FCC Exam Question: 3-25C5

The beta cutoff frequency of a bipolar transistor is the frequency at which:

A. Base current gain has increased to 0.707 of maximum.
B. Emitter current gain has decreased to 0.707 of maximum.
C. Collector current gain has decreased to 0.707.
D. Gate current gain has decreased to 0.707.
Correct Answer: B

Explanation: In a bipolar junction transistor (BJT), beta ($\beta$) is the common-emitter current gain, defined as the ratio of collector current ($I_C$) to base current ($I_B$). This gain is frequency-dependent; it remains relatively constant at lower frequencies but begins to decrease as the operating frequency increases, primarily due to internal junction capacitances. The beta cutoff frequency ($f_{\beta}$) is the specific frequency at which the common-emitter current gain ($\beta$) drops to 0.707 (or 1/√2) of its maximum low-frequency value. The 0.707 factor signifies the -3dB point, where the amplifier's power output has fallen to half its maximum. Therefore, option B is correct because it accurately describes the decrease in the common-emitter current gain (often simply referred to as "emitter current gain" in this context) to the defined 0.707 level at the cutoff frequency. Options A describes an increase, which is incorrect. Option C is vague and misses the "of maximum" qualifier. Option D refers to gate current gain, which is relevant to Field-Effect Transistors (FETs), not bipolar transistors.

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Includes Elements 1, 3, 6, 7R, 8, and 9.