Semiconductor (Diode) Lasers and Properties

In summary, the program does not work as intended because the output power is linear when it should be exponential.
  • #1
dalye13
2
0
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 0.03AV = 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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  • #2
dalye13 said:
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 0.03AV = 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.:smile:

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?
 
  • #3
berkeman said:
Welcome to PhysicsForums.:smile:

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
 

Related to Semiconductor (Diode) Lasers and Properties

1. What is a semiconductor laser?

A semiconductor laser is a type of laser that uses a semiconductor material, typically a combination of gallium, arsenic, and other elements, to produce coherent light. It is also known as a diode laser because it uses a p-n junction diode as its active medium.

2. How does a semiconductor laser work?

A semiconductor laser works by passing an electric current through the p-n junction diode, causing the electrons and holes to recombine and emit photons. These photons are then reflected back and forth between two mirrors, creating a coherent beam of light that exits through one of the mirrors.

3. What are the properties of a semiconductor laser?

Some of the key properties of a semiconductor laser include its small size, high efficiency, and ability to produce a narrow, intense beam of light. It also has a fast response time and can operate at high frequencies, making it useful for applications such as telecommunications and laser printers.

4. What are the advantages of using a semiconductor laser?

One of the main advantages of using a semiconductor laser is its compact size, which allows for easy integration into electronic devices. It also has a longer lifespan and lower power consumption compared to other types of lasers. Additionally, semiconductor lasers can be modulated and controlled with high precision, making them useful for a variety of applications.

5. What are some common applications of semiconductor lasers?

Semiconductor lasers have a wide range of applications, including telecommunications, barcode scanners, laser pointers, laser printers, and optical data storage. They are also used in medical and scientific equipment, such as laser eye surgery and spectroscopy. Semiconductor lasers are also used in military and defense technologies, such as targeting and guidance systems.

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