage dependent - as either temperature or reverse voltage increases, leakage increases (I don't know the physics behind this, but I did see it in my testing). 

Q3 in a Tonebender Mark III is leakage biased; it depends on the leakage of the C-B diode to turn it on at all. Whatever current flows through the C-B junction flows back into the transistor and turns it on.  If the C-B leakage (our bias current!) is 5uA and the gain of the transistor is 80 then the transistor would try to conduct 400uA, which across our 18k collector resistor is 7.2V (i.e. too much). We could alter the quiescent bias point by using a simple resistor from base to ground, which would steal some of the leakage current, but since the leakage is temperature dependent, the biasing of the transistor would move around as the ambient temperature changed.  To mitigate this, instead the designer used another reverse-biased diode from base to ground, the idea being that the leakages of the two diodes will track each other with temperature and keep the biasing stable.  What other diodes will work?  Schottky diodes also leak, but perhaps not the right amount (power devices like the 1N5817 leak a lot, maybe 10uA, where as BAT41s leak more like 0.01uA, not enough), but you can try them - just build it and then heat it up and watch to see if the biasing changes.

If you want to test your own transistors or diodes at home, try this: (for PNP transistors flip the power supply polarity)


(And what about silicon? Well, modern silicon transistors are designed to minimize this leakage to the level of irrelevance.)

[sevenisthenumber]

Thanks bro, thats a huge help!

[PRR]

Q3 is not "properly biased", with or without D1. (Thermal current is not likely to bring Q3 into a good amplifying zone.)

As such, small signals into Q3 will not come out.

But are the signals here small?

Q1 Q2 have a gain of about 40 and a maximum output like 1.2V peak positive, 6V peak negative.

30mV input to Q1 will deliver 1.2V out of Q2. BIG signal.

So for most guitar playing, "there are no small signals". Even weak strums will slam Q2. And Q2's output is SO hot that Q3 is slammed.

Both ways.

Without D1: Q2 output and Q3 input tries to go positive 1.2V.... only on the initial transient. But because Q3 Base can only rise 0.3V, the excess 0.9V goes to charge the 0.1uFd cap. Eventually the 0.1u cap charges-down so much that Q3 Base is held "off" ALL the time. Sound stops. (This is identical to Grid Blocking in a tube amp.)

Put D1 back in. Now positive swings charge the 0.1u cap one way, but negative swings charge the 0.1u cap the other way. On average, Q3 Base is centered near zero, though slammed +0.3V and -0.3V to cause 4Vpk or 2Vpk square-waves toward the output (then reduced by tone and volume networks).

The "ideal" part for D1 for "symmetrical" clipping might be the Base-Emitter junction of a transistor of the same transistor-type as Q3, but backward (B to E and E to B). This could possibly be a no-gain reject. ("13 Transistors!!" radios often used dud transistors as diodes.)

If dud transistors cost money and similar diodes are cheaper, a diode is fine (and often seen).

If symmetry is not needed/wanted, a different diode (or part-transistor) can be tried. I sure would try a little Silicon here. That might tickle the ear, or just cause excess compression..... hard to say, easy to try.

And in fact, this plan can't do "symmetrical" because Q2 can pull-down better than this stage can pull-up. Somewhat averaged by the large (and user-variable!) resistor in series with the 0.1u cap. As some asymmetry is "spice" for fuzz, this is part of the "flavor". As in any cooking, you may want to try different spicing.