DIY SMU output stage

[ezalys]

In Dave Erickson’s SMU design (https://www.djerickson.com/diy_smu/), he explains that the output stage is a current output amplifier. When I look at this, I see an opamp stabilizing a cascoded diamond buffer, which I think of as a unity gain voltage buffer. How do I understand this as having current output?

[David Hess]

I had the same thought when I saw his description.  I think what he meant is that the amplifier has a high impedance current output when combined with the output current shunt resistance and the current control feedback loop, which is not the same thing because some SMUs do use a high impedance current output amplifier, which his is not.

[ezalys]

Yeah when you're measuring small currents, the output impedance has to go up to compensate -- or something like that.

What SMUs use a high impedance current output amplifier? Any with schematics I could have a look at?

[David Hess]

Yeah when you're measuring small currents, the output impedance has to go up to compensate -- or something like that.

What SMUs use a high impedance current output amplifier? Any with schematics I could have a look at?

It is a common feature in transistor curve tracers for the base/gate drive amplifier.  If you are just looking for examples of voltage in current out, then look at transconductance amplifiers.

Transconductance outputs are not difficult to make, but do require a different way of thinking.

[Kleinstein]

A somewhat confusing part with the usual SMU circuits is that there are different grounds and a floating extra supply (possibly more than 1).  In many aspects it is similar (but 4 quadrants) to the floating regulator type lab supplies. So it would help to first understand the workings of such a supply.

[djerickson]

I'd like to (attempt to) clarify the DIY-SMU / K236 output stage. It's complicated, kind of mind-twisting. The amplifier board itself,  AMPIN to OCOM out (without local feedback) is voltage input and current output, so is a Transconductance amplifier. This type of amplifier configurations is pretty rare, but used occasionally. If it helps, think of an LM13700.

Why current output? The output current path is the common of the +/- 170V supplies, back through the FET drains (cascode amplifiers, common gate) to the bipolar transistor (Q3/Q4) collectors, which appear as a current output. The open-loop transconductance gain is set by R9/R26, 25 ohms, so -25A / V gain. This is high gain, considering that the max output current is 100mA, so the input V is -25V/A * 0.1A = -0.25V The common emitter transistors invert the gain.

Then local feedback is applied to reduce the amplifier gain to x-20. The feedback circuit is on the main board, buffer opamp U12B, R33 (10K) and R25 (200K). The effect of this is to convert the Transconductance amplifier into a voltage amplifier with fixed voltage gain.

I'm not 100% sure why it was built this way, but I suspect that this clever configuration provides a high voltage amplifier with only a couple of transistors and a low voltage opamp, and helps stabilize the circuit for all types of load impedances.  Electrometers, current sources, SMUs and other DC instruments use similar configurations, so I think it is useful to understand it.

Clear as mud, right? Hope this helps.

Cheers, Dave