Math Problem Statement

For the transistor circuits shown below, if the current gain of Q1 = 100 and the current gain of Q2 = 50, what is the overall gain of each circuit?

Solution

To solve the problem, let's analyze the circuits and calculate the overall current gain for each.

Understanding the problem:

  • Q1 (npn) has a current gain (β1\beta_1) of 100.
  • Q2 (pnp) has a current gain (β2\beta_2) of 50.
  • The overall current gain for a two-transistor Darlington or similar compound configuration can be approximated by: βoverall=β1β2+β1+β2\beta_{\text{overall}} = \beta_1 \cdot \beta_2 + \beta_1 + \beta_2 However, for high values of β1\beta_1 and β2\beta_2, the term β1β2\beta_1 \cdot \beta_2 dominates, so it simplifies to: βoverallβ1β2\beta_{\text{overall}} \approx \beta_1 \cdot \beta_2

Calculations for the two circuits:

  1. For the first circuit (left side):

    • Current gain of Q1 (β1\beta_1) = 100
    • Current gain of Q2 (β2\beta_2) = 50 βoverall=10050=5000\beta_{\text{overall}} = 100 \cdot 50 = 5000
  2. For the second circuit (right side):

    • The configuration and calculation method remain the same.
    • βoverall=5000\beta_{\text{overall}} = 5000

Answer:

The overall gain for each circuit is 5000.

Let me know if you'd like further clarifications or additional details! Here are five related questions to deepen your understanding:

  1. How would the overall gain change if β1\beta_1 or β2\beta_2 increased or decreased?
  2. What are the advantages of using such compound transistor configurations?
  3. How does this calculation change for circuits with additional resistors or feedback loops?
  4. Can this calculation apply to circuits with three or more transistors?
  5. What practical applications benefit from high current gain in transistor circuits?

Tip: Always check the operating conditions (like saturation and cutoff regions) to ensure the calculated gains are achievable in real circuits!

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Math Problem Analysis

Mathematical Concepts

Transistor Current Gain
Circuit Analysis

Formulas

Overall Gain (βoverall) = β1 × β2 + β1 + β2
Approximation: βoverall ≈ β1 × β2 (when β1 and β2 are large)

Theorems

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Suitable Grade Level

Undergraduate Electrical Engineering