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Old 19th Jan 2023, 5:46 pm   #1
Malcolm T
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Default Op Amp help

I need to measure at the test points on an Op Amp IC. to measure the stated voltages on the pins, Obviously the multi meter red lead goes to the IC pin but does the negative multi meter lead go to chassis ground of the PSU ?.
Never done any work on fault finding op amp circuits.
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Old 19th Jan 2023, 5:58 pm   #2
RogerEvans
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Default Re: Op Amp help

Usually op-amps are designed to work with both positive and negative supply rails, and the output can be positive or negative with respect to the zero volts of the supply. So you will quite likely find that none of the op-amp pins are connected to the supply zero.

If the op-amp is designed to work in its linear regime with negative feedback, and it is working properly, then the two input terminals will be within a few mV of each other. If the op-amp is being used as a comparator then the very large open loop gain will drive the output close to one of the supply voltages, the positive supply if in+ is greater than in- and the negative supply if the inputs are the other way around.

Best of luck,

Roger
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Old 19th Jan 2023, 6:31 pm   #3
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Default Re: Op Amp help

You've asked a very general question, so there isn't a specific answer.

Opamps have been used in an awful lot of different configurations for an awful lot of different purposes. What you need to know is what was being thought os as a reference by whoever wrote those numbers.

Quite often, opamps are run between a pair of power supplies, one positive, the other negative and their common grounded. Commonly, this sort of power arrangement uses supplies of equal but opposite voltages. There are a few opamps which used asymmetric power rail voltages, just waiting to catch you out.

Another common scheme is to use an opamp working from a single rail usually positive, with its other power pin being ground. Usually, the inputs need to be comfortably within the bounds of ground and the power supply (These ARE opamps, so there are exceptions which can go outside the power supply bounds). This means that you'll most often see arrangements so that the signal input pins are elevated from ground to stay reasonably centred in the ground-power rail voltage region.

So you need to look at the circuit to see what the designer's been doing. The voltage reference for voltage markings on the schematic might be from the power supply midpoint of a dual power supply arrangement, or it might be from whatever common reference is used for signals.

The opamp datasheet for the model of chip might help you, but then again it might not because there's a long history of using opamps in ways or for things never envisioned by their manufacturers.

So you have to look at the actual circuit and understand what's going on to be quite sure. Signal ground or chasdsis or supply midpoint are likely answers, but not guaranteed for all cases.

David
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Old 19th Jan 2023, 6:40 pm   #4
Malcolm T
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Default Re: Op Amp help

Right thanks for replies , yes back to the circuit diagram then.
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Old 19th Jan 2023, 7:04 pm   #5
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Default Re: Op Amp help

Rather than check voltages in abstract, you need to measure them in a functioning circuit unless you want to take the full-on design engineer route.
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Old 19th Jan 2023, 8:00 pm   #6
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Default Re: Op Amp help

There is a modest number of basic opamp circuits and most designers do not stray very far from them at all. You get variations of supply arrangement like single versus dual supplies ad that covers most things you are likely to come across. But it is then important to be able to spot the basic circuits so you can then realise when something different is in front of you. There are far fewer basic circuit types than there are weird types, so it's feasible to memorise and recognise the basics, but never to remember all the wild ones.

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Old 19th Jan 2023, 8:41 pm   #7
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Default Re: Op Amp help

It should be worth mentioning that not only are designs using two rails like +15v 0v and -15v, you can often get the "false 0v rail designs".
So to simplify, if there is only a say, +12v and 0v supply available, the common trick is to have a potential divider with two resistors supplying a half rail of 6 volts.
This mimicks the + 0 and - rails as such.
so what would have been +15 becomes +12 and accordingly
0v becomes 6 volts
-15 becomes 0 volts.
I hope that makes sense?
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