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•Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage Single Stage Amplifiers •Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage Hassan Aboushady University of Paris VI References • B. Razavi, “Design of Analog CMOS Integrated Circuits”, McGraw-Hill, 2001. H. Aboushady University of Paris VI 1 Single Stage Amplifiers •Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage Hassan Aboushady University of Paris VI Basic Concepts I • Amplification is an essential function in most analog circuits ! • Why do we amplify a signal ? • The signal is too small to drive a load • To overcome the noise of a subsequent stage • Amplification plays a critical role in feedback systems In this lecture: • Low frequency behavior of single stage CMOS amplifiers: • Common Source, Common Gate, Source Follower, ... • Large and small signal analysis. • We begin with a simple model and gradually add 2nd order effects • Understand basic building blocks for more complex systems. H. Aboushady University of Paris VI 2 Approximation of a nonlinear system Input-Output Characteristic of a nonlinear system ≈ α +α +α 2 + +α n ≤ ≤ y(t) 0 1x(t) 2 x (t) ... n x (t) x1 x x2 In a sufficiently narrow range: ≈ α +α y(t) 0 1x(t) α where 0 can be considered the operating (bias) point and α1 the small signal gain H. Aboushady University of Paris VI Analog Design Octagon H. Aboushady University of Paris VI 3 Single Stage Amplifiers •Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage Hassan Aboushady University of Paris VI Common Source Stage with Resistive Load = − Vout VDD RD I D µ C W 2 M1 in the saturation region: V = V − R n ox (V −V ) out DD D 2 L in TH µ nCox W 2 M1 in limit of saturation: V −V = V − R (V −V ) in1 TH DD D 2 L in1 TH ⎡ 2 ⎤ M1 in the = − µ W − − Vout Vout VDD RD nCox ⎢(Vin VTH )Vout ⎥ linear region: L ⎣ 2 ⎦ H. Aboushady University of Paris VI 4 Common Source Stage with Resistive Load R V M1 in deep V = V on = DD linear region: out DD R + R W on D 1+ µ C R (V −V ) n ox L D in TH H. Aboushady University of Paris VI Common Source Stage with Resistive Load M1 in the saturation region: µ C W V = V − R n ox (V −V )2 out DD D 2 L in TH Small signal gain: ∂V W A = out = −R µ C (V −V ) v ∂ D n ox in TH Vin L = − gm RD Same relation can be derived Small signal model for the saturation region from the small signal equivalent circuit To minimize nonlinearity, the gain equation must be a weak function of signal dependent parameters such as gm ! H. Aboushady University of Paris VI 5 Example 1 Sketch ID and gm of M1 as a function of the Vin: • M1 in the saturation region: • M1 in the linear region: 2 µ C W W ⎡ Vout ⎤ = n ox − 2 I = µ C (V −V )V − I D (Vin VTH ) D n ox ⎢ in1 TH out ⎥ 2 L L ⎣ 2 ⎦ ∂I W ∂I W g = D = µ C (V −V ) g = D = µ C V m ∂ n ox in TH m ∂ n ox out VGS L VGS L = VDD Vout + µ W − 1 nCox RD (Vin VTH ) H. Aboushady L University of Paris VI Voltage Gain of a Common Source Stage = − Av gm RD = − µ W VRD Av 2 nCox I D L I D = − µ W VRD Av 2 nCox L I D How to increase Av ? Trade-offs: • Increase W/L Greater device capacitances. • Increase VRD Limits Vout swing. • Reduce ID Greater Time Constant. H. Aboushady University of Paris VI 6 Taking Channel Length Modulation into account Calculating Av starting from the Large Signal Equations: µ C W V = V − R n ox (V −V )2 (1+ λV ) out DD D 2 L in TH out ∂V W A = out = −R µ C (V −V )(1+ λV ) v ∂ D n ox in TH out Vin L µ C W ∂V − R n ox (V −V )2 λ out D in TH ∂ 2 L Vin = − − λ Av RD gm RD I D Av λI =1/ r r R RD gm D O = − O D A = − Av gm v + λ r + R 1 RD I D O D H. Aboushady University of Paris VI Taking Channel Length Modulation into account Calculating Av starting from the Small Signal model: = − gmV1(rO // RD ) Vout V A = out = −g (r // R ) = v V m O D V1 Vin in H. Aboushady University of Paris VI 7 Example 2 Assuming M1 biased in saturation, calculate the small signal voltage gain : • I1 : Ideal current source Infinite Impedance = − Av gmrO • Intrinsic gain of a transistor: This quantity represents the maximum voltage gain that can be achieved using a single device. µ C W I = n ox (V −V )2 (1+ λV ) = I D1 2 L in TH out 1 • Constant Current: As Vin increases, Vout must decrease such that the product remains constant H. Aboushady University of Paris VI CS Stage with Current-Source Load • Both transistors operate in the saturation region: = − Av gm (rO1 // rO2 ) • The output impedance and the minimum required VDS of M2 are less strongly coupled than the value and voltage drop of a resistor. = − VDS 2,min VGS 2 VTH 2 • This value can be reduced to a few hundred millivolts by simply increasing the width of M2. •If rO2 is not sufficiently high, the length and width of M2 can be increased to achieve a smaller λ while maintaining the same overdrive voltage. •The penalty is the large capacitance introduced by M2 at the output node. •Increasing L2 while keeping W2 constant increases rO2 and hence the voltage gain, but at the cost of higher |VDS2| required to maintain M2 in saturation H. Aboushady University of Paris VI 8 CS with Source Degeneration Large Signal model: Small Signal model: ∂I ∂I ∂V G = D = D GS m ∂ ∂ ∂ Vin VGS Vin = − VGS Vin I D RS ∂V ∂I GS =1− D R ∂ ∂ S Vin Vin ∂ ⎛ ∂ ⎞ = I D − I D Gm ⎜1 RS ⎟ ∂ ∂ ⎜ ∂ ⎟ I D I D gmV1 VGS ⎝ Vin ⎠ g = G = = m ∂V m + = ()− GS Vin V1 gmV1RS Gm gm 1 RSGm g g G = m A = −G R G = m m + v m D m + 1 gm RS 1 gm RS g R A = − m D v + 1 gm RS H. Aboushady University of Paris VI CS with Source Degeneration g 1 G = m = >> ≈ m + + for RS 1/ gm Gm 1/ RS 1 gm RS 1/ gm RS ID is linearized at the cost of lower gain. Small Signal model including body effect and channel length modulation: = − − VX Iout gmV1 gmbVX rO = − + − − Iout RS gm (Vin Iout RS ) gmb ( Iout RS ) rO I g r G = out = m O m + + + Vin RS [1 (gm gmb )RS ]rO H. Aboushady University of Paris VI 9 With and Without Source Degeneration g r G = m O m + + + 1 [1 (gm gmb )RS ]rO = ≠ RS 0 RS 0 H. Aboushady University of Paris VI Estimating Gain by Inspection g R R A = − m D = − D v + + 1 gm RS 1/ gm RS Resistance seen at the Drain Gain = − Total Resistance in the Source Path Example: R A = − D v + 1/ gm1 1/ gm2 H. Aboushady University of Paris VI 10 Output Resistance of Degenerated CS = − V1 I X RS The current flowing in r : − + O I X (gm gmb )V1 = + + I X (gm gmb )RS I X = + + + VX rO[I X (gm gmb )RS I X ] I X RS = VX = + + + Rout rO[1 (gm gmb )RS ] RS I X = + + + Rout [1 (gm gmb )rO ]RS rO ≈ + + = + + Rout (gm gmb )rO RS rO Rout [1 (gm gmb )RS ]rO H. Aboushady University of Paris VI Voltage Gain of Degenerated CS The current through RS must equal that through RD: V I = I = out RD RS RD VS = − RS VS Vout RD Vout The current through rO : I = − − (g V + g V ) rO m 1 mb BS RD V R R V I = − out −[g (V +V S ) + g V S ] V = I r − out R rO m in out mb out out rO O S RD RD RD RD = −Vout − + RS + RS − RS Vout rO [gm (Vin Vout ) gmbVout ]rO Vout RD RD RD RD Vout = − gmrO RD V R + R + r + (g + g )R r H. Aboushady in D S O m mb S O University of Paris VI 11 Voltage Gain of Degenerated CS Vout = − gmrO RD + + + + Vin RD RS rO (gm gmb )RS rO + + + Vout = − gmrO RD[RS rO (gm gmb )RS rO ] + + + + + + + Vin RS [1 (gm gmb )RS ]rO RD RS rO (gm gmb )RS rO Vout = () Gm Rout // RD Vin The output resistance = + + Rout [1 (gm gmb )RS ]rO of a degenerated CS stage: I g r The Transconductance G = out = m O of a degenerated CS stage: m + + + Vin RS [1 (gm gmb )RS ]rO H. Aboushady University of Paris VI General expression to calculate Av by inspection Lemma: = − Av Gm Rout Gm : the transconductance of Rout : the output resistance the circuit when the output is of the circuit when the input shorted to grounded. voltage is set to zero. • For high voltage gain the output resistance must be high! A “buffer” is needed to drive a low-impedance load. The source follower can operate as a voltage buffer. H. Aboushady University of Paris VI 12 Single Stage Amplifiers •Basic Concepts •Common Source Stage •Source Follower •Common Gate Stage •Cascode Stage Hassan Aboushady University of Paris VI Source Follower (Common Drain) Large Signal Behavior M1 turns on in saturation: = Vout I D RS µ C W V = n ox (V −V −V )2 R out 2 L in out TH S To calculate gm : ∂V W ∂V ∂V out = µ C (V −V −V )(1− out − TH )R ∂ n ox in out TH ∂ ∂ S Vin L Vin Vin Since, VTH = VTH 0 + γ ( 2ΦF +VSB − 2ΦF ) ∂ TH ∂ TH ∂ V γ ∂ V V = V SB = SB ∂ ∂ ∂ Φ + ∂ Vin VSB Vin 2 2 F VSB Vin ∂ =η Vout ∂ V H.
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