# Voltage Follower - OPAMP

I am going through an OPAMP Voltage follower/Buffer Circuit.

I have 1 question regarding this Voltage follower circuit using OP-AMP and 2 questions in general regarding OP-AMP:

1. Let's say my Vcc rails of the OP-AMP are +5V and 0V. And I am giving a constant Vin=2V to the Non-inverting terminal of the OP-AMP and the Output terminal of the OP-AMP is fedback to the Inverting terminal without any feedback resistance.

I know, since this is a voltage follower circuit, the Output voltage follows the input voltage until the input voltage hits the saturation voltage (Vcc) of the OP-AMP.

But how does the voltages start initially?

Like, at the start, the voltage at the output terminal is 0V and 0V is at the Inverting terminal as well. Now, we give the Vcc power to the OP-AMP and the Input voltage to the non-inverting terminal. Vcc=+5V. And Vin=2V. Since the voltage at the non-inverting terminal is greater than the voltage at the inverting terminal, in this case, 2V and 0V, the Output should go to the saturation voltage, right? (Assuming this OP-AMP has high enough gain say 10000) So, the output goes to 5V and then this 5V is fedback to the inverting terminal which then makes the output go to negative saturation and hence output 0V (as -Vcc = 0V)

Where am I going wrong with the Voltage follower operation? In my understanding, the output still swings from +Vcc to -Vcc only.

I can understand the circuit with the Help of Math and the equations of Gain and the resistors of Feedback and so on. But when understanding with the intuitive working, I am misunderstanding. Please tell me how this voltage follower works at the initial start-up and how the output follows the input without any math or equations.

1. The Ideal characteristic of OP-AMP is that the differential voltage between the input terminals is 0V.

But in the real world, we give different input voltages to the terminals so as to amplify the difference between them, right?

If an ideal OP-AMP maintains the differential voltage between the input as 0V, then the OP-AMP can't amplify the differential voltage between the terminals, right?

Isn't this contradictory to the actual purpose of the OP-AMP? What am I missing here?

Last question,

1. The OP-AMP has high input impedance and Low output impedance. This means that the OP-AMP input terminals does not have any input currents drawn from the source.

And Low output impedance means, that it can produces how much ever current the load requires? Is there a limitation of how much current the OP-AMP can source?

And the output current to the load comes from the +Vcc power terminals of the OP-AMP right? Is there a limitation to the amount of current the OP-AMP can provide?

• If you have time try to read this electronics.stackexchange.com/questions/441184/…
– G36
Apr 5, 2020 at 6:55
• The op-amp output will not jump from 0V to 5V instantaneously. It will take some time. There will be a voltage "ramp" at the output. Also, as the output voltage increases from 0V towards Vcc the differential voltage seen by the amplifier input (Vdif = V+ - V- ) will be smaller and smaller. For example, at time 0s we have Vout = 0V and Vdif = (2V - 0V) = 2V but at time = 1s we have Vout = 1V at the output and Vdif = 2V -1V = 1V and at time 1.5s we have Vout=1.5V and Vdiff = 2V-1.5=0.5V.
– G36
Apr 5, 2020 at 9:16
• And because the Vdiff is getting smaller and smaller the rate of change of an output voltage will also get smaller. I should pointed out that any output voltage larger than 2V will change the sign of a Vdiff voltage. And this will cause that the output voltage will stop increases and it will start to decreases into a negative direction towards Vee.
– G36
Apr 5, 2020 at 9:17
• And the circuit will reach the equilibrium when this equation is met Vo = AOL*(Vin - V-) (only one solution is possible). And for the Voltage follower, we have $$V_0 = \frac{A_{OL}}{1 +A_{OL}} \times V_{IN}$$ Where Aol is a opamp open-loop gain
– G36
Apr 5, 2020 at 9:23