Math Problem Statement

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

Solution

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

gm=ICVBE=ICVTg_m = \frac{\partial I_C}{\partial V_{BE}} = \frac{I_C}{V_T}

where:

  • ICI_C is the collector current,
  • VTV_T is the thermal voltage, VT=kTqV_T = \frac{kT}{q},
  • kk is the Boltzmann constant (1.38×1023 J/K1.38 \times 10^{-23} \text{ J/K}),
  • qq is the charge of an electron (1.6×1019 C1.6 \times 10^{-19} \text{ C}),
  • TT is the absolute temperature in Kelvin.

Given Data:

  • Temperature T=350 KT = 350 \text{ K}
  • Base-emitter voltage VBE=0.8 VV_{BE} = 0.8 \text{ V}
  • Reverse saturation current IS=1011 mA=1014 AI_S = 10^{-11} \text{ mA} = 10^{-14} \text{ A}
  • Ideality factor η=1\eta = 1 (since the problem assumes ideal behavior)

Step 1: Calculate the Thermal Voltage (VTV_T):

VT=kTq=(1.38×1023 J/K)×350 K1.6×1019 C0.0302 VV_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 (ICI_C):

The collector current in a BJT in forward active mode is given by: IC=ISeVBEVTI_C = I_S e^{\frac{V_{BE}}{V_T}} Substituting the given values: IC=1014 A×e0.8 V0.0302 VI_C = 10^{-14} \text{ A} \times e^{\frac{0.8 \text{ V}}{0.0302 \text{ V}}}

Calculate the exponent: 0.8 V0.0302 V26.49\frac{0.8 \text{ V}}{0.0302 \text{ V}} \approx 26.49

Now, calculate the collector current: IC=1014 A×e26.491014 A×2.61×1011=2.61×103 A=2.61 mAI_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 (gmg_m):

gm=ICVT=2.61×103 A0.0302 V0.0864 A/Vg_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 A/V0.0864 \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.

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Math Problem Analysis

Mathematical Concepts

Transconductance
Bipolar Junction Transistor (BJT)
Thermal Voltage

Formulas

Transconductance formula: gm = IC / VT

Theorems

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Suitable Grade Level

Advanced College