# floyd12-25, op-amp input offset voltage

Discussion in 'Homework Help' started by PG1995, Mar 23, 2012.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

I was doing a problem about an op-amp. Here is the problem along with its solution; please have a look on it. Here is what the author says on the topic. Please help me with it. Thank you.

Regards
PG

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2. ### Ron H AAC Fanatic!

Apr 14, 2005
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What the heck is the 1 doing in the equation? It doesn't belong there.

EDIT: When Aol is a zillion, the 1 is insignificant. Imagine an amplifier with Aol=0 (I know, an absurd example). The offset would be Vio. That is not possible.

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3. ### steveb Senior Member

Jul 3, 2008
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469
The formula I'm used to seeing is the one valid for both inverting and noninverting amplifier configurations. There is not really a difference between these configurations when the input signal is zero and only the offset voltage is considered.

Vout=(1+R2/R1) Vio

For the open loop case R2=infinity and the opamps open loop gain takes over. Based on this someone might say Vout=(1+Aol) Vio is correct, but I think I agree with RonH because an open loop opamp should just multiply the voltage difference on the input by Aol. Hence, I think Vout=Aol Vio. I'd have to look it up or derive it with a particular opamp model to be sure, but off the top of my head, I'll say that's right. In the end, the 1 does not matter for any practical case. Put it in or take it out and you get the same answer.

For the buffer circuit in the book, R2=0 and the formula reduces to Vout=Vio which agrees with the formula Vout=Vcl Vio

4. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you, RonH, Steve.

So, is this formula, V_OUT(error)=A_cl(V_IO), incorrect? Please let me know. Thanks.

Best regards
PG

5. ### steveb Senior Member

Jul 3, 2008
2,433
469
Well, as I pointed out, it seems to agree with the formula I'm familiar with. The closed loop gain of a buffer is one. So, I'd say yes, it is correct. The way to be sure is to use a nonideal opamp model and derive it directly. I'm pretty sure you'll get that answer though.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
It is correct for a noninverting amplifier.
For an inverting amplifier, it is V_OUT(error)=(A_cl+1)(V_IO). The sign of the error depends on which input pin V_IO is referenced to.

7. ### steveb Senior Member

Jul 3, 2008
2,433
469
The fact that it is inverting or noninverting should not matter. Both configurations are identical when the input signals are zero. But, I guess we define closed loop gain differently, so yes, i agree.

EDIT: By the way, that's why I like the formula in terms of R2 and R1. I find it less confusing.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Nope. The input offset in an inverting amp is NOT located at the signal input. It is always between the noninverting and inverting pins. To the input offset, the circuit looks like a noninverting amplifier.
Consider a zero gain inverting amp (Rf=0, or Rin=∞). Do you think the offset will disappear?

EDIT: It appears you changed your mind while I was typing.

9. ### steveb Senior Member

Jul 3, 2008
2,433
469
Yes, sorry. You were too fast for me.

It took me 2 seconds to realize that the noninverting and inverting values of Acl are different. You must have slipped in there in those 2 seconds.

10. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

I think I understand it a little now.

1: Acl(NI) = 1 + Rf/Ri
2: Acl(I) = -Rf/Ri
3: Acl(VF) = 1

Yes, it is conceptually helpful to have the formula, V_OUT(error)=A_cl(V_IO), in terms of Rf and Ri, i.e. V_OUT(error)=(1+Rf/Ri)(V_IO). e.g. In case of open loop Rf is infinite (in other words, Rf is open) and we can use Aol instead in the formula.

Perhaps, the author should have been more careful and have stated the formulas for NI amplifier and inverting amplifier separately like this:

For NI op-amp: V_OUT(error)=A_cl(V_IO)
For Inverting op-amp: V_OUT(error)=(1+A_cl)(V_IO)

Thank you very much, Steve, RonH, for your help.

Best wishes
PG

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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Go to the noninverting and inverting op amp wikis and look at the exact equations for Vout in each case. Plug in Rf=∞ in each equation, and the open loop output offset voltage equation will be the result.

12. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
1: For NI op-amp: V_OUT(error) = A_cl(V_IO) = (1+Rf/Ri)(V_IO)
2: For Inverting op-amp: V_OUT(error) = (1+A_cl)(V_IO) = (1-Rf/Ri)(V_IO)

When Rf=∞:
#1 reduces to Aol(V_IO)
#2 reduces to (1-Aol)(V_IO)

I hope I haven't made a mistake. I think in #2 "1" can be ignored because Aol is far greater than it. Then, #2 becomes V_OUT(error) = -A_ol(V_IO).

I'm sorry if I'm still wrong. Thank you.

Regards
PG

13. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
Your equations are approximately correct for Aol>>1, but the "1" in each equation still gives the wrong answer if Aol is low. The equations should work for ANY value Aol. What if Aol were 0, 1, or 2? If I were your instructor, I MIGHT give you partial credit for your answer.
I told you in my previous post how to get a proper derivation. Are you ignoring it, or do you not understand it?

Last edited: Mar 25, 2012
14. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5

I don't understand it. Please have a look on the attachment. Thanks.

Regards
PG

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15. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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You wrote:
Not true. Aol is not in the equation. When Rf=∞, it reduces to V_OUT(error) = ∞*V_IO, which is obviously wrong.
Again, Aol is not in the equation. When Rf=∞, it reduces to V_OUT(error) = -∞*V_IO.
The answers are wrong because the original equations are approximations, assuming that Aol>>>>1.

See attachment.

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16. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5

As you and Steve pointed said that "1" doesn't belong in the formula here, now that I understand it somewhat I also agree with you.

In the question statement it isn't mentioned if the op-amp is an inverting one or the NI, so which versions of the two, Aol(V_IO) or -Aol(V_IO), should be used? It look in the given solution it was assumed that the op-amp is NI. Please let me know. Thank you.

Regards
PG

17. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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I would leave out the minus sign, because, AFAIK, the input offset is not defined as being referenced to either pin. Input offset can be positive or negative.

PG1995 likes this.
18. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

Please have a look on the attachment. It contains quite a few questions but it wouldn't have been any helpful, rather confusing, if I had made queries in parts; at least making all the queries about related topics in one turn can help you to help me. I don't need broad understanding of these topics, basic knowledge will do. I have very basic knowledge about the workings of electronic devices such as diodes etc. It would be very kind of you if you could help me with the queries. Thank you very much.

Regards
PG

PS: Higher resolution copy of the attached image can be found here: http://img402.imageshack.us/img402/5374/copyoffloydamp9original.jpg

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Last edited: Apr 12, 2012
19. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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I don't have the time or the energy to answer all these questions. Maybe someone else does.
There are so many resources available on the internet. I think you can find the answers to all these questions by doing your own research, and you will probably learn more in the process.

20. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

Thanks for the advice and giving my questions a look. Yes, you are right about the availability of the resources. But many a time those resources are hard to understand. Just take as an example a Wikipedia article. You would find a lot of complex information there. Sometimes, I do understand some of the stuff but many a time that stuff takes my confusion to higher level while solving one. But I have to agree that this is the best way to learn. But in my view the present education system doesn't really appreciate this approach, especially where I study. Besides this, I have an exam a day after tomorrow so I'm also pressed for time.

Regards
PG