I am preparing for an exam and I have a problem with the following exercise (hope my translation is good enough):

We have an op-amp circuit with gain \$G=+1V/V\$ for AC signals and \$G=0V/V\$ for DC. The op-amp has two power supplies. The input resistance (for medium-range AC signals) is \$50k\Omega\$. Assume that additional capacitors (if needed) are infinitely large.

The input signal - square wave with duty cycle \$50\%\$, has \$V_{pp}=50mV, \ f=1MHz\$ and offset \$ 5V\$.

Op-amp parameters: \$A_0=10^5V/V, \ f_{UG}=2MHz, \ I_+=I_-=2\mu A, \ SR=2V/\mu s\$

  1. Draw a diagram of the circuit, calculate every resistance
  2. On one graph, draw \$v_{in}(t)\$ and \$v_{out}(t)\$
  3. Calculate time \$t_x\$ which passes from the moment when output signal reaches \$-20mV\$ to the moment when it reaches \$+20mV\$

First question: is information about input bias currents important?

Secondly, I think that the circuit will look like this:

enter image description here

And here, for medium-range AC signals \$R_{in}=R_3=50k\Omega\$

In this circuit \$G=1+\frac{R_2}{R_1}=1V/V\$, hence \$R_1=+\infty\$

And since \$R_3\$ is a compensating resistor and it sees only \$R_2\$, then we got: \$R_2=R_3=50k\Omega\$

Because the amplitude of the input signal is equal \$50mV\$, then there is a possibility that SR effect can occur. We need to check it. \$A_{in}=A_{out}=50mV\$

I am not sure about English naming, thus:

\$f_{UG} \ - \ \text{Unity Gain, open loop} \\ f_{CL} \ - \ \text{-3dB point, closed loop} \$

And we have:



\$A_{out}\cdot2\pi\cdot f_{CL}=50mV\cdot2\pi\cdot2MHz\approx 0.63V/\mu s < SR=2V/\mu s\$

There will be no SR effect.

Rise time:



Hence, \$t_r\$ is the answer for question no. 3.

And the graph, I think it will look like this

enter image description here

Is my solution correct?

  • \$\begingroup\$ I think your graph is slightly off. For the increasing curve, subtract 25 mV, and for the decreasing one, add 25 mV. Your circuit is a highpass filter, and you shouldn't be seeing any constant level besides the 0 level. \$\endgroup\$
    – hcabral
    Commented Jul 11, 2018 at 0:56

1 Answer 1


Your calculations look correct to me, I don't see why the exact value of the input bias current would be relevant in this case. However I'm just a student myself.

Good luck with your exams!


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