Optics 082 · Polarization, anisotropy, and modulation

Electro-Optic Modulator

How can a voltage across a crystal control the brightness of a light beam? Through the Pockels effect: an applied voltage changes the crystal's refractive index in proportion, and that shifts the light's phase. At the half-wave voltage the shift reaches half a wave. Between polarizers the phase change becomes a brightness change, and a bias voltage sets the operating point so the output rises and falls with the signal voltage.

A compensated pair of Pockels crystals in 3D. The half-wave voltage follows from the electrode geometry and the electro-optic tensor; follow the complex polarization and the RC and transit-time response, freeze the waveform and compute harmonics from the samples.

Geometry-derived half-wave voltage

V_\pi

$0.50$

Selected detector power

P_D/P_0

$50\%$

Assumed RC amplitude cutoff

f_{\rm RC}

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Electro-Optic Modulator

Source sinusoid peak amplitude

V_{\rm ac}

100\,\mathrm V

RF drive frequency

f

40\,\mathrm{MHz}

Bias as fraction of half-wave voltage

b

0.50\,V_\pi

Length of each matched crystal

L

20\,\mathrm{mm}

Driven crystal electrode gap

d

2\,\mathrm{mm}

Assumed driver series resistance

R

50\,\Omega

Analyzer: zero removed, one inserted

a

1

Prepared axis from lab Y toward Z

\alpha

45^\circ

Source RF probe phase

\varphi

90^\circ

The crystal turns voltage into phase; the analyzer decides whether that phase becomes a change in brightness. Slide the analyzer out, change the cell geometry to tune the half-wave voltage, and shift the bias to find where the output oscillates at twice the drive frequency. Capture the detector waveform before looking at its harmonics.

The crystal is a teaching material with properties fixed at 600 nm, the compensation and analyzer are ideal, and the drive circuit is an assumed lumped circuit. Commercial bandwidth, thermal and acoustic effects are not modeled, and nothing is calibrated against experiment. Transverse dimensions are enlarged sixfold.

Physics tutorial

Audit the physics of Electro-Optic Modulator

BackgroundAn applied voltage changes the refractive index seen by each of two polarization components. Their relative delay is $\Gamma=\pi\frac{V_{\rm eff}}{V_\pi}$ . That usually turns the polarization into an ellipse instead of rotating a straight line. An analyzer then turns this phase difference into power, part absorbed and part collected.

Why it mattersHow can a voltage across a crystal control the brightness of a light beam?

Start with the essentials

Focus question: How can a voltage across a crystal control the brightness of a light beam?
One-sentence intuition: With a crossed analyzer, the transmitted power is $T_D=\sin^2(2\alpha)\sin^2\!\frac{\Gamma}{2}$ . Take the analyzer out and the lossless crystal pair still changes the polarization, but the power stays constant. If the input lies along either eigenaxis, the phase difference cannot modulate the intensity at all.

Core mathematical model

From the electro-optic tensor to half-wave voltage

\begin{aligned}\Delta n_e&=-\tfrac12n_e^3r_{33}V/d\\\Delta n_o&=-\tfrac12n_o^3r_{13}V/d\\V_\pi&=\frac{\lambda_0 d}{L(n_e^3r_{33}-n_o^3r_{13})}\end{aligned}

Only one crystal is driven, so there is no push-pull factor of two. The passive crystal has the same length with its ordinary and extraordinary axes swapped, which cancels the natural retardance at zero voltage. The material constants come from an MIT problem set and are fixed at six hundred nanometres. They illustrate the physics and are not a calibrated fit to RF material data.

A relative phase, not a mechanical rotation

\begin{aligned}\mathbf E_{\rm pair}&=\begin{pmatrix}e^{-i\Gamma/2}&0\\0&e^{i\Gamma/2}\end{pmatrix}\binom{\cos\alpha}{\sin\alpha}\\T_D&=\sin^2(2\alpha)\sin^2(\Gamma/2)\\A+T_D&=1\end{aligned}

The two components lie along laboratory Y and Z, and the time dependence is a negative imaginary exponential. The common phase is removed, and the relative phase is the ordinary phase minus the extraordinary phase. The power formula holds only with the crossed analyzer. Without it, the total power is one and nothing is absorbed.

Electrical filtering and optical flight time

\begin{aligned}H_{\rm RC}&=\frac{1}{1+i2\pi fRC}\\H_{\rm opt}&=H_{\rm RC}\,\operatorname{sinc}(\pi f\tau_a)\\&\quad\cdot e^{-i2\pi f(\tau_a/2+\tau_{\rm after})}\\C&=\epsilon_0\epsilon_r\frac{Lw}{d}+C_p\end{aligned}

Sinc means the sine divided by its argument. The transit through the active crystal uses an assumed common group index. The remaining delay covers the passive crystal and the actual air path to the sensor. The field is imposed, uniform and lumped; the model does not solve a traveling-wave electrode or acoustic resonances. The DC bias is not attenuated. The peak amplitude of the sine wave is defined at the ideal voltage source, before the resistor.

Common difficulties

Does the light pulse at the drive frequency?

Typical misconceptionA sinusoidal voltage always produces a sinusoidal optical signal at the same frequency.

Better mental modelTransmission through the analyzer is a nonlinear function of voltage. At zero bias, positive and negative voltages give the same power, so the main signal is at twice the drive frequency. At quadrature bias, a small drive gives a strong fundamental. The harmonic bars analyze the detected intensity, not sidebands of the optical frequency.

Run the experiment

01
Scene 1: Analyzer out: constant power
Start with the analyzer out. Drag the waveform probe and compare the polarization ellipses, drawn at equal scale: the power stays at one.
What to observe: The beam runs along laboratory X. In the driven crystal, light travels along the crystal y axis; crystal z points up along laboratory Y, and crystal x lies along laboratory Z. A positive voltage makes the lower electrode positive, so the electric field points up. The passive crystal has the same length, is turned a right angle about the beam, and has no applied field.
02
Scene 2: Quadrature: one pulse per cycle
Insert the analyzer. Compare quadrature bias with null bias, acquire a waveform for each, and compare the fundamental with the second harmonic.
What to observe: The input power is defined just after the source polarizer. The analyzer stays crossed with that polarizer’s axis, and when you remove it, it slides out of the beam. The analyzer absorbs the power it rejects, and the opaque detector stops the transmitted beam. The light has a fixed wavelength of six hundred nanometres and is shown in false color.
03
Scene 3: Null bias: frequency doubling
Double the electrode gap or shorten both matched crystals: the crystals in the scene and the half-wave voltage change together. Then raise the frequency or the series resistance to see the amplitude drop and the phase lag.
What to observe: The selected RF phase refers to the source clock at the moment of detection. The electric-field arrows in the cell show the electrode voltage at that instant. The optical ellipse follows an earlier voltage, averaged over the transit time, because that is the voltage that acted on the detected light. The field markers compare matching states along the beam; they are not a snapshot of both crystals at one instant. The white tips run on a separate, slowed-down clock for the optical carrier.