AW: Re: AW: Transistor matching

[Haible_Juergen#Tel2743]

> > Just a stupid question about Moog's Transistor matching circuit:
> > Do we really need these current sources ??
> > I had to select a few transistors recently, and I just used a
> > bipolar 15V supply and two 150k resistors. Ok, 150k from
> > 15V - Vbe is not an ideal current source, but two identical
> > transistors should give the same Vbe at the same 150k
> > resistors as well, no? Or did I miss something?
>
> The point is that the Vbe is directly connected to the Ic. However, there 
is
> also some bulk resistances which has to be compensated for etc. so you can 

> _not_
> rely on any external voltages in that sence... the Ic current source is
> therefor
> necessary to get a correct Ic current. The Vbe can now be measured more
> accuratly.

Who needs an exact current of 100.00 uA ??
The only important thing is that you measure the Vbe of *all*
transistors at the *same* current.
The voltage drop over the 150k will determine the current.
It will be approx. 14.4 V, and it will vary 2 millivolts, if you match
transistors for 2 mV.
If you had two identical transistors, there would be no error at all
with the 2-resistor-circuit. If you have matched two transistors that
are 2mV apart, say 600mV and 598mV, the two currents would
be 14.400V/150kOhm = 96uA and 14.402V/150kOhm = 96.01uA.
Should not have a great effect on Vbe.

> The current source isn't expensive, so WHY cut it out?

I build it once, on veroboard. Searching it somewhere in the
closet or basement takes me longer than soldering two
resistors to a transistor socket. AND the two-resistor
method works bipolar, i.e. one socket for both NPN
and PNP: Base is tied to gnd, Collector and Emitter are
connected to adjacent rails with 150k resistors.
So you measure at a *nearly* constant current, and at
a collector voltage of 0.6V. If you plug in a PNP, just
turn it upside down, the circuit is symmetrical!

JH.