Problem 75 An input voltage can be differentiated in time by the R following op amp differentiator circuit: (a) Using the method of node voltages, prove V; that the following relationship exists between the input and output voltages: dvi v.(t) = –RC dt (b) Using R = 250 k2, C = 10 µF, and vi(t) = 12*r(t) volt where r(t) is the unit ramp input. Hint: r(t) = t*u(t) where u(t) is the unit step input. In other words, vi(t) = 12*t for t>0 but is equal to zero for t<0.

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**Problem 75**

An input voltage can be differentiated in time by the following op amp differentiator circuit:

![Differentiator Circuit Diagram](differentiator_circuit.png)

**(a)** Using the method of node voltages, prove that the following relationship exists between the input and output voltages:

\[ v_o(t) = -RC \frac{dv_i}{dt} \]

**(b)** Using \( R = 250 \, \text{k}\Omega \), \( C = 10 \, \mu\text{F} \), and \( v_i(t) = 12 r(t) \) volt where \( r(t) \) is the unit ramp input.

*Hint: \( r(t) = t \ast u(t) \) where \( u(t) \) is the unit step input. In other words, \( v_i(t) = 12 \ast t \) for \( t > 0 \) but is equal to zero for \( t < 0 \).*

**Explanation of the Circuit Diagram:**

The diagram depicts an operational amplifier (op amp) differentiator circuit. It includes:
- An input voltage source \( v_i \) connected to a capacitor \( C \).
- The capacitor \( C \) connects to the inverting input of the op amp.
- A feedback resistor \( R \) is connected from the output to the inverting input.
- The non-inverting input is grounded.
- The output voltage is denoted as \( v_o \). 

This configuration allows the output voltage to be proportional to the rate of change of the input voltage, thus differentiating the input signal.
Transcribed Image Text:**Problem 75** An input voltage can be differentiated in time by the following op amp differentiator circuit: ![Differentiator Circuit Diagram](differentiator_circuit.png) **(a)** Using the method of node voltages, prove that the following relationship exists between the input and output voltages: \[ v_o(t) = -RC \frac{dv_i}{dt} \] **(b)** Using \( R = 250 \, \text{k}\Omega \), \( C = 10 \, \mu\text{F} \), and \( v_i(t) = 12 r(t) \) volt where \( r(t) \) is the unit ramp input. *Hint: \( r(t) = t \ast u(t) \) where \( u(t) \) is the unit step input. In other words, \( v_i(t) = 12 \ast t \) for \( t > 0 \) but is equal to zero for \( t < 0 \).* **Explanation of the Circuit Diagram:** The diagram depicts an operational amplifier (op amp) differentiator circuit. It includes: - An input voltage source \( v_i \) connected to a capacitor \( C \). - The capacitor \( C \) connects to the inverting input of the op amp. - A feedback resistor \( R \) is connected from the output to the inverting input. - The non-inverting input is grounded. - The output voltage is denoted as \( v_o \). This configuration allows the output voltage to be proportional to the rate of change of the input voltage, thus differentiating the input signal.
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