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?
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ATrue
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BFalse
Answer
Correct Answer: False
Explanation
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:
Write impedances: ZL = jωL, ZR = R.High-pass output across L: magnitude rises with frequency, approaching 1 as ω → ∞.Low-pass output across R: magnitude falls with frequency beyond the break ω = R/L.Hence, “output across R” for an RL differentiator is incorrect.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