Question

Problem 2 (25 points) The threshold current of a semiconductor laser is a critical electrical current required for the semiconductor laser to start emitting laser light. Above the threshold current, the output power of a semiconductor laser increases as the electrical current increases. For a semiconductor laser around the wavelength 1.31 micron, the threshold current is 6 mA. When the electrical current increases from 20 mA to 40 mA, the optical output laser increases as thie ron the threshold current power increases from 28 m W to 68 mW. What is the slope efficiency of this laser? (10 points) (a) (b) If the laser is biased at 30 mA and modulated with a harmonic varying electric current of amplitude of 20 mA, what is the modulation amplitude of the optical output power?

Answer 1

Sol.

a)

Δ1 Ith 20mA 40mA Current

From the above diagram it is clear that the power output from the laser diode will increase almost linearly after the threshold current.

We can now calculate the slope efficiency by the formula:

Generally,

Po

in this case,

( change in current is taken after the threshold current )

slopeAI

68mW - 28mW 40 stope 40m.A- 20m.A 20 2.0 mW/mA

b)

lin 50mA 20mA 30mA 20mA 10mA OmA time

Let the modulating current be a sinusoidal signal with peak amplitude of 20mA

the input current can be expressed as :

lin = 30 + 20 sin(2π ft 27l

As the laser diode is biased at 30mA, the peak input current will be ( when the input current is maximum):

$I_1=30mA+20mA=50mA$

similarly, in the negative half cycle,

Io = 30mA-20mA 10mA

now, we can calculate the power output at the maximum and minimum current values using the graph in part a of this question.

using the formula

Po

i) we have,

stope-' th
where
1150mA
,
10 10mA
,  $I_{th}=6mA$

therefore,

$P_1=\eta_{slope} \times (I_1-I_{th}) = 2.0\times(50-6)=88mW$

similarly,

$P_0=\eta_{slope} \times (I_0-I_{th}) = 2.0\times(10-6)=8mW$

The optical modulation amplitude (OMA) is calculated using its definition, i.e OMA is the difference between the power output when the input is 'on' and the output power when input is 'off'

In our case, when the input is 'on' we get $P_1$ and when the input is 'off' we get $P_0$

therefore, $OMA=P_1-P_0=88-8=80mW$

Hence we have obtained the required value $OMA=80mW$.