I made a small black box temperature compensated current source with one of those LT 2 terminal regulators and then simply used a multimeter on volts. If you want I can try to trace out the schematic of what I dead bugged back in the day.
It is amasing idea. Can You share schematic etc.?
I cracked the chip trying to pry it off the case to see which one it was. But I owe you a circuit! My main question is, which error are you willing to accept?
These 2 terminal current regulators normally have a positive temperature coefficient, so the trick is to grab a couple of small signal diodes and find one with roughly the same or opposite behaviour and put this one in your feedback network in a way that counter-acts the current increase. The stereotype implementation of this can be found in the LM234 datasheet. But I was just looking over at the newer 2 terminal regulators, e.g. LT3082 and LT3092, and they seem to be a lot less temperature dependent, so one of those should do the trick. Between 0 and 50°C you don't seem to have more than 0.005 µA drift on the internal reference current source so that's more than sufficiently stable for most applications I think, and I doubt you're planning to do arctic exploration with this thing.
So lets assume we run it of two 9V batteries, of which we lose 2V to the dropout. So we use a measurement current of 100 mA, because that makes for easy math. So we would then realistically be able to go from 150 Ohm down to about 10 mOhm (1 mV) on the average meter. So lets take the LT3092 (
http://cds.linear.com/docs/en/datasheet/3092fc.pdf ) with the suggested LT1634 shunt reference, as shown on page 18 of the LT3092 datasheet. The LT1634 has a change of about 0.5 mV over the desired temperature range (Page 6 of the LT1634 datasheet -
http://cds.linear.com/docs/en/datasheet/1634ff.pdf ), at 10 µA however the current has no real effect on the voltage, so we can throw that variation of 5 nA in the trash bin. The LT3092's output current in this configuration is simply Reference Voltage / Shunt resistor. So, neglecting the temperature increase of the 12.5 Ohm resistor we'd need, we could calculate the effect of the temperature variation now quite easily: 0.5 mV / 125 Ohm = 40 µA. So on the 150 Ohm measurement (15V) this would create an error of 6 mV, and on the 10 mOhm we're down to 0.4 µV, so I think we're safe if we modify that circuit slightly to take the temperature increase of the resistor into account. I'll draw something up after lunch.
*Back after lunch*
We could either come up with a complicated feedback circuit that acts and talks like a resistor based on the reference, but then we're stuck with using an OTA, and it's the equivalent of shooting a mosquito with an ICBM. (Though, as a side note, I do approve of that.) So the easier solution is to take it into account in the resistor and minimize the resistor heating by using multiple in parallel. Loads of 24.5 to 24.9 resistors available right here! e.g. if we take
http://www.farnell.com/datasheets/2111203.pdf you have about a 0.12 Ohm resistance change over the desired temperature range. Now since we're using it in parallel we have 12.45 Ohm + (0 - 0.062 Ohm) + +/- 1%. This means we go from 100.4 mA to 99.9 mA roughly, or an error of 0.5 mA, which in turn is about 5 µV error for 10 mOhm or 0.5%. And on a reading of 15V for 150 Ohm we get 0.075V error (also 0.5%). So I think we're quit safe in just taking this circuit with a small adjustment and a 1W thick film 24.9 Ohm 1% resistor with 100 ppm/°C.
So I'd like to propose the following circuit, I took the 1N5817 from memory, so not sure if it'd be really suitable for this. But I think it was 1A with about a 0.45V forward drop. The trimmer is optional, but I highly recommend it, might have to take a slightly higher value (since the bias current is 100 µA). The 18V zener with a power NPN is as input protection against higher voltages. Think this is a simple enough approach that should be sufficient. If you care more about accuracy it's a good idea to invest in a good handheld LCR meter I think?
Do tell if you see any mistakes, haven't done a quick ballpark calculation like this in quite a while.
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