Question 5 of 7: Longitudinal LiNbO$_3$ Pockels-cell modulator — half-wave voltage, $R$ and $C$, and drive power

Question 5: Longitudinal LiNbO$_3$ Pockels-cell modulator — half-wave voltage, $R$ and $C$, and drive power (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. A longitudinal (field parallel to the beam) lithium-niobate Pockels cell, with the crystal constants taken from page 1 of the paper.

Find. A labelled sketch with the required external optics and drive, the half-wave voltage, the crystal capacitance and the load resistance that gives the stated bandwidth, and the electrical drive power.

(a) Arrangement and operation

The beam passes in sequence through a polarizer, the electro-optic crystal, and an analyser crossed with respect to the polarizer. The polarizer is oriented at 45° to the crystal’s induced principal axes so that it launches equal components on the fast and slow axes. Transparent ring (or thin-film) electrodes on the two end faces apply the modulating voltage along the direction of propagation — that is what makes the geometry longitudinal — and the drive source feeds them through a load resistor $R$, the crystal itself providing the parallel-plate capacitance $C$.

Applying a voltage changes the refractive index linearly through the Pockels effect, $n = n_{o} - \tfrac{1}{2}r n_{o}^{3}\mathcal{E}$ (the relation printed on page 1), so the two polarisation components accumulate a relative phase retardation $\Gamma = 2\pi r n_{o}^{3} V/\lambda$ that depends only on the applied voltage, not on the crystal length, because the longer path and the weaker field cancel exactly. The analyser converts that retardation into a transmitted intensity $T = \sin^{2}(\Gamma/2)$: at $V = 0$ the light is blocked and at $V = V_{\pi}$ it is fully transmitted, which is precisely the on/off behaviour wanted for digital amplitude pulse modulation. For analogue use a quarter-wave plate is added to bias the cell to the half-transmission point, where the response is most nearly linear.

(b) to (d) — the numbers

Approach. The half-wave voltage follows from the retardation expression; the capacitance is the parallel-plate value of the crystal itself; $R$ then follows from the specified $RC$ bandwidth; and the drive power is the power the source delivers to $R$ when swinging a sinusoid of amplitude $V_{\pi}$.

1. Half-wave voltage. Setting the retardation to $\pi$ in $\Gamma = 2\pi r n_{o}^{3}V/\lambda$ gives the standard longitudinal result $$V_{\pi} = \frac{\lambda}{2\,n_{o}^{3}\,r} = \frac{624\times10^{-9}}{2\,(2.30)^{3}(30\times10^{-12})} = \frac{624\times10^{-9}}{7.300\times10^{-10}}$$ $$\boxed{V_{\pi} = 855\ \text{V}}$$ Note it does not depend on $L$ or $D$: in the longitudinal geometry the field is $V/L$ and the interaction length is $L$, so the geometry cancels. This high drive voltage is the well-known drawback of longitudinal cells, and the reason integrated modulators use the transverse geometry, where $V_{\pi}$ is scaled down by the aspect ratio $d/L$.
2. Electrode area and crystal capacitance. The end faces are circular of diameter 20 mm, so $$A = \frac{\pi D^{2}}{4} = \frac{\pi (0.020)^{2}}{4} = 3.142\times10^{-4}\ \text{m}^{2}$$ and, treating the crystal as a parallel-plate capacitor whose plate separation is its own length, $$C = \frac{\varepsilon_{0}\varepsilon_{r}A}{L} = \frac{(8.854\times10^{-12})(32)(3.142\times10^{-4})}{0.030}$$ $$\boxed{C = 2.97\ \text{pF}}$$
3. Load resistance for the stated bandwidth. The modulator is a single-pole $RC$ network, so $f_{3\text{dB}} = 1/(2\pi RC)$ and $$R = \frac{1}{2\pi f_{3\text{dB}} C} = \frac{1}{2\pi (1.0\times10^{6})(2.967\times10^{-12})}$$ $$\boxed{R = 53.6\ \text{k}\Omega}$$ A larger $R$ would give more voltage swing for a given drive current but would slow the cell; this value is the one that lands the pole exactly at 1.0 MHz.
4. Drive power. A sinusoidal supply of peak amplitude $V_{\pi}$ (the minimum needed to reach full transmission on each cycle) has r.m.s. value $V_{\pi}/\sqrt{2}$, and the power is dissipated in the load resistance: $$P = \frac{V_{\text{rms}}^{2}}{R} = \frac{V_{\pi}^{2}}{2R} = \frac{(854.8)^{2}}{2(53.64\times10^{3})}$$ $$\boxed{P = 6.8\ \text{W}}$$ Nearly 7 W to modulate a milliwatt-class HeNe beam at only 1 MHz is a striking figure, and it is the honest reason bulk longitudinal Pockels cells are used for Q-switching and pulse gating rather than for continuous high-rate modulation.