I spent quite a while yesterday and the day before messing with this. I used a circuit posted by Kassutronics which is the same as the above except the top resistor was 4.7k.
All that’s going on there is that resistor plus the diode form a simple voltage reference, ~600 mV, for the collector voltage. Base is at ground and emitters are a little above -12 V. You put a reference transistor on one side and leave it there, and put transistors to check on the other side and measure voltage between the two emitters. Then you pair off transistors that give nearly the same measurement.
At first I built the matching circuit on a solderless breadboard and spent some time trying out an idea for expediting the process and generally trying to get consistent and sensible results. Ended up deciding the breadboard was not the way to go, especially after discovering the reading would change if I moved the jumper that ran from base to ground from one spot on the ground rail to another! So I built the circuit on stripboard, abandoned trying to improve the method, and just got on with it. Resistors were matched to 0.1%, 100.5k each, though if my CircuitLab simulation is correct, a 0.1k difference corresponds to only a 24 µV change in the measurement and I can only measure to precision of 0.1 mV.
With the stripboard, results seemed to be consistent and stable at the ~0.2 mV level. Out of 23 transistors tested all were within 2.5 mV of my reference transistor (and all read higher than the reference, apparently I picked a very atypical transistor for my reference) and I was able to pair off 20 of them with measurements differing by ≤ 0.1 mV.
(What you’re measuring is a voltage difference ∆V corresponding to ∆I/R where ∆I is the difference between the two emitter currents and R is 100k (or 100.5k in my case). Each current is about 11.4 V / 100.5k ~ 114 µA. Measuring ∆V to 0.1 mV corresponds to measuring ∆I to ~1 nA, or a relative difference in emitter currents of about 1 part in 100,000. Not bad for a multimeter and a piece of stripboard. Note, though, that the emitter voltages also differ, so the underlying quantity of interest, I_S, isn’t directly proportional to the emitter current. Per the CircuitLab simulation, it looks as though an emitter current difference of 1 part in 100,000 corresponds to a difference in I_S of something like half a percent.)
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