# why opamp input track

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#### mona123

##### Member level 5 I am still not clear what is the physical mechanism that makes opamp inputs track each other? we say it is because opamp has high gain. can someone explain better. how high a gain is needed? thanks.

#### crutschow The inputs of an op amp are differential, that is, the output voltage is multiplied by the difference voltage between the two inputs.
The gain is how much the output voltage is multiplied by the difference voltage on the input.
The voltage difference at the input is only low if the op amp is configured with negative feedback from the output to the (-) input (either as an inverting amp or as a non-inverting amp).
Here's a more detailed discussion of that.

#### D.A.(Tony)Stewart With a linear gain of >1 million, the output can only be linear by negative feedback to force in differential input near zero.

#### mona123

##### Member level 5 thanks sunny , can you explain a little better? what do you mean by linear gain?

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thanks crutschow, so are you saying input difference is low due to negative feedback and is not dependent on open loop opamp gain? that's where I am confused. so if I make a single stage or two stage opamp it won't matter?

#### vGoodtimes The output is given by:
Vout = Av*(Vnoninverting - Vinverting) = Av*Vdiff

This means Vdiff = Vout/Av

When Av is 10, Vdiff will be Vout/10. Eg, for 10V of output, Vdiff is 1V.
When Av is 100000, Vdiff will be Vout/10000. Eg, for 10V of output, Vdiff is 0.0001V (100uV)

As gain increases, the Vdiff decreases. For easy analysis, the inputs are assumed to track. Further analysis of the circuit will show if the gain is suitably high for this approximation to work.

• mona123

### mona123

Points: 2

#### LvW Perhaps it helps to analyze what happens after switch-on the opamp power supplies:

Example: Supply voltages +/- Vs=+/-10V; non-inverting gain stage with gain of "+2". That means: Feedback factor k=0.5.

1.) Apply at t=0 an input voltage Vin=1V. The opamp is not yet working in its linear range and the output will be either at Vs=10V or Vs=-10V (lets assume +10V)
2.) The voltage at the inverting terminal will be 0.5Vs=5V>Vin=1V. Hence, the voltage at the inverting terminal dominates (is larger) and the output voltage will change in the direction to -Vs.
3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inv. terminal a feedback voltage of +0.99980004V .
4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region and produces an ouput of nearly +2V: (1-0.99980004)*1E4=0.00019996*1E4=1.9996001V.
5.) That means: We have an equilibrium because the output voltage has a value which exactly meets the closed-loop gain Vout=Vin*[1E4/(1+0.5*E4)]=1.9996001 V.
6.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.0002V) that in can be neglected (assumed to be zero for calculations) in many cases.

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• mona123

### mona123

Points: 2

#### abhishekgrover

##### Newbie level 5 Physical mechanism that makes opamp input track each other:
Opamp inputs track each other only in the case of negative feedback. In the other two configurations, positive feedback and open loop ,there is no virtual short between the two inputs of opamp. So one should first make this clear that this property of inputs tracking each other or virtual short between the two input terminals is a characteristic of the negative feedback configuration. This theory of negative feedback was originally developed for control systems and then after some time it was used in analog circuit design. To make you understand the physical mechanism I would first start with the negative feedback theory. Consider the figure below: In negative feedback we are subtracting y (output) from x (input) and the result e is then given as input to the amplifier. The amplifier amplifies e and the result multiplied by the feedback element is then subtracted from input. When the input signal increases, error e also increases, correspondingly the output of the amplifier also increases and therefore y also increases. Now this increased value is then subtracted from x . When the input signal decreases, error e also decreases, correspondingly the output of the amplifier also decreases and therefore y also decreases. Now this decreased value is subtracted from x . Thus we can see that the negative feedback is working to minimise the error. This property of minimising the error between the inputs is a characteristic of the negative feedback. Now to understand we use the opamp in place of the amplifier A block and the feedback network is replaced by the R2 and R1 network. To understand the physical mechanism one should also know what are the blocks inside the opamp. An opamp basically consists of three stages: differential amplifier at the input, an intermediate amplifier for gain and an output amplifier as the output stage. The differential amplifier acts like a subtractor of the negative feedback configuration shown above , subtracting the output of the feedback network from the input, thus minimising the error between the two and this is how the two inputs of the opamp track each other. Although for calculations we do take that the two terminals are virtual short but there is a potential difference between the two inputs as the opamp we use is not an ideal voltage amplifier. As the specifications like voltage gain(infinite for ideal), input impedance(infinite for ideal) of the opamp we are using moves closer to that of the ideal voltage amplifier the potential difference between the two terminals decreases and we can say that the two terminals are virtually short.

• mona123

### mona123

Points: 2

#### crutschow ........................
thanks crutschow, so are you saying input difference is low due to negative feedback and is not dependent on open loop opamp gain? that's where I am confused. so if I make a single stage or two stage opamp it won't matter?
For the input difference to be small, the op amp must be operating in a linear mode with negative feedback.
However the amount of voltage difference between the two inputs for a given output voltage is determined by the op amp open loop gain.

#### mona123

##### Member level 5 Perhaps it helps to analyze what happens after switch-on the opamp power supplies:

Example: Supply voltages +/- Vs=+/-10V; non-inverting gain stage with gain of "+2". That means: Feedback factor k=0.5.

1.) Apply at t=0 an input voltage Vin=1V. The opamp is not yet working in its linear range and the output will be either at Vs=10V or Vs=-10V (lets assume +10V)
2.) The voltage at the inverting terminal will be 0.5Vs=5V>Vin=1V. Hence, the voltage at the inverting terminal dominates (is larger) and the output voltage will change in the direction to -Vs.
3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inv. terminal a feedback voltage of +0.99980004V .
4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region and produces an ouput of nearly +2V: (1-0.99980004)*1E4=0.00019996*1E4=1.9996001V.
5.) That means: We have an equilibrium because the output voltage has a value which exactly meets the closed-loop gain Vout=Vin*[1E4/(1+0.5*E4)]=1.9996001 V.
6.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.0002V) that in can be neglected (assumed to be zero for calculations) in many cases.

#### D.A.(Tony)Stewart With a linear gain of >1 million, the output can only be linear by negative feedback to force in differential input near zero.

By linear gain it is meant that the output is in the linear range ( not saturated to rail) and the input is in the linear range ( defined by CM range limits) and the output is proportional to the input.
The open loop gain is what I meant to say.

• mona123

Points: 2