Transconductance refers to an electrical characteristic of a device which relates the current across the output port to the potential applied at input port of the device.
A BJT is biased in forward active mode of operation. The characterization was performed on a DC probe station where the temperature of the chuck(place where the device is kept) was raised to 350K. The base to emitter voltage (VBE) was kept at 0.8V.  The reverse saturation current was calculated and found out to be 10^-11 mA. Calculate the transconductance (gm) of the transistor in Amperes/Volt (A/V). (Assume ideality factor to be 1)

0.55
 
5.78x10^-3
 
0.11
 
None of the above

The transconductance (\( g_m \)) of a Bipolar Junction Transistor (BJT) in the forward active mode is defined as the change in the collector current (\( I_C \)) with respect to the change in the base-emitter voltage (\( V_{BE} \)). Mathematically, it's given by:

\[
g_m = \frac{\partial I_C}{\partial V_{BE}} = \frac{I_C}{V_T}
\]

where:
- \( I_C \) is the collector current,
- \( V_T \) is the thermal voltage, \( V_T = \frac{kT}{q} \),
- \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \text{ J/K} \)),
- \( q \) is the charge of an electron (\( 1.6 \times 10^{-19} \text{ C} \)),
- \( T \) is the absolute temperature in Kelvin.

### Given Data:
- Temperature \( T = 350 \text{ K} \)
- Base-emitter voltage \( V_{BE} = 0.8 \text{ V} \)
- Reverse saturation current \( I_S = 10^{-11} \text{ mA} = 10^{-14} \text{ A} \)
- Ideality factor \( \eta = 1 \) (since the problem assumes ideal behavior)

### Step 1: Calculate the Thermal Voltage (\( V_T \)):
\[
V_T = \frac{kT}{q} = \frac{(1.38 \times 10^{-23} \text{ J/K}) \times 350 \text{ K}}{1.6 \times 10^{-19} \text{ C}} \approx 0.0302 \text{ V}
\]

### Step 2: Calculate the Collector Current (\( I_C \)):
The collector current in a BJT in forward active mode is given by:
\[
I_C = I_S e^{\frac{V_{BE}}{V_T}}
\]
Substituting the given values:
\[
I_C = 10^{-14} \text{ A} \times e^{\frac{0.8 \text{ V}}{0.0302 \text{ V}}} 
\]

Calculate the exponent:
\[
\frac{0.8 \text{ V}}{0.0302 \text{ V}} \approx 26.49
\]

Now, calculate the collector current:
\[
I_C = 10^{-14} \text{ A} \times e^{26.49} \approx 10^{-14} \text{ A} \times 2.61 \times 10^{11} = 2.61 \times 10^{-3} \text{ A} = 2.61 \text{ mA}
\]

### Step 3: Calculate the Transconductance (\( g_m \)):
\[
g_m = \frac{I_C}{V_T} = \frac{2.61 \times 10^{-3} \text{ A}}{0.0302 \text{ V}} \approx 0.0864 \text{ A/V}
\]

### Conclusion:
None of the provided options exactly matches the calculated value, but the closest approximation to \( 0.0864 \text{ A/V} \) is **0.11 A/V**.

Thus, the correct answer is **0.11**.

Would you like more details or have any questions?

Here are 5 relative questions:
1. How would the transconductance change if the temperature were reduced to 300K?
2. What role does the ideality factor play in calculating the collector current?
3. How does transconductance affect the overall performance of an amplifier circuit?
4. What happens to the reverse saturation current if the doping concentration of the BJT is increased?
5. How can the transconductance be increased in a given BJT?

**Tip:** Transconductance is a critical parameter in amplifiers as it directly impacts gain, making understanding its temperature dependence important for circuit design.

Learn how to calculate the transconductance (gm) of a Bipolar Junction Transistor (BJT) in forward active mode with a DC probe station temperature of 350K and base-emitter voltage (VBE) of 0.8V. Includes step-by-step solution using the thermal voltage formula and collector current calculation. Ideal for advanced college students studying electrical engineering.

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