Electro-Optic Modulation

Table of Contents:

  1. Introduction
  2. Amplitude Modulation
  3. Phase Modulation


Knowing that the electro-optic effect changes the index ellipsoid of a medium, we can then see that the propagation of light through this medium would also be affected. This is one of the most common applications of the electro-optic effect: to change the propagation of an incident light beam. To begin this explanation, we can consider a z-cut KDP plate with an electric field applied parallel to z. KDP has crystal symmetry $\bar{4}2m$, which, with light propagating along the z-axis, we found in the previous section has birefringence given as:

\begin{equation} n_{y’} – n_{x’} = n_0^3r_{63}E_z \end{equation}

If the thickness of the plate is given as “d”, the phase retardation of the plate is then given by:

\begin{equation} \Gamma = \frac{\omega}{c}(n_{y’}-n_{x’})d = \frac{2\pi}{\lambda}n_0^3r_{63}V \end{equation}

where “V” is the voltage applied to the crystal and is given by $V = E_zd$. Because the amount of delay induced by the electro-optic modulator is proportional to the applied voltage, the electro-optic modulator is essentially a variable waveplate! As such, it can be used to convert any initial polarization state into any other through control of the voltage. The voltage that yields a total phase delay of $\Gamma = \pi$ is known as the “half-wave voltage”, and is given by:

\begin{equation} V_{\pi} = \frac{\lambda}{2n_0^3r_{63}} \end{equation}

where “$\lambda$” is the wavelength of the light propagating through the crystal. Note that the half-wave voltage is proportional to the wavelength of the incident light and inversely proportional to the electro-optic coefficient.


The key to understanding amplitude modulation is to consider how one may change the amplitude of the light using an electro-optic modulator and some polarizers. If the electro-optic modulator has a sinusoidal input voltage, such that the modulator effectively “turns on and off” for certain periods of time, then how might the changing polarization be affected by polarizers? If you can understand this, then you have the essence of electro-optic modulators! If not, no worries, I will explain exactly how they work here. Consider operation of the electro-optic modulator at the half-wave voltage. If you place an input polarizer parallel to the x-axis of the crystal without the presence of an electric field, then the retardation through the modulation plate would be $\pi$, which means that the output polarization of the field is now oriented along the y-axis. If you place another polarizer at the output that is parallel to the y-axis, then during the times when $\Gamma = \pi$, i.e. when $V = V_{\pi}$, then the output field would propagate through the polarizer unaffected. Conversely, when $\Gamma = 0$, i.e. when $V = 0$, then all of the light would be blocked and nothing would emerge from the polarizer. Thus, we have amplitude modulation! The total transmission is then given as:

\begin{equation} T = sin^2(\frac{\Gamma}{2}) = sin^2(\frac{\pi}{2}\frac{V}{V_\pi}) \label{transmit}\end{equation}

The figure below shows the transmission vs voltage for the cross polarized electro-optic modulator.

The transmission vs voltage plot for a crossed polarized electro-optic modulator. Photo from: https://www.newport.com/n/electro-optic-modulator-faqs

In order to maintain operation in the linear regime, a more typical setup uses a bias with a fixed retardation of $\Gamma = \frac{\pi}{2}$, which can be introduced either by applying a bias voltage to the crystal of $V = \frac{1}{2}V_{\pi}$ or by using a quarter wave plate that would introduce the same amount of delay. The axes of the quarter wave plate would be aligned with the crystal in the electro-optic modulator, which we can see makes sense because the quarter wave plate is essentially “finishing the job” of the electro-optic modulator. This then allows the sinusoidal voltage applied to the crystal to be much smaller such that $V = \frac{1}{2}V_{\pi}$ and the modulation would be much more sensitive to the applied signal.

We can state this mathematically by taking the total retardation as:

\begin{equation} \Gamma = \frac{1}{2}\pi + \Gamma_msin\omega_mt \end{equation}

where the retardation from either the voltage bias or the quarter-wave plate is the “$\frac{1}{2}\pi$ component. Then, using Equation \eqref{transmit}, we have:

\begin{align} \frac{I_0}{I_i} & = sin^2(\frac{1}{4}\pi + \frac{1}{2}\Gamma_msin\omega_mt) \\ & = \frac{1}{2}[1 + sin(\Gamma_msin\omega_m{t})] \label{amp_trans} \end{align}

Looking at these equations, we can see that the second term in \eqref{amp_trans} would oscillate with distortions and would contain a non-trivial number of higher order harmonics. Here we can see the need for the bias: with the bias, the modulation voltage can be decreased significantly and we can make the approximation that $\Gamma << 1$ such that we can rewrite the equation as:

\begin{equation} \frac{I_0}{I_i} \approx \frac{1}{2}(1+\Gamma_msin\omega_mt) \end{equation}


If you can understand amplitude modulation, then phase modulation will be a walk in the park! In the case of phase modulation, instead of using a polarizer that is aligned to the normal $x$-axis, we would use a polarizer aligned with the $x’$ axis of the crystal. As one may expect, then, the light would not have any amplitude changes as it propagates along the $x’$-axis but it would have a phase change. In this case, if the electric field is applied along the z-direction, the output phase would be changed by:

\begin{equation}\Delta\phi_{x’} = -\frac{\omega{d}}{c}\Delta{n_{x’}} \end{equation}

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