EOM

EOM Kazan

Company Qubig produces eom PS3D-BC

$\delta \nu =FSR \frac{}{2\pi\lambda}\delta V$ 

Parameter	MIN	TYP	MAX	units
Bandwidth	DC		200	MHz
Maximum voltage			500	V
$V_\pi \, DC$			1	kV

EOM Calgary

We bought a 10 GHZ phase modulator NIR-MPX80 from iX-Blue. The material is lithium niobate, characteristics are

Parameter	MIN	TYP	MAX	units
Wavelength	780	850	890	nm
Insertion loss		3.5	4.5	dB
Electro-optic bandwidth	10	12		GHz
$V_\pi \, RF_{50kHz}$		4	5	V

Absolute maximum ratings:

Parameter	MAX	units
RF input power	28	dBm
Optical input power	14	dBm

How to find modulation depth from these parameters?

Let's assume we apply the optical wave with frequency $\omega_0$ and amplitude $E_{in}$ and modulate the EO crystal with RF frequency $\omega_{rf}$ The output of EOM is described by the following expression:

$E_{out}= E_{in} e^{i\omega_o t +\beta \cos (\omega_{rf}t+\phi) } = E_{in} e^{i\omega_o t} \left( J_{0}(\beta) +\sum_{k=1} J_{k}(\beta) e^{ik\omega_{rf}} + (-1)^{k} J_{-k}(\beta) e^{-ik\omega_{rf}} \right)$

Thus the maximum intensity of the first sideband is reached at $\beta \approx 1.82$, $E_{1}/E_{in} \approx 0.33$.

Thus by varying applied rf power and finding maximum it is possible to translate it into modulation depth index.