How does the feedback work in an inverting op amp amplifier?

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Heidi said:
......................

If I understand the purpose of that circuit correctly, I'm curious, "How did you know the voltage at the inverting terminal would approach zero with no feedback in advance?"
Click to expand...
I didn't know in advance. I took the easy way and just increased the size of the capacitor until the Bode plot showed no low-frequency roll-off above about 1 Hz. If you reduce the size of the capacitor you will start to see a low frequency rolloff below some low frequency.
For calculation purposes you can determine what RC value is required to have a roll-off at 1Hz much greater than the open-loop DC gain.
For example, with 100F and 1KΩ the rolloff at 1Hz is -115.9dB.
This is well below the DC gain (Aol) of the op amp which is set at 100k (100dB).
 
Please forgive my stubbornness, crutschow, I'd still like to make 100% sure of one thing: When there is no feedback at the inverting terminal, an op amp's output is obtained by the "product of its open-loop gain and its voltage difference between input terminals" (assume neglecting the input offset voltage or any other factors). Therefore, before we can accept the output voltage as a representation of the open-loop gain happily, we also have to make sure (in this case) the inverting input voltage is approaching zero. Am I right?

( Aol*(Vp-Vn)=Aol*(1-0)=Aol
and
Aol*(Vp-Vn)=Vout
So Aol=Vout )
 
Hi,

I think you might do well to look at a few solid cases. For example, with Vp=0 and Vn=1mv, 2mv, 5mv, 10mv, etc., maybe even less, like Vn=1uv, 2uv, 5uv, etc. If the output can no longer support the feedback for proper linear operation then it saturates in an attempt to satisfy the feedback requirement for linear operation.
 
Sorry, MrAl, I don't understand the meaning of the paragraph "if the output can no longer ..." Did you want me to look at

or this

or something else?
 
You are right. As I noted, with a 100F and 1kΩ low-pass filter in the feedback the feed back is reduced by 115.9db (a factor of 1/628k) at 1Hz. Thus for a gain of 100dB (100,000) the feedback voltage is still a factor of 15.9dB (a factor of 6) below the input voltage.
 
You are talking about a transient AC measurement where saturation can happen.
The open loop Bode-plot test is a linear model AC simulation where actual signal levels are of no consequence or concern.
 
crutschow said:
You are talking about a transient AC measurement where saturation can happen.
The open loop Bode-plot test is a linear model AC simulation where actual signal levels are of no consequence or concern.
Click to expand...

Hi,

Yup
 
Heidi said:
Sorry, MrAl, I don't understand the meaning of the paragraph "if the output can no longer ..." Did you want me to look at
View attachment 91055
or this
View attachment 91056
or something else?
Click to expand...


Hello,

When we use a real op amp, the output can only go to certain levels, and even when we use an ideal op amp there are certain things that happen that help to explain the operation of the op amp circuit.
For example, with 1uv input differential and a gain of 1000 the output is only 1mv. With an input of 1v the output would pin at one rail, and that would prevent the proper feedback from getting to the inverting input because it might require 100v or even 1000v to get the right feedback to keep it in the linear mode.

What is it exactly that you are having a problem with though? Can you explain it carefully?
 
I have problem understanding the "if ..." statement in your post #123, it seems so "technical" to me. But that's fine, I can understand your explanations above quite well.

I still have many annoying questions in my head, such as "in the recent example, why was the inverting input voltage going to zero while no feedback?" But I think I'd better keep going and learning more (perhaps especially my Englisg ability) instead of keeping asking and asking (or it might become an "endless loop"). So I decided to postpone my exploration about op amps until I start the learning of the internal circuitry.

Thank you all!
 
Hello again! I know that this op amp feedback topic perhaps is too 'old' and getting too boring, but I did some experiments and calculation and I would like you see if there's anything incorrect.

For the ideal op amp "opamp" in LTspice, I set up an equivalent model which is similar to MrAl's in post #49 except that it didn't have the voltage clipping function.


For the equivalent circuit on the right-hand side, the gain Av is equal to

Av= 100K / (1 + jw*R1*C1) = Aol / (1 +jw*R1*C1), (Aol indicates the open-loop dc gain of an op amp)

which has a -3dB frequency f3db = 1/2*pi*R1*C1 and a dc gain of 100K.

According to the definition of gain bandwidth product GBW

GBW = Aol * f3db = Aol/(2*pi*R1*C1) ---- (1)

hence R1*C1 = Aol/(2*pi*GBW) ------ (2)

For GBW=1MHz, Aol=100K, C1=1uF, we have R1=15915.49Ω, and f3db=10Hz.

------------------------------------------------------------------------------------------------
From the GBW definition (1), we see that the product of the dc gain and -3dB frequency is a constant which is 10^6 in this case.
For GBW=1Mhz and C1=1uF but now Aol is 1k, then R1=159.1549Ω from (2), f3db=1KHz from (1), and Aol*f3db=10^6 as well.

-----------------------------------------------------------------------------------------------
For non-inverting configuration, a transient analysis shows they have the same output.


... Sorry, I'll be back later.
 

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Hi,

Just a quick note here...

If you want to follow the LT Spice version i think you should use the op amp model where we use a differential voltage to single ended current (with gain) stage as the basic op amp gain stage and with R and C in parallel on the output to ground. If R is made equal to 1 ohm then C is even easier to calculate:
C=Aol/(GBW*2*pi)

So the gain stage output is:
Iout=(vp-vn)*Aol

and vp is the non inverting input and vn the inverting input, and R and C are in parallel with the output of this stage.
 
Hi MrAl,

Did you want me to use the circuit like the one on the left hand side below where G1 is a voltage-controlled-current-source instead of a voltage-controlled-voltage-source, and R is connected to ground?


For the left circuit, let Vd=vp-vn in which Vd, vp, vn are all in phasor forms.

Vout1/Vd = Aol*[R/(s*C)] / [R+1/(s*C)] = Aol*R / (1+R*C*s)
in which the -3dB frequency
f3db = 1/2*pi*R*C,
the dc voltage gain = Aol*R,
therefore GBW=Aol/(2*pi*C),
or C=Aol/(GBW*2*pi).

Although the current flowing out of G1 is
(vp-vn)*Aol,
the voltage Vout1=Vout2.

But I have a feeling that I didn't draw the left circuit the way you want me to. Did I?
 
Hi,

I think that looks right. I suggested that so your results would agree with the LT Spice op amp you were using previously.
 
MrAl said:
Hi,

I think that looks right. I suggested that so your results would agree with the LT Spice op amp you were using previously.
Click to expand...
But I thought the two circuits in post #133 are equivalent if I considered their input differential voltages (vp-vn), output voltages or -3dB bandwidths. Besides, the (VCCS, R) pair is the Norton equivalent of the (VCVS, R) pair if R=1Ω. The results in post #130 using the VCVS version approximated those the ideal op amp in LT Spice resulted.

Or did I misunderstand your point?

Thank you.
 
Thank you very much, cowboybob.

In that Analog Engineer's Pocket Reference, page No.41 "op amp bandwidth":

the author defined GBW using "closed loop gain", but next in a plot, they used "open loop gain" vs frequency:

I remember seeing the same thing in another TI paper. Do you know what the possible reason is?

There's one more thing, the circuit below is also from TI( **broken link removed** ), it seems to used to determine the open-loop gain of a op amp, do you know why they use (V+ /2) as one of the input supply, not (-V+)? How would you analyze that circuit? `Thank you!
 


Hello,

Yes that is true if R=1, but if R is not equal to 1 then they are different. So if you use the second type then just keep R=1. The reason is R also acts as a gain in one of them but not the other.
 
Gain x BW is the definition of GBW. The plot of this for all possible gains is shown in the open-loop gain versus frequency, for which the GBW is a constant for typical op amps.
If you draw a horizontal line at a particular closed-loop gain, then where it intercepts the open-loop plot is the circuit bandwidth at that gain.
I don't quite understand your confusion on this. (?)

The bottom op amp plus input is biased at 1/2 V+ since it is being powered by a signal supply and, in general, you want to bias the op amp at the middle of the total supply voltage for proper operation. It's the same as powering the op amp from (V+/2) and (-V+/2) and grounding the plus input.
 
Sorry, my confusion is why some people define GBW using closed-loop gain while some using open-loop gain.

To solve this mystery, I tried to find the voltage transfer functions for both an inverting and non-inverting op amp configurations and their corner frequencies. For a certain kind of op amps, I got a conclusion:
(the closed-loop dc gain)*(bandwidth in that closed-loop condition)
=(the open-loop dc gain of an op amp itself)*(open-loop bandwidth of that op amp)
= a constant for that op amp, which is why we have a GBW value for an op amp

I'm not sure if that is a general rule, hope that you could correct some of the mistakes. Here's what I did:


crutschow said:
If you draw a horizontal line at a particular closed-loop gain, then where it intercepts the open-loop plot is the circuit bandwidth at that gain
Click to expand...
Could you please show me how to prove that statement?

Although a simulation agrees with the prediction:

(Vout2 represents the closed-loop gain variation with frequency, Vout1 represents the open-loop gain variation of that op amp model with frequency.)

I'd still like to know how we obtain such a powerful tool. I tried the following approach to get a provement, but failed.

Ac = the closed-loop dc gain of some op amp configuration
fc = the bandwidth of that closed-loop configuration
Ao = the open-loop dc gain of the op amp
fo = the open-loop bandwidth of the op amp

What I know is that the two rectangular areas are equal:
Ao*fo = Ac*fc
and the two curves have the same slope: -20dB/decade.

If I can show that a=b or a/b=1, the statement is proved:
a=fo*10^(Ao/20)
b=fc*10^(Ac/20)

a/b=(fo/fc)*10^((Ao-Ac)/20)
but it doesn't seem to be equal to unity ...
 

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op amp works simply by using as much gain as available to ensure the differential input is zero.

Since the inverting input has two sources the original input and the output via passive components, both must add up to zero or whatever voltage is on (+) side.

The attenuation from the output to (-) is what controls the gain and ends up being just the Z impedance ratio of the -Zfb/Zin(-) or in the simplest case -Rf/Rin
 
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