I have some minor questions about two different op amp circuits.

Circuit Problem #1:


I am asked to determine Vout with the following conditions: R1 = 1 kΩ, R2 = 0 Ω, RL = 1 Ω, and Vin = 43.5 V.

I understand how to do the analysis, but the trick in this question is R2 = 0 Ω. My thinking is that this is just a short, so V- = 0 V (it's grounded). But assuming ideal op amp rules, V+ is also 0 V, yet it has Vin. So I'm lost. The lack of a resistor is a problem, but I'm not 100% sure why.

The solutions state the following: When R2 = 0 Ω, then V+ ≠ V- , so the op amp is not ideal and it'll be in saturation.

My questions:

1. Why does zero resistance at R2 cause V+ ≠ V-?
2. How is it known that the op amp will be saturated? I am not given the (not shown) supplied voltages, so I have no way of knowing whether it's saturated or not.

Circuit Problem #2:


I am to determine V1, which is referenced to the ground.

I have done the analysis on this circuit and I have determined that at the node on the left side of the branch containing the 1 MΩ resistor, the nodal voltage (with respect to the ground, of course) is 27 V. Yet in the solutions, it simply states that V1 = 27 V. What about the 1 MΩ resistor? Am I forgetting something basic or is this a mistake? And if it is a mistake, what's the correct answer?

 

If there's no current through the 1M resistor, then there's no voltage drop across it, and the voltage at both ends is the same.

 

You've largely answered your own questions. If R2 is 0Ω, the what voltage will the V- input be at. What voltage will the V+ input be at? Are they the same? If not, then V+ != V-, does it. Whenever the inputs are not the same (by more than a few millivolts in most cases) the opamp will saturate. As you indicated you don't know what the supply voltages are (or how close the opamp can get to the supplies), about the best you can do is describe in words what the output of the opamp will be in terms of the supply voltages, such as "the output will be as close to the positive supply voltage as it can get" or something like that.

 

Whenever the inputs are not the same (by more than a few millivolts in most cases) the opamp will saturate.

This is the main part I don't understand, though. Is there some rule about op amps that states this?

 

If there's no current through the 1M resistor, then there's no voltage drop across it, and the voltage at both ends is the same.

Why would there be no current through that resisitor?

 

This is the main part I don't understand, though. Is there some rule about op amps that states this?

Yes, it is the very large open loop gain. Let's say the difference is 1mV or .001 Volts. Now the open loop gain of the opamp is maybe 100,000. So we multiply 1 mV * 100,000 = 100V We suspect that this value is to high for a solid state opamp and we understand this to mean that the output goes as close to the supply rail as it can go. If you make the difference -0.001 volts then it will go as close to the negative rail as possible. BTW I suspect that this example was incorrectly copied from somewhere because it is not a useful configuration and the feedback resistor has no apparent purpose when R2=0

 

This is the main part I don't understand, though. Is there some rule about op amps that states this?

Remember, the open loop gain of the opamp is very high. For an ideal opamp the open loop gain is infinite.

 

Why would there be no current through that resisitor?

The right hand end of that resistor is not connected. Hence no current flows through that resistor.

 

This is the main part I don't understand, though. Is there some rule about op amps that states this?

It sounds like you are trying to apply a bunch of rules about opamps without understand what and opamp is and where those rules came from. An opamp is a very high gain differential amplifier in which the output voltage is equal to a very large gain factor, typically several hundred thousand to sometimes over ten million, multiplied by the difference between the V+ and V- inputs. For simplicity, let's say that the gain is one million and that the opamp is powered from +/-10V supplies. This means that for the output to be equal to 10V only requires a voltage difference at the inputs of 10uV. Even with a very low gain of just 100,000 would only require a differential input voltage of 0.1mV. But once the output hits the limits imposed by the supply rails, it can't go any further even if the input voltage differential is higher than that -- that is called saturation.

 

The right hand end of that resistor is not connected. Hence no current flows through that resistor.

The problem states that V1 "is referenced to the ground." So current would flow through 1 MΩ resistor, no?

Edit: I think my interpretation of the circuit drawing is wrong. I assumed that a not-shown ground is connected to where V1 is written (thereby extending the branch with the 1 MΩ resistor to the ground), but in actual fact, there is no ground from that branch. So what you're saying, I think, is that V1 is across the 250 Ω resistor to the ground. Is that correct?

 

It sounds like you are trying to apply a bunch of rules about opamps without understand what and opamp is and where those rules came from. An opamp is a very high gain differential amplifier in which the output voltage is equal to a very large gain factor, typically several hundred thousand to sometimes over ten million, multiplied by the difference between the V+ and V- inputs. For simplicity, let's say that the gain is one million and that the opamp is powered from +/-10V supplies. This means that for the output to be equal to 10V only requires a voltage difference at the inputs of 10uV. Even with a very low gain of just 100,000 would only require a differential input voltage of 0.1mV. But once the output hits the limits imposed by the supply rails, it can't go any further even if the input voltage differential is higher than that -- that is called saturation.

That makes perfect sense, thanks! I glossed over this part of my book. Rereading coupled with your explantation helped. Thanks once again!

 

No, current cannot flow through 1 MΩ resistor because there is no closed loop for the current to flow (one end of that resistor is left open).

 

No, current cannot flow through 1 MΩ resistor because there is no closed loop for the current to flow (one end of that resistor is left open).

Thanks. I made an edit to my previous comment before reading this. I think I misinterpreted this circuit drawing because of the way I've seen it drawn in my book before.

 

I think, is that V1 is across the 250 Ω resistor to the ground. Is that correct?

Yes, exactly.

 

There is current flow in the 1 Meg resistor the moment you connect the meter or oscilloscope probe to it, and the meter or probe has a ground connection. Then the 1 Meg and the test equipment form a voltage divider.

So, if you used a DMM with 1 Megohm input impedance, the voltage would be one half that of the OP Amp's output.

Now all you need to do is calculate the voltage as read on a simpson 260 meter and an oscilloscope.

Shunt your virtual equipment with an appropriate resistor to simulate.

You would need to figure out the full scale voltage of the simpson-260 you would use and then calculate the appropriate 20,000 ohms per volt resistance used on that scale, to shunt the virtual meter.

With no external connection or virtual perfect test equipment connected, the full op amp output is at V1.