Lecture 13 Small Signal Model

Microelectronic Devices and Circuits Small-Signal Operation MOSFET Small-Signal Model - Summary IG = 0 Kn (V −V ) 2 GS TN ID = 2 (1+ λ VDS) Transconductance: • • • Since gate is insulated from channel by gate-oxide input resistance of transistor is infinite. Small-signal parameters are controlled by the Q-point. For the same operating point, MOSFET has lower transconductance and an output resistance that is similar to the BJT. gm = 2I D = 2K I n D VGS−VTN Output resistance: ro = 1 = 1+λ VDS ≅ 1 go λ ID λ ID Amplification factor for lVDS<<1: 2Kn µ f = gmro = 1+λ VDS ≅ 1 λ ID λ ID MOSFET Small-Signal Operation Body Effect in Four-terminal MOSFETs Drain current depends on threshold voltage which in turn depends on vSB. Back-gate transconductance is: gmb= ∂iD ∂i =− D ∂vBS Q-point ∂vSB Q-point # % gmb=− % % $ &# & ∂iD (%∂VTN ( (% ( ∂VTN(' %$ ∂vSB (' =−(−gmη)= gmη Q-point 0 < η < 3 is called the back-gate transconductance parameter. bulk terminal is a reverse-biased diode. Hence, no conductance from the bulk terminal to other terminals. MOSFET Small-Signal Operation Small-Signal Model for PMOS Transistor • For a PMOS transistor vSG = VGG − vgg iD = I D − id • Positive signal voltage vgg reduces source-gate voltage of the PMOS transistor causing decrease in total current exiting the drain, equivalent to an increase in the signal current entering the drain. • The NMOS and PMOS small-signal models are the same! Common-Source Amplifiers Small-Signal Analysis - ac Equivalent Circuit • ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources are ac ground. Common-Source Amplifiers Small-Signal Equivalent Circuit • Input voltage is applied to the gate terminal • Output signal appears at the drain terminal • Source is common to both input and output signals Thus circuit is termed a Common-Source (C-S) Amplifier. • The terminal gain of the C-S amplifier is the gain from the gate terminal to the drain terminal AvtCE = vd = −gm RL vg RL = ro RD R3 Common-Source Amplifiers Input Resistance and Signal-Source Gain RiG = Define RiG as the input resistance looking into the base of the transistor. Rin is the resistance presented to vi vg = RG ii Rin = RI + RG The signal source voltage gain is: ACS v = RG vo vo v g = AvtCS = RI + RG vi v g vi " R % CS G Av = −gm RL $ ' # RI + RG & Common-Source Amplifiers “Rule of Thumb” Design Estimate % CS " R CS G Av = −gm RL $ ' ≅ Avt #RI + RG & Typically: AvtCS = −gm RL ro >>RD and R3 >>RD RL = ro RD R3 AvCS ≅−gm RD = − I D RD " VGS −VTN % $ ' # & 2 I D RD represents the voltage dropped across drain resistor RD V with VGS −VTN = 1 V I D RD = DD 2 A typical design point is ∴ AvCS ≅−VDD Our rule-of-thumb estimate for the C-S amplifier: the voltage gain equals the power supply voltage. Note that this is 10 times smaller than that for the BJT! Common-Source Amplifiers Voltage Gain Example • Problem: Calculate voltage gain, input resistance and maximum input signal level for a common-source amplifier with a specified Q-point • Given data: Kn = 0.50 mA/V2, VTN = 1 V, • λ = 0.0133V-1, Q-point is (0.241 mA, 3.81 V) • Assumptions: Transistor is in the active region. Signals are low enough to be considered small signals. • Analysis: gm = 2K n I D( 1+λ VDS) = 0.503 mS RG = R1 R2 = 892 kΩ ro = λ −1 +VDS = 328 kΩ ID RL = or RD R3 = 17.1 kΩ Common-Source Amplifiers Voltage Gain Example (cont.) gm = 0.503 mS ro = 328 kΩ RG = 892 kΩ RL = 17.1 kΩ & # R # 892kΩ & CS Av = −gm RL % G ( = −0.503mS(17.1kΩ) % ( = −8.60 (0.999) = −8.59 $1kΩ + 892kΩ' $ RI + RG' & & # R # R Rin = RI + RG = 893 kΩ vgs = vi % G ( → vi % G ( ≤ 0.2 (VGS −VTN) $ RI + RG ' $RI + RG ' VGS −VTN ≅ 2I D = 0.982 V Kn Check the rule-of-thumb estimate: # 893kΩ & ∴ vi ≤ 0.2 (0.982V) % ( = 0.197 V $ 892kΩ' ACSv ≅−VDD = −12 V (ballpark estimate) C-E and C-S Amplifiers Output Resistance C-E and C-S Amplifiers Output Resistance (cont.) Apply test source vx and find ix (with vi = 0) vbe = 0 → gm vbe = 0 v ∴ Rout = x = RC ro ix Rout ≅ RC for ro >>RC vgs = 0 → gm vgs = 0 vx = R r D o ix Rout ≅ RD for ro >>RD ∴ Rout = For comparable bias points, output resistances of C-S and C-E amplifiers are similar. BJT and FET Small-Signal Model Summary Common-Emitter / Common-Source Amplifiers Summary Amplifier Power Dissipation Static power dissipation in amplifiers is found from their dc equivalent circuits. (a) Total power dissipated in BJT: PD = VCE IC + VBE IB Total power supplied is: PS = VCC ( IC + I2 ) (b) Total power dissipated in MOSFET: PD = VDS ID Total power supplied is: PS = VDD ( ID + I2 ) Amplifier Signal Range vCE = VCE −Vm sin ω t where Vm is the output signal. Active region operation requires vCE ≥ vBE So: Vm ≤ VCE −VBE Also: vRc ()t = I C RC −Vm sinω t ≥ 0 ∴Vm ≤ min % &I C RC , (VCE −VBE) (' Similarly for MOSFETs: VM ≤min #%$I D RD,(VDS−(VGS−VTN))(' &