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Question

What is the relationship between the common-emitter current gain ($\beta$) and the common-base current gain ($\alpha$) of a bipolar junction transistor (BJT)?

The correct answer is
$\beta = \frac{\alpha}{1 - \alpha}$

BJT Current Gain Relationship: Alpha ($\alpha$) and Beta ($\beta$)

Understanding BJT Current Gains

A bipolar junction transistor (BJT) has two primary current gain parameters:
  • Common-Base Current Gain ($\alpha$): Defined as the ratio of collector current ($I_C$) to emitter current ($I_E$). It is typically close to 1 (e.g., 0.95 to 0.99). $\alpha = \frac{I_C}{I_E}$
  • Common-Emitter Current Gain ($\beta$): Defined as the ratio of collector current ($I_C$) to base current ($I_B$). It is typically much larger than 1 (e.g., 20 to 200). $\beta = \frac{I_C}{I_B}$

Deriving the Relationship

The relationship between these gains can be derived using Kirchhoff's Current Law at the emitter node: $I_E = I_C + I_B$ Using the definition of $\alpha$, we can write $I_C$ in terms of $I_E$: $I_C = \alpha \cdot I_E$ Substitute this into the KCL equation: $I_E = (\alpha \cdot I_E) + I_B$ Rearrange to solve for $I_B$: $I_B = I_E - (\alpha \cdot I_E)$ $I_B = I_E (1 - \alpha)$ Now, use the definition of $\beta$: $\beta = \frac{I_C}{I_B}$ Substitute $I_C = \alpha \cdot I_E$ and $I_B = I_E (1 - \alpha)$: $\beta = \frac{\alpha \cdot I_E}{I_E (1 - \alpha)}$ Cancel out $I_E$: $\beta = \frac{\alpha}{1 - \alpha}$

Final Relationship

The relationship between the common-emitter current gain ($\beta$) and the common-base current gain ($\alpha$) is: $\beta = \frac{\alpha}{1 - \alpha}$
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Important Questions from Configuration of BJT

  1. BC147 is the transistor used for:

  2. Which of the following is NOT true for a common collector transistor?

  3. The other name for the common collector amplifier is -

  4. Match List I with List II:

    List I

    (Bias Configuration of BJT)

    List II

    (Stability factor equation)

    (A)Fixed Bias Configuration(I)S(V BE ) = \(\rm −\frac{\beta/R_E}{\beta+R_{TH}/R_E}\)
    (B)Emitter Bias Configuration(II)S(V BE ) = −β/R E
    (C)Voltage Divider Configuration(III)S(V BE ) =  \(\rm −\frac{\beta/R_C}{\beta+R_E/R_C}\)
    (D)Feedback Bias Configuration(IV)S(V BE ) =  \(\rm −\frac{\beta/R_E}{\beta+R_B/R_E}\)

    Choose the correct answer from the options given below :

  5. During normal working of transistor as amplifier, the emitter junction is _______.

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