question on op amp circuit

Discussion in 'Analog & Mixed-Signal Design' started by debdut123, Sep 27, 2016.

1. debdut123 Thread Starter New Member

Aug 30, 2016
10
1
Hello, I had a question in op amp?
If we provide equal dc voltages to an op amp it will produce some voltage 'Veq'.
The op amp has a finite gain. My understanding is, if we use this circuit in a negative feedback loop of another circuit X, its input voltages come closer to each other.
However they do not necessary become exactly equal. I think, they will become equal when the circuit X requires the op amp output voltage to be Veq for stabilization.
In other cases, when inputs are not exactly equal, what is this input difference called? Is it input offset voltage?

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2. #12 Expert

Nov 30, 2010
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Offset voltage inherent in the (not perfect) differential input stage of the amplifier, offset voltage developed by the input bias current flowing through the resistors of the external circuit, and the very tiny difference caused by the fact that the op-amp gain is finite.

It seems to me that, If an op-amp has an open loop gain of 1,000,000, the input voltage difference caused by this factor must be equal to 1/1,000,000 of the output voltage. I'm sure there is math for this, which I don't know, but the logic is good.

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3. debdut123 Thread Starter New Member

Aug 30, 2016
10
1
Thank you #12 for the reply.
In my circuit, I have not considered mismatches between the input transistors. I have simulated through process variations only and the input voltages of the opamp in feedback circuit seems to change.
I am writing a report on it. Shall I call the input voltage difference the input offset voltage because of finite gain of the op amp?

4. #12 Expert

Nov 30, 2010
16,704
7,354
The input voltage offset due to mismatch of the input transistors is a characteristic of the amplifier chip. It is noted on all datasheets for op-amps and comparators. It is called, "Vio" (Voltage input offset). What you seem to be examining is the offset caused by finite gain. There must be a tiny difference in the input voltages to cause the output to change. This is related to the gain of the chip. You must also consider the input bias current passing through the external resistors in your circuit. Any further comments need the circuit you are using. You can post a schematic.

5. debdut123 Thread Starter New Member

Aug 30, 2016
10
1
I have attached an image.
I am trying to make the two currents equal through the op amp. All the resistors can be thought equal.
Currents are not exactly equal because of a small offset between the op amp's inputs.
This is due to the finite gain of the op amp.

NOTE: Very sorry!! I incorrectly inverted the op amp terminals.

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6. #12 Expert

Nov 30, 2010
16,704
7,354
I don't understand. What method are you using? What are your measurements? Where is the error? What are the voltages.? What size are the resistors? Which op-amp are you using?

I'm good with op-amps, but I'm not going to guess 11 variables, the method you're using, what you're trying to accomplish, and try to give you an answer.

Last edited: Sep 27, 2016
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