can instrumentation amplifier measure beyond rails? - Page 1
Forum section: electronics / beginners

author: exe
author_rank: Unknown
timestamp: March 20, 2020, 03:42:34 pm
content: Hello!I'm designing a power supply with a digital control and read-back. I want to measure output voltage which ranges between 0 and -18V relative to the ground (that's because control part is floating around positive output). Can I measure output voltage without a resistor divider? That's because divider would put some current loading, which I'd like to avoid. I could use a current-to-voltage converter, but it loads the output too. Desired output signal is 0...+2.5V.So, how else I could sense the output voltage?It's not that essential, I can live with a small current error, but it would be great if we could avoid it. Loading within, say, 100-200uA is fine. So, in the worst case I'll use a resistor divider with high values of resistors.

author: chris_leyson
author_rank: Super Contributor
timestamp: March 20, 2020, 04:30:33 pm
content: Use a 10:1 voltage divider, 180k/20k for example followed by a 16bit ADC. I recently built a two channel A-D converter using a MAX1416 in differential mode with an LM4040 2V5 reference.and 1% resistors. Once calibrated for full scale or near full scale with a 5 1/2 digit meter it's accurate to +/- 1mV. Not bad for a $5 part.

author: Zero999
author_rank: Super Contributor
timestamp: March 20, 2020, 07:52:22 pm
content: Some amplifiers can go beyond the rails, but not by that much.If it's acceptable for the output voltage to be 2.5V, when the input is 0V and 0V, when the input is -18V, then all that's needed is a few resistors. The values don't have to be massively high for a current of <100µA.

author: David Hess
author_rank: Super Contributor
timestamp: March 21, 2020, 05:57:13 am
content: Quote from: exe on March 20, 2020, 03:42:34 pmThose are the two common options; either divide the inputs down or use inverting (current to voltage) amplifiers.The easiest solution is to extend the supply voltage to extend the input voltage range so that a non-inverting input with the lowest possible input bias current can be used. But another solution I have seen in some old HP power supplies is to compensate for the current drawn by any output dividers by adding it back in via the current error amplifier.But now that I think about it, there is a way to do exactly what you suggest.Varactor diode input operational amplifiersnot only have very low input bias currents but also have common mode input voltage ranges which exceed their supply voltage by 100s or even 1000s of volts limited only by transformer isolation. Whether such a design is economically practical is a whole different question.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 21, 2020, 02:24:35 pm
content: If you have digital control and readout, why does the loading matter? It's trivial math to subtract the calculated divider current.

author: exe
author_rank: Unknown
timestamp: March 26, 2020, 07:20:21 pm
content: Thank you very much for all your suggestions, now I understand my options.Quote from: David Hess on March 21, 2020, 05:57:13 amDo you know the name of a model that does that? I'd like to understand how it works. Motivation: I'd like to put shunt resistor before bjt pass transistor (to have lower output impedance and for the sake of education). Problem is, base current creates an error. I wonder if I can easily compensate for that.

author: David Hess
author_rank: Super Contributor
timestamp: March 28, 2020, 04:01:12 am
content: Quote from: exe on March 26, 2020, 07:20:21 pmIt would have been in their bench power supplies from the 1970s to 1980s. I puzzled over the schematic for a while.

author: Zero999
author_rank: Super Contributor
timestamp: March 28, 2020, 03:14:59 pm
content: Quote from: exe on March 26, 2020, 07:20:21 pmNo, put the resistor after the transistor and use negative feedback to reduce the output impedance.

author: David Hess
author_rank: Super Contributor
timestamp: March 28, 2020, 06:01:18 pm
content: Quote from: Zero999 on March 28, 2020, 03:14:59 pmAnother reason to put the current shunt on the output of an emitter follower is that by taking AC feedback before the resistor, the phase margin of the amplifier is increased allowing a lower ESR output capacitor.