P-N Junction Diode
Ideal Diode Equation (Shockley)
I_D = I_S \left(e^{V_D / \eta V_T} - 1\right)
| Symbol | Description | Unit |
|---|---|---|
| I_D | Diode current | A |
| I_S | Reverse saturation current | A |
| V_D | Forward voltage across diode | V |
| \eta | Ideality factor (1 for ideal, 2 for recombination-dominant) | dimensionless |
| V_T | Thermal voltage = kT/q ≈ 26 mV at 300 K | V |
Worked example
A silicon diode with I_S = 1 nA and η = 1 is forward-biased at V_D = 0.65 V at 300 K. Find I_D.
Given: I_S=1×10⁻⁹ A, V_D=0.65 V, η=1, V_T=0.02585 V
- Exponent = V_D/(η·V_T) = 0.65/0.02585 = 25.14
- e^{25.14} ≈ 8.1×10^{10}
- I_D = 1×10⁻⁹ × (8.1×10^{10} − 1) ≈ 1×10⁻⁹ × 8.1×10^{10}
- I_D ≈ 81 mA
Answer: I_D ≈ 81 mA
Thermal Voltage
V_T = \frac{kT}{q}
| Symbol | Description | Unit |
|---|---|---|
| k | Boltzmann constant = 1.38×10⁻²³ J/K | J/K |
| T | Absolute temperature | K |
| q | Electron charge = 1.6×10⁻¹⁹ C | C |
Worked example
Find V_T at T = 400 K.
Given: k=1.38×10⁻²³, T=400, q=1.6×10⁻¹⁹
- V_T = (1.38×10⁻²³ × 400) / (1.6×10⁻¹⁹)
- = 5.52×10⁻²¹ / 1.6×10⁻¹⁹
- = 0.03450 V = 34.5 mV
Answer: V_T = 34.5 mV at 400 K
Diode Small-Signal Resistance
r_d = \frac{\eta V_T}{I_D}
| Symbol | Description | Unit |
|---|---|---|
| r_d | Dynamic (small-signal) diode resistance | Ω |
| I_D | DC bias current | A |
Worked example
Find the small-signal resistance of a diode biased at I_D = 2 mA, η = 1, T = 300 K.
Given: I_D=2×10⁻³ A, η=1, V_T=26 mV
- r_d = η·V_T / I_D
- r_d = 1 × 0.026 / (2×10⁻³)
- r_d = 0.026 / 0.002 = 13 Ω
Answer: r_d = 13 Ω
Built-in (Contact) Potential
V_0 = \frac{kT}{q} \ln\!\left(\frac{N_A N_D}{n_i^2}\right)
| Symbol | Description | Unit |
|---|---|---|
| V_0 | Built-in potential | V |
| N_A, N_D | Acceptor and donor concentrations | cm⁻³ |
| n_i | Intrinsic carrier concentration (1.5×10¹⁰ cm⁻³ for Si at 300 K) | cm⁻³ |
Worked example
Find V_0 for a silicon p-n junction with N_A=10¹⁶ cm⁻³ and N_D=10¹⁵ cm⁻³ at 300 K.
Given: N_A=10¹⁶, N_D=10¹⁵, n_i=1.5×10¹⁰, V_T=0.026 V
- V_0 = 0.026 × ln(10¹⁶ × 10¹⁵ / (1.5×10¹⁰)²)
- = 0.026 × ln(10³¹ / 2.25×10²⁰)
- = 0.026 × ln(4.44×10¹⁰)
- = 0.026 × 24.22 = 0.630 V
Answer: V_0 ≈ 0.63 V
Depletion Width
W = \sqrt{\frac{2\varepsilon_s (V_0 - V_A)}{q}\left(\frac{1}{N_A} + \frac{1}{N_D}\right)}
| Symbol | Description | Unit |
|---|---|---|
| W | Total depletion width | m |
| \varepsilon_s | Semiconductor permittivity (11.7ε₀ for Si) | F/m |
| V_A | Applied voltage (positive forward, negative reverse) | V |
Worked example
Find depletion width for the above junction (V_0=0.63 V) under 5 V reverse bias.
Given: ε_s=11.7×8.85×10⁻¹²=1.035×10⁻¹⁰ F/m, V_0=0.63, V_A=−5, N_A=10²² m⁻³, N_D=10²¹ m⁻³, q=1.6×10⁻¹⁹
- V_0 − V_A = 0.63 − (−5) = 5.63 V
- 1/N_A + 1/N_D = 10⁻²² + 10⁻²¹ = 11×10⁻²²
- W = √(2×1.035×10⁻¹⁰ × 5.63 / 1.6×10⁻¹⁹ × 11×10⁻²²)
- = √(2×1.035×10⁻¹⁰×5.63×11×10⁻²² / 1.6×10⁻¹⁹)
- = √(1.285×10⁻³⁰ / 1.6×10⁻¹⁹) = √(8.03×10⁻¹²) ≈ 2.83 μm
Answer: W ≈ 2.83 μm
BJT — DC Analysis
BJT Current Relationships
I_C = \beta I_B, \quad I_E = I_C + I_B = (1+\beta) I_B, \quad \alpha = \frac{\beta}{1+\beta}
| Symbol | Description | Unit |
|---|---|---|
| \beta | Common-emitter current gain (h_FE) | dimensionless |
| \alpha | Common-base current gain | dimensionless |
| I_B, I_C, I_E | Base, collector, emitter currents | A |
Worked example
A BJT has β = 120 and I_B = 25 μA. Find I_C, I_E, and α.
Given: β=120, I_B=25×10⁻⁶ A
- I_C = β·I_B = 120 × 25×10⁻⁶ = 3 mA
- I_E = I_C + I_B = 3×10⁻³ + 25×10⁻⁶ = 3.025 mA
- α = β/(1+β) = 120/121 = 0.9917
Answer: I_C=3 mA, I_E=3.025 mA, α=0.9917
Voltage-Divider Bias — Thevenin Base Voltage
V_{TH} = V_{CC} \cdot \frac{R_2}{R_1 + R_2}, \quad R_{TH} = R_1 \| R_2
| Symbol | Description | Unit |
|---|---|---|
| V_{TH} | Thevenin equivalent base voltage | V |
| R_1, R_2 | Voltage-divider resistors | Ω |
| V_{CC} | Supply voltage | V |
Worked example
Find I_C for a voltage-divider biased NPN BJT: V_CC=12V, R1=47 kΩ, R2=10 kΩ, R_E=1 kΩ, β=100, V_BE=0.7V.
Given: V_CC=12, R1=47k, R2=10k, R_E=1k, β=100, V_BE=0.7
- V_TH = 12 × 10/(47+10) = 120/57 = 2.105 V
- R_TH = 47k||10k = 470k/57k = 8.246 kΩ
- KVL: V_TH = I_B·R_TH + V_BE + (1+β)I_B·R_E
- 2.105 = I_B(8246 + 0.7 + 101×1000) → I_B = (2.105−0.7)/(8246+101000) = 1.405/109246
- I_B = 12.86 μA; I_C = 100×12.86μA = 1.286 mA
Answer: I_C ≈ 1.29 mA
BJT Q-Point Stability Factor
S = \frac{1 + \beta}{1 + \beta \cdot R_E / (R_E + R_B)}
| Symbol | Description | Unit |
|---|---|---|
| S | Stability factor (lower is more stable) | dimensionless |
| R_E | Emitter resistor | Ω |
| R_B | Effective base resistance | Ω |
Worked example
Find S for β=100, R_B=8.246 kΩ, R_E=1 kΩ.
Given: β=100, R_B=8246 Ω, R_E=1000 Ω
- Numerator = 1+β = 101
- Ratio R_E/(R_E+R_B) = 1000/9246 = 0.1082
- Denominator = 1 + 100×0.1082 = 1 + 10.82 = 11.82
- S = 101/11.82 = 8.54
Answer: S ≈ 8.54 (good stability; S=1 is ideal)
BJT Small-Signal Model
Transconductance g_m
g_m = \frac{I_C}{V_T}
| Symbol | Description | Unit |
|---|---|---|
| g_m | Transconductance | A/V (S) |
| I_C | Quiescent collector current | A |
Worked example
Find g_m for a BJT biased at I_C = 2 mA at 300 K.
Given: I_C=2×10⁻³ A, V_T=0.026 V
- g_m = I_C / V_T = 2×10⁻³ / 0.026
- g_m = 76.9 mA/V
Answer: g_m = 76.9 mA/V
Small-Signal Input Resistance r_π
r_\pi = \frac{\beta}{g_m} = \frac{V_T}{I_B}
| Symbol | Description | Unit |
|---|---|---|
| r_\pi | Base–emitter small-signal resistance | Ω |
Worked example
Find r_π for β=150, I_C=3 mA at 300 K.
Given: β=150, I_C=3 mA, V_T=26 mV
- g_m = I_C/V_T = 3×10⁻³/0.026 = 115.4 mA/V
- r_π = β/g_m = 150/0.1154 = 1300 Ω = 1.3 kΩ
Answer: r_π = 1.3 kΩ
CE Amplifier Voltage Gain (with bypass capacitor)
A_v = -g_m R_C = -\frac{\beta R_C}{r_\pi}
| Symbol | Description | Unit |
|---|---|---|
| R_C | Collector resistance | Ω |
| A_v | Voltage gain (negative → inverting) | dimensionless |
Worked example
Find voltage gain for CE amplifier: I_C=1 mA, R_C=4.7 kΩ, β=100 at 300 K.
Given: I_C=1 mA, R_C=4700 Ω, β=100
- g_m = 1×10⁻³/0.026 = 38.46 mA/V
- A_v = −g_m × R_C = −0.03846 × 4700
- A_v = −180.8
Answer: A_v ≈ −181 (inverting, ~45 dB)
Early Effect — Output Resistance r_o
r_o = \frac{|V_A| + |V_{CE}|}{I_C} \approx \frac{|V_A|}{I_C}
| Symbol | Description | Unit |
|---|---|---|
| V_A | Early voltage (typically 50–200 V for BJT) | V |
| r_o | Output resistance | Ω |
Worked example
Find r_o for a BJT with V_A = 80 V biased at I_C = 2 mA.
Given: V_A=80 V, I_C=2×10⁻³ A
- r_o ≈ V_A / I_C
- r_o = 80 / (2×10⁻³)
- r_o = 40,000 Ω = 40 kΩ
Answer: r_o = 40 kΩ
MOSFET — DC Analysis
NMOS Threshold Voltage
V_{tn} = V_{FB} + 2\phi_F + \frac{\sqrt{2\varepsilon_s q N_A (2\phi_F + V_{SB})}}{C_{ox}}
| Symbol | Description | Unit |
|---|---|---|
| V_{FB} | Flat-band voltage | V |
| \phi_F | Fermi potential = (kT/q)ln(N_A/n_i) | V |
| C_{ox} | Gate oxide capacitance per unit area | F/m² |
| V_{SB} | Source-body voltage | V |
Worked example
An NMOS has V_FB=−1 V, 2φ_F=0.7 V, Q_dep_max/C_ox=0.5 V, V_SB=0 V. Find V_tn.
Given: V_FB=−1, 2φ_F=0.7, √(2εqN_A·2φ_F)/C_ox=0.5
- V_tn = V_FB + 2φ_F + √(2εqN_A·2φ_F)/C_ox
- V_tn = −1 + 0.7 + 0.5
- V_tn = 0.2 V
Answer: V_tn = 0.2 V
NMOS Drain Current — Triode Region
I_D = \mu_n C_{ox} \frac{W}{L} \left[(V_{GS} - V_{tn})V_{DS} - \frac{V_{DS}^2}{2}\right]
| Symbol | Description | Unit |
|---|---|---|
| \mu_n C_{ox} | Process transconductance parameter k'_n | A/V² |
| W/L | Transistor aspect ratio | dimensionless |
| V_{GS} | Gate-source voltage | V |
| V_{DS} | Drain-source voltage | V |
| V_{tn} | Threshold voltage | V |
Worked example
NMOS: k'_n=200 μA/V², W/L=10, V_tn=0.5 V, V_GS=1.5 V, V_DS=0.3 V. Find I_D (check region first).
Given: k'_n=200×10⁻⁶, W/L=10, V_tn=0.5, V_GS=1.5, V_DS=0.3
- V_GS−V_tn = 1.5−0.5 = 1.0 V. Since V_DS=0.3 < 1.0 V, device is in triode.
- I_D = 200×10⁻⁶ × 10 × [(1.0)(0.3) − (0.3)²/2]
- = 2×10⁻³ × [0.30 − 0.045]
- = 2×10⁻³ × 0.255 = 0.51 mA
Answer: I_D = 0.51 mA (triode region)
NMOS Drain Current — Saturation Region
I_D = \frac{1}{2} \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{tn})^2 (1 + \lambda V_{DS})
| Symbol | Description | Unit |
|---|---|---|
| \lambda | Channel-length modulation parameter | V⁻¹ |
Worked example
Same NMOS as above but V_DS=2 V, λ=0.02 V⁻¹. Find I_D in saturation.
Given: k'_n=200 μA/V², W/L=10, V_GS−V_tn=1.0 V, V_DS=2 V, λ=0.02
- Check: V_DS=2 > V_GS−V_tn=1 ✓ (saturation)
- I_D = 0.5 × 200×10⁻⁶ × 10 × (1.0)² × (1 + 0.02×2)
- = 10⁻³ × 1 × 1.04
- = 1.04 mA
Answer: I_D = 1.04 mA
MOSFET Transconductance g_m
g_m = \sqrt{2 \mu_n C_{ox} \frac{W}{L} I_D} = \frac{2I_D}{V_{GS} - V_{tn}}
| Symbol | Description | Unit |
|---|---|---|
| g_m | MOSFET transconductance | A/V |
Worked example
Find g_m for the NMOS in saturation above (I_D=1 mA, V_GS−V_tn=1 V, k'_n W/L=2 mA/V²).
Given: I_D=1×10⁻³ A, V_GS−V_tn=1 V
- Method 1: g_m = 2I_D/(V_GS−V_tn) = 2×10⁻³/1 = 2 mA/V
- Method 2: g_m = √(2×k'_n(W/L)×I_D) = √(2×2×10⁻³×10⁻³) = √(4×10⁻⁶) = 2×10⁻³ A/V ✓
Answer: g_m = 2 mA/V
MOSFET Amplifier Configurations
Common-Source Amplifier Gain
A_v = -g_m (R_D \| r_o)
| Symbol | Description | Unit |
|---|---|---|
| R_D | Drain resistance | Ω |
| r_o | MOSFET output resistance = 1/(λI_D) | Ω |
Worked example
CS amplifier: g_m=2 mA/V, R_D=10 kΩ, r_o=50 kΩ. Find A_v.
Given: g_m=2×10⁻³, R_D=10k, r_o=50k
- R_D||r_o = (10k × 50k)/(10k+50k) = 500k/60k = 8.33 kΩ
- A_v = −g_m × 8.33×10³ = −2×10⁻³ × 8333 = −16.67
Answer: A_v ≈ −16.7
Source Follower (Common-Drain) Gain
A_v = \frac{g_m R_S}{1 + g_m R_S}
| Symbol | Description | Unit |
|---|---|---|
| R_S | Source resistance | Ω |
Worked example
Source follower: g_m=5 mA/V, R_S=2 kΩ. Find A_v.
Given: g_m=5×10⁻³, R_S=2000
- g_m R_S = 5×10⁻³ × 2000 = 10
- A_v = 10/(1+10) = 10/11 = 0.909
Answer: A_v ≈ 0.91 (slightly less than unity, non-inverting)
Special Diodes and Breakdown
Zener Voltage Regulation
V_{out} = V_Z, \quad I_Z = \frac{V_{in} - V_Z}{R_S} - I_L
| Symbol | Description | Unit |
|---|---|---|
| V_Z | Zener breakdown voltage | V |
| R_S | Series current-limiting resistor | Ω |
| I_L | Load current | A |
Worked example
A Zener regulator: V_in=15 V, V_Z=9 V, R_S=300 Ω, R_L=1.5 kΩ. Find I_Z.
Given: V_in=15, V_Z=9, R_S=300, R_L=1500
- Total current from supply: I_S = (V_in − V_Z)/R_S = (15−9)/300 = 6/300 = 20 mA
- Load current: I_L = V_Z/R_L = 9/1500 = 6 mA
- I_Z = I_S − I_L = 20 − 6 = 14 mA
Answer: I_Z = 14 mA
Diode Junction Capacitance
C_j = \frac{C_{j0}}{\left(1 - V_A/V_0\right)^m}
| Symbol | Description | Unit |
|---|---|---|
| C_{j0} | Zero-bias junction capacitance | F |
| m | Grading coefficient (0.5 for abrupt, 0.33 for graded) | dimensionless |
| V_A | Applied voltage (negative for reverse) | V |
Worked example
Find C_j for abrupt junction: C_j0=10 pF, V_0=0.7 V, V_A=−3 V.
Given: C_j0=10×10⁻¹², V_0=0.7, V_A=−3, m=0.5
- 1 − V_A/V_0 = 1 − (−3)/0.7 = 1 + 4.286 = 5.286
- (5.286)^{0.5} = 2.299
- C_j = 10 pF / 2.299 = 4.35 pF
Answer: C_j ≈ 4.35 pF
Frequency Response of Amplifiers
Unity-Gain Bandwidth Product
f_T = \frac{g_m}{2\pi(C_{\pi} + C_{\mu})}
| Symbol | Description | Unit |
|---|---|---|
| f_T | Transition frequency (unity current gain) | Hz |
| C_\pi | Base-emitter capacitance | F |
| C_\mu | Base-collector (Miller) capacitance | F |
Worked example
BJT: g_m=40 mA/V, C_π=20 pF, C_μ=2 pF. Find f_T.
Given: g_m=0.04 A/V, C_π=20×10⁻¹², C_μ=2×10⁻¹²
- f_T = g_m / [2π(C_π+C_μ)]
- = 0.04 / [2π × 22×10⁻¹²]
- = 0.04 / (1.382×10⁻¹⁰)
- = 289.4 MHz
Answer: f_T ≈ 289 MHz
Miller Capacitance
C_{Miller} = C_{\mu}(1 + g_m R_C)
| Symbol | Description | Unit |
|---|---|---|
| C_{Miller} | Effective input capacitance due to Miller effect | F |
Worked example
Find Miller capacitance for C_μ=2 pF, g_m=40 mA/V, R_C=5 kΩ.
Given: C_μ=2×10⁻¹², g_m=0.04, R_C=5000
- g_m R_C = 0.04 × 5000 = 200
- C_Miller = 2×10⁻¹² × (1+200) = 2×10⁻¹² × 201
- = 402 pF
Answer: C_Miller = 402 pF
Quick reference
| Formula | Expression |
|---|---|
| Shockley Diode Equation | I_D = I_S(e^{V_D/\eta V_T}-1) |
| Thermal Voltage | V_T = kT/q \approx 26\,\text{mV at 300 K} |
| Diode Small-Signal Resistance | r_d = \eta V_T / I_D |
| Built-in Potential | V_0 = V_T \ln(N_A N_D/n_i^2) |
| BJT Currents | I_C = \beta I_B,\; I_E=(1+\beta)I_B |
| BJT Transconductance | g_m = I_C/V_T |
| BJT Small-Signal r_π | r_\pi = \beta/g_m |
| CE Voltage Gain | A_v = -g_m R_C |
| Early Voltage r_o | r_o \approx V_A/I_C |
| NMOS Triode Current | I_D = k_n(W/L)[(V_{GS}-V_t)V_{DS}-V_{DS}^2/2] |
| NMOS Saturation Current | I_D = (k_n W/2L)(V_{GS}-V_t)^2 |
| MOSFET g_m | g_m = 2I_D/(V_{GS}-V_t) |
| CS Amplifier Gain | A_v = -g_m(R_D\|r_o) |
| Miller Capacitance | C_M = C_\mu(1+g_m R_C) |
| Transition Frequency | f_T = g_m/[2\pi(C_\pi+C_\mu)] |
Exam tips
- GATE regularly asks you to determine the region of operation (triode vs. saturation) before calculating MOSFET drain current — always check V_DS against V_GS−V_tn first.
- In BJT small-signal problems, examiners penalise missing the Early effect (r_o) when the question states a non-zero λ or V_A; always include r_o in the output resistance.
- For voltage-divider bias calculations, the stiff-bias approximation (ignoring loading) is only valid when β·R_E >> R_B; verify this condition or use the exact Thevenin method.
- Zener regulator questions frequently require you to verify that I_Z stays above I_Z(min) under maximum load — calculate both extremes of load current.
- Miller effect questions often hide the high-frequency bandwidth limitation; calculate C_Miller and then upper −3dB frequency f_H = 1/(2πR_sig C_Miller).
- When a problem specifies body effect (V_SB ≠ 0), the threshold voltage increases for NMOS — always recalculate V_tn before finding I_D.