BJT Biasing and DC Analysis
Voltage Divider Bias — Base Voltage
V_B = V_{CC} \cdot \frac{R_2}{R_1 + R_2}
| Symbol | Description | Unit |
|---|---|---|
| V_B | Base voltage | V |
| V_{CC} | Supply voltage | V |
| R_1 | Upper bias resistor | Ω |
| R_2 | Lower bias resistor | Ω |
Worked example
Find V_B for a voltage-divider bias circuit with V_CC = 12 V, R1 = 47 kΩ, R2 = 10 kΩ.
Given: V_CC = 12 V, R1 = 47 kΩ, R2 = 10 kΩ
- V_B = 12 × (10 / (47 + 10))
- V_B = 12 × (10 / 57)
- V_B = 12 × 0.1754
- V_B = 2.105 V
Answer: V_B ≈ 2.1 V
Emitter Current
I_E = \frac{V_E}{R_E} = \frac{V_B - V_{BE}}{R_E}
| Symbol | Description | Unit |
|---|---|---|
| I_E | Emitter current | A |
| V_{BE} | Base-emitter voltage (≈ 0.7 V for Si) | V |
| R_E | Emitter resistor | Ω |
Worked example
With V_B = 2.1 V and R_E = 1 kΩ, find I_E.
Given: V_B = 2.1 V, V_BE = 0.7 V, R_E = 1 kΩ
- V_E = V_B - V_BE = 2.1 - 0.7 = 1.4 V
- I_E = V_E / R_E = 1.4 / 1000
- I_E = 0.0014 A
Answer: I_E = 1.4 mA
Collector-Emitter Voltage
V_{CE} = V_{CC} - I_C(R_C + R_E)
| Symbol | Description | Unit |
|---|---|---|
| V_{CE} | Collector-emitter voltage | V |
| I_C | Collector current (≈ I_E for large β) | A |
| R_C | Collector resistor | Ω |
Worked example
Find V_CE with V_CC = 12 V, I_C = 1.4 mA, R_C = 3.3 kΩ, R_E = 1 kΩ.
Given: V_CC = 12 V, I_C = 1.4 mA, R_C = 3.3 kΩ, R_E = 1 kΩ
- V_CE = 12 - 0.0014 × (3300 + 1000)
- V_CE = 12 - 0.0014 × 4300
- V_CE = 12 - 6.02
- V_CE = 5.98 V
Answer: V_CE ≈ 6 V
Transconductance
g_m = \frac{I_C}{V_T} = \frac{I_C}{26\text{ mV}}
| Symbol | Description | Unit |
|---|---|---|
| g_m | Transconductance | A/V (S) |
| I_C | DC collector current | A |
| V_T | Thermal voltage (26 mV at 300 K) | V |
Worked example
Find g_m for a BJT biased at I_C = 1.4 mA.
Given: I_C = 1.4 mA, V_T = 26 mV
- g_m = I_C / V_T
- g_m = 0.0014 / 0.026
- g_m = 0.05385 A/V
Answer: g_m ≈ 53.8 mA/V
Small-Signal BJT Amplifier Parameters
Small-Signal Input Resistance r_π
r_\pi = \frac{\beta}{g_m}
| Symbol | Description | Unit |
|---|---|---|
| r_\pi | Small-signal base-emitter resistance | Ω |
| \beta | Current gain (h_FE) | dimensionless |
Worked example
Find r_π for a BJT with β = 120 biased at I_C = 1.4 mA.
Given: β = 120, g_m = 53.8 mA/V
- r_π = β / g_m
- r_π = 120 / 0.0538
- r_π = 2230 Ω
Answer: r_π ≈ 2.23 kΩ
Common-Emitter Voltage Gain
A_v = -g_m \cdot R_C'
| Symbol | Description | Unit |
|---|---|---|
| A_v | Voltage gain | V/V |
| R_C' | Effective collector resistance (R_C ∥ R_L) | Ω |
Worked example
Find A_v for g_m = 53.8 mA/V, R_C = 3.3 kΩ, R_L = 10 kΩ.
Given: g_m = 53.8 mA/V, R_C = 3.3 kΩ, R_L = 10 kΩ
- R_C' = R_C ∥ R_L = (3300 × 10000) / (3300 + 10000)
- R_C' = 33000000 / 13300 = 2481 Ω
- A_v = -0.0538 × 2481
- A_v = -133.5
Answer: A_v ≈ -133.5 V/V
Common-Emitter Input Resistance
R_{in} = R_1 \| R_2 \| r_\pi
| Symbol | Description | Unit |
|---|---|---|
| R_{in} | Total input resistance seen by source | Ω |
Worked example
Find R_in with R1 = 47 kΩ, R2 = 10 kΩ, r_π = 2.23 kΩ.
Given: R1 = 47 kΩ, R2 = 10 kΩ, r_π = 2.23 kΩ
- R1 ∥ R2 = (47000 × 10000) / (57000) = 8246 Ω
- R_in = 8246 ∥ 2230
- R_in = (8246 × 2230) / (8246 + 2230)
- R_in = 18388580 / 10476 = 1755 Ω
Answer: R_in ≈ 1.76 kΩ
MOSFET Biasing and Parameters
MOSFET Drain Current (Saturation)
I_D = \frac{\mu_n C_{ox} W}{2L}(V_{GS} - V_{th})^2 = \frac{k_n}{2}(V_{GS} - V_{th})^2
| Symbol | Description | Unit |
|---|---|---|
| I_D | Drain current | A |
| k_n | Process transconductance parameter (μ_n C_ox W/L) | A/V² |
| V_{GS} | Gate-source voltage | V |
| V_{th} | Threshold voltage | V |
Worked example
Find I_D for an NMOS with k_n = 0.5 mA/V², V_GS = 3 V, V_th = 1 V.
Given: k_n = 0.5 mA/V², V_GS = 3 V, V_th = 1 V
- V_GS - V_th = 3 - 1 = 2 V
- I_D = (0.5e-3 / 2) × (2)²
- I_D = 0.25e-3 × 4
- I_D = 1 mA
Answer: I_D = 1 mA
MOSFET Transconductance
g_m = \sqrt{2 k_n I_D} = k_n(V_{GS} - V_{th})
| Symbol | Description | Unit |
|---|---|---|
| g_m | MOSFET transconductance | A/V |
Worked example
Find g_m for k_n = 0.5 mA/V², I_D = 1 mA.
Given: k_n = 0.5 mA/V², I_D = 1 mA
- g_m = √(2 × 0.5e-3 × 1e-3)
- g_m = √(1e-6)
- g_m = 1 mA/V
Answer: g_m = 1 mA/V
MOSFET Output Resistance
r_o = \frac{V_A}{I_D} = \frac{1}{\lambda I_D}
| Symbol | Description | Unit |
|---|---|---|
| r_o | Output resistance | Ω |
| V_A | Early voltage | V |
| \lambda | Channel-length modulation parameter | V⁻¹ |
Worked example
Find r_o with V_A = 50 V and I_D = 1 mA.
Given: V_A = 50 V, I_D = 1 mA
- r_o = V_A / I_D
- r_o = 50 / 0.001
- r_o = 50000 Ω
Answer: r_o = 50 kΩ
Op-Amp Ideal and Practical Configurations
Inverting Amplifier Gain
A_v = -\frac{R_f}{R_1}
| Symbol | Description | Unit |
|---|---|---|
| R_f | Feedback resistor | Ω |
| R_1 | Input resistor | Ω |
Worked example
Find gain for an inverting op-amp with R_f = 100 kΩ, R_1 = 10 kΩ.
Given: R_f = 100 kΩ, R_1 = 10 kΩ
- A_v = -R_f / R_1
- A_v = -100000 / 10000
- A_v = -10
Answer: A_v = -10 V/V (gain of 10, inverting)
Non-Inverting Amplifier Gain
A_v = 1 + \frac{R_f}{R_1}
Worked example
Find gain for a non-inverting op-amp with R_f = 47 kΩ, R_1 = 4.7 kΩ.
Given: R_f = 47 kΩ, R_1 = 4.7 kΩ
- A_v = 1 + (47000 / 4700)
- A_v = 1 + 10
- A_v = 11
Answer: A_v = 11 V/V
Summing Amplifier Output
V_{out} = -R_f \left(\frac{V_1}{R_1} + \frac{V_2}{R_2}\right)
| Symbol | Description | Unit |
|---|---|---|
| V_1, V_2 | Input voltages | V |
| R_1, R_2 | Input resistors | Ω |
Worked example
Find V_out for V1 = 2 V, V2 = 3 V, R1 = R2 = 10 kΩ, Rf = 10 kΩ.
Given: V1 = 2 V, V2 = 3 V, R1 = R2 = Rf = 10 kΩ
- V_out = -10000 × (2/10000 + 3/10000)
- V_out = -10000 × (0.0002 + 0.0003)
- V_out = -10000 × 0.0005
- V_out = -5 V
Answer: V_out = -5 V
Slew Rate
SR = \frac{dV_{out}}{dt}\bigg|_{max} = \frac{I_{bias}}{C_{comp}}
| Symbol | Description | Unit |
|---|---|---|
| SR | Slew rate | V/μs |
| I_{bias} | Internal bias current | A |
| C_{comp} | Internal compensation capacitor | F |
Worked example
An op-amp has SR = 1 V/μs. Find max frequency for undistorted output at V_p = 5 V.
Given: SR = 1 V/μs = 1×10⁶ V/s, V_p = 5 V
- SR = 2π f_max V_p
- f_max = SR / (2π V_p)
- f_max = 1×10⁶ / (2π × 5)
- f_max = 1×10⁶ / 31.416
- f_max = 31831 Hz
Answer: f_max ≈ 31.8 kHz
Feedback Amplifiers
Closed-Loop Gain with Feedback
A_f = \frac{A}{1 + A\beta}
| Symbol | Description | Unit |
|---|---|---|
| A_f | Closed-loop gain | V/V |
| A | Open-loop gain | V/V |
| \beta | Feedback factor | dimensionless |
Worked example
Find A_f with A = 10000, β = 0.01.
Given: A = 10000, β = 0.01
- 1 + Aβ = 1 + 10000 × 0.01 = 1 + 100 = 101
- A_f = 10000 / 101
- A_f = 99.01
Answer: A_f ≈ 99 V/V ≈ 1/β for large loop gain
Loop Gain
T = A \cdot \beta
| Symbol | Description | Unit |
|---|---|---|
| T | Loop gain (Barkhausen denominator) | dimensionless |
Worked example
Find loop gain for A = 10000 and β = 0.01.
Given: A = 10000, β = 0.01
- T = A × β = 10000 × 0.01
- T = 100
Answer: T = 100 (40 dB loop gain)
Bandwidth Extension by Feedback
f_{Hf} = f_H(1 + A\beta)
| Symbol | Description | Unit |
|---|---|---|
| f_{Hf} | Closed-loop upper cutoff frequency | Hz |
| f_H | Open-loop upper cutoff frequency | Hz |
Worked example
Amplifier has f_H = 1 kHz, A = 10000, β = 0.01. Find closed-loop bandwidth.
Given: f_H = 1000 Hz, Aβ = 100
- f_Hf = f_H × (1 + Aβ)
- f_Hf = 1000 × (1 + 100)
- f_Hf = 1000 × 101
- f_Hf = 101000 Hz
Answer: f_Hf = 101 kHz
Oscillators
RC Phase-Shift Oscillator Frequency
f_o = \frac{1}{2\pi RC\sqrt{6}}
| Symbol | Description | Unit |
|---|---|---|
| f_o | Oscillation frequency | Hz |
| R | Resistance per stage | Ω |
| C | Capacitance per stage | F |
Worked example
Find f_o for an RC phase-shift oscillator with R = 10 kΩ, C = 10 nF.
Given: R = 10 kΩ, C = 10 nF
- f_o = 1 / (2π × 10000 × 10e-9 × √6)
- √6 = 2.449
- Denominator = 2π × 10000 × 10e-9 × 2.449
- = 2π × 2.449e-4 = 1.539e-3
- f_o = 1 / 1.539e-3 = 649.7 Hz
Answer: f_o ≈ 650 Hz
Wien Bridge Oscillator Frequency
f_o = \frac{1}{2\pi RC}
Worked example
Find oscillation frequency for R = 15.9 kΩ, C = 10 nF.
Given: R = 15.9 kΩ, C = 10 nF
- f_o = 1 / (2π × 15900 × 10e-9)
- = 1 / (2π × 1.59e-4)
- = 1 / 9.99e-4
- = 1001 Hz
Answer: f_o ≈ 1 kHz
Hartley Oscillator Frequency
f_o = \frac{1}{2\pi\sqrt{(L_1+L_2)C}}
| Symbol | Description | Unit |
|---|---|---|
| L_1, L_2 | Inductors in the tank circuit | H |
| C | Tank capacitor | F |
Worked example
Find f_o for L1 = 1 mH, L2 = 1 mH, C = 100 pF.
Given: L1 = L2 = 1 mH, C = 100 pF
- L_total = L1 + L2 = 2 mH = 2e-3 H
- f_o = 1 / (2π × √(2e-3 × 100e-12))
- = 1 / (2π × √(2e-13))
- = 1 / (2π × 4.472e-7)
- = 1 / 2.81e-6 = 355.9 kHz
Answer: f_o ≈ 356 kHz
Frequency Response and Filters
3 dB Cutoff Frequency (RC Low-Pass)
f_c = \frac{1}{2\pi RC}
| Symbol | Description | Unit |
|---|---|---|
| f_c | Lower/upper 3 dB frequency | Hz |
Worked example
Find f_c for R = 1 kΩ, C = 0.1 μF.
Given: R = 1000 Ω, C = 0.1 μF = 0.1e-6 F
- f_c = 1 / (2π × 1000 × 0.1e-6)
- = 1 / (2π × 1e-4)
- = 1 / 6.283e-4
- = 1591.5 Hz
Answer: f_c ≈ 1.59 kHz
Gain-Bandwidth Product
GBW = A_v \cdot BW = A_0 \cdot f_H
| Symbol | Description | Unit |
|---|---|---|
| GBW | Gain-bandwidth product (constant for single-pole op-amps) | Hz |
| A_0 | DC open-loop gain | V/V |
| f_H | Open-loop 3 dB frequency | Hz |
Worked example
An op-amp has GBW = 1 MHz. Find open-loop gain at f = 10 kHz.
Given: GBW = 1 MHz, f = 10 kHz
- A_v(at f) = GBW / f
- A_v = 1×10⁶ / 10×10³
- A_v = 100
Answer: A_v = 100 V/V = 40 dB at 10 kHz
Power Amplifiers and Rectifiers
Class-A Amplifier Efficiency
\eta = \frac{P_{out(ac)}}{P_{dc}} \times 100\%,\quad \eta_{max} = 25\%
| Symbol | Description | Unit |
|---|---|---|
| \eta | Collector efficiency | % |
| P_{out(ac)} | AC output power | W |
| P_{dc} | DC input power from supply | W |
Worked example
Class-A amp: V_CC = 12 V, I_CQ = 50 mA, V_out(peak) = 5 V, I_out(peak) = 45 mA. Find η.
Given: V_CC = 12 V, I_CQ = 50 mA, V_p = 5 V, I_p = 45 mA
- P_dc = V_CC × I_CQ = 12 × 0.05 = 0.6 W
- P_out = (V_p × I_p) / 2 = (5 × 0.045) / 2 = 0.1125 W
- η = (0.1125 / 0.6) × 100
- η = 18.75%
Answer: η ≈ 18.75%
Class-B Push-Pull Max Efficiency
\eta_{max} = \frac{\pi}{4} \approx 78.5\%
Worked example
Confirm Class-B max efficiency derivation for V_CC = 15 V, R_L = 8 Ω at full swing.
Given: V_CC = 15 V, R_L = 8 Ω, V_p = V_CC = 15 V (ideal)
- P_out(max) = V_CC² / (2 R_L) = 225 / 16 = 14.06 W
- P_dc = (2 V_CC) / (π R_L) × V_CC — derived as 2V_CC² / (π R_L)
- P_dc = 2 × 225 / (π × 8) = 450 / 25.13 = 17.9 W
- η = 14.06 / 17.9 = 0.785 = 78.5%
Answer: η_max = 78.5% = π/4
Full-Wave Rectifier Ripple Factor
\gamma = \frac{V_{r(rms)}}{V_{dc}} = \frac{1}{4\sqrt{3}fCR_L}
| Symbol | Description | Unit |
|---|---|---|
| \gamma | Ripple factor | dimensionless |
| f | Supply frequency | Hz |
| C | Filter capacitor | F |
| R_L | Load resistance | Ω |
Worked example
Find ripple factor for a full-wave rectifier with C = 1000 μF, R_L = 100 Ω, f = 50 Hz.
Given: C = 1000e-6 F, R_L = 100 Ω, f = 50 Hz
- γ = 1 / (4√3 × 50 × 1000e-6 × 100)
- 4√3 = 6.928
- Denominator = 6.928 × 50 × 0.1 = 34.64
- γ = 1 / 34.64 = 0.0289
Answer: γ ≈ 0.029 (2.9%)
Quick reference
| Formula | Expression |
|---|---|
| Voltage Divider Bias V_B | V_B = V_CC × R2/(R1+R2) |
| BJT Transconductance | g_m = I_C / V_T |
| Small-Signal r_π | r_π = β / g_m |
| CE Voltage Gain | A_v = -g_m × R_C' |
| MOSFET Drain Current (Sat) | I_D = (k_n/2)(V_GS - V_th)² |
| MOSFET g_m | g_m = √(2 k_n I_D) |
| Inverting Op-Amp Gain | A_v = -R_f / R_1 |
| Non-Inverting Op-Amp Gain | A_v = 1 + R_f/R_1 |
| Closed-Loop Gain | A_f = A / (1 + Aβ) |
| Slew Rate Limit Frequency | f_max = SR / (2π V_p) |
| Wien Bridge Oscillator | f_o = 1/(2πRC) |
| RC Phase-Shift Oscillator | f_o = 1/(2πRC√6) |
| Gain-Bandwidth Product | GBW = A_v × BW |
| Class-B Efficiency | η_max = π/4 ≈ 78.5% |
| Ripple Factor (FWR) | γ = 1/(4√3 f C R_L) |
Exam tips
- GATE examiners frequently test whether you choose r_e or r_π model correctly — r_π = β/g_m while r_e = 1/g_m; confusing them costs marks.
- For op-amp slew rate questions, always check whether the output frequency exceeds SR/(2πV_p) before assuming linear operation.
- In feedback amplifier questions, identify the topology (series-shunt, shunt-series) first — the affected input/output resistance formulas differ by a factor of (1+Aβ).
- Oscillator questions often ask for minimum gain condition: for Wien bridge A_v ≥ 3; for RC phase-shift A_v ≥ 29.
- MOSFET saturation condition is V_DS ≥ V_GS - V_th — verify this before applying the square-law I_D formula.
- Power amplifier efficiency questions in ESE distinguish between power delivered to load and total DC input power — use both in the ratio.