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.Thank you, RonH, Steve.
So, is this formula, V_OUT(error)=A_cl(V_IO), incorrect? Please let me know. Thanks.
Best regards
PG
It is correct for a noninverting amplifier.Thank you, RonH, Steve.
So, is this formula, V_OUT(error)=A_cl(V_IO), incorrect? Please let me know. Thanks.
Best regards
PG
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.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.
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.The fact that it is inverting or noninverting should not matter. Both configurations are identical when the input signals are zero.
Yes, sorry. You were too fast for me.EDIT: It appears you changed your mind while I was typing.
1: For NI op-amp: V_OUT(error) = A_cl(V_IO) = (1+Rf/Ri)(V_IO)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.
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.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
Thank you for the reply.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?
Not true. Aol is not in the equation. When Rf=∞, it reduces to V_OUT(error) = ∞*V_IO, which is obviously wrong.1: For NI op-amp: V_OUT(error) = A_cl(V_IO) = (1+Rf/Ri)(V_IO)
When Rf=∞:
#1 reduces to Aol(V_IO)
Again, Aol is not in the equation. When Rf=∞, it reduces to V_OUT(error) = -∞*V_IO.2: For Inverting op-amp: V_OUT(error) = (1+A_cl)(V_IO) = (1-Rf/Ri)(V_IO)
When Rf=∞:
#2 reduces to (1-Aol)(V_IO)
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.Thank you for the reply.
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
I don't have the time or the energy to answer all these questions. Maybe someone else does.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
HiI 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.
by Duane Benson
by Duane Benson
by Jake Hertz
by Jake Hertz