Man I don’t understand

That's apparent.
You seem to not understand how negative feedback works (did you study that?)
Let's try again.
Due to the high gain of the op amp with negative feedback, it will do whatever it can to keep the (-) input essentially identical to the (+) input (which in this case is at ground potential, making the (-) input a virtual ground).
Ideally the current into either op amp input terminal is zero so that doesn't enter into the equation.

The means the input current through R2 is the same as if R2 was going to ground, so that's how you calculate its current.

Due to negative feedback the op amp must then adjust the output voltage so that the current through R1 is equal to the current through R2 to keep the (-) input at virtual ground voltage.
Depending upon the relative value of the two resistors, this output voltage magnitude can be less or greater than the input voltage.
(Hint: what does the relative output voltage polarity compared to the input have to be so that the above conditions are satisfied?)

 

I reversed the resistors. Normally we'd number them from left to right, but your schematic doesn't do that. I meant that Vin is across R2.

View attachment 232061
What is the voltage across R1?

KCL (Kirchoff's Current Law). There are 3 branches associated with pin 2 (AKA V-). The sum of the current into and out of that node must sum to zero. Reread Post #11. Call the current in R1, I(R1). call the current in R2, I(R2). Call the current into the minus input of the op amp Iin. Now KCL says:

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I(R!) + I(R2) + Iin = 0, but we can consider Iin to be effectively 0, because it is very much less than either of the other two currents -- think femtoamps. So​

I(R1) = -I(R2) now we want to write the voltages in terms allowed by Ohms law where a current can be replaced the voltage drop across the resistor divided by the value of the resistor. So​

I(R1) = (Vout - Virtual_GND)/R1 & I(R2) = (Vin - Virtual_GND)/R2 substitute into the result above from KCL​

Vout = -(R1/R2)*Vin​

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Don't forget the negative sign. If Vin is above ground the Vout will be below ground and vice versa. That is how the inverting opamp configuration works. Where does Vout come from if it does not come from Vin you might ask. It comes from the power supply terminal which are not shown in the diagram and must be assumed. An immutable law of Physics says you can't get (or make) something from nothing. the amplifer is not really passing anything from the input to the out -- it's more like it is making a much bigger copy of what it sees on the input.

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KCL (Kirchoff's Current Law).

Why are you quoting my post? Did I somehow seem confused to you??

 

Why are you quoting my post? Did I somehow seem confused to you??

I didn't see your post and I typically take a while to write one and I sometimes edit it after the initial posting. I would not think that you would be confused about much of anything. My responses were all directed at the TS, none of it was meant for you.

 

I didn't see your post

You quoted from it (?).

 

You quoted from it (?).

I'm sorry I thought I was replying to the TS.

 

Hi everyone, we’ve just started looking at Op Amps in my University Physics Course and in particular negative feedback. i have been given a rule that in the circuit below because V+ is connected to ground then V- and V+ will both be equal to 0v. If this is the case then what is the point in the circuit, it seems to me like the current just flows through the resistor at the top and no voltage gets amplified. If someone could explain this to me I would be very grateful.
Thanks Everyone

Your question is very reasonable. If you have zero volts into an amplifier then how can you have some actual voltage like 1v or 2v or 3v coming out. If the gain is 10 and you have 1v input then you get 10v output, very simple, and that is because 1v times 10 is 10 volts. If you have 0 volts input however then 0v times 10 is still 0 volts, so how can we have any output at all with 0v input.

The answer is that the input is not really zero volts it is a small voltage that gets amplified by a very large gain. We often call it zero volts because it makes the approximate calculation simpler in theory. We first have to abandon the question of why zero volts input produces a non zero output though, and follow with a different theory where we assume that the input is only at the very input, that is, Vin not one or both of the inputs to the op amp itself (inverting or non inverting). Once we do that, we can use the 'theory' that the output is Vin times the gain of the CIRCUIT, not the gain of the op amp itself. The gain of the circuit we calculate using other means, and for this simpler circuit is it just the feedback resistor resistance divided by the input resistor resistance, which then is just R1/R2 here.
So using Vin as the input ONLY and the gain as G=R1/R2, we can calculate the output as:
Vout=Vin*G

and that's the end of it.

Now if you would like to know how this actually works because we are still assuming that both inputs to the op amp are zero and we only get an output out of the op amp itself, the answer again is that the inputs are not really both zero, it just makes the above calculation easier to think of it that way. IN reality, the gain of the op amp is very very high and the input at the inverting terminal is very very low but not zero, so a very small input times a very large gain is still a reasonable output.
For example, if we had R1=1k and R2=1k then the gain using the above simpler calculation is -1 (remember it's inverting so it is 1 but negative). So if we input 2v then we get -2v output. Simple.
But in reality what is happening is if the gain of the OP AMP ITSELF (not the gain of the circuit) is 100, then the input at the inverting terminal is actually 0.019607843137255 volts, and that times 100 is of course 1.9607843137255 and that times -1 for inverting is -1.9607843137255 volts.
So you see the output is not exactly the theoretical -2v it is a tiny bit different, and the input it not zero it is a tiny bit different. That is what makes this kind of circuit appear to work as the simpler calculation we did far above.
Also, note that as the internal gain of the op amp goes higher, to say 1000, we get an output of
-1.996007984031936 volts, and notice that is even closer to the theoretical -2 volts.
As the internal op amp gain is raised higher, the output gets closer and closer to the theoretical -2 volts and that is how we can see that the simpler result far above is a reasonable approximation.

So in the end you have to consider the circuit gain, not the internal op amp gain, if you want to calculate the output using the simpler theory. If you want more exact results, then you have to resort to using either a voltage controlled voltage source or a voltage controlled current source to calculate a more exact output result.

So that's how you get a non zero output from a 'seemingly' zero input, because the input at the op amp itself is not really zero it is slightly different, and that tiny difference is what makes an op amp work.

You will also see the theoretical calculation done by calculating the input current assuming both inputs are at 0v, and then using that to calculate the current through the feedback resistor, and since both inputs are at 0v then the output must be the same as that across the feedback resistor. That is the simpler theoretical solution which again assumes that the input to the circuit is the main featuer not the input to the op amp really because we assume they are both non zero and thus the input to the inverting terminal is really nonsensical, yet it works to help make the calculation simpler.

View attachment 232060

 

I had the same question, the answer is there is gain. Think through it.

 

Hi everyone, we’ve just started looking at Op Amps in my University Physics Course and in particular negative feedback. i have been given a rule that in the circuit below because V+ is connected to ground then V- and V+ will both be equal to 0v. If this is the case then what is the point in the circuit, it seems to me like the current just flows through the resistor at the top and no voltage gets amplified. If someone could explain this to me I would be very grateful.
Thanks EveryoneView attachment 232060

@JackBidds

Title: Understanding Basic Electronics, 1st Ed.
Publisher: The American Radio Relay League
ISBN: 0-87259-398-3

and our favorite:

 

I had the same question, the answer is there is gain.

But the gain can be -1 or smaller.

 

Of course it can be - anything inverted is negative gain, if you’re a poor engineer, even transistor can have a net loss in amplitude. A log amplifier often curtail gain... the point is there’s gain and high input impedance.

 

the point is there’s gain and high input impedance.

Don't see how that's really pertinent to the TS's original question about negative feedback in an op amp (?).

 

I'm addressing this...

as far as negative feedback, you guys have it covered... but read Harold Black.

i have been given a rule that in the circuit below because V+ is connected to ground then V- and V+ will both be equal to 0v. If this is the case then what is the point in the circuit, it seems to me like the current just flows through the resistor at the top and no voltage gets amplified.

 

I'm addressing this...

as far as negative feedback, you guys have it covered... but read Harold Black.

I know sometimes we call it "negative gain" but really i think the better choice of words for the description the way it plays out in real life is "signed gain". Sometimes we ignore the gain sign so we might call that "unsigned gain". So really either inverting or non inverting has a gain, like 1, 2, 5, 10, etc., but sometimes we have to specify the sign.