Then could you explain in more detail what you mean by "through the op amp's output" in this sentence:

"The only paths the 1uA current into the inverting input can flow is through the op amp's output or through the load R."

If I'm looking at the same circuit you are, I see 3 terminals on the triangle representing the opamp. If 1 μA flows into the + input, it would seem that it must exit through the only other 2 terminals, namely the - input or the output terminal, or it must exit through the unshown power supply.

In order for it to flow through the resistor R, it must first pass through the ammeter, mustn't it? Or, it would have to reach R via the - input.

I don't understand how it can flow through R without also exiting the opamp's output terminal.

I am talking about the current into the inverting input.

 

I keep hammering on this, but it's falling on deaf ears.
As I said, and The Electrician reiterated, the power supplies are implicit, but necessary for the circuit to operate.
THE INPUT CURRENT FLOWS THROUGH ONE OF THE POWER SUPPLIES (IN THIS CASE, THE NEGATIVE ONE). IT DOES NOT FLOW THROUGH THE OUTPUT.

Please pardon the shouting.

 

I keep hammering on this, but it's falling on deaf ears.
As I said, and The Electrician reiterated, the power supplies are implicit, but necessary for the circuit to operate.
THE INPUT CURRENT FLOWS THROUGH ONE OF THE POWER SUPPLIES (IN THIS CASE, THE NEGATIVE ONE). IT DOES NOT FLOW THROUGH THE OUTPUT.

Please pardon the shouting.

Does it passes through the output?

 

One might possibly propose that save for a non-ideal input bias requirement the op-amp is in all other respects "ideal".

Given this proposition, one might then argue that both inputs require identical bias current. Suppose the 1uA alluded to in the question is the required bias current for the two identical inputs.

Would one not then have a 1uA current flow in the feedback path to provide the bias for the -ve input?

Agreed - the question makes no mention of this. It's unfortunately a poor proposed question and leads to confusion and frustration. Perhaps we should be shouting at the teacher rather than one another on the forum.

 

Does it passes through the output?

I've said three times already:

From post #14:

I realize that there is no other path in the schematic posted, but an op amp will not work without a power supply, and that's where the input current will go.

From post #16:

If the op amp is powered, then the input current will flow through the negative supply back to GND. If the op amp is not powered, then there can be no input current.

From post #22:

THE INPUT CURRENT FLOWS THROUGH ONE OF THE POWER SUPPLIES (IN THIS CASE, THE NEGATIVE ONE). IT DOES NOT FLOW THROUGH THE OUTPUT.

I don't know how else to say it.

 

One might possibly propose that save for a non-ideal input bias requirement the op-amp is in all other respects "ideal".

Given this proposition, one might then argue that both inputs require identical bias current. Suppose the 1uA alluded to in the question is the required bias current for the two identical inputs.

Would one not then have a 1uA current flow in the feedback path to provide the bias for the -ve input?

Agreed - the question makes no mention of this. It's unfortunately a poor proposed question and leads to confusion and frustration. Perhaps we should be shouting at the teacher rather than one another on the forum.

It is a very poorly thought out question.
Input offset current is a real "feature" of real op amps, and I assume this question is aimed at this phenomenon.

 

One might possibly propose that save for a non-ideal input bias requirement the op-amp is in all other respects "ideal".

Given this proposition, one might then argue that both inputs require identical bias current. Suppose the 1uA alluded to in the question is the required bias current for the two identical inputs.

Would one not then have a 1uA current flow in the feedback path to provide the bias for the -ve input?

Agreed - the question makes no mention of this. It's unfortunately a poor proposed question and leads to confusion and frustration. Perhaps we should be shouting at the teacher rather than one another on the forum.

That is how universities (and schools in general) work. They ask the question in an easy way but if you analyze the problem more thoroughly you might find that the question makes no sense.

The argument here is that Ron H says the bias current is provided by the power supply and I say is provided by the op amp's output. I agree that the bias current is provided by the power supply but to reach the inverting input, in this example, it has to pass through the op amp's output or through the load.

 

That is how universities (and schools in general) work. They ask the question in an easy way but if you analyze the problem more thoroughly you might find that the question makes no sense.

The argument here is that Ron H says the bias current is provided by the power supply and I say is provided by the op amp's output. I agree that the bias current is provided by the power supply but to reach the inverting input, in this example, it has to pass through the op amp's output or through the load.

I agree with that, but there was no mention of inverting input current in the statement of the problem.

 

It is a very poorly thought out question.
Input offset current is a real "feature" of real op amps, and I assume this question is aimed at this phenomenon.

The marks of the question are low to consider the input bias current offset. This might not make sense to you but for students it is fine because they solve it as the teacher shows them. Most of them care how to pass the exam and not to learn what is really going on.

 

The marks of the question are low to consider the input bias current offset. This might not make sense to you but for students it is fine because they solve it as the teacher shows them. Most of them care how to pass the exam and not to learn what is really going on.

The question directly addresses input offset current (implicitly). In the problem statement, no mention was made of inverting input current.

 

I agree with that, but there was no mention of inverting input current in the statement of the problem.

You are right Ron H, I thought the question said into the inverting input. I have read it again and it says into the non inverting input.

I am sorry for the argument, that is the reason I was saying it has to pass through the output or the load.

Nevertheless, I think the question should say into the inverting input because the value of Rin is not given in the question. Also, Rin should be in series with the non inverting input to affect the output due to the bias current.

 

I will be sure to pass on your grievances to the Professor next time I see him.

 

Here is a question, and sorry if it's a stupid one.

Real Op amps have an input resistance. Is this the same value at each input terminal?

 

You usually talk about differential input resistance because that's the kind of signal you apply to the op amp.
It's between both terminals:

http://en.wikipedia.org/wiki/File:Op-Amp_Internal.svg

 

I will be sure to pass on your grievances to the Professor next time I see him.

Yes, that is a good idea to make things clear.