Op-amp - input voltage

op-amp --- input voltage

Hello,
I find op-amps very fun! And I understood mostly everything I was offered to study. But here is the thing.

Why does the voltage has to be the same at the input leads?

To be more specific.

Lets say we have a inverting op-amp. And we have an imput Vin at -ive lead.

+ive lead is grounded. Why are both then at 0 potential? Why can't -ive be at Vin and +ive at 0? Whats going on there?

Does this has something to do with differential amplifier ?

An op amp has really high gain. One common one, the 741, has an open loop gain of 200,000. That means that even with a 1mV difference between the two inputs, the output would be 200V!

The only way you can have a stable, reasonable output voltage level is by using negative feedback. The entire point of negative feedback, whether used in an inverting or non-inverting configuration, is that it uses the output to adjust one of the inputs to bring them close to equal.

One thing must be said here: the op amp is not sentient, and does not "try" or "think" about this at all. As an abstract analogy, consider a ball in a wok. No matter which spot in the wok you place the ball, it will always roll towards the center. It's not because the ball "looks for" or "seeks out" that spot, but is because of the wok's geometry.

Consider an inverting op amp with Rin = Rf for simplicity, so the closed loop gain is 1. Vin = 0V, so Vout = 0V as well.

What happens when we change the input voltage? Increase it to 1V.

One helpful way to think about Rf and Rin is that they form a voltage divider. The instant you put 1V on there, the output is till 0V. Calculate with the voltage divider formula, and the voltage at the inverting input is 0.5V. Because the inverting input is higher than the non-inverting, it will force the output to go down.

However, once the output starts changing, that changes the output of your voltage divider too. You will have to modify the voltage divider formula to take into account having neither of the ends connected to ground, which is an interesting exercise in algebra.

As the output goes down, the voltage divider output (the voltage at the inverting input) goes down too. Once the output reaches -1V, the voltage at the inverting input is 0V, the same as the non-inverting input.

If the output goes lower, say to -1.1V, the voltage at the inverting input goes lower than 0V. Because the inverting input is now lower than the non-inverting, it forces the output back up.

Does that help?

One helpful way to think about Rf and Rin is that they form a voltage divider. The instant you put 1V on there, the output is till 0V. Calculate with the voltage divider formula, and the voltage at the inverting input is 0.5V. Because the inverting input is higher than the non-inverting, it will force the output to go down.

I understood everything up to this part.

Because the inverting input is higher than the non-inverting, it will force the output to go down.

Can you explain this further, please?

The two inputs of a diff amp are named based on how the output responds to them.

At the non-inverting input, the output has the same polarity as the input. If you increase the input, the output does the same and goes up.

At the inverting input, the output has the opposite polarity as the input. If you increase the input, the output goes down.

That's how they are designed to respond.

An op amp is just a differential amplifier with a monstrous amount of open loop gain.

EDIT: Perhaps I did say it a little ambiguously in the first post.

If you hold the on-inverting input at the same voltage, and increase the inverting input, the output voltage decreases.

If you hold the inverting input at the same voltage, and increase the non-inverting input, the output voltage increases.

The two inputs of a diff amp are named based on how the output responds to them.

At the non-inverting input, the output has the same polarity as the input. If you increase the input, the output does the same and goes up.

At the inverting input, the output has the opposite polarity as the input. If you increase the input, the output goes down.

That's how they are designed to respond.

An op amp is just a differential amplifier with a monstrous amount of open loop gain.

Yes yes I understand that. Can we make a few things clear?

I understand that if you increase the input at the inverting lead, you will feedback through feedback resistor 180 out of phase voltage, resulting in negative feedback.

But what does this have to do with non-inverting lead, it is just grounded, sitting there...

Mr. Jiggy? :D

Redbelly98
Staff Emeritus
Homework Helper

Yes yes I understand that. Can we make a few things clear?

I understand that if you increase the input at the inverting lead, you will feedback through feedback resistor 180 out of phase voltage, resulting in negative feedback.

But what does this have to do with non-inverting lead, it is just grounded, sitting there...
I don't know what circuit Jiggy-Ninja had in mind, but here is one with a grounded non-inverting input:

The important features of the op-amp are:

1. The extremely high open loop gain:
$$V{out} = \begin{cases} +V_{cc} & \text{if } & & A \cdot (V_+ - V_-) \ \ge \ +V_{cc} \\ A \cdot (V_+ - V_-) & \text{if} & -V_{cc} \ \le & A \cdot (V_+ - V_-) \ \le \ +V_{cc} \\ -V_{cc} & \text{if } & & A \cdot (V_+ - V_-) \ \le \ -V_{cc} \\ \end{cases}$$
where $A$ is generally somewhere between 100,000 to 1,000,000.
2. There is negligible current entering or leaving the two inputs.
3. There is feedback from the output to the inverting input.

Let's think about what happens if V+ and V- are not roughly equal. Let's take Vcc=15V, and assume that (1) the input to the circuit is well within the range ±Vcc, and (2) R1=R2.

If V+ and V- are not roughly equal, there are two possibilities: either V+>V- or V+<V- by a significant amount. We'll consider both.

I. V+>V- by a significant amount.
According to feature (1) descirbed above, this mean Vout will be equal to +Vcc=15V. Since R1=R2, then V- is simply the average of the input and 15V. But we assume the input is well above -Vcc=-15V, so this means V- is positive, and therefore greater than V+=0V. This contradicts the statement that V+>V- by a significant amount, so V+ cannot be significantly greater than V-.

II. V+<V- by a significant amount.
By a similar argument, we arrive at a contradictory statement, so V+ cannot be significantly less than V-.

Since V+ can be neither significantly greater or less than V-, V+ must be approximately equal to V-. (Note, this no longer holds if the input voltage is either below -Vcc or above +Vcc in this example with R1=R2).​

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I don't know what circuit Jiggy-Ninja had in mind, but here is one with a grounded non-inverting input:

The important features of the op-amp are:

1. The extremely high open loop gain:
$$V{out} = \begin{cases} +V_{cc} & \text{if } & & A \cdot (V_+ - V_-) \ \ge \ +V_{cc} \\ A \cdot (V_+ - V_-) & \text{if} & -V_{cc} \ \le & A \cdot (V_+ - V_-) \ \le \ +V_{cc} \\ -V_{cc} & \text{if } & & A \cdot (V_+ - V_-) \ \le \ -V_{cc} \\ \end{cases}$$
where $A$ is generally somewhere between 100,000 to 1,000,000.
2. There is negligible current entering or leaving the two inputs.
3. There is feedback from the output to the inverting input.

Let's think about what happens if V+ and V- are not roughly equal. Let's take Vcc=15V, and assume that (1) the input to the circuit is well within the range ±Vcc, and (2) R1=R2.

If V+ and V- are not roughly equal, there are two possibilities: either V+>V- or V+<V- by a significant amount. We'll consider both.

I. V+>V- by a significant amount.
According to feature (1) descirbed above, this mean Vout will be equal to +Vcc=15V. Since R1=R2, then V- is simply the average of the input and 15V. But we assume the input is well above -Vcc=-15V, so this means V- is positive, and therefore greater than V+=0V. This contradicts the statement that V+>V- by a significant amount, so V+ cannot be significantly greater than V-.

II. V+<V- by a significant amount.
By a similar argument, we arrive at a contradictory statement, so V+ cannot be significantly less than V-.

Since V+ can be neither significantly greater or less than V-, V+ must be approximately equal to V-. (Note, this no longer holds if the input voltage is either below -Vcc or above +Vcc in this example with R1=R2).​

Thank you for your in-depth reply. I think I understand. I will have to work out few things still.

Thank thank you very much, I have a lot to work with now.

And thank you Jiggy, for your previous post, although you were absent from further discussion.​

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sophiecentaur
Gold Member
2020 Award

Neither of the two input signals needs to be 'grounded' exactly. All that is necessary is that they should be biased to some voltage within the operating range of the amplifier (see spec). Usually, there will be some feedback and the inputs pins will not be connected directly to a voltage source so this 'knife edge' sensitivity is not relevant. A 'Virtual Earth' Amp circuit ensures that the -input is maintained as near to the +input (which can have any value - not just 0V) as the gain of the op amp will allow.

You could also imagine the two inputs connected to two ends of a transformer winding, (with some bias voltage applied to the centre tap on the winding). There would then be no 'reference' input at all - the amp would just be looking at the potential difference between the + and - inputs.

Neither of the two input signals needs to be 'grounded' exactly. All that is necessary is that they should be biased to some voltage within the operating range of the amplifier (see spec). Usually, there will be some feedback and the inputs pins will not be connected directly to a voltage source so this 'knife edge' sensitivity is not relevant. A 'Virtual Earth' Amp circuit ensures that the -input is maintained as near to the +input as the gain of the op amp will allow.

You could also imagine the two inputs connected to two ends of a transformer winding, (with some bias voltage applied to the centre tap on the winding). There would then be no 'reference' input at all - the amp would just be looking at the potential difference between the + and - inputs.

Yes but this concept of virtual earth, needs time to settle in. I am not analysing mega complicated circuits, I am just interested in theory of op-amps. That is why the input was bothering me.

AlephZero
Homework Helper

Just remember that the two inputs of an op amp are not necessarily always at the same voltage. If you connected a battery directly between the inputs, nothing bad would happen and you wouldn't fry the chip.

Op amps are often used in circuits that are designed so the inputs stay at the same voltage, because there is some feedback from the output to the input that makes the IC work as a linear amplifier.

The "virtual earth" idea is quite simple. If you connect one input to the "real" earth, and the two inputs have the same voltage, the other one is forced to stay at 0V. The input current to the op amp is so small you can ignore it, so you have a point in the circuit that is always at 0V (earthed) but not physically connected to earth - hence the word "virtual".

Just remember that the two inputs of an op amp are not necessarily always at the same voltage. If you connected a battery directly between the inputs, nothing bad would happen and you wouldn't fry the chip.

Op amps are often used in circuits that are designed so the inputs stay at the same voltage, because there is some feedback from the output to the input that makes the IC work as a linear amplifier.

The "virtual earth" idea is quite simple. If you connect one input to the "real" earth, and the two inputs have the same voltage, the other one is forced to stay at 0V. The input current to the op amp is so small you can ignore it, so you have a point in the circuit that is always at 0V (earthed) but not physically connected to earth - hence the word "virtual".

Because of the infinite impedance and all. I get it :) Thanks

In a nutshell from a engineering point of view, if we have a feedback path, output will drive or draw current, through that path, until voltages on both input leads are approximately the same(cannot be the same because of offset voltage)

Redbelly98
Staff Emeritus
Homework Helper

In a nutshell from a engineering point of view, if we have a feedback path, output will drive or draw current, through that path, until voltages on both input leads are approximately the same(cannot be the same because of offset voltage)
Yes, well said. Provided that feedback is to the inverting input.

Yes, well said. Provided that feedback is to the inverting input.

Cool. I am off to learn applications.

sophiecentaur