The circuit shown is a differentiator, where the operational amplifier (op-amp) configuration, along with the resistor $R$ and capacitor $C$, responds to changes in the input voltage $V_{in}$.

Given Information:

• Input $V_{in}$: A step function that jumps from 0 V to 5 V at $t = 0$.
• Resistor $R = 56 \, \text{k}\Omega$.
• Capacitor $C = 0.022 \, \mu\text{F}$.

Objective:

Compute the output voltage $V_{out}$.

Solution:

For a differentiator circuit, the output $V_{out}$ is given by the derivative of the input voltage $V_{in}$: $V_{out} = -RC \frac{dV_{in}}{dt}$

1. Determine $\frac{dV_{in}}{dt}$: Since $V_{in}$ is a step function that goes from 0 V to 5 V at $t = 0$, the derivative $\frac{dV_{in}}{dt}$ can be approximated as a very large spike at $t = 0$ (infinite slope in ideal conditions).

2. Calculate $V_{out}$ immediately after the step: For a step input from 0 to 5 V, $\frac{dV_{in}}{dt}$ at $t = 0$ can be considered as an impulse. Thus: $V_{out} = -RC \cdot \delta(t) \cdot 5 \, \text{V}$

3. Magnitude Calculation: Substituting the values $R = 56 \, \text{k}\Omega$ and $C = 0.022 \, \mu\text{F}$: $V_{out} = -5 \cdot 56 \times 10^3 \times 0.022 \times 10^{-6}$ $V_{out} = -5 \cdot 1.232 = -6.16 \, \text{V}$

Thus, the output voltage $V_{out}$ will produce a negative spike of approximately $-6.16$ V when the step input is applied.

Would you like further details or have any questions?

1. How does a differentiator circuit respond to other types of input signals?
2. What would happen if the input was a sine wave instead of a step function?
3. How does the choice of $R$ and $C$ values affect the output voltage?
4. What are practical limitations of using an ideal differentiator circuit?
5. How can this circuit be modified to limit high-frequency noise?

Tip: In ideal conditions, a differentiator's output is proportional to the rate of change of the input signal.