Feedback be dammed, here is why the inputs of an op amp are assumed to be "equal or very close" in many cases.

Take a traditional op amp with a +/- 15 volt power supply. We will look at it when the output is NOT in saturation, or the output is in the range of
-15V <= Vout <= +15V

The output of an op amp is given by the equation:

Vout = Av * (V+ - V-)
where Vout is the output voltage, V+ the + input, V- the - input, and Av is the open loop gain.

Now even with an old fashioned 741 type the open loop gain is 50,000. So let us see what is going on at the input for the largest signal the amp can make:

Vout = +15V = 50,000 * (V+ - V-)

(V+ - V-) = 15V / 50,000 = 0.3 mV

So the largest signal the amp can make is produced by a voltage difference of only 0.0003 volts.

Hey, that's damn close to zero so let's just assume it really is zero and get some work done.

I like your approach, but how this approach will differ if you assume that the op amp in saturation (No feedback)

 

In a closed loop application, the output is fed back to the inverting input so as the output goes up / down, the potential on the inverting input goes down / up. That reduces the potential difference between the non-inverting input and the inverting input.
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Is the closed loop is including the positive feedback also?

 

I like your approach, but how this approach will differ if you assume that the op amp in saturation (No feedback)

In saturation you assume that the output of the amp is at one of its extreme limits (usually assumed to be one of the power rails, but often it is a volt or two inside the rails).

 

Is the closed loop is including the positive feedback also?

While positive feedback is technically "closed loop", it operates so as to keep the output in saturation at one extreme or the other, so you cannot assume that the differential input voltage is zero or even small.

 

In saturation you assume that the output of the amp is at one of its extreme limits (usually assumed to be one of the power rails, but often it is a volt or two inside the rails).

Hey WBahn,

I'm sorry if I argue too much. But I think he is already assumed that Vout equal one of the extreme limit which is 15V

 

While positive feedback is technically "closed loop", it operates so as to keep the output in saturation at one extreme or the other, so you cannot assume that the differential input voltage is zero or even small.

For the negative feedback: assume the input at V-=1V and the input at V+ = 0V and the power supply for that op amp = +/-15V. At that moment(at t = 0) the op amp will produce A(V+ - V-) V, since the max output is +/-15 then the voltage at V- will be 1 - 15 --> -14 (Based on the feedback), then the op amp will produce(at t = 1) --> A(0- -14) which will give the max voltage which i is 15V ! . Based on that the op amp will saturate ! Is this true?

 

For the negative feedback: assume the input at V-=1V and the input at V+ = 0V and the power supply for that op amp = +/-15V. At that moment(at t = 0) the op amp will produce A(V+ - V-) V, since the max output is +/-15 then the voltage at V- will be 1 - 15 --> -14 (Based on the feedback), then the op amp will produce(at t = 1) --> A(0- -14) which will give the max voltage which i is 15V ! . Based on that the op amp will saturate ! Is this true?

There is no way to discuss what the voltage at V- will be because you haven't give a circuit, so there is no way to determine how the voltage at the output will affect the voltage at the inputs.

 

There is no way to discuss what the voltage at V- will be because you haven't give a circuit, so there is no way to determine how the voltage at the output will affect the voltage at the inputs.

Oh, it's my mistake. Just assume an inverting amplifier circuit , where the Gain is equal to - R_f/R_in.

 

So let's take a very simplistic view of how this works. Let's assuming that Rin = Rf = 1kΩ. Let's assume that, initially, the input and output voltages are at 0 V which means that the V- is at 0 V and everyone is happy.

Now change Vin to 2 V. There is now 2 V across 2 kΩ and 1 mA of current flows resulting in a voltage of 1 V at V-. In response to this, the output starts heading to the negative rail (let's say that that is -15 V). But as it gets to -1 V the voltage at V- goes to 0 V and the amplifier is no longer trying to drive the output to the negative rail. Let's assume the gain of the opamp is a measly 1000. When the output of the opamp is -0.9 V the voltage at V- will be 0.05 V giving a differential voltage of -50 mV, so the output will still be driving toward -50 V (of course it will saturate at -15 V). But what about when the output of the opamp reaches -0.998 V? Now the voltage at V- will be 1 mV so the differential input voltage will be -1 mV and the output will be driving toward -1 V, which it is very close to already. The end result is that it will settle into an output voltage just shy (in magnitude) of -1 V because there needs to be a bit of an error in order to give a differential signal of just under (in magnitude) -1 mV in order to result in the final output voltage.