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Energy consumption in optical modulators for interconnects

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Abstract

We analyze energy consumption in optical modulators operated in depletion and intended for low-power interconnect applications. We include dynamic dissipation from charging modulator capacitance and net energy consumption from absorption and photocurrent, both in reverse and small forward bias. We show that dynamic dissipation can be independent of static bias, though only with specific kinds of bias circuits. We derive simple expressions for the effects of photocurrent on energy consumption, valid in both reverse and small forward bias. Though electroabsorption modulators with large reverse bias have substantial energy penalties from photocurrent dissipation, we argue that modulator diodes with thin depletion regions and operating in small reverse and/or forward bias could have little or no such photocurrent energy penalty, even conceivably being more energy-efficient than an ideal loss-less modulator.

©2012 Optical Society of America

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Figures (4)

Fig. 1
Fig. 1 (a) Example drive circuit for a p-i-n modulator diode. (b) Equivalent circuit.
Fig. 2
Fig. 2 Diode band diagrams at (a) zero bias; (b) reverse bias V TOT ; (c) forward bias V F .
Fig. 3
Fig. 3 Diode band structures for (a) indirect gap materials and (b) heterostructures.
Fig. 4
Fig. 4 Example drive circuit for a forward-biased modulator diode.

Equations (21)

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Δ E CES =(1/2)C[ ( V B + V DD ) 2 V B 2 ]=(1/2)C V DD 2 +C V DD V B
Δ E R =C 0 V DD ( V DD V OUT ) d V OUT =(1/2)C V DD 2
Δ E DISS =2Δ E R =C V DD 2
Δ E bit =(1/4)C V DD 2
δ E diss =ω+e V TOT
E PCA ( V TOT )= E abs ( V TOT )+ V TOT Q PC = E abs ( V TOT )[ 1+ V TOT /( ω/e ) ]
β= E bittrans / E bittot
E bittrans =| E 1 E 0 |/2
E outo ( V TOT )=[ 1η( V TOT ) ] E ino
E abs ( V TOT )=η( V TOT ) E ino
E bittrans =(1/2)| E outo ( V DD + V B ) E outo ( V B ) |= E ino | η hi η lo |/2
E bittot =(1/2){ [ E PCA ( V DD + V B )+ E outo ( V DD + V B ) ]+[ E PCA ( V B )+ E outo ( V B ) ] }
E bittot = E ino [ 2+ η hi μ hi + η lo μ lo ]/2
β= | η hi η lo | 2+ η hi μ hi + η lo μ lo
β=ΔT/2
β 1 η lo 2+ μ hi + η lo μ lo
β1/( 2+ μ hi )
β 1 η hi 2+ η hi μ hi + μ lo
β1/( 2+ μ lo )
β( 1 η hi )/( 2+ μ lo )
E bit =(1/4)C V DD 2 + E bittrans /β
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