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14th July 2016, 07:03 #1
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determining the opamp stability state
Hi i have attached an image of a very simple opamp. I am learning the basics of opamp.
The 1st image we have the opamp amplifying the difference of the two inputs
and therefore the output voltage, X = 0.99 which is depicted in figure 2. I was taught (just a simple example as such) that now if you take the difference of the two inputs 1V 0.99V = 0.01. This difference multiplied by gain => 0.01 * 99 which results in an output of 0.99 again. And state will always remain the same and this is known to be the opamp stable point. I can understand that the opamp inputs are almost equal which is wat feedback is used for. This example is very simple.
But say i have 3V in the noninverting terminal and 2V in the inverting terminal and gain is 4. So at first (32) *4 = 4V. So now this 4V output is given to the noninverting terminal.
So next it will be (34) * 4 = 4. (4 now to the non inverting terminal) and so
Next it will be (3(4))*4 = 28V. And say out power supply can supply only to 15V So now output will be 15V and will be fed back. And so (315)*4 =48V... This example keeps changing the output. How can i bring it to a stable state like in the first example shown. Like to find the stable point for this example in a intuitive way without any equation. Can someone pls help!!! Thank you!!! :)

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14th July 2016, 08:24 #2
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Re: determining the opamp stability state
From your first example you state the two op amp inputs are almost equal at the stable point, so just apply that reasoning to the second case.
When 3V is applied to the noninverting input, the output will start to increase until the inverting input voltage essentially equals the noninverting input voltage. Thus the output voltage equals the input voltage at the stable point.
Don't know where you came up with a gain of 4 for the op amp.
Op amps typically have an openloop gain of over a 100,000.
And how does 4V from the output get to the noninverting terminal? It goes to the inverting terminal.
The noninverting terminal is fixed at the 5V input voltage.Zapper
Curmudgeon Elektroniker

14th July 2016, 09:18 #3
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Re: determining the opamp stability state
Hi,
But say i have 3V in the noninverting terminal and 2V in the inverting terminal and gain is 4. So at first (32) *4 = 4V. So now this 4V output is given to the noninverting terminal.
So next it will be (34) * 4 = 4. (4 now to the non inverting terminal) and so
Next it will be (3(4))*4 = 28V. And say out power supply can supply only to 15V So now output will be 15V and will be fed back. And so (315)*4 =48V... This example keeps changing the output. How can i bring it to a stable state like in the first example shown.
You are thinking in "steps". Steps in time and steps in voltage.
But this is not true with analog systems.
When IN+ = 3V, IN = 2V and gain = 4 then you imagine the output jumps from 2V (because output = IN) to 4V.
But it doesn´t jump. It moves. Every OPAMP has a limited slew rate. For your example let´s assume it is 1V/us.
so imagine 10 ns of time later..
(now you may say: this 10ns is a step, too! True. But while the voltage moves continously, not in steps, we just take a snapshot 10ns later of the voltage to see what happens. You may vary the timng of your snapshots... but the result will be the same..)
Within the 10ns and a slew rate of 1V/us the voltage moves 0.01V
Now the voltages are 3V, 2.01V > (tries to target) 3.96V
10ns later: 3V, 2.02V > 3.92V
10ns later: 3V, 2.03V > 3.88V
and so on...
after a while:3V, 2.38V > 2.48V
10ns later: 3V, 2.39V > 2.44V
10ns later: 3V, 2.40V > 2.40V (hit the target. Stable)
(This is a simplified view!)
Klaus
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14th July 2016, 10:35 #4
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Re: determining the opamp stability state
The original post is showing that a negative feedback loop with pure delay (your assumed step delay) but no bandwidth limitation won't be stable.
Any real amplifier and respectively any feedback loop has however bandwidth limitations (poles in terms of linear system analysis). A loop ruled by a single dominant pole (the case of typical unity gain compensated OP) will be generally stable.
A negative feedback with delay plus slew rate limitation (a nonlinear system, by the way) will show limit cycles and is not stable.

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14th July 2016, 12:57 #5

14th July 2016, 13:30 #6
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Re: determining the opamp stability state
I presume pure delay plus slew rate limitation without a pole is unstable. But it's a theoretical case.

15th July 2016, 12:40 #7
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Re: determining the opamp stability state
Thank you all for the reply!!! I knew that opamp gain is generally very high. But i found the gain mentioned to be 4 on the following link
http://www.sanfoundry.com/linearint...configuration/
and i'm sorry i mistyped it. Yes the 4V goes to the inverting terminal and the noninverting is fixed.Last edited by preethi19; 15th July 2016 at 12:50.

15th July 2016, 12:50 #8
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Re: determining the opamp stability state
Yes  the contents of the link is somewhat confusing.
In the headline, they speak about "openloop"  and they show an ampfier symbol (triangle) with a gain of 4.
Certainly, this symbolizes an opamp with a negative feedback and a closedloop gain of 4 (but they should mention it).
My recommendation: Never blindly trust any information found in the internet.

15th July 2016, 15:08 #9
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Re: determining the opamp stability state
Permit me to remind the very basics:
1. An ideal opamp has infinite gain. This number does not play in the gain of may practical circuit.
2. The two inputs (inverting and noninverting) are always at virtual ground (what is that!)
3. All practical circuits using opamp used as an amplifier use feedback.
4. The gain of an opamp with feedback depends on the resistors in the network and not on the actual gain (which is anyway very large) of the opamp.

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15th July 2016, 16:47 #10
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