Unity Gain Buffer Initial Conditions

Question

^{simulate this circuit – Schematic created using CircuitLab}

A typical way of analysing the unity gain buffer is:

Where A is Open Loop Gain

Vout = A(Vin - Vp) and Vp = Vout therefore

Vout = A(Vin - Vout)

Vout = AVin - AVout

Vout + AVout = AVin

Vout (1+A) = AVin

Vout / Vin = A / (1+A)

if A >> then 1+A ~ A

Vout / Vin = A/A = 1

I would like to get an understanding of how this works from initial conditions ie step response

At t= -1 Vin, Vp and Vout = 0

At t= 0 Vin = 1V and as Vp = 0, Vout = A(1-0) = A

As Vout = A then Vp = A such that Vout = A(1 - A) = A - A^2

Subbing again gives Vout = A(1 - (A-A^2) = A(1-A+A^2)

This is obviously just going to run away so must be incorrect, perhaps it is my initial condition assumption that is wrong?

Alternatively using the simplified model of an op amp:

I could assume that at t=0 Vp = Vin therefore Vout = Vin

However Vin = A(Vin - Vin) is a contradiction

Obviously this is a well known circuit so I must have a fundamental misunderstanding but what have I got wrong?

*Edit

Just to clarify the purpose in asking,

I'm trying to get an understanding of how negative feedback works. The standard analysis at the start of the question is effectively time independent and I would like to know how the output rises with time

Answer 1

When you are considering a step response, you need to include the time delays at the different locations for the answer to be reasonable. One way to model this is to add an RC filter at the output of the op-amp. The other thing that you haven't accounted for is that the op-amp can't actually drive the output to greater than the positive supply voltage.

Taken in combination, and assuming Vs = 5V, this means after the positive input steps from 0 to 1, the output of the op-amp would follow a curve something like this, ignoring feedback for now:

OK, now what happens if we do include feedback? Remember that the negative input to the op-amp is the same node as the op-amp output. Once the output starts to get close to 1 V, the op-amp will begin to drive less current. In particular, once \$V_{out} > 1\text{ V} - \frac{5\text{ V}}{A}\$, the output stops driving high as hard as possible. This doesn't mean it flattens out immediately, but the endpoint/goal of the RC settling starts to drop. I will call this goal point Vfinal and I've plotted it alongside Vout in the plot below:

Zooming in to the time when Vfinal drops:

Here is the schematic for the above plots (though I actually just simulated them in python):

^{simulate this circuit – Schematic created using CircuitLab}

Python to generate the above plots, if you are interested:

from matplotlib import pyplot as plt
import numpy as np
a = 1000
vp = 1.
vs = 5.
ts = np.linspace(0, 0.3, 10000)
v_rc = 5.0 * (1.0 - np.exp(-ts))
# plt.plot(ts, v_rc)
# plt.axis([0, 8, 0, 5])
# plt.xlabel('time (# time constants)')
# plt.ylabel('voltage')
# plt.show()

vout = 0.
vfinals = [0.]
vouts = [0.]
for i, t in enumerate(ts[1:], 1):
    vfinal = a * (vp - vout)
    vfinal = np.clip(vfinal, 0., 5.)
    vfinals.append(vfinal)
    dt = t - ts[i - 1]
    dv = (vfinal - vout) * dt
    vout += dv
    vouts.append(vout)

plt.plot(ts, vouts, '-', ts, vfinals, '--')
# plt.axis([0.21, 0.24, 0, 5.1])
plt.axis([0.22, 0.233, 0.98, 1.02])
plt.xlabel('time (# time constants)')
plt.ylabel('voltage')
plt.legend(['Vout', 'Vfinal'])
plt.show()

Answer 2

I'm trying to get an understanding of how negative feedback works.

Think of it like a control system and an error amplifier.

You set the "demand" on the non-inverting input and the negative feedback causes the op-amp output to rapidly change to a voltage that makes the input error minimal. In other words, the inverting input is made to be nearly the same voltage as the "demand".

Because the open loop gain is massive at low frequencies, the amount of error required to "drive" the control system into stabilization is quite small.

You can, if you want, regard the op-amp as a kind of motorized control system: -

When you grasp this, replace the motor and position measurement potentiometer with a wire link and you have a simple op-amp buffer. It's easy after that!

Picture from this q and a.

Answer 3

In addition to Justin's answer here's a mathematical way of looking at it:

Using Vout = A(Vp - Vn)

The incorrect assumption that I was making was that Vout == Vp == Vn which is wrong as: Vp - Vn = 0, which is a contradiction as this would make Vout = 0

Looking as Justin's answer we can see that Vout (and therefore Vn) start at 0v and rises with time until it flattens at a value near Vp (Which we now know cant be Vp)

So as a conjecture, is it possible that for some small value d and a given value of Vp that Vout and Vn both equal (Vp - d):

(Vp - d) = A(Vp - (Vp - d)) = A(Vp - Vp + d) = Ad,

Giving:

Vp = Ad +d = d(A+1)

Therefore:

d = Vp / A+1

Testing this with values: Vp = 1V, A = 1000

d = 1 / 1001 ~ 9.99E-4

1 - 9.99E-4 = 1000(1 - (1 - 9.99E-4))

1 - 9.99E-4 = 1000(9.99E-4)

0.999 = 0.999 - conjecture proven

Adding to this, in the case of increasing or decreasing Vp (in unsaturated operation)

if Vp increases then the output will rise as Vp - Vn has increased until Vout (and therefore Vn) reaches (Vp - d)

if Vp decreases then Vp - Vn will have decreased meaning Vout will decrease in turn reducing Vn until Vn = (Vp-d)

Hopefully this explanation helps others as the combination of this and Justin's' answer makes it far easier form me to understand how negative feedback works as opposed to what it does.