# operational amplifier circuits etc.

Discussion in 'Mathematics and Physics' started by PG1995, Apr 19, 2014.

1. ### NorthGuyWell-Known Member

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2mv is not particularly small offset. You get 0.185V/A, so 2mV will introduce approximately 10mA of error. At 2A reading, it's 0.5%. I think it's fine.

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2. ### PG1995Active Member

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Let's say that the input offset voltage is +2mv then what does it mean? I understand that the +2mV voltage should be applied to the op-amp to make its Vout zero but which input terminal should be grounded and which terminal should be applied with +2mV? Should we ground inverting terminal or non-inverting terminal? This might be useful to understand what I'm asking.

The same goes for the offset -2mV. Which input terminal should be grounded?

Suppose, the input voltage on the non-inverting input terminal is 1V which means we expect voltage of 5V on the output for gain of 5, and the offset is 2mV. The output voltage will be 5-10mV=4.99V. Right?

What would be the output voltage for the same setting if the offset were instead -2mV? Thank you.

Regards
PG

3. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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PG:

You missed the point entirely,I think. The offset voltage is "referred to the input". So, it is effectively measured by putting the OP amp in a buffer configuration and measuring the output. with the +input grounded. So, with a gain of +1, input offset = output offset.

Now, you can now figure out where to put the 2 mV.

BUT your thinking about it ALL wrong because if the input offset voltage max is +-2 mV, you have to be able to inject that range if you want to compensate. In reality the differential input is not 100% identical.

Some OP amps have a potentiometer that could be used to provide the offset compensation which is strongly affected by temperature.

Referred to input, remember.

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5. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

That's about right yes. If we have 1v on the input and a gain of 5 and there is 2mv input offset, then the 'real' input looks like 1.002v so the output is 1.002*5=5.010 volts. Likewise, if the input offset is -2mv then the 'real' input is 0.998v so we get a lower output.

If you were to compensate on the input you would add or subtract 2mv on the input. This means you have a choice of which terminal you could use to compensate if you have -2mv available or +2mv available as needed. If you only have +2mv available then your choice of input terminal depends on the actual offset of the op amp itself. This means in theory if you only have +2mv available (not -2mv) then you must set the circuit up such that you can switch input terminals if needed.

The simplest view of the input offset is as a small battery with voltage 2mv, but it can be plus and minus. This battery can be in series with either the inverting terminal or the non inverting terminal. If the offset is positive aiding the input and it's on the non inverting terminal, then the output voltage will be higher than with no offset. If it is negative then the output will be lower than with no offset. If it is viewed as being on the inverting terminal, then it has the opposite effect.

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6. ### PG1995Active Member

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Thank you, KISS, MrAl.

I don't think that I have it right as KISS also pointed out above so let's go thru the basics.

First let's see that if my basic understanding is along the right lines.

Let's say input offset for the given op-amp in this picture is +2mV. It means that differential voltage of +2V should be applied at the input to make Vout=0V.

In the picture although the non-inverting input is grounded, the op-amp still erroneously 'think' that there is an input voltage of -2mV present at the input which makes Vout=-2mV.

Therefore, it needs actual Vin=+2mV at the non-inverting input to make Vout=0V because -2mV+2mV=0V.

Best wishes
PG

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7. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Yes.

To clarify the input offset voltage has a min and max range. So, if it's specified as +-2mV, it could actually wind up as 0, dependent on temperature.

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8. ### PG1995Active Member

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Thank you, KISS.

So, I had it correct though I'm also waiting for MrAl's comments. I'm still confused about two points.

Perhaps, to be precise, you should have written |input offset| = |output offset|. Please let me know if I have it correct. I understand that there isn't really a big difference but to me it might indicate a loophole in my understanding.

In this picture from my previous post, Vout was confirmed to be -2mV. The input offset voltage is said to be differential voltage. The differential voltage is Vdiff = Vin - Vf where Vf is feedback voltage. In that buffer circuit input offset voltage was taken to be 2 mV, i.e. Vdiff=+2mV. The Vf was -2mV. It gives Vdiff=Vin-Vf => Vin=+2mV+2mV=4mV. The Vin for that circuit should have been +4mV and not +2mV. Where am I going wrong? I believe that input offset voltage isn't really the differential voltage I'm familiar with.

Please also help me with this next step. Thank you.

Regards
PG

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9. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Input offset = output offset.

Agreed, for this case.

input offset voltage = (The inverting input) - (the non-inverting input)

We call it differential, because it's not referenced to ground.

In general, you can test an OP amp when it's not used as a comparator by looking at this voltage.

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10. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Here is a diagram of the five possible basic cases of input offset which should make this more clear.

I've calculated the first three cases, you should calculate the last two.

Note that the "input offset" is shown as a small equivalent battery external to the op amp, but it's really inside. The calculations are the same however.
In each case the drawing of the op amp is a drawing where the op amp has zero input offset so that we can show it externally.

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11. ### PG1995Active Member

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Thank you, KISS, MrAl.

You are essentially saying that Vdiff=Vinv - Vnoninv. I had always thought that Vdiff=Vin-Vf where Vin could be on either non-inverting input or inverting input. In your case, the order of subtraction is fixed but in my case the order depends upon Vin. Please let me know if my way is correct.

Please have a look here. Thank you.

Best wishes
PG

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12. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

That's very good. I am happy that you now understand input offset voltage.

Another good point that KISS brought up is that this test voltage we have been working with was 2mv, because that was the spec of the op amp. But that is the max spec, so it could be less than that...anywhere from 0v to 2mv plus or minus. So a circuit built to help null this out would have to be adjustable, and it would have to be adjusted in the actual circuit being used.

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13. ### PG1995Active Member

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Thank you, MrAl.

But what's the correct equation for differential voltage? Is it Vdiff = Vinv - Vnoninv, or Vdiff = Vin - Vf? Thanks.

Regards
PG

14. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Not sure what you mean exactly.

The differential voltage between the two INPUT terminals is vp-vn. But internally it would be vp-vn+vos.
vp being non inverting terminal voltage to ground, vn inverting terminal voltage, vos input offset voltage and vos can be plus or minus.

15. ### PG1995Active Member

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Thanks for letting me know.

So, differential voltage is always voltage at non-inverting input minus voltage at inverting input.

16. ### MrAlWell-Known MemberMost Helpful Member

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Hi,

Yes, does that sound surprising?

17. ### PG1995Active Member

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No, it doesn't!

That was a note to self for reference in future.

18. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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Sorry, I goofed with my previous explanation, I got a bit dyslexic: order of the subtraction was backwards

One thing that either may confuse you or help you is suppose A is the non-inverting input and B is the the inverting one. The OP amp takes (A-B) and multiplies it by a huge number called the open loop gain.

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19. ### PG1995Active Member

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Hi again

We discussed four cases here about how input offset voltage can affect the operation of an op-amp. If a datasheet gives input offset to be 2mV then it might be anywhere between 0-(+/-)2mV. But how do we know that if the input offset is on non-inverting input or inverting input? Perhaps, we don't know it either. Thanks.

Regards
PG

20. ### KeepItSimpleStupidWell-Known MemberMost Helpful Member

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You don't know. That's why Mr. Al said that you have to design your compensation for the +- worst case. You may have to design your compensation to be a function of temperature.

Electrometers have a Zero check/Zero correct function primarily because Vos is very temperature sensitive. In a current to voltage converter that measures pA, Vos is a big error term.

There is a special op amp that's called an auto-zero OP amp, but it has other issues.

Digital pots have been used for this function because trimpots are virtually gone from instrumentation. There is another device that eludes me right now, but it;s essentially an electrically programmable resistor which can be programmed to just about any value, It doesn't have the bits/accuracy problems, but can't be programmed forever and the process takes a while.

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21. ### PG1995Active Member

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Hi MrAl

Remember this circuit we were discussing. I wanted to ask you which op-amp would be good choice between LM358 and OP290 for the given circuit. Both are single supply but neither is rail-to-rail. I remember that you told me that such an op-amp could go down to the ground but it won't go up completely to the positive rail. For example, if +5V supply is used then the op-amp can touch the ground but it won't go above, say, 4V.

Please note that OP290 has very low input offset voltage compared to LM358. Thanks.