# Non-inverting op-amp circuit with gain = 1

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: 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:

$f_{CL}=\frac{f_{UG}}{G}=f_{UG}=2MHz$

So:

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

There will be no SR effect.

Rise time:

$t_r=\frac{0.35}{f_{CL}}=175ns$

$10\%\cdot50mV=5mV$

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

And the graph, I think it will look like this Is my solution correct?

• 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. Jul 11 '18 at 0:56