RL differentiator topology: In an RL differentiating circuit (first-order high-pass using RL), the output voltage is taken across the resistor. True or false?

Introduction / Context:
For RL networks, which node yields differentiation (high-pass, edge emphasis) and which yields integration (low-pass) depends on whether the output is across the inductor or the resistor. This affects timing circuits, snubbers, and pulse-shaping.

Given Data / Assumptions:

• Series RL network driven by a voltage source.
• Seeking differentiator behavior (high-pass).
• Assume small τ = L/R relative to input edge times for strong differentiation.

Concept / Approach:

Taking the output across the inductor gives a high-pass transfer H(jω) = jωL / (R + jωL). Output across the resistor yields the complementary low-pass H(jω) = R / (R + jωL). Therefore, an RL differentiator uses the inductor node, not the resistor node.

Step-by-Step Solution:

Verification / Alternative check:

Apply a step input. The inductor node shows a sharp spike (dV/dt response) that decays with τ = L/R, consistent with differentiation. The resistor node shows a conventional exponential toward the step value (integration/low-pass).

Why Other Options Are Wrong:

• Choosing “True” reverses the RL roles compared with RC networks and leads to an unintended low-pass output.

Common Pitfalls:

Assuming duality with RC without swapping which element the output is taken across; the dual network places the output across the reactive component for RL differentiators.

Final Answer:

False