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ππ = ππ ππΌπ·πΆ Noise current density, In β΄ πππππ π(πππ )π βππ‘ = πΌππππ π(πππ ) ππ Noise power density, Vn2 πΌππππ π(πππ ) = β2ππΌπ·πΆ π΅ = πΌπ βπ΅ Also white noise πππππ π(πππ )π‘ππ‘ππ = βπππππ π(πππ )1 2 + πππππ π(πππ )2 2 π πΊ = [π 1 β₯ π 2 ] Where Vn is rms noise in 1Hz (noise density) Shot noise πππππ π(πππ ) = β4πππ΅π = ππ βπ΅ No dependence on Thermal/Johnson frequency ο¨ white noise RD R1 Rsig 1 1 π1 = β πΆ1 (π π ππ + π πΊ ) πΆ1 π πΊ VS 1 πΆ2 (π πΏ + π ππ» ) = [ππ β₯ π π· ] π2 = Vdd π ππ» C2 Vout C1 R2 RS CS RL Burst (βpopcornβ) noise Flicker noise (1/f) Also LF noise π(πππ )ππ’ππ π‘ = πππ = 1 ππ β πΎ2 π(πππ )πππππππ rds2 πΌπ = βπΎ1 π βπ΅ π 1 + ππ π π ππ1 = πΆπ π π πΆπ π π Combining above graphs: For ΟS0<Ο1 For ΟS0>Ο1 LF noise 1 Hence AV set by π΄π = βππ1 [ππ1 β₯ ] ππ2 transistor size ππ1 and is =β ππ2 independent of bias circuit (π/πΏ)1 =β (π/πΏ)2 Gate-drain connected load i.e. diode connected LF MOSFET performance dictated by C1, C2 and CS πΌπ βπ΅ π 2 1+( ) ππ ππ0 = Emitter coupled transistors responds only to different signals VO2 VO2 Uses transistor in active region as a load to maintain gain and have low power Common mode input, VCM = Vi1 = Vi2 Before: π΄πΆπ1 β β After: π΄π = βππ1 [ππ1 β₯ ππ2 ] πΌπΆ = πΌπππ = π πΆ 2π πΈπΈ REE is the current source small signal resistance Keep low ππΆπΆ β ππ΅πΈ π Input common mode resistance, π πππ = ππ + 2π πΈπΈ (1 + ππ ππ ) (Same applies to FETs) If Io1 through Q1 has W/L, then Io2 = 2Io1 for Q2 that has 2W/L (half the current, half the area) Cascode π ππ’π‘ β (1 + ππ ππ2 )ππ3 ο¨ massive CMRR Differential mode input, Vi1 = - Vi2 Currents can be set by adjusting W/L ratio β can replace Rs with MOS Qs MOS cascode current sources increase Rout Differential gain, Ad = 2Ad1 Transistor biased in active region However, decreased Rout ο¨ emitter degeneration 2. Wilson current mirror πΌπππ πΌππ’π‘ β 2 1+ π½ ro effect decreased (Rout increased) Single ended differential gain, π΄π1 = βππ1 π πΆ BJT and FETs require large resistances Solutions: 3. Widlar current source Current mirror with RE on transistor 2 emitter Keep high Figure of merit for op-amp, Common Mode Rejection Ratio, π΄π πΆππ π = = ππ π πΈπΈ β« 10000(= 80ππ΅) π΄πΆπ 1. Current mirror πΌππ’π‘ = πΌπππ ππΆπΈ (1 + ) 2 ππ΄ 1+ π½ Due to finite Ξ² 2 Input differential resistance, π ππ = 2ππ 2 πΌπππ = πΌππ’π‘ (1 + ) π½ Due to finite ro If Ξ² and ro = β ο¨ Iout = Iref