Question 5 of 7: Longitudinal Pockels-Cell Modulator (LiNbO 3 )

Question 5: Longitudinal Pockels-Cell Modulator (LiNbO₃) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) modulator schematic and operating principle, (b) the half-wave voltage $V_\pi$, (c) the drive resistance $R$ and crystal capacitance $C$, (d) the sinusoidal drive power.

Part (a) — configuration and operating principle

In the longitudinal Pockels configuration the modulating electric field is applied parallel to the direction of light propagation, through ring or transparent electrodes on the two end faces of the crystal, so the light and the field share the same path length $L$ through the crystal. A polarizer ahead of the crystal sets the input polarization at $45^{\circ}$ to the crystal's induced fast/slow birefringent axes; the applied voltage induces a linear (Pockels) birefringence proportional to $E$, retarding one polarization component relative to the other; an analyzer (typically crossed with the polarizer) after the crystal then converts this voltage-dependent phase retardation into an intensity modulation at the output detector.

Approach (b–d). For the longitudinal geometry the interaction length and the electrode spacing are the same length $L$, so $L$ cancels out of the retardation formula and $V_\pi$ depends only on $\lambda$, $n_o$ and $r$. The capacitance follows from the crystal's end-face area as a parallel-plate capacitor of separation $L$, and $R$ then follows from the stated RC time constant.

1. Part (b) — half-wave voltage. For a longitudinal modulator the field-induced phase retardation is $\Delta\phi=(2\pi/\lambda)\,n_o^3\,r\,V$ (the crystal length $L$ cancels between the optical path length and the electrode spacing $E=V/L$). Setting $\Delta\phi=\pi$ for the half-wave point, $$V_\pi=\frac{\lambda}{2n_o^3r}=\frac{624\times10^{-9}}{2(2.30)^3(30\times10^{-12})} =\boxed{855\ \text{V}}.$$
2. Part (c) — R and C. The crystal's end-face area (diameter 20 mm) is $A=\pi(D/2)^2=\pi(0.010)^2=3.14\times10^{-4}\ \text{m}^2$, and treating the crystal as a parallel-plate capacitor of separation $L=30$ mm, $$C=\frac{\varepsilon_0\varepsilon_rA}{L}=\frac{(8.854\times10^{-12})(32)(3.14\times10^{-4})}{0.030} =\boxed{2.97\ \text{pF}}.$$ The RC-limited bandwidth condition $f_{3dB}=1/(2\pi RC)$ then fixes the load resistance: $$R=\frac{1}{2\pi f_{3dB}C}=\frac{1}{2\pi(1.0\times10^6)(2.97\times10^{-12})} =\boxed{53.6\ \text{k}\Omega}.$$
3. Part (d) — sinusoidal drive power. Driving the crystal with a sinusoid of peak amplitude $V_\pi$ across the load resistor $R$ dissipates an average power $$P=\frac{V_{rms}^2}{R}=\frac{(V_\pi/\sqrt2)^2}{R}=\frac{(855/\sqrt2)^2}{53.6\times10^3} =\boxed{6.81\ \text{W}}.$$