Can someone explain the transistor piggyback theory?

[MattAnonymous]

I read something recently about piggybacking a Si transistor to get a more germanium sound.  Can you supply a schematic, an explanation, and/or how/why this works?

Thanks,
Matt

[Gringo]

Make a search for "miss piggy", there are a couple of (long) topics about this, and check brett's site for schems:

http://www.members.optusnet.com.au/~jethro.dog/gallery.html

Hope it helps.

[R.G.]

I originally posted the idea that you could lower the current gain of a silicon bipolar transistor by putting a second transistor in parallel with it, but with a disconnected collector.

The idea is that the current into the base-emitter junction of a bipolar is a single valued-function of its voltage - that is, one and only one current value results from each value of voltage, although the function may not be linear.

By putting another identical junction in parallel with the "working" one, the parallel (I use the term "dummy") junction will have the same voltage as the working one, and should in consequence, eat as much current. That, I reasoned, should leave only half the base current for the working junction, and so the overall transistor would have equivalently half the current gain.

That turned out to be correct in concept, wrong in detail. I forgot the internal rbe of the transistor junction. What a bipolar transistor base-emitter junction does that a diode junction does not is to have an equivalent nonlinear resistor arising from the transistor action. This internal resistor causes the working junction to have a higher voltage at the same current as the dummy junction; looked at the other way, the dummy junction eats more current than the working one at the same voltage.

That forces the dummy junction to eat more than half the base current, so the current gain is lowered by more than half - sometimes a lot more than half.

Brett's work showed that putting an external resistor from the dummy junction to the emitter of the working junction could make the dummy junction eat a variable amount of the input base current, effectively letting you tune the current gain of a piggybacked transistor.

And that's how that works.

[Jason Stout]

IMO, That is one for the archives!

[Siemen]

Hi, I'm new to this forum (and this is my first message).
I realise that the last post dates from 2004, but I just wanted to clear something up (since this technique is much spoken of on this forum).

How is it possible that this "dummy" transistor eats up any current at all if the collector is left floating?!
I'm an EE student (in Belgium) and I'm looking at these characteristics and I just can't figure out how there can be any current at all...
(http://www.pling.org.uk/cs/dadimg/bjtcharacteristics.gif)

The emittor current that is controlled by the base current, should be originating from the collector, but it's left open. Is IE coming from IB then?
Also, what's the value of VCE (on the characteristic abovce) if C is left floating? And how can I deduce IE if C is left open?

To make a long story short: what happens to the equations when C is left floating? Is it: IE=IB?

Greets from a stompbox noobie (but one that's hungry for more)

[Siemen]

Also: couldn't you use a diode instead of a transistor?

[anchovie]

Also: couldn't you use a diode instead of a transistor?

What a bipolar transistor base-emitter junction does that a diode junction does not is to have an equivalent nonlinear resistor arising from the transistor action.

[amptramp]

How is it possible that this "dummy" transistor eats up any current at all if the collector is left floating?!
I'm an EE student (in Belgium) and I'm looking at these characteristics and I just can't figure out how there can be any current at all...
(http://www.pling.org.uk/cs/dadimg/bjtcharacteristics.gif)

The emittor current that is controlled by the base current, should be originating from the collector, but it's left open. Is IE coming from IB then?
Also, what's the value of VCE (on the characteristic abovce) if C is left floating? And how can I deduce IE if C is left open?

To make a long story short: what happens to the equations when C is left floating? Is it: IE=IB?

Greets from a stompbox noobie (but one that's hungry for more)

With a disconnected collector, Ie = Ib.  The transistor acts as a diode and if you use the same type transistor as the one that is connected, you get the same junction geometry and doping.  If the collector is open, the voltage would be that of its neighbour in the junction, Vb.  If you only have two connections, the current into one is the current out of the other.  There is no place for electrons to be "consumed".

[CynicalMan]

Heh. I just had this idea a few days ago and was about to try it out this weekend. Guess I should learn to Google first. Still, good to know that it works.

[PRR]

> How is it possible that this "dummy" transistor eats up any current at all if the collector is left floating?!

A frequent bias technique is to force a semi-fixed current Ib toward the Base. The Collector current is then, obviously, hFE*Ib. If Ib is 1uA and hFe is 100, Ic is 100uA and Ie is 101uA. "A" below.

If this 100:1 ratio is inconvenient, we could generally re-design the bias or load. However some people think hFE is a Magic Parameter, and wish to change it.

R.G.'s hasty thought was that an additional transistor BE junction bridged across the original BE junction would divert half the current and cut gain in half. On further thought, he realized that Vbe (same in both transistors) is influenced by C-E current; or if CE current is zero, by Ib alone. So the bridged transistor steals almost all the intended Base current. "Gimmicked hFE" is very-very low. "B" below. (The 1000Meg passes nearly-no current; just to stop SPICE complaining about unused legs.)

There is another connection: collector to base. This is a very common well-known connection. Ideally the "hFE" is Unity. (A wee bit higher since the "in use" transistor's collector voltage is not-small compared to Early Voltage.) "C" below.



A third connection MAY be the "half hFE" solution: Unused collector to V+.

[Brymus]

Heh. I just had this idea a few days ago and was about to try it out this weekend. Guess I should learn to Google first. Still, good to know that it works.

Did it work for you ?
I tried Miss Piggy version 3 and it wouldnt bias with 2N5088s,I havent tried replacing the 10K's between emitters with 25K pots yet (version 5)

[CynicalMan]

I only got as far as simming it. It seemed tricky to get just right, though.

[EATyourGuitar]

there are lots of devi circuits that have transistors piggybacked with all 3 legs of both transistors used. she does also have some circuits where the collector is disconnected but its not a piggyback, just a diode junction.

[brett]

Hi

I only got as far as simming it. It seemed tricky to get just right, though.

Because the files describing BJTs in SPICE are quite simple, and oriented the usual things done with BJTs, they don't simulate piggybacks well.
Empirically, we know that for most stages a piggyback without the inter-emitter resistor (Ree) give very low hFE (2 to 10 ?), while an Ree of 47k usually has a small effect. Values around 2 to 10k give the 80 to 50% reduction in hFE desired in many vintage circuits (e.g. for using 2N3906s or 2N5088s for Q1 in a fuzzface).
cheers

[pinkjimiphoton]

sorry to bump a zombie,
but what about this?
i stumbled across it working on a weird old "parallel" fuzz from an ovation guitar preamp, working out how to switch between two different transistors.

in my case, both are ge... but it seems to be the OPPOSITE of the effect rg describes here.

https://www.diystompboxes.com/smfforum/index.php?topic=121953.0