Need help with understanding SMU output stage - Page 1
Forum section: electronics / beginners

author: DanielDlouhy
author_rank: Contributor
timestamp: January 15, 2024, 09:14:32 am
content: Hello everybody,I want to make my own simple SMU. Power supply board and main board are not a problem to design for me, but I am not really sure about the output amplifier. I got lost, when trying to design my own. I decided to look for inspiration on the internet and I found the well known Dave Erickson's SMU. I would like to modify and use the output amplifier in my unit.Can someone please help me understand the working principle of this Amplifier? I am completely lost with this one.I know that Dave Erickson said, that it takes ±10 V and amplifies it to ±150 V, but I don't know how. Even simulations didn't help, I wasn't able to make it do anything. In my unit I need it to take ±10 V and output ±100V and up to 500mA. Or should I use a different design for my use? Pardon my probably stupid question.I added a schematic, but everything else is on Dave's page -https://www.djerickson.com/diy_smu/Thank you in advance. Daniel

author: kimballa
author_rank: Contributor
timestamp: January 15, 2024, 05:52:42 pm
content: The top and bottom rails of the amplifier stage are marked as +/-170V.The amplifier stage itself is a class B transistor amplifier; there is some more stuff going on in there than I can understand quickly, but I think the main principle is fairly straightforward - take a "small" signal (+/-10V) and amplify it onto the larger power rails.Somewhere else in the SMU, it is generating that split 340V bus - there's probably another schematic with a transformer that's stepping up those power lines.

author: DanielDlouhy
author_rank: Contributor
timestamp: January 15, 2024, 07:40:21 pm
content: I can see the class B amp, but that would have voltage gain of 1. And in this design it has voltage gain of 15, how? How does the circuitry around it make it have voltage gain of 15?In all of my simulations it just has gain of 1, for some reason.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 15, 2024, 07:46:35 pm
content: The circuit can be a bit confusing,as the supply for the main regulator part and the power stage are separate and they don't have a common ground / link.The amplifier part is bit similar to the common lab supplies with a floating regulator, just for both polarities.The output transitors work in a kind of common emitter configuration, not as an emitter follower. So the transistors provide voltage gain.

author: Yuu
author_rank: Regular Contributor
timestamp: January 15, 2024, 08:39:54 pm
content: Quote from: Kleinstein on January 15, 2024, 07:46:35 pmWhere do you see transistors in a common emitter configuration?Q7/Q8, Q3/Q4, and Q10/Q9 all look like they're in a emitter follower configuration. The only thing I see that looks like it might provide gain are transistors Q11/Q12 which look like they're in some common base config.On this topic, where would you learn more about circuits like this? On amazon I really only find two books on power amplifiers. I own one of them. I haven't seen a topology like this. There are also some power amplifiers I've seen at work that are mysterious.I get the feeling power amplifier design is something esoteric that is handed down from senior engineers and there's not much writing on it. I'd love to be proven wrong.

author: Kleinstein
author_rank: Super Contributor
timestamp: January 15, 2024, 09:32:49 pm
content: To learn / understand the circuit, it helps to first have a reduced version, so without the extra MOFETs (they are there for high voltage) and also without the short circuit protection.The circuit as shown is already simpler than most SMUs, as it does not include a 2nd set of transformer taps (e.g. a +-15 V supply).Q3 and Q4 are working in a kind of emiiter circuit, as they get the control (base) signal from a source relative to there emitter. The main point of the circuit one may easy overlook is that the supply to the input side has it's ground / reference point linked to the emitter side of Q3 / Q4. The ground of the +-150 V supply is the putput. So it is a bit tricky to have these votlages as lables and not as visible voltage sources.This type of amplifier is also sometimes found in high power audio amplifiers, though not very often. One may find it in class H types as the step to the extra supply taps is relatively easy in this configuration.A first step may be to looks at the common Harrison configuration for lab supplies, that uses a floating regulator. The idea it the same, just only 1 polarity.

author: dobsonr741
author_rank: Frequent Contributor
timestamp: January 15, 2024, 10:52:15 pm
content: The Keithley schematics,https://www.djerickson.com/diy_smu/files/Keithley_236_237_Source_Measure_Unit_SMU_Service_Manual_and_schematics.pdfDavid linked on his page, section 4.2.2 Output Stage and 4.2.3 V/I Feedback has a nice summary.

author: Yuu
author_rank: Regular Contributor
timestamp: January 15, 2024, 11:26:23 pm
content: Quote from: Kleinstein on January 15, 2024, 09:32:49 pmIt's kind of hard to see the labels in the picture. I think I got the wrong idea from an initial look. I assumed the input was the AMPIN on pin 5 of J3 coming into the left side of R10. I'm used to audio power amps and op-amp-like circuits and a lot of that looked like emitter followers I see on output stages. Will try to understand your comment more later when I have time. Thank you.Quote from: dobsonr741 on January 15, 2024, 10:52:15 pmThanks. Will take a look at this.

author: dobsonr741
author_rank: Frequent Contributor
timestamp: January 16, 2024, 01:30:51 am
content: I'm curious why this topology survived 30 years - i.e. "grounding" what normally would be the output and go the difficulty of the floating mid-supply reference for the high voltage rail.I can think of one pro and con - I wonder if I'm on the right track:Pro: current measurement is at GND potential, avoiding the extreme common mode voltage on the shuntCon: there is no PSSR - deviation of the floating supply will influence the output directly.

author: DanielDlouhy
author_rank: Contributor
timestamp: January 16, 2024, 12:43:12 pm
content: I was looking in the schematic of the Keithley 236 and I saw that some transistors connected like this. (see picture) Why is it like this? If I understand it correctly it works like a diode, so why don't just use a diode then? And what is the purpose of that capacitor?

author: Kleinstein
author_rank: Super Contributor
timestamp: January 16, 2024, 02:43:44 pm
content: Some meters use the BE junction as a kind of 5-9 V zener diode with relatively low leakage. Normal zener diodes are lower tolerance and may be more robust / lower noise, but also higher leakage at some 2-4V. A transistor can also make an intermediate speed low leakage diode. The true low leakage diodes are slow recovery and the faster diodes are usually really fast and high leakage. There are very few small diodes in between with low leakage and a trr in the 50-200 ns range.