Electro-Optic Modulators

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This is a continuation from the previous tutorial - Pockels Effect.

The index changes induced by the Pockels effect can be utilized to construct a variety of electro-optic modulators, in either bulk or waveguide structures. An electro-optically induced rotation of principal axes is not required for the functioning of an electro-optic modulator though it often accompanies the index changes. However, the directions of the principal axes in the presence of an applied electric field, whether rotated or not, have to be taken into consideration in the design and operation of an electro-optic modulator.

In this tutorial, we consider the operation principles of basic electro-optic modulators. Although some of the concepts, such as transverse phase modulation, that are considered in this tutorial for bulk devices can be directly applied to guided-wave devices, specific guided-wave electro-optic devices are discussed in the next tutorial.

Phase Modulators

The phase of an optical wave can be electro-optically modulated. For this type of application, the optical wave is linearly polarized in a direction that is parallel to one of the principal axes, $$\hat{X},\hat{Y},$$ or $$\hat{Z}$$, of the crystal in the presence of a modulation field.

The preferred choice is a principal axis that has a large electro-optically induced index change but remains in a fixed direction as the magnitude of the modulation electric field varies. In LiNbO3, this can be accomplished by applying the electric field along the $$z$$ axis, as shown in figure 6-2 below.

In this case, $$\hat{X}=\hat{x},\hat{Y}=\hat{y},$$ and $$\hat{Z}=\hat{z}$$, as discussed earlier.

There are two possible arrangements: transverse modulation, where the optical wave propagates in a direction perpendicular to the modulation field, as shown in figure 6-2(a), and longitudinal modulation, where the modulation field is parallel to the direction of optical wave propagation, as shown in figure 6-2(b).

Transverse phase modulators

We first consider the situation of the transverse phase modulator shown in figure 6-2(a), where the optical wave propagates in the $$X$$ direction. In this case, the optical wave can be polarized in either the $$Z$$ or $$Y$$ direction. If it is linearly polarized in the $$Z$$ direction, its space and time dependence can be written as

$\tag{6-37}\mathbf{E}(X,t)=\hat{Z}\mathcal{E}\exp(\text{i}k^ZX-\text{i}\omega{t})=\hat{Z}\mathcal{E}\exp(\text{i}\varphi_Z-\text{i}\omega{t})$

For propagation over a crystal of length $$l$$, the total phase shift is

$\tag{6-38}\varphi_Z=k^Zl=\frac{\omega}{c}n_Zl=\frac{\omega}{c}\left(n_el-\frac{n_e^3}{2}r_{33}E_{0z}l\right)=\frac{\omega}{c}\left(n_el-\frac{n_e^3}{2}r_{33}V\frac{l}{d}\right)$

where $$V=E_{0z}d$$ is the voltage applied to the modulator shown in figure 6-2(a).

For a sinusoidal modulation of a frequency $$f=\Omega/2\pi$$, the modulation voltage can be written as

$\tag{6-39}V(t)=V_{\text{pk}}\sin\Omega{t}$

which has a peak value of $$V_{\text{pk}}$$. The optical field at the output plane, $$X=l$$, of the crystal is

$\tag{6-40}\mathbf{E}(l,t)=\hat{Z}\mathcal{E}\text{e}^{\text{i}\omega{n_e}l/c}\exp[-\text{i}(\omega{t}+\varphi_{\text{pk}}\sin\Omega{t})]$

where

$\tag{6-41}\varphi_{\text{pk}}=\frac{\omega}{c}\frac{n_e^3}{2}r_{33}V_{\text{pk}}\frac{l}{d}=\frac{\pi n_e^3}{\lambda}r_{33}V_{\text{pk}}\frac{l}{d}$

is the peak phase shift known as the phase modulation depth for the $$Z$$-polarized optical field. Using the Bessel-function identities

$\tag{6-42}\exp(-\text{i}\varphi_{\text{pk}}\sin\Omega{t})=\sum_{q=-\infty}^{\infty}J_q(\varphi_{\text{pk}})\text{e}^{-\text{i}q\Omega{t}}$

and $$J_{-q}=(-1)^qJ_q$$, we find that

$\tag{6-43}\mathbf{E}(l,t)=\hat{Z}\mathcal{E}\text{e}^{\text{i}\omega n_el/c}\left\{J_0(\varphi_{\text{pk}})\text{e}^{-\text{i}\omega t}+\sum_{q=1}^{\infty}J_q(\varphi_{\text{pk}})[\text{e}^{-\text{i}(\omega+q\Omega)t}+(-1)^q\text{e}^{-\text{i}(\omega-q\Omega)t}]\right\}$

Thus, a series of side bands at the harmonics of the modulation frequency are generated on both high- and low-frequency sides of the optical carrier frequency by the sinusoidal phase modulation.

If the optical field is instead linearly polarized in the $$Y$$ direction, the phase shift after propagation through the crystal is

$\tag{6-44}\varphi_Y=k^Yl=\frac{\omega}{c}n_Yl=\frac{\omega}{c}\left(n_ol-\frac{n_o^3}{2}r_{13}E_{0z}l\right)=\frac{\omega}{c}\left(n_ol-\frac{n_o^3}{2}r_{13}V\frac{l}{d}\right)$

The phase modulation depth is then

$\tag{6-45}\varphi_{\text{pk}}=\frac{\omega}{c}\frac{n_o^3}{2}r_{13}V_{\text{pk}}\frac{l}{d}=\frac{\pi n_o^3}{\lambda}r_{13}V_{\text{pk}}\frac{l}{d}$

for the modulation voltage given in (6-39). Since $$n_o\approx{n_e}$$ but $$r_{33}\approx3.6r_{13}$$, it can be seen by comparison of (6-45) with (6-41) that for a desired modulation depth, the modulation voltage required for a $$Y$$-polarized optical wave is about 3.6 times that for a $$Z$$-polarized wave.

Longitudinal phase modulators

For the longitudinal phase modulator shown in figure 6-2(b), an optical wave of any polarization in the $$XY$$ plane will experience the same amount of phase shift because $$n_X=n_Y$$. For a crystal of length $$l$$ as shown in figure 6-2(b), we have

$\tag{6-46}\varphi_{X}=\varphi_{Y}=\frac{\omega}{c}\left(n_ol-\frac{n_o^3}{2}r_{13}E_{0z}l\right)=\frac{\omega}{c}\left(n_ol-\frac{n_o^3}{2}r_{13}V\right)$

where $$V=E_{0z}l$$ for the longitudinal modulator. Therefore, with a sinusoidal modulation voltage as given in (6-39), the modulation depth of the longitudinal phase modulator is

$\tag{6-47}\varphi_{\text{pk}}=\frac{\omega}{c}\frac{n_o^3}{2}r_{13}V_{\text{pk}}=\frac{\pi n_o^3}{\lambda}r_{13}V_{\text{pk}}$

which is independent of crystal length $$l$$.

It is seen that the voltage required for a given modulation depth is independent of the physical dimensions of the modulator in the case of longitudinal modulator, while it is proportional to $$d/l$$ in the case of transverse modulation. One advantage of transverse modulation is that the required modulation voltage can be substantially lowered by reducing the $$d/l$$ dimensional ratio of a transverse modulator. Another advantage is that the electrodes of a transverse modulator can be made with standard techniques and can be patterned if desired, while those of a longitudinal modulator have to be made of transparent conductors that can be very difficult, if not impossible, to fabricate in the dimensions of a typical optical waveguide. However, if a large input and output aperture is desired such that $$d/l\gt1$$, it becomes advantageous to use longitudinal modulation rather than transverse modulation.

The relative advantages and disadvantages of transverse versus longitudinal modulation discussed above also hold true for the polarization and intensity modulators discussed in the following.

Example 6-2

As a practical example, consider the LiNbO3 transverse and longitudinal phase modulators shown in figure 6-2(a) and (b), respectively, where the modulation voltage is applied along the $$z$$ axis of the crystal. Find the required voltage $$V_{\text{pk}}$$ for a phase modulation depth of $$\varphi_{\text{pk}}=\pi$$ at $$\lambda$$ =1 μm for optical waves of different polarizations.

We first consider the transverse modulator shown in figure 6-2(a). Using (6-41) and the parameters of $$r_{13}$$ = 8.6 pm V-1, $$r_{33}$$ = 30.8 pm V-1, $$n_e$$ = 2.159, and $$n_o$$ = 2.238 for LiNbO3, the peak voltage required to have $$\varphi_{\text{pk}}=\pi$$ for a $$Z$$-polarized wave is found to be

$V_{\text{pk}}=\frac{\lambda}{n_e^3r_{33}}\frac{d}{l}=\frac{1\times10^{-6}}{2.159^3\times30.8\times10^{-12}}\frac{d}{l}V=3.23\frac{d}{l}kV$

Using (6-45), we find that the required peak voltage for a $$Y$$-polarized wave is

$V_{\text{pk}}=\frac{\lambda}{n_o^3r_{13}}\frac{d}{l}=\frac{1\times10^{-6}}{2.238^3\times8.6\times10^{-12}}\frac{d}{l}V=10.4\frac{d}{l}kV$

For a bulk modulator where $$d$$ and $$l$$ are generally of the same order of magnitude, the required modulation voltage is on the order of kilovolts. However, for a waveguide modulator of typical waveguide dimensions, $$d/l$$ is of the order $$10^{-3}$$. For example, in a transverse waveguide modulator that has dimensions of $$d$$ = 5 μm and $$l$$ = 5 mm, the peak voltage required is reduced to 3.23 and 10.4 V for $$Z$$- and $$Y$$-polarized waves, respectively.

For the longitudinal modulator shown in figure 6-2(b), the optical wave is polarized in the $$XY$$ plane. From (6-47), we find the peak voltage required for $$\varphi_{\text{pk}}=\pi$$ is always

$V_{\text{pk}}=\frac{\lambda}{n_o^3r_{13}}=\frac{1\times10^{-6}}{2.238^3\times8.6\times10^{-12}}\text{V}=10.4\text{ kV}$

Polarization Modulators

In the operation of an electro-optic polarization modulator, the optical wave is not linearly polarized in a direction that is parallel to any of the principal axes in the presence of the modulation field. The optical field can be decomposed into two linearly polarized normal modes. If the two normal modes see different field-induced indices of refraction, there is a field-dependent phase retardation between the two modes. The polarization of the optical wave at the output of the crystal can then be controlled by the modulation field.

The LiNbO3 transverse modulator discussed above becomes a polarization modulator if the input optical field polarized in the $$YZ$$ plane is parallel to neither $$\hat{Y}$$ nor $$\hat{Z}$$:

$\tag{6-48}\mathbf{E}(0,t)=(\hat{Y}\mathcal{E}_Y+\hat{Z}\mathcal{E}_Z)\text{e}^{-\text{i}\omega{t}}$

where $$\mathcal{E}_Y\ne0$$ and $$\mathcal{E}_Z\ne0$$, as is shown in figure 6-3(a) below.

At the output, we have

$\tag{6-49}\mathbf{E}(l,t)=\left(\hat{Y}\mathcal{E}_Y\text{e}^{\text{i}k^Yl}+\hat{Z}\mathcal{E}_Z\text{e}^{\text{i}k^Zl}\right)\text{e}^{-\text{i}\omega{t}}=\left(\hat{Y}\mathcal{E}_Y\text{e}^{\text{i}\Delta\varphi}+\hat{Z}\mathcal{E}_Z\right)\text{e}^{\text{i}k^Zl-\text{i}\omega{t}}$

where

$\tag{6-50}\Delta\varphi=(k^Y-k^Z)l$

is the phase retardation between the $$Y$$ and $$Z$$ components. Using (6-25) [from the Pockels Effect tutorial], we have

$\tag{6-51}\Delta\varphi=\frac{\pi}{\lambda}\left[2(n_o-n_e)l+(n_e^3r_{33}-n_o^3r_{13})V\frac{l}{d}\right]$

The intrinsic birefringence of a uniaxial LiNbO3 crystal causes a voltage-independent background phase retardation of $$\Delta\varphi_0=2\pi(n_o-n_e)l/\lambda$$ at $$\text{V}=0$$ in the absence of an applied field. For a given crystal length, the background phase retardation is fixed, and the function of the device cannot be varied if no modulation field is applied. By properly choosing the value of $$l$$, the device can function as a quarter-wave or half-wave plate.

An applied field causes an additional voltage-dependent phase retardation. The output polarization state of a given input optical wave with nonvanishing $$Y$$- and $$Z$$-field components can be varied by varying the modulation voltage. Thus, the device functions as a voltage-controlled polarization modulator.

For proper operation of the device as a voltage-controlled polarization modulator, the background phase retardation can be compensated by a fixed bias voltage. To find the compensation bias voltage, we note that any phase retardation that is an integral multiple of $$2\pi$$ has no net effect on the polarization of the optical wave at the output of the device. Therefore, the net effect of the background phase retardation can be evaluated by expressing $$\Delta\varphi_0$$ as

$\tag{6-52}\Delta\varphi_0=\frac{2\pi}{\lambda}(n_o-n_e)l=2m\pi+\frac{2\pi}{\lambda}(n_o-n_e)\Delta{l}$

where $$m$$ is a properly chosen integer for

$\tag{6-53}-\frac{\lambda}{2(n_o-n_e)}\lt\Delta{l}=l-\frac{m\lambda}{n_o-n_e}\le\frac{\lambda}{2(n_o-n_e)}$

so that

$\tag{6-54}-\pi\lt\Delta\varphi_0-2m\pi=\frac{2\pi}{\lambda}(n_o-n_e)\Delta{l}\le\pi$

The polarization of the output wave is actually only determined by the following differential phase retardation:

$\tag{6-55}\Delta\varphi-2m\pi=\frac{2\pi}{\lambda}(n_o-n_e)\Delta{l}+\frac{\pi}{\lambda}(n_e^3r_{33}-n_o^3r_{13})V\frac{l}{d}$

A fixed bias voltage for compensation of the background phase retardation can then be chosen as

$\tag{6-56}V_b=\frac{2(n_e-n_o)}{n_e^3r_{33}-n_o^3r_{13}}\frac{d}{l}\Delta{l}=\frac{2(n_e-n_o)}{\lambda}\Delta{l}V_\pi$

where

$\tag{6-57}V_\pi=\frac{\lambda}{n_e^3r_{33}-n_o^3r_{13}}\frac{d}{l}$

is the half-wave voltage, which can also be denoted as $$V_{\lambda/2}$$. The voltage-controlled phase retardation can then be recast in the following form:

$\tag{6-58}\Delta\varphi-2m\pi=\frac{V-V_b}{V_\pi}\pi$

From (6-53) and (6-56), we find that the bias voltage can always be chosen within a range of $$-V_\pi\le{V_b}\lt{V_\pi}$$. At $$V-V_b=\pm{V_\pi}$$, the device functions as a half-wave plate that has a phase retardation of $$\Delta\varphi=2m\pi+\pi$$. At $$V-V_b=\pm{V_\pi}/2$$, the device functions as a quarter-wave plate with a phase retardation of $$\Delta\varphi=2m\pi\pm\pi/2$$. Therefore, the quarter-wave voltage $$V_{\pi/2}$$, or $$V_{\lambda/4}$$, if half that of the half-wave voltage, both measured with respect to the bias point.

The background phase retardation contributed by the intrinsic birefringence is the major drawback of the LiNbO3 transverse polarization modulator discussed here. Although it can be compensated by a bias voltage that falls within the range of $$\pm{V_\pi}$$, the requirement of such a DC bias voltage complicates the operation of the device, particularly when it is modulated at a high frequency. Because the bias voltage depends on the length and the refractive indices of the device, it is susceptible to changes in the operating condition, such as temperature variations caused by operation of the device. Furthermore, the bias voltage is also a function of optical wavelength because $$\Delta{l}$$ varies as the optical wavelength varies. In practice, the bias voltage has to be carefully adjusted for each individual device in a given operating condition due to small variations in device length and refractive indices.

Example 6-3

LiNbO3 transverse polarization modulator for $$\lambda$$ = 1 μm as shown in figure 6-3(a) has dimensions of $$d$$ = 5 μm and $$l$$ = 5 mm. Find the half-wave voltage $$V_\pi$$ and the bias voltage $$V_b$$ required for compensating the background phase retardation from the intrinsic birefringence of LiNbO3. If the length varies by $$\pm$$5 μm due to fabrication errors or changes in the operating condition, what are the changes in $$V_\pi$$ and $$V_b$$, respectively?

Using parameters of LiNbO3 given in table 6-2 in the Pockels Effect tutorial, we find from (6-57) equation that $$V_\pi\approx4.68\text{ V}$$. With $$n_o-n_e=0.079$$ at $$\lambda$$ = 1 μm found from the data in table 6-2, we find that $$\Delta{l}=0$$ with $$m$$ = 395 for $$l$$ = 5 mm. Therefore, $$V_b$$ = 0 from (6-56).

A length variation of $$\pm$$5 μm amounts to a change of $$\pm$$0.1% in the total length. From (6-57), we find that it causes a change of only about $$\mp$$0.1% in $$V_\pi$$. However, it results in $$\Delta{l}=\pm5$$ μm, also with $$m$$ = 395. From (6-56), we find that the required compensation bias voltage is $$V_b\approx\mp3.70\text{ V}$$. We therefore see that a small variation in the length of the device causes a similarly small change in $$V_\pi$$, but it can lead to a large change in the compensation bias voltage.

The LiNbO3 longitudinal modulator shown in 6-2(b) cannot function as a polarization modulator because $$n_X=n_Y$$. Instead, we consider the GaAs longitudinal modulator shown in figure 6-3(b). In this case, the principal axes and their corresponding indices of refraction in the presence of a modulation field are those given in (6-35) and (6-36) [refer to the Pockels Effect tutorial], respectively. The optical wave to be modulated propagates in the $$Z$$ direction and has both $$X$$ and $$Y$$ field components. At the input end, it can be written as

$\tag{6-59}\mathbf{E}(0,t)=(\hat{X}\mathcal{E}_X+\hat{Y}\mathcal{E}_Y)\text{e}^{-\text{i}\omega{t}}$

After propagating through the crystal, the optical field is

$\tag{6-60}\mathbf{E}(l,t)=\left(\hat{X}\mathcal{E}_X\text{e}^{\text{i}k^Xl}+\hat{Y}\mathcal{E}_Y\text{e}^{\text{i}k^Yl}\right)\text{e}^{-\text{i}\omega{t}}=\left(\hat{X}\mathcal{E}_X+\hat{Y}\mathcal{E}_Y\text{e}^{\text{i}\Delta\varphi}\right)\text{e}^{\text{i}k^Xl-\text{i}\omega{t}}$

where

$\tag{6-61}\Delta\varphi=\left(k^Y-k^X\right)l$

is the phase retardation between the $$Y$$ and $$X$$ components of the optical field. Using (6-36) [refer to the Pockels Effect tutorial] and the fact that $$V=E_{0z}l$$ for the longitudinal modulator, we have

$\tag{6-62}\Delta\varphi=\frac{2\pi}{\lambda}n_o^3r_{41}V=\frac{V}{V_\pi}\pi$

where the half-wave voltage is

$\tag{6-63}V_\pi=\frac{\lambda}{2n_o^3r_{41}}$

It can be seen from (6-62) that no bias voltage is needed for this GaAs modulator because both $$x$$ and $$y$$ axes are ordinary axes in the absence of an applied field. However, because of the longitudinal modulator scheme, $$V_\pi$$ is a constant independent of both dimensions $$l$$ and $$d$$. Therefore, in comparison with the LiNbO3 transverse modulator, the advantage of this modulator in regarding to no bias voltage is completely offset by the disadvantage due to its longitudinal modulation scheme. In a GaAs transverse polarization modulator, both problems are eliminated.

Example 6-4

For $$\lambda$$ = 1 μm, the parameters for GaAs in table 6-2 [refer to the Pockels Effect tutorial] yield $$V_\pi\approx9.72\text{ kV}$$ from (6-63), which is independent of the dimension of the GaAs longitudinal modulator. Though no bias voltage is needed because GaAs has no intrinsic birefringence, this half-wave voltage cannot be reduced by varying the dimensions of the modulator. For $$\Delta\varphi$$ to vary in the range between 0 and $$\pi$$, the modulation voltage has to be varied between 0 and 9.72 $$\text{kV}$$.

Amplitude Modulators

An electro-optic amplitude modulator can be constructed by simply placing a polarization modulator between a polarizer at the input end and another, often referred to as an analyzer, at the output end. Usually, the axis of the polarizer and that of the analyzer are arranged to be orthogonally crossed, although other arrangements are possible.

Figure 6-4 shows a typical setup for a GaAs longitudinal amplitude modulator. In this arrangement, the polarizer ensures that the input optical wave is linearly polarized in the $$y$$ direction while the analyzer passes only the $$x$$ component of the optical wave at the output end. The input field is $$\mathbf{E}(0,t)=\hat{y}\mathcal{E}\text{e}^{-\text{i}\omega{t}}$$, which can be written in the form of (6-59) with $$\mathcal{E}_X=\mathcal{E}_Y=\mathcal{E}/\sqrt{2}$$. Then, from (6-60), the field at the output end of the crystal is

$\tag{6-64}\mathbf{E}(l,t)=\frac{\mathcal{E}}{\sqrt{2}}(\hat{X}+\hat{Y}\text{e}^{\text{i}\Delta\varphi})\text{e}^{\text{i}k^Xl-\text{i}\omega{t}}=\frac{\mathcal{E}}{2}\left[\hat{x}(1-\text{e}^{\text{i}\Delta\varphi})+\hat{y}(1+\text{e}^{\text{i}\Delta\varphi})\right]\text{e}^{\text{i}k^Xl-\text{i}\omega{t}}$

where $$\Delta\varphi$$ is the same as that given in (6-61). Because the analyzer passes only the $$x$$ component of the optical field, the transmittance of the amplitude modulator is

$\tag{6-65}T=\frac{I_\text{out}}{I_\text{in}}=\sin^2\frac{\Delta\varphi}{2}=\frac{1}{2}(1-\cos\Delta\varphi)$

A similar result is obtained for an amplitude modulator constructed by placing any other polarization modulator, such as the LiNbO3 transverse polarization modulator shown in figure 6-3(a), between a pair of properly oriented polarizer and analyzer.

Because $$\Delta\varphi$$ varies linearly with applied voltage $$V$$, the amplitude modulator would have a linear response if its transmittance $$T$$ varied linearly with $$\Delta\varphi$$. It can be seen from (6-65) that this is not generally true. However, for small variations of $$\Delta\varphi$$, it is approximately true near $$\Delta\varphi=\pi/2$$, as can be seen by substituting

$\tag{6-66}\Delta\varphi=\frac{\pi}{2}+\delta\varphi$

in (6-65) to get

$\tag{6-67}T=\frac{1}{2}(1+\sin\delta\varphi)\approx\frac{1}{2}(1+\delta\varphi)$

for $$|\delta\varphi|\ll\pi$$. By setting the operating point at a bias phase retardation of $$\Delta\varphi=\pi/2$$, the device has a linear small-signal response, as shown in figure 6-5. Then, with a voltage such as that given in (6-39), the output intensity will be sinusoidally modulated. This bias phase retardation can be obtained either by operating the device with a fixed bias voltage of $$V_\text{b}=V_{\pi/2}$$, as shown in figure 6-4(a), or by inserting a properly oriented quarter-wave plate between the modulator crystal and the analyzer to introduce an extra fixed phase retardation of $$\pi/2$$, as shown in figure 6-4(b).

For large-signal applications, the response of the amplitude modulator is nonlinear. The device is then often used as an electro-optically controlled ON-OFF modulator. In these types of applications. a bias is neither useful nor necessary.

The next part continues with the guided-wave electro-optic modulators tutorial