Output voltage and output current of an Opamp

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Output voltage and output current of an OP-Amp

Taking the OP-Amp as an ideal OP-Amp, the voltage at terminal 1 is 0 with respect to ground.

The voltage at B is $V_B = - 0.5 V$

$i_{in}$ is from A to B.

There is another 5 mA current going from C to B.

I think both of these current should go towards D. Then, no current is coming out of the OP-Amp, i.e. the direction of $i_o$ is opposite to what is given in the question.Since, $i_in$ is very very less than 5mA, approximately 5 mA goes through 50 $\Omega$ resistor. So, the voltage drop across this resistor is 0.25 V. Since, the current is towards D, $V_o = -0.5V -0.25V = -0.75 V$So, the correct option is (c).

Is this correct?

 

[Charles Link]

Yes, this is correct. :)

 

[Pushoam]

Thanks.

 

[LvW]

Yes, this is correct. :)

Charles - please, can you explain ? Why do you think, that (c) is correct?
Pushoam has started his calculation with an assumption (ideal opamp) that is NOT in accordance with the task description (A=200).

 

[Charles Link]

Charles - please, can you explain ? Why do you think, that (c) is correct?
Pushoam has started his calculation with an assumption (ideal opamp) that is NOT in accordance with the task description (A=200).

Perhaps an oversight on my part. I spotted the $A=200$ also, but I think (maybe incorrectly) that that is a typo. Usually Op-Amps have very high gain, e.g. $A=200,000$ and higher. The assumption is made that the inverting terminal is a virtual ground. A gain of only 200 would likely make that assumption somewhat inaccurate. $\\$ Additional comment: I believe the purpose of the exercise is to teach students to calculate ideal op-amp configurations by hand. A hand calculation of the non-ideal A=200 would be somewhat difficult. I'm not sure what is presently in use for that=when I was in school in 1980, a program called SPICE would have computed the result.

 

[LvW]

Yes - without any doubt, a gain calculation based on an open-loop gain of only A=200 would be somewhat more involved.
However, based on Black`s classical feedback formula, it is not a big problem.
On the other hand, perhaps you are right and there was a typing error - and A=200k would be the envisaged number,
But - who knows?

 

[gneill]

A by-hand analysis would not be too difficult. The op-amp is replaced by a suitable input resistance and a controlled voltage source with a gain of 200. Two essential nodes means that nodal analysis will be easy enough to do by hand.

 

[Charles Link]

A by-hand analysis would not be too difficult. The op-amp is replaced by a suitable input resistance and a controlled voltage source with a gain of 200. Two essential nodes means that nodal analysis will be easy enough to do by hand.

I do think though, in the multiple choice question, they were simply looking for the answer for an ideal op-amp. I do think the gain of $A=200$ was mislabeled.

 

[Pushoam]

But, in the question, input impedance is also given.
But, I don't know any other way to calculate it.
But, from the given options, doesn't it become clear whether one has to idea op - amp approximation or not?

 

[Charles Link]

But, in the question, input impedance is also given.
But, I don't know any other way to calculate it.
But, from the given options, doesn't it become clear whether one has to idea op - amp approximation or not?

In general, an ideal op-amp is well known to require a very large gain factor. $A=10^6$ is not uncommon for a commercial op-amp. I do think this was probably a misprint, because learning how to compute the case of the ideal op-amp is much more important than doing the non-ideal case with a low gain such as $A=200$.

 

[gneill]

I doubt that it was a misprint as both the gain and input impedance of the amplifier were given, and both values were much less that typical of a modern, garden variety op-amp. The question has the benefit of demonstrating how close to ideal performance you can get with even modest gain and input impedance, thanks to the feedback loop. The early, discrete component op-amps (both tube and transistor based) had characteristics akin to these specs.

Recognizing that you can use an ideal approximation to choose amongst the given multiple choice answers is commendable. Being able to justify that approximation is an even better thing

 

[Tom.G]

I believe there is a bigger problem than Ideal vs Non-ideal op-amp. The 50KΩ feedback resistor needs 5V -0.5V across it to balance the I_D (+5% for the input current). That puts 5V -0.5V volts at the junction of the 100Ω and the 50Ω resistors. However the problem asks for the voltage at V_O, not the voltage at the 100Ω resistor

 

[Charles Link]

I believe there is a bigger problem than Ideal vs Non-ideal op-amp. The 50KΩ feedback resistor needs 5V across it to balance the I_D (+5% for the input current). That puts 5V voltage at the junction of the 100Ω and the 50Ω resistors. However the problem asks for the voltage at V_O, not the voltage at the 100Ω resistor

Please check your arithmetic. It's -.5 volts, not 5 Volts. I believe the OP got a reasonably accurate answer.

 

[Tom.G]

Please check your arithmetic. It's -.5 volts, not 5 Volts.

Oops! You're right. Thanks. Am correcting the post. (That's what happens when I don't write things down.)