# Semiconductor (Diode) Lasers and Properties

• dalye13
dalye13
Homework Statement
How does a change in a diode laser's cavity length affect the output power, slope efficiency and threshold current? how are they all related?
Relevant Equations
(P/P_o) = exp(gL)
Slope Efficiency = dP/VdI
P = SE*(P-P_th) = SE*V*(I-I_th) (where P=VI) = SE*(E/q)*(I-I_th) (Where E=hf)
Code:
print ('Calculate threshold, power, slope efficiency for different lengths of SC Laser')

g = 510 # The gain of the laser, arbitrary value of 510 m^-1 was picked

I = np.linspace(0, 0.03,5) #DRIVE CURRENT; 100 values of current, 'I', between 0A and 0.03A

V = 1.8 #INPUT VOLTAGE; arbitrary value of 1.8V was picked

l = np.linspace(0.000001, 0.001, 5) #LENGTH; 10 values of 'l' between 1 micron and 1mm

p_out = {}
print(l, I)

for x in range(len(l)): #for each element in 'l', create a list of P_out
p_out[x] = []
for y in range(len(I)): #For each P_out, populate with 'P' formula below at every 'I' value in the 'I' array
p = (math.exp(g*l[x])) * V * (I[y]-I_th) #Power formula exp(gl)*V*(I-I_th)
p_out[x].append(p) #appending p_out of x with p from above

print(p_out)

fig, ax = plt.subplots()
for x in range(len(l)):
ax.plot(I, p_out[x])
#print (math.exp(g*l[x]))

#ax.set_xscale('log')
plt.savefig('PowervsCurrentfordifferentLengths.png', dpi =600)
plt.xlabel('Current, I [A]')
plt.ylabel('Optical Power, P [mW]')
plt.ylim([0,0.25]) #limits on x axis
plt.title('Optical Power Vs Current for 10 values of L between 1 micron and 1 mm')
plt.show

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Homework Statement:: How does a change in a diode laser's cavity length affect the output power, slope efficiency and threshold current? how are they all related?
Relevant Equations:: (P/P_o) = exp(gL)
Slope Efficiency = dP/VdI
P = SE*(P-P_th) = SE*V*(I-I_th) (where P=VI) = SE*(E/q)*(I-I_th) (Where E=hf)

Code:
print ('Calculate threshold, power, slope efficiency for different lengths of SC Laser')

g = 510 # The gain of the laser, arbitrary value of 510 m^-1 was picked

I = np.linspace(0, 0.03,5) #DRIVE CURRENT; 100 values of current, 'I', between 0A and 0.03A

V = 1.8 #INPUT VOLTAGE; arbitrary value of 1.8V was picked

l = np.linspace(0.000001, 0.001, 5) #LENGTH; 10 values of 'l' between 1 micron and 1mm

p_out = {}
print(l, I)

for x in range(len(l)): #for each element in 'l', create a list of P_out
p_out[x] = []
for y in range(len(I)): #For each P_out, populate with 'P' formula below at every 'I' value in the 'I' array
p = (math.exp(g*l[x])) * V * (I[y]-I_th) #Power formula exp(gl)*V*(I-I_th)
p_out[x].append(p) #appending p_out of x with p from above

print(p_out)

fig, ax = plt.subplots()
for x in range(len(l)):
ax.plot(I, p_out[x])
#print (math.exp(g*l[x]))

#ax.set_xscale('log')
plt.savefig('PowervsCurrentfordifferentLengths.png', dpi =600)
plt.xlabel('Current, I [A]')
plt.ylabel('Optical Power, P [mW]')
plt.ylim([0,0.25]) #limits on x axis
plt.title('Optical Power Vs Current for 10 values of L between 1 micron and 1 mm')
plt.show
Welcome to PhysicsForums. So does your program work? What is your summary of what you have found? Are you able to answer the questions in the problem statement?

dalye13
Welcome to PhysicsForums. So does your program work? What is your summary of what you have found? Are you able to answer the questions in the problem statement?

Thanks!
No not really, I thought the expression for P=exp{gL}*V*(I-I_th) was along the right path, but when I run the code, it produces a very linear graph when its meant to be exponential. I've tried many combinations of inputs to see if my units were off (probably still are) but nothing seemed to work... now granted the code could also be wrong but I believe its okay, so that means I messed up in the theory setting up the problem to put into the programme... Because if the curve is not exponential then there's no way of me finding the threshold current from the graph... overall I've really confused myself with everything hahahaha