The Light-Emitting Diode (LED) — Question 7

Section 41-9 · Physics pages 6964–6965

Question 7 Carrier storage limits the optical modulation bandwidth Small excess population changes obey dδN/dt + δN/τ = δI/e, with constant τ = 10.0 ns. Optical modulation is proportional to δN. Use phasors proportional to eiωt. ωτ |H| Schematic; labels, not drawing lengths, specify the data. (a) Derive the normalized modulation transfer function and its minus-3-dB frequency. (b) At 30.0 MHz, calculate the amplitude ratio and phase lag.
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Question 7 - Solution Carrier storage limits the optical modulation bandwidth ωτ |H| (a) Method and calculation (iω + 1/τ) eN = eI/e ⇒H(ω) = 1/(1 + iωτ) . |H| = 1/ p 1 + ω2τ 2 ⇒f3dB = 1/(2πτ) = 15.92 MHz . (b) Interpretation and further calculation ωτ = 2π(30 × 106)(10−8) = 1.885 ⇒|H| = .4686 . arg H = −tan−1(1.885) = −62.05◦. Physical check This first-order result neglects electrical parasitics and assumes linear recombination about the bias point. A nonlinear rate law requires the differential lifetime there.

Question 7

Carrier storage limits the optical modulation bandwidth

Small excess population changes obey dδN/dt+δN/τ=δI/ed\delta N/dt+\delta N/\tau=\delta I/e, with constant τ=10.0ns\tau=10.0\,\mathrm{ns}. Optical modulation is proportional to δN\delta N. Use phasors proportional to eiωte^{i\omega t}.

Diagram shown in the original SVG worksheet.

  1. Derive the normalized modulation transfer function and its minus-3-dB frequency.

  2. At 30.0 MHz, calculate the amplitude ratio and phase lag.

Question 7 - Solution

Carrier storage limits the optical modulation bandwidth

Diagram shown in the original SVG worksheet.

(a) Method and calculation

(iω+1/τ)Ñ=Ĩ/e⇒H(ω)=1/(1+iωτ).(i\omega+1/\tau)\widetilde N=\widetilde I/e\Rightarrow\boxed{H(\omega)=1/(1+i\omega\tau)}.|H|=1/1+ω2τ2⇒f3dB=1/(2πτ)=15.92MHz.|H|=1/\sqrt{1+\omega^2\tau^2}\Rightarrow\boxed{f_{3\mathrm{dB}}=1/(2\pi\tau)=15.92\,\mathrm{MHz}}.

(b) Interpretation and further calculation

ωτ=2π(30×106)(10−8)=1.885⇒|H|=.4686.\omega\tau=2\pi(30\times10^6)(10^{-8})=1.885\Rightarrow\boxed{|H|=.4686}.arg⁡H=−tan⁡−1(1.885)=−62.05∘.\boxed{\arg H=-\tan^{-1}(1.885)=-62.05^\circ}.

Physical check

This first-order result neglects electrical parasitics and assumes linear recombination about the bias point. A nonlinear rate law requires the differential lifetime there.