Question:
The slope efficiency of a laser diode is \(0.5\) W/A, and the output optical power at a current of \(100\) mA is \(30\) mW. Assuming piece-wise linear characteristics of the laser diode, the threshold current of the laser is ________ mA.
The slope efficiency of a laser diode is \(0.5\) W/A, and the output optical power at a current of \(100\) mA is \(30\) mW. Assuming piece-wise linear characteristics of the laser diode, the threshold current of the laser is ________ mA.
Show Hint
Write the power-current line as \(P=\eta_s(I-I_{th})\) using the given slope efficiency, plug in the one known operating point, and solve for the current-axis intercept.
Updated On: Jul 22, 2026
- \(0\)
- \(20\)
- \(40\)
- \(60\)
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The Correct Option is C
Solution and Explanation
Step 1: Recall the piece-wise linear model of a laser diode.
Below the threshold current \(I_{th}\), a laser diode gives essentially no coherent optical output. Above threshold, the output power rises linearly with current at a constant rate called the slope efficiency, \(\eta_s\). So
\[ P = \eta_s (I - I_{th}) \quad \text{for } I \geq I_{th} \]
Step 2: Keep the units consistent.
The slope efficiency is \(\eta_s = 0.5\) W/A. Since \(1\) W/A is the same ratio as \(1\) mW/mA (both numerator and denominator scale by \(1000\)), we can write \(\eta_s = 0.5\) mW/mA and work entirely in mW and mA.
Step 3: Substitute the known operating point.
At \(I=100\) mA, \(P=30\) mW, so
\[ 30 = 0.5(100 - I_{th}) \]
Step 4: Solve for the threshold current.
\[ \frac{30}{0.5} = 100 - I_{th} \]
\[ 60 = 100 - I_{th} \]
\[ I_{th} = 100 - 60 = 40 \text{ mA} \]
Step 5: Check every option directly in the model, and see why the rest fail.
Option (A) \(I_{th}=0\) would mean \(P=0.5\times100=50\) mW at \(100\) mA, not the given \(30\) mW, so it fails. Option (B) \(I_{th}=20\) mA gives \(P=0.5(100-20)=40\) mW, not \(30\) mW, so it fails. Option (D) \(I_{th}=60\) mA gives \(P=0.5(100-60)=20\) mW, not \(30\) mW, so it fails too. Only \(I_{th}=40\) mA gives \(P=0.5(100-40)=30\) mW, matching the data exactly.
Final Answer:
The threshold current is \(40\) mA. \[ \boxed{I_{th} = 40 \text{ mA}} \]
Below the threshold current \(I_{th}\), a laser diode gives essentially no coherent optical output. Above threshold, the output power rises linearly with current at a constant rate called the slope efficiency, \(\eta_s\). So
\[ P = \eta_s (I - I_{th}) \quad \text{for } I \geq I_{th} \]
Step 2: Keep the units consistent.
The slope efficiency is \(\eta_s = 0.5\) W/A. Since \(1\) W/A is the same ratio as \(1\) mW/mA (both numerator and denominator scale by \(1000\)), we can write \(\eta_s = 0.5\) mW/mA and work entirely in mW and mA.
Step 3: Substitute the known operating point.
At \(I=100\) mA, \(P=30\) mW, so
\[ 30 = 0.5(100 - I_{th}) \]
Step 4: Solve for the threshold current.
\[ \frac{30}{0.5} = 100 - I_{th} \]
\[ 60 = 100 - I_{th} \]
\[ I_{th} = 100 - 60 = 40 \text{ mA} \]
Step 5: Check every option directly in the model, and see why the rest fail.
Option (A) \(I_{th}=0\) would mean \(P=0.5\times100=50\) mW at \(100\) mA, not the given \(30\) mW, so it fails. Option (B) \(I_{th}=20\) mA gives \(P=0.5(100-20)=40\) mW, not \(30\) mW, so it fails. Option (D) \(I_{th}=60\) mA gives \(P=0.5(100-60)=20\) mW, not \(30\) mW, so it fails too. Only \(I_{th}=40\) mA gives \(P=0.5(100-40)=30\) mW, matching the data exactly.
Final Answer:
The threshold current is \(40\) mA. \[ \boxed{I_{th} = 40 \text{ mA}} \]
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