JEE Mains Physics · Semiconductor Electronics

Transistors and the CE Amplifier

A transistor's emitter current splits into a small base current and a large collector current; their ratios α and β, and the common-emitter gains built from β, are what the questions ask for.

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Why this matters

Eighteen PYQs, thirteen of them multiple choice: five from 2021, nine from 2022 and four from 2023. Transistors have since left the JEE Main syllabus, and the bank agrees: there has been no transistor question since 2023, the last one from April 2023. Nine are about structure, α, β and the operating regions; nine are about the common-emitter amplifier, finding β from a characteristic and then the voltage or power gain. The page is kept for revising the older papers.

Concept 1 of 2: Transistor structure, α and β

A transistor is three layers, n-p-n or p-n-p. The emitter is packed with carriers and pushes them into a very thin, lightly doped base. Few of them recombine there; almost all sweep on into the collector. So the collector current is nearly the whole emitter current, and the small base current controls a much larger collector current.

Definition

Emitter: heavily doped, supplies the carriers. Base: very thin and lightly doped. Collector: the largest region, moderately doped.
Two diodes joined back to back do not make a transistor: their common middle region is far too thick, so the carriers recombine there.
Currents: $I_E = I_B + I_C$ . $\alpha = I_C/I_E$ is just under 1; $\beta = I_C/I_B$ is tens to hundreds.
$\alpha = \dfrac{\beta}{1 + \beta}$ and $\beta = \dfrac{\alpha}{1 - \alpha}$ . The same holds for changes: $\Delta I_B = \Delta I_E - \Delta I_C$ .
Active region (emitter-base forward biased, collector-base reverse biased): the transistor amplifies.
Cut-off (both junctions reverse biased) and saturation (both forward biased): the transistor is a switch, off and on.
An n-p-n transistor carries more current than a p-n-p one, because electrons move more easily than holes.
An oscillator is an amplifier with positive feedback: part of the output is fed back to the input in phase.

Transistor currents

I_E = I_B + I_C, \qquad \alpha = \frac{\beta}{1 + \beta}, \qquad \beta = \frac{\alpha}{1 - \alpha}

Worked example

In a transistor the emitter current is 5 mA and the base current is 100 μA. Find the collector current, α and β.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 10 · Q8Moderate

Example 1 · Semiconductor Electronics · Transistors and the CE Amplifier

The correct relation between

\alpha

(ratio of collector current to emitter current) and

\beta

(ratio of collector current to base current) of a transistor is:

α is below 1, β is large

α compares the collector current with the larger emitter current, so it is just under 1. β compares it with the small base current, so it is tens or hundreds. A relation that gives α above 1 has the two swapped.

The emitter current is the sum

I_E = I_B + I_C. From a change in emitter and collector current, the base change is their difference, and β is the collector change divided by that difference.

A switch uses cut-off and saturation

The active region is for amplifying. A transistor used as a switch is driven between cut-off (off) and saturation (on).

Drill 8 more on transistor structure, α and β

Concept 2 of 2: Gains of a common-emitter amplifier

In the common-emitter circuit a small signal on the base changes the base current a little and the collector current β times as much. That collector change flows through the load resistor, so the output voltage changes far more than the input did. The voltage gain is β scaled by the ratio of load resistance to input resistance.

Definition

Input resistance: $r_i = \dfrac{\Delta V_{BE}}{\Delta I_B}$ at constant $V_{CE}$ .
Current gain: $\beta = \dfrac{\Delta I_C}{\Delta I_B}$ , the slope of the transfer characteristic ( $I_C$ against $I_B$ ).
Voltage gain: $A_V = \beta\,\dfrac{R_L}{r_i}$ . Output voltage: $v_{out} = A_V v_{in}$ , 180° out of phase with the input.
Power gain: $A_P = \beta A_V = \beta^{2}\,\dfrac{R_L}{r_i}$ .
d.c. working point: $I_C = \beta I_B$ and $V_{CE} = V_{CC} - I_C R_C$ .
Keep the units straight: a change in mA divided by a change in μA is a factor of 1000.

Common-emitter gains

\beta = \frac{\Delta I_C}{\Delta I_B}, \quad r_i = \frac{\Delta V_{BE}}{\Delta I_B}, \quad A_V = \beta\,\frac{R_L}{r_i}, \quad A_P = \beta A_V

Worked example

In a common-emitter amplifier, a 20 mV change in base-emitter voltage changes the base current by 40 μA and the collector current by 4 mA. The load is 3 kΩ. Find the input resistance, β, the voltage gain and the power gain.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q23Moderate

Example 2 · Semiconductor Electronics · Transistors and the CE Amplifier

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of

10mV

is added to the base-emitter voltage, the base current changes by

10\mu A

and the collector current changes by

1.5\text{ }mA

. The load resistance is

5k\Omega

. The voltage gain of the transistor will be

mA over μA is a factor of a thousand

β from a 3 mA change against a 25 μA change is 120, not 0.12. Convert both to the same unit before dividing.

Power gain has β twice

Power gain is current gain times voltage gain, β × A_V = β² R_L ÷ r_i. Using β once gives the voltage gain again.

Use the input resistance, not the base resistor

The voltage gain uses r_i, the transistor's own input resistance. When the question gives a separate input resistance, use it rather than the base resistor R_B; R_B stands in for r_i only when nothing else on the input side is given.

Drill 8 more on gains of a common-emitter amplifier

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 (2)

Transistor structure, α and β
Transistor currents
$I_E = I_B + I_C, \qquad \alpha = \frac{\beta}{1 + \beta}, \qquad \beta = \frac{\alpha}{1 - \alpha}$
Gains of a common-emitter amplifier
Common-emitter gains
$\beta = \frac{\Delta I_C}{\Delta I_B}, \quad r_i = \frac{\Delta V_{BE}}{\Delta I_B}, \quad A_V = \beta\,\frac{R_L}{r_i}, \quad A_P = \beta A_V$

Watch out for (6)

α is below 1, β is large
→ Transistor structure, α and β
The emitter current is the sum
→ Transistor structure, α and β
A switch uses cut-off and saturation
→ Transistor structure, α and β
mA over μA is a factor of a thousand
→ Gains of a common-emitter amplifier
Power gain has β twice
→ Gains of a common-emitter amplifier
Use the input resistance, not the base resistor
→ Gains of a common-emitter amplifier

Test yourself on Semiconductor Electronics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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