I was solving a problem related to voltage follower in Fitzgerald's *Basic Electrical Engineering*, 5th ed., pages 458-459. The op-amp characteristics are (respectively open loop gain, open loop bandwidth, input resistance):

$$\begin{align}
&A\_0=10^5\\
&f\_h=10\ \mathrm{Hz}\\
&R\_s=10\ \mathrm{k\Omega}\\
\end{align}$$

If the input voltage is:

$$v\_s(t)=20\cos(0.5\text{M} \times t)^\*$$

Then what is the output, measured across the load R = 1 kΩ? The book says it should be 20 V, which is obvious since closed loop gain is (very, very nearly) 1, but lagging the input by 4.6°.

Why does this lag take place? How should I redraw the circuit to realise this? The only thing that I could conclude is that the equivalent voltage controlled voltage source is:

$$A(v\_s-v\_o)=A[20\ \mathrm{V}\angle0 -20\ \mathrm{V}\angle(-\theta)], \text{where} -\theta \text{ is the lag.}$$

[![Voltage follower](https://i.stack.imgur.com/7l0MH.png)](https://i.stack.imgur.com/7l0MH.png)