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MHT-CET Physics · Semiconductor Devices

The Transistor and the Common-Emitter Amplifier

In a transistor the emitter current splits into a small base current and a large collector current, I_E = I_B + I_C; the ratios α = I_C/I_E and β = I_C/I_B describe it, and in common-emitter mode it amplifies with a gain of β R_L/R_in and a 180° phase reversal.

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Why this matters

29 PYQs, none HARD — steady marks for anyone who keeps α and β apart. Three shapes: the current ratios and their identities, the common-emitter amplifier's voltage and power gain and its phase, and how an n-p-n transistor differs from a p-n-p one.

Concept 1 of 3: Current Ratios α and β

Almost every carrier the emitter sends reaches the collector; a few recombine in the thin base. So α = I_C/I_E is just under 1, and β = I_C/I_B, collector current per base current, is large. Every relation between them comes from I_E = I_B + I_C.

Definition

  • IE=IB+ICI_E = I_B + I_C; α=ICIE<1\alpha = \dfrac{I_C}{I_E} < 1; β=ICIB>1\beta = \dfrac{I_C}{I_B} > 1.
  • β=α1−α\beta = \dfrac{\alpha}{1 - \alpha}, α=β1+β\alpha = \dfrac{\beta}{1 + \beta}, 1α−1β=1\dfrac{1}{\alpha} - \dfrac{1}{\beta} = 1, β−ααβ=1\dfrac{\beta - \alpha}{\alpha\beta} = 1.
  • '80% of the emitted electrons reach the collector' means α=0.8\alpha = 0.8, so β=4\beta = 4.
  • ΔIC=β ΔIB\Delta I_C = \beta\,\Delta I_B.

Current gains

α=ICIE,β=ICIB,β=α1−α\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}, \quad \beta = \frac{\alpha}{1 - \alpha}

Worked example

A transistor has IE=10I_E = 10 mA and IB=0.2I_B = 0.2 mA. Find ICI_C, α and β.
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q26Easy

Example 1 · Semiconductor Devices · Transistors — BJT, CE Amplifier, and Gain

For a common emitter transistor, if ICIE=0.95\frac{I_{C}}{I_{E}}= 0.95, then the current gain is

Calling the collector fraction β

'90% reach the collector' is α = 0.9, and β is 0.90.1=9\frac{0.9}{0.1} = 9 — not 90. The options offer α and β both ways round.
Drill 12 more on current ratios α and β

Concept 2 of 3: The Common-Emitter Amplifier

The input goes in between base and emitter, the output comes out between collector and emitter. A small change in base current makes a β-times larger change in collector current, and the load turns that into a large voltage — upside down, because more collector current means more drop across the load and less voltage at the collector.

Definition

  • Voltage gain AV=β RLRinA_V = \beta\,\dfrac{R_L}{R_{\text{in}}}; power gain =β×AV=β2RLRin= \beta \times A_V = \beta^2\dfrac{R_L}{R_{\text{in}}}; so power gain / voltage gain =β= \beta.
  • Given the voltage gain: power gain =AV2RinRout= A_V^2\dfrac{R_{\text{in}}}{R_{\text{out}}}.
  • Output is 180∘180^\circ (π\pi) out of phase: Vi=2cos⁡(ωt+ϕ)V_i = 2\cos(\omega t + \phi) with gain 126 gives 252cos⁡(ωt+ϕ+π)252\cos(\omega t + \phi + \pi).
  • Base–emitter junction forward biased, collector–base reverse biased.
  • Saturation: IC,sat=VCCRCI_{C,\text{sat}} = \dfrac{V_{CC}}{R_C}, least base current IC,satβ\dfrac{I_{C,\text{sat}}}{\beta}.

CE amplifier

AV=β RLRin,AP=β AVA_V = \beta\,\frac{R_L}{R_{\text{in}}}, \qquad A_P = \beta\,A_V

Worked example

β = 50, RL=4 kΩR_L = 4\,\text{k}\Omega, Rin=1 kΩR_{\text{in}} = 1\,\text{k}\Omega. Voltage gain, power gain, and the output for a 5 mV input?
Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q44Easy

Example 2 · Semiconductor Devices · Transistors — BJT, CE Amplifier, and Gain

In common emitter transistor amplifier, load resistance is 6.5 kΩ6.5\ \text{k}\Omega and input resistance is 1.3 kΩ1.3\ \text{k}\Omega. If current gain is 78, the voltage gain is

Squaring β in the voltage gain

Voltage gain has ONE factor of β; power gain has two. 78×6.51.3=39078 \times \frac{6.5}{1.3} = 390 is the voltage gain; 782×578^2 \times 5 would be the power gain.
Drill 13 more on the common-emitter amplifier

Concept 3 of 3: n-p-n and p-n-p Transistors

Both work the same way; only the carriers swap. An n-p-n emitter pushes electrons into a p-type base; a p-n-p emitter pushes holes into an n-type base. Seen as two diodes, the base is the shared middle.

Definition

  • n-p-n: emitter injects ELECTRONS into the base; p-n-p: emitter injects HOLES.
  • Two-diode picture of an n-p-n: both diodes have their p-side (anode) at the base, pointing out to E and to C.
  • In either, the emitter–base junction is forward biased in normal operation.

Currents

IE=IB+ICI_E = I_B + I_C

Worked example

What does the emitter inject into the base in a p-n-p transistor, and in an n-p-n?
Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q8Easy

Example 3 · Semiconductor Devices · Transistors — BJT, CE Amplifier, and Gain

Which one of the operations of n-p-n transistor differs from that of p-n-p transistor?

Drawing the diodes pointing inward

An n-p-n's base is p-type, so the diode arrows point OUT from the base to E and C. Arrows pointing into the base describe a p-n-p.
Drill 1 more on n-p-n and p-n-p transistors

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Current Ratios α and β

    Current gains

    α=ICIE,β=ICIB,β=α1−α\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}, \quad \beta = \frac{\alpha}{1 - \alpha}
  • The Common-Emitter Amplifier

    CE amplifier

    AV=β RLRin,AP=β AVA_V = \beta\,\frac{R_L}{R_{\text{in}}}, \qquad A_P = \beta\,A_V
  • n-p-n and p-n-p Transistors

    Currents

    IE=IB+ICI_E = I_B + I_C

Watch out for (3)

Test yourself on Semiconductor Devices

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