Hi there.

In the attached question you will see part c) asks the reader to explain what happens when the non-inverting terminal allows a current of 1uA into the op-amp, and to answer what new current will be through the ammeter.

I am stuck on this. I dont want any worked solutions, just an idea would be handy.

Cheers,
Fraser

 

oh by the way, I cut out the bit that said the ammeter has a resistance of 5ohm, hence the reason for putting it inside the feedback loop.

 

Ask yourself what happens if Vin is a zero resistance source. What are the relevances of input current and the value of Rin?

 

I don't know what you're asking me.

In an ideal op-amp, your 5V would all be across the input resistor because the op amp has infinite input resistance, and no current would go into the opamp.

But here, some current does go into the op-amp and I've not come across any equations/abstractions to deal with it before.

 

The issue for those trying to understand the problem is that you haven't given the complete picture.

You only supply part (c) of a problem which refers to earlier questions.

I presume this is a circuit which converts an input current value Iin to a related voltage drop Vin across a shunt resistor Rin, and then translates this voltage to a 'usable' current indication on the ammeter A.

Also a minor additional point - Since Rin develops the voltage Vin we probably need to know its value or at least the relative level of Iin with respect to the 1uA leakage current. For instance, if you are measuring an input current of 50uA you would need Rin to be 100kΩ to develop a voltage of 5V (say). The fact that the anticipated 'ideal' input voltage will be degraded by the 1uA current leakage into the op-amp is possibly an important consideration in the overall problem solution. It would be a different story if you were measuring a current of 10A via a shunt Rin of 0.5Ω.

I hope the ammeter isn't a moving coil type - the accuracy on those isn't all that great.

 

I think for 3 marks Vin is applied externally and Rin just exists there.

The question asks you to calculate the current through R and then add 1uA to that current because that 1uA flows through the ammeter too.

It asks to find the percentage error as to see that as the output current increases, the error due to the 1uA, is decreased.

However, posting the whole question will make sure you get the right answer.

 

Since the problem states that you must take into account the opamp's input current to the non-inverting input then I would think that is likely that the value for Rin has been given earlier in the problem.

Can you provide us with any additional information that is missing from this excerpted problem statement that you may have omitted? For example, without the 5 ohm ammeter resistance that you almost withheld the voltage drop across the meter would have been difficult to determine. Is there more information that you can supply?


hgmjr

 

I think for 3 marks Vin is applied externally and Rin just exists there.

The question asks you to calculate the current through R and then add 1uA to that current because that 1uA flows through the ammeter too.

In a real op amp, input current flows to one of the power supplies, not to the output. This op amp has no explicit power supplies, but I'm guessing that the answer will not assume that the input current flows to the output.

It asks to find the percentage error as to see that as the output current increases, the error due to the 1uA, is decreased.

However, posting the whole question will make sure you get the right answer.

 

I think for 3 marks Vin is applied externally and Rin just exists there.

I think this is the most likely scenario.

Thank you all for looking at the question. I double checked and there is definitely no other information, R(in) is not given so presumably it is just there to give a finite input resistance to the circuit.

mik3, you say that because an extra 1uA flows into the op-amp, that same amount will flow out. Can I ask why that is so?

 

here is the complete question in case anyone is curious. it's the first out of the six.

 

My best guess is that the intent was for Rin to be in series with Vin instead of in parallel. Otherwise, input bias current would have no effect on the output.

 

My best guess is that the intent was for Rin to be in series with Vin instead of in parallel. Otherwise, input bias current would have no effect on the output.

I was thinking the same thing. They're showing a parallel resistance such as one would expect to be part of a Norton equivalent for the impedance of the source, but they are showing it with a voltage source (apparently, since they designate the source Vin, and not Iin) rather than with a current source. ???

There are some other parts of the problem that I think are ill-posed.

For example, in part d, they ask "is it likely that the op-amp will be able to provide the required current through the 10Ω resistor?". Well, of course it will, since they said earlier "assume the op-amp is ideal"; an ideal op-amp can provide any amount of current. Then, they say "What is a typical short current for a 741 op-amp?". Apparently, they want the student to no longer consider the op-amp ideal, even though they didn't say that outright.

If I were the student, I would try to cover the bases and say something like, "if the opamp is ideal, then it can supply the current through the 10Ω resistor, but it's a real opamp, most likely not."

 

Rin does not affect the answer of the question.

Assuming the op amp can supply the current required by the resistor in each case plus the 1uA input current, then for both cases the ammeter will measure an error current of 1uA. This is because the feedback loop will force the voltage across R to equal Vin. Thus, the current through R does not depend on the ammeter's resistance anymore. Thus the current through R will be Vin/R in both cases. However, the ammeter will measure the Vin/R current plus the 1uA current flowing into the inverting input. Therefore, the error current will be always 1uA. The voltage across the ammeter is not the same for both cases but it depends on Vin/R and the ammeter's resistance.

 

Rin does not affect the answer of the question.

Assuming the op amp can supply the current required by the resistor in each case plus the 1uA input current, then for both cases the ammeter will measure an error current of 1uA. This is because the feedback loop will force the voltage across R to equal Vin. Thus, the current through R does not depend on the ammeter's resistance anymore. Thus the current through R will be Vin/R in both cases. However, the ammeter will measure the Vin/R current plus the 1uA current flowing into the inverting input. Therefore, the error current will be always 1uA. The voltage across the ammeter is not the same for both cases but it depends on Vin/R and the ammeter's resistance.

Can you give an example if an op amp (ideal or otherwise) in which the input current flows to the output?
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.

 

Can you give an example if an op amp (ideal or otherwise) in which the input current flows to the output?
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.

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

Assuming the 1uA is flowing through the op amp's output, then the ammeter would measure that 1uA.

If the 1uA is flowing through R, then the voltage across R would change (drop in this case). The feedback loop would try to correct this change in the voltage across R and this will cause a 1uA extra current to the Vin/R current to flow through the ammeter. Thus, the ammeter would measure a 1uA error current again.

As it is seen, in both cases the ammeter would measure the 1uA error current. Thus, it doesn't matter whether the 1uA current is flowing through the op amp's output or through R.

If the op amp is a single supply one, then the only way the 1uA can flow into the inverting input is through the op amp's output.

If the op amp is powered by a dual power supply, then it is possible for the 1uA current to flow through R.

 

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.

 

Thanks for the input guys, excuse the pun.

I spoke to my tutor who has a copy of the answers to these tutorial questions, and he spent a while looking at it and said the question was a bit "artificial".

I think mik3 is correct in this case. And considering it's only worth three marks, I don't think they are expecting us to solve it at a transistor level.

Thanks for your help

 

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

Are you assuming that the 1μA into the non-inverting input is accompanied by a similar 1μA into the inverting input? The problem said no such thing, and it's not necessary for the two inputs of an opamp to have the same bias currents.

Furthermore, there must be some other path for current flow, else where will the .5 amp into R come from when the input is 5 volts? From the unshown, but necessarily implied power supply, apparently.

 

Are you assuming that the 1μA into the non-inverting input is accompanied by a similar 1μA into the inverting input? The problem said no such thing, and it's not necessary for the two inputs of an opamp to have the same bias currents.

Furthermore, there must be some other path for current flow, else where will the .5 amp into R come from when the input is 5 volts? From the unshown, but necessarily implied power supply, apparently.

I am not assuming that the current into the non-inverting input is 1uA, just ignore it since it does not affect the output voltage in this case.

I know there is a power supply from where the power to the op amp and R is provided.

 

I am not assuming that the current into the non-inverting input is 1uA, just ignore it since it does not affect the output voltage in this case.

I know there is a power supply from where the power to the op amp and R is provided.

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.