Question

A BJT in CE mode has \(\beta = 120\) and collector current \(I_C = 2.4\text{ mA}\). The small-signal transconductance at room temperature (\(V_T \approx 25\text{ mV}\)) is approximately

• A. 0.024 S
• B. 0.048 S
• C. 0.096 S
• D. 0.12 S

Hint:

The transistor current gain parameter (\(\beta = 120\)) is extra information not needed to solve this problem! BJT transconductance is unique because it depends only on the collector current and thermal voltage (\(g_m = \frac{I_C}{V_T}\)).

Answer 1

The Correct Option is C

Concept: The small-signal transconductance (\(g_m\)) of a Bipolar Junction Transistor (BJT) measures the sensitivity of the collector current output response relative to changes in the base-emitter input voltage. In the Common Emitter (CE) configuration, transconductance depends directly on the DC bias collector current (\(I_C\)) and the thermal voltage (\(V_T\)): \[ g_m = \frac{I_C}{V_T} \] where \(V_T = \frac{\bar{k}T}{q}\) is the thermal voltage, which is approximately equal to 25 mV (or 26 mV) at standard room temperature.

Step 1: Identify the given values and convert units to standard form. The parameters provided are:
• DC Collector current (\(I_C\)) = \(2.4\text{ mA} = 2.4 \times 10^{-3}\text{ A}\)
• Thermal voltage (\(V_T\)) = \(25\text{ mV} = 25 \times 10^{-3}\text{ V}\)
• Current gain factor (\(\beta\)) = 120

Step 2: Calculate the small-signal transconductance (\(g_m\)). Substitute the values into the transconductance formula: \[ g_m = \frac{2.4 \times 10^{-3}\text{ A}}{25 \times 10^{-3}\text{ V}} \] The \(10^{-3}\) scale terms in the numerator and denominator cancel out directly: \[ g_m = \frac{2.4}{25}\text{ Siemens (S)} \] To evaluate this division easily, multiply both the numerator and the denominator by 4: \[ g_m = \frac{2.4 \times 4}{25 \times 4} = \frac{9.6}{100} = 0.096\text{ S} \] Thus, the small-signal transconductance of the BJT is equal to 0.096 S, matching option (C).