Transistor and Amplifier
Premium coaching notes on transistor structure, p-n-p and n-p-n action, CB/CE/CC configurations, current gains, characteristics, amplifier action, switching, oscillator basics, numericals and PYQs.
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1. Transistor Structure
A junction transistor is a three-layer, two-junction semiconductor device. It has three terminals: emitter E, base B and collector C.
Emitter
Emitter is heavily doped and injects charge carriers into the base.
Base
Base is very thin and lightly doped. Only a small fraction of carriers recombine in the base.
Collector
Collector is moderately doped and physically larger. It collects most carriers coming from emitter.
2. PNP and NPN Transistor Action
For normal transistor action, emitter-base junction is forward biased and collector-base junction is reverse biased.
NPN Transistor
Electrons are majority carriers. Electrons are injected from n-type emitter into the thin p-type base. A small part recombines in the base producing IB, while most electrons reach the collector and produce IC.
PNP Transistor
Holes are majority carriers. In p-n-p transistor, holes are injected from emitter into base. A small percentage recombines in base and produces base current. Most carriers reach collector and produce collector current.
3. CB, CE and CC Configurations
A transistor can be connected in three basic configurations depending on which terminal is common to input and output circuits.
| Configuration | Input Terminals | Output Terminals | Main Feature |
|---|---|---|---|
| Common-base (CB) | Emitter-base | Collector-base | Low input resistance, voltage gain possible, current gain less than 1. |
| Common-emitter (CE) | Base-emitter | Collector-emitter | High current and voltage gain; output is phase reversed. |
| Common-collector (CC) | Base-collector | Emitter-collector | High input resistance; used as emitter follower. |
4. Current Gain and Relation between α and β
Common-base Current Gain α
Since IC is slightly less than IE, α is less than 1. Typical value is 0.95 to 0.99.
Common-emitter Current Gain β
β is large because a small base current controls a much larger collector current.
5. CE Transistor Characteristics
In common-emitter configuration, base-emitter side is input and collector-emitter side is output.
CE Input Characteristic
Graph between base current IB and base-emitter voltage VBE at constant VCE.
CE Output Characteristic
Graph between collector current IC and collector-emitter voltage VCE at constant base current IB.
6. Transistor as Amplifier
A transistor amplifier uses a small input signal to control a larger output signal. In active region, small change in input current produces a large change in collector current.
1. Common-base PNP Amplifier
Input is applied in the emitter-base circuit and output is taken from the collector-base circuit. Current gain is less than 1, but voltage gain may be high.
2. Common-base NPN Amplifier
Base is common to input and output. This configuration has low input resistance and high output resistance.
3. Common-emitter PNP Amplifier
Emitter is common to input and output. It gives current gain, voltage gain and power gain; polarities are according to p-n-p operation.
4. Common-emitter NPN Amplifier
Input is applied between base and emitter; output is taken between collector and emitter. CE amplifier gives high current gain, high voltage gain and 180° phase reversal.
Gains
Current gain Ai = ΔIout / ΔIin. Voltage gain Av = ΔVout / ΔVin. Power gain = Ai × Av.
Resistances
Input resistance ri = ΔVin / ΔIin. Output resistance ro = ΔVout / ΔIout.
Phase Reversal
In CE amplifier, output voltage is 180° out of phase with the input signal.
7. Transistor as Switch and Logic Applications
A transistor can behave like an electronic switch. It is OFF in cut-off region and ON in saturation region.
| Region | Condition | Switch State | Logic Meaning |
|---|---|---|---|
| Cut-off | IB ≈ 0, IC ≈ 0 | OFF | No collector current flows. |
| Active | Collector current controlled by base current | Amplifier mode | Used for signal amplification. |
| Saturation | Both junctions effectively forward biased | ON | Maximum collector current flows. |
8. Transistor Oscillator
A transistor oscillator converts DC power into AC oscillations. It requires an amplifier, positive feedback and a frequency-selective tank circuit.
Feedback
A fraction of output is fed back to input in the correct phase to compensate energy losses.
Tank Circuit
Inductor L and capacitor C exchange energy and decide the frequency of oscillation.
Barkhausen Condition
In simple words: loop gain must be 1 and feedback must be in phase with input.
Sustained Oscillations
Oscillations continue when energy supplied by transistor equals energy lost in the circuit.
9. Numericals
Numerical 1: Relation between IE, IB and IC
Question: In a transistor, IB = 40 μA and IC = 4 mA. Find IE.
Given: IB = 0.04 mA, IC = 4 mA
Formula: IE = IB + IC
Substitution: IE = 0.04 + 4
Calculation: IE = 4.04 mA
Final Answer: IE = 4.04 mA
Exam Tip: Convert μA to mA before adding.
Numerical 2: β from Currents
Question: If ΔIC = 2 mA and ΔIB = 20 μA, find β.
Given: ΔIC = 2 mA, ΔIB = 0.02 mA
Formula: β = ΔIC / ΔIB
Substitution: β = 2 / 0.02
Calculation: β = 100
Final Answer: β = 100
Exam Tip: β has no unit.
Numerical 3: α from β
Question: Find α if β = 49.
Given: β = 49
Formula: α = β / (1 + β)
Substitution: α = 49 / 50
Calculation: α = 0.98
Final Answer: α = 0.98
Exam Tip: α is always less than 1.
Numerical 4: β from α
Question: A transistor has α = 0.96. Find β.
Given: α = 0.96
Formula: β = α / (1 - α)
Substitution: β = 0.96 / 0.04
Calculation: β = 24
Final Answer: β = 24
Exam Tip: A small change in α near 1 causes a large change in β.
Numerical 5: Voltage Gain and Power Gain
Question: An amplifier has current gain 50 and voltage gain 20. Find power gain.
Given: Ai = 50, Av = 20
Formula: Power gain = Ai × Av
Substitution: Power gain = 50 × 20
Calculation: Power gain = 1000
Final Answer: Power gain = 1000
Exam Tip: Power gain is often expressed as a pure ratio.
Numerical 6: Transconductance
Question: In a CE transistor, ΔIC = 4 mA for ΔVBE = 20 mV. Find gm.
Given: ΔIC = 0.004 A, ΔVBE = 0.020 V
Formula: gm = ΔIC / ΔVBE
Substitution: gm = 0.004 / 0.020
Calculation: gm = 0.2 siemens
Final Answer: gm = 0.2 S
Exam Tip: Use ampere and volt for transconductance.
10. PYQs and Exam-Style Questions
Older CBSE Pattern
Question: Why is the emitter-base junction forward biased and collector-base junction reverse biased in normal transistor action?
Solution: Forward bias at emitter-base junction injects majority carriers from emitter into base. Reverse bias at collector-base junction sweeps most carriers into collector.
Final Answer: This biasing allows small base current to control large collector current.
Exam Tip: Write both junction biasing conditions clearly.
NEET Style
Question: In a transistor, IE = 10 mA and IC = 9.8 mA. Find IB.
Given: IE = 10 mA, IC = 9.8 mA
Formula: IE = IB + IC
Substitution: IB = 10 - 9.8
Calculation: IB = 0.2 mA
Final Answer: IB = 0.2 mA
Exam Tip: Base current is much smaller than collector current.
JEE Main Style
Question: If α = 0.98, calculate β.
Given: α = 0.98
Formula: β = α / (1 - α)
Substitution: β = 0.98 / 0.02
Calculation: β = 49
Final Answer: β = 49
Exam Tip: Never use β = 1 / α.
JEE Advanced Style
Question: A CE amplifier gives output voltage opposite in phase to input voltage. Explain why.
Solution: When base current increases, collector current increases. Larger collector current causes larger voltage drop across collector resistor, so collector voltage decreases. Therefore output variation is opposite to input variation.
Final Answer: CE amplifier gives 180° phase reversal.
Exam Tip: Explain using voltage drop across RC.
IB Physics Style
Question: State the condition for a transistor to act as a closed switch.
Solution: A transistor acts as a closed switch when it is driven into saturation region by sufficient base current.
Final Answer: ON condition: saturation region.
Exam Tip: Cut-off means OFF, saturation means ON.
IGCSE Style
Question: Name the three terminals of a transistor.
Solution: The three terminals are emitter, base and collector.
Final Answer: Emitter, base, collector.
Exam Tip: Use E, B, C labels in circuit diagrams.
A-Level Style
Question: A transistor switch controls a relay. Why is a small base current sufficient?
Solution: In a transistor, small base current controls a much larger collector current because β is large in CE mode.
Final Answer: Current amplification allows small input current to control large load current.
Exam Tip: Mention β or current gain in switching explanation.
11. Quick Revision
Core Relations
- IE = IB + IC
- α = ΔIC / ΔIE
- β = ΔIC / ΔIB
α and β
- α = β / (1 + β)
- β = α / (1 - α)
- α is less than 1; β is usually large.
Biasing
- Emitter-base junction: forward biased.
- Collector-base junction: reverse biased.
- CE amplifier output has phase reversal.
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