Analog Circuits

Syllabus

Diode circuits (clipping, clamping, rectifiers); BJT and MOSFET amplifiers (biasing, AC coupling, small signal analysis, frequency response); current mirrors and differential amplifiers; op-amp circuits (amplifiers, summers, differentiators, integrators, active filters, Schmitt triggers, oscillators).

Key concepts and formulas

Diode circuits. A clipper removes part of a waveform above or below a reference; a clamper shifts the DC level so the waveform's peak sits at a reference (a negative clamper on a ±V_m sine gives 0 to −2V_m). Rectifier averages: half-wave V_m/π, full-wave 2V_m/π.

Small-signal models.

• BJT: g_m = I_C/V_T, r_π = β/g_m, r_o = V_A/I_C.
• MOSFET: g_m = 2I_D/(V_GS − V_T), r_o = 1/(λI_D).

Amplifier configurations (with load R_L′ = R_C ∥ R_L, ignoring r_o):

Frequency response. Coupling and bypass capacitors set the lower cut-off; internal capacitances set the upper cut-off. The Miller effect: a capacitance C between input and output of an inverting stage of gain −A appears at the input as C(1 + A). Gain-bandwidth product is constant for a single-pole amplifier.

Current mirror. Two matched transistors with the same V_BE (or V_GS) copy a reference current: I_out = I_ref × (W/L)₂/(W/L)₁ for MOSFETs; for BJTs, I_out = I_ref/(1 + 2/β).

Differential amplifier. Differential gain A_d = g_mR_D (differential output); common-mode gain A_cm ≈ −R_D/(2R_SS) per side. CMRR = |A_d/A_cm|, in dB 20 log₁₀ CMRR. A large tail resistance (current source) raises the CMRR.

Op-amp circuits. Inverting gain −R_f/R₁, non-inverting 1 + R_f/R₁; summer v_o = −R_fΣ(vᵢ/Rᵢ); integrator −(1/RC)∫v dt; differentiator −RC dv/dt. Schmitt trigger (inverting, output ±V_sat, feedback divider R₁ from output to + input and R₂ from + input to ground): thresholds ±V_satR₂/(R₁ + R₂), hysteresis width 2V_satR₂/(R₁ + R₂). Oscillators: Barkhausen criterion |Aβ| = 1, phase 0°; Wien bridge f = 1/(2πRC) with gain 3; RC phase shift (3 sections) f = 1/(2πRC√6) with gain 29.

Worked example

A BJT is biased at I_C = 1 mA with β = 100 and V_T = 25 mV: find g_m and r_π, and the CE gain with R_C = R_L = 10 kΩ. Find the Miller input capacitance for 2 pF across a stage of gain −50. Then find the CMRR in dB for A_d = 100 and A_cm = 0.01, and the frequency of a Wien bridge oscillator with R = 10 kΩ and C = 10 nF.

import math

IC, beta, VT = 1e-3, 100, 0.025
gm = IC / VT
print(f"gm = {gm * 1000:.0f} mA/V, r_pi = {beta / gm / 1000:.2f} kohm")
RL = 10e3 * 10e3 / (10e3 + 10e3)
print(f"CE gain = {-gm * RL:.0f}")
print(f"Miller input capacitance = {2 * (1 + 50):.0f} pF")
print(f"CMRR = {20 * math.log10(100 / 0.01):.0f} dB")
print(f"Wien bridge f = {1 / (2 * math.pi * 10e3 * 10e-9):.2f} Hz")

gm = 40 mA/V, r_pi = 2.50 kohm
CE gain = -200
Miller input capacitance = 102 pF
CMRR = 80 dB
Wien bridge f = 1591.55 Hz

Practice questions