How does the negative feedback amplifier below work?

[Jy158654]

Homework Statement

How does the difference between V- and V+ reduce and eventually become zero?

Relevant Equations

G = G0/(1+G0)

The book says that " The op-amp multiplies the potential difference between V+ and V- by the open-loop voltage gain to produce Vout."
So my first question is, since we say we assume open-loop voltage gain is infinite, so does that mean the Vout produce will be infinite too?

Then, the book says that "As Vout is 'fed back' to V-, the value of V- increases and this reduces the difference between V- and V+. Very quickly this difference becomes zero again and Vout = Vin." The problem is, I have checked some websites and all of their photo of negative feedback amplifier consists of resistors to attenuate and take some fraction of the voltage to the V-, but the photo shown in this book doesn't consist any of the resistors, so if Vout = infinite, doesn't that mean V- in this example also has infinite value without any attenuation??

And also, this example says that the initial input, Vin was +0.1V, and Vout/V- was 0V, so how does the increase in V- result to reduce the difference between V- and V+ and becomes zero in difference at the end? I am a little confused over this part here.

Hope there's someone who can help me in my confusion. Thank you.

 

[Jy158654]

Addition: The closed loop voltage gain i checked from the websites is this, which consists of feedback fraction, which formula is correct between this formula and the formula shown in my main post's photo?

 

[DaveE]

The formulas are the same when β=1, which is the case in your first schematic. β is the fraction of the output voltage that is applied at the - input, which in the first case is β=1. 100% of the output is applied to the - input.

 

[Jy158654]

The formulas are the same when β=1, which is the case in your first schematic. β is the fraction of the output voltage that is applied at the - input, which in the first case is β=1. 100% of the output is applied to the - input.

Since we assume infinite open loop voltage gain, then the Vout hence the V- is infinite too in this case??

 

[DaveE]

Infinities are hard to deal with (as you describe). I think it is better to think really, really big, but finite. In the real world, op-amp gains are something like 100,000. If you are mathematically inclined, think of the open loop gain A approaching inifinity (A→∞), not equal to infinity (A=∞).

Perhaps it would help if you look for a stable solution with low gain, like A=10. Then repeat it for A=100, 1000, ... You will see the trend is for the output and the - input to equal the + input, with a small error. That error will decrease as the open loop gain, A, increases.

At this point you should just assume that the circuit will "know" how to find that stable point where V₀=A(V₊ - V_-). You aren't wrong to wonder about that, it's a significant question for more complex circuits. You will get to that a little later in your studies.

 

[Tom.G]

Homework Statement:: How does the difference between V- and V+ reduce and eventually become zero?
Relevant Equations:: G = G0/(1+G0)

So my first question is, since we say we assume open-loop voltage gain is infinite, so does that mean the Vout produce will be infinite too?

V_OUT can only go as far as the supply voltages, less about a volt for internal overhead in the OP-Amp.

Now since an op-amp by definition has an "almost infinite" gain, any small difference between the inputs will cause a large change in output voltage.

Just to make things a little easier to follow with some numbers attached, let us assume the OP-Amp gain=100.

With 0.5V applied at V+ input, the output will head toward 50V. But that can't happen because the supply voltage is only 6V. Since the output is directly connected to V- input, that's where V- input is headed.

As V- approaches the 0.5V on V+, you will see that the two input pins are approaching the same value. Of course when they are the same value, the output will be Zero. Which is not exactly what we are after.

When the V- input reaches +0.495V there is a difference of 0.005V between the inputs. With the amplifier gain of 100, that 0.005V is amplified by 100, to give an output of 0.5V.

Yea, that 0.495V is not exactly matched to the 0.5V input. That's because the OP-Amp gain is kinda low^{see also Note 1 below}. In the real world, OP-Amps often have a gain of 100 000 or more. If you do the math you will see that a real world OP-Amp would have an input difference of 5microvolts rather than 5millivolts.

Note 1: Actually, I took a shortcut in the math here to make it easier to follow. The correction factor for the input difference and for the output is (100/101), as you correctly showed with G0/1+G0 in your first post.

Hope this helps!

Cheers,
Tom

 

[Jy158654]

With 0.5V applied at V+ input, the output will head toward 50V. But that can't happen because the supply voltage is only 6V. Since the output is directly connected to V- input, that's where V- input is headed.

As V- approaches the 0.5V on V+, you will see that the two input pins are approaching the same value. Of course when they are the same value, the output will be Zero. Which is not exactly what we are after.

When the V- input reaches +0.495V there is a difference of 0.005V between the inputs. With the amplifier gain of 100, that 0.005V is amplified by 100, to give an output of 0.5V.

Hello thanks for your help! Does that mean that when the output is 50V(as in your example), the output is "slowly fed back" to the V- input? Because I have always thought that the V- will immediately become 50V as well when the output becomes 50V.

And also, does that mean that the input difference usually have, let's say 5 microvolts difference, which is so small and unsigficant, and so we assume that the difference between V+ and V- is zero? Thank you!

 

[DaveE]

And also, does that mean that the input difference usually have, let's say 5 microvolts difference, which is so small and unsigficant, and so we assume that the difference between V+ and V- is zero? Thank you!

Yes, exactly. That's why one of the simplifications for "ideal" op-amps is infinite gain, even though, as you said, that doesn't really make sense. Then you can assume that the difference is zero, instead of so close to zero that you don't care.

The speed that feedback happens can be assumed to be very, very, very fast (nearly infinite) for analog circuits. In the real world, it will depend on the type of components in the feedback path. We will sometimes put things like capacitors in the circuit to intentionally slow it down. However, don't worry about that yet, it's a little too advanced, and doesn't actually change the feedback concept that you are learning now.