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

  • \$\begingroup\$ Post a picture of the problem from the book because how you have interpreted it misses vital information. \$\endgroup\$
    – Andy aka
    Commented Oct 9, 2022 at 9:44
  • \$\begingroup\$ This can be solved using a Bode plot and knowledge of the phase response of single pole low-pass systems. \$\endgroup\$
    – user57037
    Commented Oct 10, 2022 at 5:57

2 Answers 2


The open loop gain and bandwidth define a gain Bode plot. Since the bandwidth is given as 10 Hz, we can assume there is a single pole in the Bode plot at 10 Hz and the slope is -20 dB per decade. Here is the plot.

Note that a gain of \$ 10^5 \$ is 100 dB. And a gain of 1 is 0 dB.

enter image description here

The effect of feedback is to reduce the DC gain and increase the bandwidth. The Bode plot provides a simple way to locate the new bandwidth, because the gain decreases at 20 dB per decade of frequency.

This type of plot can be hand sketched also.

For a single pole low-pass response, the output phase offset, \$ \theta \$, can be calculated using this formula: (found here)

$$\theta = -arctan(2 \times \pi \times f \times \tau)$$

\$ \tau \$ is the time constant of the response, and f is the frequency in Hz. We can find \$ \tau \$ for the closed-loop amplifier using the low-pass bandwidth. For the closed-loop amplifier, the bandwidth is 1 MHz. First we convert that to radians per second: \$ 1 MHz \times 2 \times \pi = 6,280,000 \$ rad/second

\$ \tau \$ is the reciprocal of that. \$ \tau = \left( \frac {1} {6,280,000} \right) = 159 ns \$

Now we can calculate phase lag, \$ \theta \$, at 80 kHz for the closed loop amplifier.

$$ \theta = -atan(2 \times \pi \times 80,000 \times 159 \times 10^9) = -0.0798 radians $$
$$ -0.0798 \times \frac {180}{\pi} = -4.57 degrees $$

All images were created by me.


Why does this lag take place?

Remember that op-amp are (1st approximation) low pass filter ... Aop is first order function.
The corner at a "low" frequency (10 Hz -> 1000 Hz).

enter image description here

Now, 20V @ 80 kHz, are you sure that "slew-rate" is ok for your op-amp?
See my answer at this post.

Why does this lag take place? How should I redraw the circuit to realise this?

Here is a Maple sheet for calculating the true answer (opamp gain first order).
NB: the op-amp output impedance not taken into account, so R is not useful.
Vm, Vp, Vo are voltage of minus input, plus input, and Vout output.

enter image description here

The last answer is the phase.

  • \$\begingroup\$ What's that input cap value?? ( 1f ) \$\endgroup\$
    – Nedd
    Commented Oct 9, 2022 at 9:07
  • \$\begingroup\$ One femtoFarad (0.001 pF), just to show some "wiring" input capacitance. Be aware that 20 V is "big" ... \$\endgroup\$
    – Antonio51
    Commented Oct 9, 2022 at 9:18

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