EE105 – Fall 2014 Microelectronic Devices and Circuits

Prof. Ming C. Wu [email protected] 511 Sutardja Dai Hall (SDH)

Lecture13-Small Signal Model-MOSFET 1

Small-Signal Operation MOSFET Small-Signal Model - Summary

IG = 0

Kn 2 ID = (VGS −VTN ) (1+ λVDS ) 2

Transconductance:

2ID gm = = 2KnID • Since gate is insulated from VGS −VTN channel by gate-oxide input resistance of transistor is infinite. Output resistance: • Small-signal parameters are controlled by the Q-point. 1 1+λVDS 1 ro = = ≅ • For the same operating point, go λID λID MOSFET has lower

transconductance and an output Amplification factor for lVDS<<1: resistance that is similar to the BJT. 1+λV 2K µ =g r = DS ≅ 1 n f m o λ I λID D

Lecture13-Small Signal Model-MOSFET 2

1 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:

∂i ∂i g = D =− D mb ∂vBS Q-point ∂vSB Q-point # &# & % ∂iD (% ∂VTN ( gmb =−% (% ( =−(−gmη)=gmη % (% ( $ ∂VTN '$ ∂vSB ' 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.

Lecture13-Small Signal Model-MOSFET 3

MOSFET Small-Signal Operation Small-Signal Model for PMOS Transistor

• For a PMOS transistor

vSG = VGG − vgg

iD = ID −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!

Lecture13-Small Signal Model-MOSFET 4

2 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.

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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

CE vd Avt = = −gm RL RL = ro RD R3 vg

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3 Common-Source Amplifiers Input Resistance and Signal-Source Gain

vg Define RiG as the input resistance RiG = = RG looking into the base of the transistor. ii

Rin is the resistance presented to vi Rin = RI + RG

The signal source voltage gain is: CS vo vo vg CS RG Av = = = Avt vi vg vi RI + RG " % CS RG Av = −gm RL $ ' # RI + RG &

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Common-Source Amplifiers “Rule of Thumb” Design Estimate

" % CS RG CS CS Av = −gm RL $ ' ≅ Avt Avt = −gm RL RL = ro RD R3 # RI + RG & I R Typically: r >> R and R >> R ACS ≅ −g R = − D D o D 3 D v m D "V −V % $ GS TN ' # 2 &

ID RD represents the voltage dropped across drain resistor RD V A typical design point is I R = DD with V −V =1 V D D 2 GS TN

CS ∴ Av ≅ −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!

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4 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

2 • Given data: Kn = 0.50 mA/V , 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: −1 λ +VDS gm = 2KnID (1+ λVDS ) = 0.503 mS ro = = 328 kΩ ID

RG = R1 R2 = 892 kΩ RL = ro RD R3 =17.1 kΩ

Lecture13-Small Signal Model-MOSFET 9

Common-Source Amplifiers Voltage Gain Example (cont.)

gm = 0.503 mS ro = 328 kΩ RG = 892 kΩ RL =17.1 kΩ # & CS RG # 892kΩ & Av = −gm RL % ( = −0.503mS(17.1kΩ)% ( = −8.60(0.999) = −8.59 $ RI + RG ' $1kΩ+892kΩ' # & # & RG RG Rin = RI + RG = 893 kΩ vgs = vi % ( → vi % ( ≤ 0.2(VGS −VTN ) $ RI + RG ' $ RI + RG '

2ID # 893kΩ & VGS −VTN ≅ = 0.982 V ∴ vi ≤ 0.2(0.982V)% ( = 0.197 V Kn $892kΩ' CS Check the rule-of-thumb estimate: Av ≅ −VDD = −12 V (ballpark estimate)

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5 C-E and C-S Amplifiers Output Resistance

Lecture13-Small Signal Model-MOSFET 11

C-E and C-S Amplifiers Output Resistance (cont.)

Apply test source vx and find ix (with vi = 0) vgs = 0 → gmvgs = 0 vbe = 0 → gmvbe = 0 v v ∴R = x = R r ∴R = x = R r out D o out C o ix ix Rout ≅ RD for ro >> RD Rout ≅ RC for ro >> RC

For comparable bias points, output resistances of C-S and C-E amplifiers are similar.

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6 BJT and FET Small-Signal Model Summary

Lecture13-Small Signal Model-MOSFET 13

Common-Emitter / Common-Source Amplifiers Summary

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7 Amplifier Power Dissipation

Static power dissipation in amplifiers is found from their dc equivalent circuits.

(a) Total power dissipated in BJT: (b) Total power dissipated in MOSFET:

PD = VCE IC + VBE IB PD = VDS ID Total power supplied is: Total power supplied is: PS = VDD ( ID + I2 ) PS = VCC ( IC + I2 )

Lecture13-Small Signal Model-MOSFET 15

Amplifier Signal Range

vCE = VCE −Vm sinωt where Vm is the output signal. Active region operation Similarly for MOSFETs:

requires vCE ≥ vBE So: Vm ≤ VCE −VBE V ≤min#I R ,(V −(V −V ))& Also: v t = I R −V sinωt ≥ 0 M $% D D DS GS TN '( Rc ( ) C C m % ' ∴Vm ≤ min&IC RC,(VCE −VBE )(

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