Transistor as Amplifier and Switch

Master BJT transistor working, CE amplifier, and digital switching for JEE Physics with smartphone processor analogy

13 min read

Prerequisites

Before studying this topic, make sure you understand:

The Hook: What’s Inside Your Smartphone Processor?

Connect: Smartphone Brain → Transistors

Your smartphone processor contains over 15 BILLION transistors in an area smaller than your fingernail! Each transistor:

  • Switches ON/OFF billions of times per second
  • Acts as a tiny amplifier or digital switch
  • Consumes less power than a grain of sand heating up

Mind-blowing scale:

  • iPhone 14 Pro chip: 16 billion transistors
  • All transistors switching → performing trillion operations/second
  • If each transistor was 1 cm, the chip would be size of Manhattan!

How does a simple sandwich of p-n-p or n-p-n semiconductors create the brain of modern civilization? Let’s decode the most important invention of the 20th century!

Interactive Demo

Visualize transistor amplification and switching action:

The Core Concept: What is a Transistor?

Transistor = Transfer Resistor

Name origin:

  • Transfer + Resistor = Transistor
  • Small current in input transfers to large current in output
  • Acts like current-controlled resistor

Two p-n Junctions Back-to-Back

Bipolar Junction Transistor (BJT) has two types:

1. npn Transistor:

  C (Collector)
  |
  n
  |
  p ←— B (Base)
  |
  n
  |
  E (Emitter)

2. pnp Transistor:

  C (Collector)
  |
  p
  |
  n ←— B (Base)
  |
  p
  |
  E (Emitter)

Three regions:

  • Emitter (E): Heavily doped, emits charge carriers
  • Base (B): Very thin (~1 µm), lightly doped
  • Collector (C): Moderately doped, collects carriers
Why Two p-n Junctions?

Single p-n junction (diode):

  • Current flows one way
  • Can’t amplify

Two junctions (transistor):

  • Base-Emitter (BE): Forward biased (allows current)
  • Base-Collector (BC): Reverse biased (normally blocks)
  • Magic: Carriers from emitter cross thin base into collector
  • Small base current controls large collector current!

Think of it like:

  • Water dam with small control valve (base)
  • Tiny adjustment → huge water flow change (collector)
  • Current amplification!

Transistor Action (npn)

Biasing for Active Mode

Proper biasing:

  • BE junction: Forward biased (VBE ≈ 0.7V)
  • BC junction: Reverse biased (VCB > 0)

Current Flow Mechanism

Step 1: Electrons injected from emitter (heavily doped n-type)

Step 2: Enter thin p-type base

  • Some recombine with holes → small base current $I_B$
  • Most (>95%) continue through thin base

Step 3: Reach reverse-biased BC junction

  • Collector electric field pulls them in
  • Creates collector current $I_C$

Result:

$$\boxed{I_E = I_B + I_C}$$

where:

  • $I_E$ = emitter current (largest)
  • $I_C$ = collector current (~99% of $I_E$)
  • $I_B$ = base current (~1% of $I_E$)

Current Gain Parameters

1. Current Gain (β or $h_{FE}$):

$$\boxed{\beta = \frac{I_C}{I_B}}$$

Typical values: β = 50 to 300

This is the amplification factor!

2. Alpha (α):

$$\boxed{\alpha = \frac{I_C}{I_E}}$$

Typical: α ≈ 0.95 to 0.99

Relationship:

$$\boxed{\beta = \frac{\alpha}{1-\alpha}}$$ $$\boxed{\alpha = \frac{\beta}{\beta + 1}}$$
The Amplification Magic

Example: β = 100

Input: Base current $I_B = 10$ µA (tiny!)

Output: Collector current $I_C = \beta I_B = 100 \times 10 = 1000$ µA = 1 mA

Amplification = 100×!

Power perspective:

  • Small base power controls large collector power
  • Like tiny steering wheel controlling massive truck
  • This is how transistors amplify signals!

In your smartphone:

  • Microphone signal: ~millivolts
  • After transistor amplification: volts
  • You can hear music clearly through speakers!

Transistor Configurations

1. Common Emitter (CE) - Most Important

Circuit:

      VCC
       |
       RC
       |
   C ──┴── Vout

Input ──┤  npn
       │
   E ──┬── GND
       |
       RE

Characteristics:

  • Input: Between base and emitter
  • Output: Between collector and emitter
  • Emitter is common to both

Properties:

ParameterValue
Current gainHigh (β)
Voltage gainHigh (~100)
Power gainVery high
Phase shift180° (inverted)
Input resistanceMedium (~1 kΩ)
Output resistanceHigh (~50 kΩ)

Use: Voltage amplifiers (most common!)

2. Common Collector (CC) - Emitter Follower

Properties:

  • Voltage gain ≈ 1 (no amplification)
  • Current gain = High
  • Output follows input (no inversion)
  • Low output impedance

Use: Impedance matching, buffer

3. Common Base (CB)

Properties:

  • Current gain < 1 (no current amplification)
  • Voltage gain = High
  • No phase inversion
  • Very low input impedance

Use: High-frequency applications, RF amplifiers

For JEE: Focus on Common Emitter - most important and frequently asked!

Common Emitter Amplifier

Circuit Analysis

DC Analysis (Biasing):

Base voltage:

$$V_B = \frac{R_2}{R_1 + R_2} V_{CC}$$

Emitter voltage:

$$V_E = V_B - V_{BE}$$

where $V_{BE} \approx 0.7$ V for Si

Emitter current:

$$I_E = \frac{V_E}{R_E}$$

Collector current:

$$I_C \approx I_E$$

(since α ≈ 1)

Collector voltage:

$$V_C = V_{CC} - I_C R_C$$

AC Analysis (Small Signal):

Voltage gain:

$$\boxed{A_v = -\frac{R_C}{r_e}}$$

where:

  • $r_e$ = dynamic emitter resistance ≈ $\frac{26 \text{ mV}}{I_E}$ at 300K
  • Negative sign indicates phase inversion

Alternative form:

$$\boxed{A_v = -\frac{\beta R_C}{R_i}}$$

Current gain:

$$\boxed{A_i = \beta}$$

Power gain:

$$\boxed{A_p = A_v \times A_i = \beta \times \frac{R_C}{r_e}}$$

Input and Output Characteristics

Input characteristic (VBE vs IB):

  • Similar to diode curve
  • Exponential rise after 0.7V
  • Controls base current

Output characteristic (VCE vs IC):

  • Shows three regions:

1. Cut-off region: $I_B = 0$, $I_C \approx 0$ (OFF)

2. Active region: $I_C = \beta I_B$ (linear amplification)

3. Saturation region: $V_{CE} < 0.2$ V (fully ON)

Smartphone Audio Amplifier

When you play music on your phone:

Input: Microphone or audio file → tiny AC signal (~mV)

Transistor amplifier:

  • Operating point set in active region
  • Small input signal varies base current
  • Collector current varies by β times
  • Output voltage varies across $R_C$

Output: Amplified signal (~V) drives speaker

Multiple stages: Typical phone has 2-3 amplifier stages

  • Each stage amplifies by ~100×
  • Total gain: $100 \times 100 = 10,000×$!
  • Millivolt input → Volt-level output → Audible sound!

Transistor as a Switch

Digital Switching

Transistor can work in two extreme states:

1. OFF State (Cut-off):

  • $I_B = 0$
  • $I_C = 0$
  • $V_{CE} = V_{CC}$ (maximum)
  • Acts like open switch

2. ON State (Saturation):

  • $I_B$ > minimum value
  • $I_C = I_{C,sat} = \frac{V_{CC}}{R_C}$
  • $V_{CE,sat} \approx 0.2$ V (minimum)
  • Acts like closed switch

Switching Circuit

      VCC
       |
       RC
       |
   C ──┴── Vout

Input ──┤  npn
       │
   E ──┴── GND

Operation:

Input LOW (0V):

  • $I_B = 0$
  • Transistor OFF
  • $V_{out} = V_{CC}$ (HIGH)

Input HIGH (5V):

  • $I_B > 0$
  • Transistor ON (saturated)
  • $V_{out} \approx 0$ V (LOW)

Notice: Output is inverted from input!

This is a NOT gate (inverter)!

Calculating Base Resistor

To ensure saturation:

$$I_B > \frac{I_{C,sat}}{\beta}$$

Base current:

$$I_B = \frac{V_{in} - V_{BE}}{R_B}$$

For saturation:

$$\boxed{R_B < \frac{(V_{in} - V_{BE}) \beta}{I_{C,sat}}}$$

Typically choose: $R_B = \frac{(V_{in} - V_{BE}) \beta}{2 I_{C,sat}}$ (provides margin)

Your Processor's Logic Gates

Every logic gate in your processor is made of transistors as switches!

NAND gate: 4 transistors NOR gate: 4 transistors Complex CPU: Billions of transistors switching!

Example - Intel Core i9:

  • 16 billion transistors
  • Each switches billions of times/second
  • Switching speed: < 100 picoseconds (10⁻¹⁰ s)!
  • All performing binary logic: 0s and 1s

Power consumption:

  • Each transistor uses ~femtojoules per switch
  • 16 billion × billions of switches/sec = ~100 watts
  • This is why CPUs need cooling fans!

From transistors to Instagram: Binary switches → Logic gates → Processors → Smartphones → Your entire digital life!

Important Formulas Summary

Current Relations

$$\boxed{I_E = I_B + I_C}$$ $$\boxed{\beta = \frac{I_C}{I_B}} \quad \text{(50-300)}$$ $$\boxed{\alpha = \frac{I_C}{I_E}} \quad \text{(0.95-0.99)}$$ $$\boxed{\alpha = \frac{\beta}{\beta+1}}$$

CE Amplifier

$$\boxed{A_v = -\frac{R_C}{r_e}} \quad \text{where } r_e = \frac{26 \text{ mV}}{I_E}$$ $$\boxed{A_i = \beta}$$ $$\boxed{A_p = A_v \times A_i}$$

Switching

$$\boxed{I_{C,sat} = \frac{V_{CC}}{R_C}}$$ $$\boxed{V_{CE,sat} \approx 0.2 \text{ V}}$$

Memory Tricks & Patterns

Mnemonic for npn vs pnp

“npn - Not Pointing iN” - Arrow points OUT “pnp - Pointing iN Permanently” - Arrow points IN

(Arrow is on emitter)

Current Relationship

“Every Base and Collector makes Emitter”

$$I_E = I_B + I_C$$

Beta Memory

“Beta is Big, Alpha Almost 1”

  • β = 50-300 (large)
  • α = 0.95-0.99 (close to 1)

CE Amplifier Sign

“CE Causes inversion” - Common Emitter has negative voltage gain (180° phase shift)

Transistor Regions

“Cut-off is Closed, Saturation is Short”

  • Cut-off: Like open circuit (no current)
  • Saturation: Like short circuit (full current)
  • Active: In between (amplification)

Pattern Recognition

  1. Current magnitudes:

    $$I_E > I_C >> I_B$$

    Typical: If $I_B = 10$ µA, then $I_C \approx 1$ mA, $I_E \approx 1.01$ mA

  2. Voltage drops:

    • $V_{BE} \approx 0.7$ V (forward-biased junction)
    • $V_{CE,sat} \approx 0.2$ V (in saturation)
    • $V_{CE,cutoff} \approx V_{CC}$ (in cut-off)

  3. Dynamic resistance:

    $$r_e(\Omega) = \frac{26 \text{ mV}}{I_E(\text{mA})}$$

    At $I_E = 1$ mA: $r_e = 26$ Ω

When to Use This

Decision Tree

Transistor as amplifier when:

  • Need voltage/current/power gain
  • Analog signal processing
  • Operating in active region
  • Small signal variations around DC operating point

Transistor as switch when:

  • Digital circuits (logic gates)
  • ON/OFF control (LED, relay)
  • Operating in saturation (ON) or cut-off (OFF)
  • No intermediate states needed

For problem solving:

Given circuit, find gain: → Identify configuration (CE/CB/CC) → Use appropriate gain formulas

Given switching circuit: → Check if transistor in saturation or cut-off → Calculate currents and voltages accordingly

Current amplification: → Use $\beta = I_C/I_B$

Common Mistakes to Avoid

Trap #1: Confusing npn and pnp

Wrong: Using npn formulas for pnp circuit

Correct:

  • npn: Conventional current flows C→E, base current is positive
  • pnp: Conventional current flows E→C, all polarities reversed

Easy identification:

  • Arrow on emitter pointing OUT → npn
  • Arrow pointing IN → pnp

For JEE: Most problems use npn (standard convention)

Trap #2: Forgetting VBE Drop

Wrong: “If base voltage is 2V, emitter voltage is 2V”

Correct:

$$V_E = V_B - V_{BE} = 2 - 0.7 = 1.3 \text{ V}$$

Always subtract 0.7V for Si transistor (0.3V for Ge)

Common JEE error: Calculating emitter current without accounting for $V_{BE}$

Trap #3: Assuming Linear Operation in Saturation

Wrong: “In saturation, $I_C = \beta I_B$”

Correct:

  • Active region: $I_C = \beta I_B$ ✓
  • Saturation: $I_C < \beta I_B$ (limited by $R_C$)

In saturation:

$$I_C = I_{C,sat} = \frac{V_{CC} - V_{CE,sat}}{R_C} \approx \frac{V_{CC}}{R_C}$$

Not dependent on β!

JEE trap: “Find IC when transistor is saturated” - don’t use β formula!

Trap #4: Phase Inversion Confusion

Wrong: “All amplifiers invert signal”

Correct:

  • CE amplifier: 180° phase shift (inverted)
  • CC amplifier: 0° phase shift (no inversion)
  • CB amplifier: 0° phase shift (no inversion)

Remember: Only Common Emitter inverts!

Negative sign in formula:

$$A_v = -\frac{R_C}{r_e}$$

The minus means inversion!

Practice Problems

Level 1: Foundation (NCERT/Basic)

Problem 1.1

A transistor has $\beta = 100$. If base current is 50 µA, find: (a) Collector current (b) Emitter current

Solution:

(a) Collector current:

$$I_C = \beta I_B = 100 \times 50 \times 10^{-6}$$ $$I_C = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$$

(b) Emitter current:

$$I_E = I_B + I_C = 0.05 + 5 = 5.05 \text{ mA}$$

Answer: (a) 5 mA, (b) 5.05 mA

Insight: Emitter current only slightly larger than collector current!

Problem 1.2

For the above transistor, find α.

Solution:

$$\alpha = \frac{\beta}{\beta + 1} = \frac{100}{100 + 1} = \frac{100}{101} = 0.99$$

Or directly:

$$\alpha = \frac{I_C}{I_E} = \frac{5}{5.05} = 0.99$$

Answer: α = 0.99

Insight: α very close to 1 for good transistor!

Level 2: JEE Main

Problem 2.1

A CE amplifier has $R_C = 2$ kΩ and transistor operating at $I_E = 2$ mA. Find voltage gain.

Solution:

Dynamic emitter resistance:

$$r_e = \frac{26 \text{ mV}}{I_E(\text{mA})} = \frac{26}{2} = 13 \text{ Ω}$$

Voltage gain:

$$A_v = -\frac{R_C}{r_e} = -\frac{2000}{13} = -154$$

Answer: $A_v = -154$ (magnitude 154, inverted phase)

Insight: Negative sign means 180° phase shift!

Problem 2.2

A transistor switch has $V_{CC} = 12$V, $R_C = 1$ kΩ. Find: (a) Collector current in saturation (b) Minimum base current needed if β = 80

Solution:

(a) Saturation collector current:

$$I_{C,sat} = \frac{V_{CC}}{R_C} = \frac{12}{1000} = 0.012 \text{ A} = 12 \text{ mA}$$

(b) Minimum base current:

For saturation:

$$I_B > \frac{I_{C,sat}}{\beta} = \frac{12}{80} = 0.15 \text{ mA}$$

Practical design (2× margin):

$$I_B = \frac{2 \times 12}{80} = 0.3 \text{ mA}$$

Answer:

  • (a) 12 mA
  • (b) Minimum 0.15 mA, recommended 0.3 mA

Insight: Provide extra base current to ensure hard saturation!

Problem 2.3

In a CE amplifier, $V_{CC} = 12$V, $R_C = 3$ kΩ, $I_C = 2$ mA. Find: (a) $V_C$ (b) $V_{CE}$ if $V_E = 1$V

Solution:

(a) Collector voltage:

$$V_C = V_{CC} - I_C R_C$$ $$V_C = 12 - (2 \times 10^{-3})(3 \times 10^3)$$ $$V_C = 12 - 6 = 6 \text{ V}$$

(b) Collector-emitter voltage:

$$V_{CE} = V_C - V_E = 6 - 1 = 5 \text{ V}$$

Answer: (a) 6V, (b) 5V

Check: $V_{CE} = 5$V > 0.2V → Transistor in active region ✓

Level 3: JEE Advanced

Problem 3.1

A CE amplifier has voltage divider bias: $R_1 = 10$ kΩ, $R_2 = 2.2$ kΩ, $R_E = 1$ kΩ, $R_C = 2.2$ kΩ, $V_{CC} = 12$V. Find: (a) Operating point $(I_C, V_{CE})$ (b) Voltage gain

Solution:

(a) DC analysis:

Base voltage:

$$V_B = \frac{R_2}{R_1 + R_2} V_{CC} = \frac{2.2}{10 + 2.2} \times 12$$ $$V_B = \frac{2.2}{12.2} \times 12 = 2.16 \text{ V}$$

Emitter voltage:

$$V_E = V_B - V_{BE} = 2.16 - 0.7 = 1.46 \text{ V}$$

Emitter current:

$$I_E = \frac{V_E}{R_E} = \frac{1.46}{1000} = 1.46 \text{ mA}$$

Collector current:

$$I_C \approx I_E = 1.46 \text{ mA}$$

Collector voltage:

$$V_C = V_{CC} - I_C R_C = 12 - (1.46 \times 2.2)$$ $$V_C = 12 - 3.21 = 8.79 \text{ V}$$

VCE:

$$V_{CE} = V_C - V_E = 8.79 - 1.46 = 7.33 \text{ V}$$

Operating point: $(1.46 \text{ mA}, 7.33 \text{ V})$

(b) AC analysis:

$$r_e = \frac{26}{1.46} = 17.8 \text{ Ω}$$ $$A_v = -\frac{R_C}{r_e} = -\frac{2200}{17.8} = -124$$

Answer:

  • (a) Q-point: (1.46 mA, 7.33V)
  • (b) $A_v = -124$

Insight: Well-designed bias point near middle of load line for maximum swing!

Problem 3.2

Show that for a transistor, $\beta = \alpha/(1-\alpha)$.

Solution:

Starting from definitions:

$$\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}$$

Also: $I_E = I_B + I_C$

$$\alpha = \frac{I_C}{I_B + I_C} = \frac{I_C/I_B}{I_B/I_B + I_C/I_B}$$ $$\alpha = \frac{\beta}{1 + \beta}$$

Solving for β:

$$\alpha(1 + \beta) = \beta$$ $$\alpha + \alpha\beta = \beta$$ $$\alpha = \beta - \alpha\beta = \beta(1 - \alpha)$$ $$\boxed{\beta = \frac{\alpha}{1-\alpha}}$$

Check: If $\alpha = 0.99$:

$$\beta = \frac{0.99}{1-0.99} = \frac{0.99}{0.01} = 99$$

✓

Proved!

Insight: Small change in α causes large change in β!

  • α: 0.98 → 0.99 (1% change)
  • β: 49 → 99 (100% change!)

This is why β varies widely between transistors!

Quick Revision Box

ParameterFormulaTypical Value
Current gain (β)$I_C/I_B$50-300
Alpha (α)$I_C/I_E$0.95-0.99
$V_{BE}$-0.7V (Si)
$V_{CE,sat}$-0.2V
CE voltage gain$-R_C/r_e$-100 to -500
$r_e$$26\text{mV}/I_E$10-50Ω

Key relation: $I_E = I_B + I_C$, $\beta = \alpha/(1-\alpha)$

JEE Strategy: High-Yield Points

What JEE Loves to Test

  1. Current relations - Most common!

    • $I_E = I_B + I_C$
    • $\beta = I_C/I_B$
    • $\alpha = I_C/I_E$
    • Conversion: $\beta = \alpha/(1-\alpha)$

  2. Transistor configurations:

    • Identify CE/CB/CC from circuit
    • Remember: CE inverts, others don’t
    • CE most commonly asked!

  3. DC biasing calculations:

    • Always use $V_{BE} = 0.7$V
    • $V_E = V_B - 0.7$
    • $I_E = V_E/R_E$
    • $I_C \approx I_E$

  4. CE amplifier gain:

    • $r_e = 26\text{mV}/I_E(\text{mA})$
    • $A_v = -R_C/r_e$
    • Negative sign = phase inversion

  5. Switching operation:

    • Cut-off: $I_B = 0$, $V_{CE} = V_{CC}$
    • Saturation: $V_{CE} = 0.2$V, $I_C = V_{CC}/R_C$
    • Don’t use $\beta$ in saturation!

  6. Conceptual questions:

    • “Why base thin?” → Most carriers reach collector
    • “Why emitter heavily doped?” → Inject many carriers
    • “Active vs saturation?” → Check $V_{CE}$ value

Time-saving tricks:

  • For β = 100: $I_C \approx 100 I_B$
  • For $I_E$: $I_E \approx I_C$ (α ≈ 1)
  • In saturation: $V_{CE} \approx 0$ (short circuit)
  • In cut-off: $I_C \approx 0$ (open circuit)

Within Electronic Devices

Connected Chapters

Real-world Applications

  • Amplifiers - Audio, RF, instrumentation
  • Switches - Digital logic, power control
  • Oscillators - Signal generation
  • Processors - Billions in CPU/GPU
  • Memory - RAM, Flash storage

Teacher’s Summary

Key Takeaways

  1. Transistor = Two p-n junctions back-to-back (npn or pnp)

    • Emitter (heavily doped) → Base (thin) → Collector (moderate)
    • BE forward biased, BC reverse biased for active operation

  2. Current amplification: $\beta = I_C/I_B$ (typically 50-300)

    • Small base current controls large collector current
    • $I_E = I_B + I_C$ (Kirchhoff’s law)
    • $\alpha = I_C/I_E \approx 0.99$ (most carriers reach collector)

  3. As amplifier (Active region):

    • Common Emitter most important
    • Voltage gain: $A_v = -R_C/r_e$ where $r_e = 26\text{mV}/I_E$
    • Phase inverted (180°)
    • Used in audio, RF, instrumentation

  4. As switch (Cut-off/Saturation):

    • Cut-off: $I_B = 0$ → transistor OFF → $V_{CE} = V_{CC}$
    • Saturation: $I_B$ large → transistor ON → $V_{CE} \approx 0.2$V
    • Foundation of digital logic gates

  5. 15 billion transistors in your smartphone processor - each a tiny amplifier or switch, working together to create the digital world!

“From a simple sandwich of doped silicon to the brain of civilization - the transistor is humanity’s most important invention after the wheel!”