Help! Anyone explain some transistor physics please?

[markeebee]

I was playing around with an Electra/Trotsky thing on a breadboard.  With a low input cap, I plugged in a BC550.  Nice and jangly.  Then substituted a 2N3904, and it was a bit more hairy.  Liked them both, so put them both on the breadboard at the same time so's I could could do a quick swap'n'listen.

While they were both in the board, connected in parallel, the sound was exactly the same as the BC550 on it's own.  When I pulled the base of the of the 550, I got the sound of the 3904, unsurprisingly.  I put a switch on the base of the 550 and it functions like a kinda boost......boosting the output when only one trannie (the 3904) is selected.

So, the scheme is like this:



My question is.....what dictates that the sound of them in parallel should be the same as the 550?  As the 3904 has more gain, I kinda thought that it would boss the parallel pair. I've searched the web but can't find an explanation anywhere - can somebody point me in the right direction?  I have a feeling that the answer might involve some mind-squishingly arcane theory and will go right over my head but I'd like to have a go at understanding.


EDIT
Whoops, got the trannie names mixed up in the last paragraph. Correct now.

[R.G.]

I suspect small differences in Vbe lets the lowest one dominate when they're in parallel.

[CynicalMan]

Someone may swing around and correct me, but my guess is that it has to do with the forward voltages of the base-emitter diodes. When you have two diodes in parallel, the one with the lower forward voltage will take nearly all of the current because it will limit the voltage to its own forward voltage, which means that the other diode won't conduct (much). So, if the forward voltage of the BC550's base-emitter diode is lower that that of the 2N3904, it will provide most of the current gain because there will be very little current into the 2N3904's base.

Edit: Aww, RG beat me. 

[PRR]

Measure the emitter current in both/all configurations. In this plan the (total) current defines gain. Not transistor type, though transistors with very different Beta will give somewhat different currents.

Or better: measure the collector voltage. This (by inference) tells if emitter current changes (collector and emitter currents are equal within 1%), but also if you find 2V or 8V at collector (with 9V supply) you know you are very close/past clipping strong signals, which is another kind of change.

I do suspect the large-area '550 has lower Vbe/Ie and thus "steals" current from the '3904.

[markeebee]

Oops, just reviewing this and I realised that I hadn't said thanks.

Thanks, y'all, much appreciated.