Chap. 5 Field-effect transistors (FET)Widely used in VLSI

used in some analog amplifiers - output stage of power amplifers (may have good thermal characteristics if designed properly)

n-channel or p-channel structure

FET - voltage controlled device

BJT - current controlled device

Physical structure of a n-channel device: Typically L = 0.35 to 10 m, W = 2 to 500 m, and the thickness of the oxide layer is in the range of 0.02 to 0.1 m.

MOSFETsMOS - metal oxide semicondutor structure (original devices had metal gates, now they are silicon)

NMOS - n-channel MOSFET

PMOS - p-channel MOSFET

CMOS - complementary MOS, both n-channel and p-channel devices used in conjuction with each other (most popular in ICs)

MESFET - metal semiconductor structure, used in high-speed GaAs devices

JFET - junction FET, early type of FET

Cross section of a CMOS integrated circuit. Note that the PMOS transistor is formed in a separate n-type region, known as an n well. CMOS

If VGS > VT (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to vGS - Vt.

Symbols and conventionsn-channel

several slightly different symbols

(source is often connected to the substrate which is usually grounded)+VDS-+ VGS -drainsourcegate

Symbols and conventionsp-channel

several slightly different symbols

(source is often connected to VDD)+VDS-+ VGS -drainsourcegate

An n-channel MOSFET with VGS and VDS applied and with the normal directions of current flow indicated. Output characteristics (n-channel)(linear)+VDS-

Input characteristics (n-channel)+VDS-ID = K(VGS-VT)2

Summary of MOSFET behavior (n-channel)VGS > VT (threshold voltage) for the device to be on

VDS > VGS - VT for device to be in saturation region

ID = K(VGS-VT)2

Enhancement mode device, VT > 0

Depletion mode device, VT < 0 (conducts with VGS = 0)

Comparison of BJT and FETFETvoltage controlled

VGS > VT for device to be on

operates in saturation region (amplifier);VDS > VGS - VT

ID = K(VGS-VT)2

BJTcurrent controlled

VBE 0.7 V for device to be on

operates in linear region (amplifier); BE junction forward biased, BC junction reversed biased

IC = bIB

MOSFET aspect ratioID = K(VGS-VT)2

K = transconductance parameter

K = 1/2 K' (W/L)

K' = mnCox, where mn is the mobility of electrons, and Cox is the capacitance of the oxide

W/L is the aspect ratio, W is the width of the gate, L is the length of the gate.

ID W/L

Prob 5.41(a)Given: VT = 2V, K = (1/2) .5 mA/V2

(a) Find V1

Use, ID = K(VGS-VT)2

10uA = (1/2) .5 (VGS - 2)2

Solve for VGS

VGS = 2.2V

V1 = - 2.2VV1VGS -+IDIG = 0n channel

Prob 5.41(b)Given: VT = 2V, K = (1/2) .5 mA/V2

(b) Find V2

Use, ID = K(VGS-VT)2

10uA = (1/2) .5 (VGS - 2)2

Solve for VGS

VGS = 2.2V

V2 = VGS = 2.2VV2VGS -+IDIG = 0n channel

Prob 5.41(f)Given: VT = 2V, K = (1/2) .5 mA/V2(f) Find VGSEquate current in load and transistor

Current in transistor: ID = K(VGS-VT)2Current in resistor: I = (5 - VGS) /100KEquate currents(5 - VGS) /100K = (1/2) .5 (VGS - 2)2

Solve for VGSVGS = 2.33V

VGS -+IDIG = 0n channel

5.4 MOSFETS at DCDC problemFind ID, and VGS, and VDS

VGS = 5VVGS > VT, so device is on

Assume device is in saturationID = K(VGS-VT)2ID = (0.05 mA/V2)(5-1)2ID = 0.8 mA

VDS = VDD - ID RDVDS = 10 - (0.8)6VDS = 5.2VVT = 1VK = 0.05 mA/V2(typical values)+ VGS -+VDS-IDIDIG = 0

General DC problem+ VGS -+VDS-IDIG = 0DC problemFind ID, and VGS

Assume device is in saturationID = K(VGS-VT)2ID = K(5 - ID RS -VT)218ID 2 - 25 ID + 8 = 0

Solve for ID, use quadratic formulaID = 0.89mA, 0.5mA, which is correct?

For ID = 0.89mA, VGS = 5 - (0.89)6 = - 0.34VFor ID = 0.5mA, VGS = 5 - (05)6 = 2V

Only for ID = 0.5mA, is transistor on!VT = 1V, K = 0.5 mA/V2

DC problem: two FETs in seriesFind VIf devices are identical

IDIG = 0n channelIG = 0VdevicedeviceV =VDD/2 = 2.5V

5.5 MOSFET as an amplifiergdsn channelac modeldsgvgs+-Ro = 1/slope of the output characteristics.dsgvgs+-SPICE modelRo

TransconductanceUseful relation: gm = 2 K IDTransconductance = gm = dID/dVGS

= d [K(VGS-VT)2]/dVGS

= 2 K(VGS-VT)

Prob. 5.86(a) Find the resistance of an enhancement loadRinRin = resistance of current source || Ro

resistance of current source = voltage across current source / current in current source

resistance of current source = vgs / gmvgs = 1/gm

Replace current source by a resistor of resistance 1/gmac model+

V

-Igds

Prob. 5.86(a) Find the resistance of an enhancement loadOften,Ro >> 1/gm

Prob. 5.86(b) To raise the resistance of the transistor by a factor of 3, what must be done?R 1/gm

= 1 / 2 K ID

= [1/2 ] [1/K] [ 1/ ID]

= [1/2 ] [1/ 1/2 K W/L ] [ 1/ ID] Decrease ID by a factor of 9Decrease W by a factor of 9Increase L by a factor of 9

5.7 Integrated Circuit MOSFET amplifiersResistors take up too much space on an integrated ciruit (IC)

Use transistors as loadsTypical amplifierDC analysis

Equate current in Q1 and load

I in Q1 = I in load

K(VGS-VT)2 = I in load

IDID

ac analysis of MOSFET amplifiersac circuitRin =

Rout = Rload || Rodsgvgs+-RinRout

ac analysis of MOSFET amplifiersAi = iout / iin =

Av = vout/vin = -gmvgs(Ro || Rload) / vgs = -gm(Ro || Rload) dsgvgs+-vout+-iin = 0-gmvgs

Transistor loads: depletion loadIV+-Depletion loadVGS = 0R = Ro || resistance of current source with 0 magnitude

= Ro ||

= RoRo = |VA| / I

Resistance iscurrent dependent

CMOS ampIrefIQ2 and Q3 form a p-channel current mirror load for Q1

Q4 and Q3 establish Iref

I = Iref due to current mirror

Given:|VT| = 1V, |VA| = 50Vp-channel mpCox = 20mA/V2n-channel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm

CMOS amp: powerGiven:|VT| = 1V, |VA| = 50Vp-channel mpCox = 20mA/V2n-channel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm

Find Total power consumed

Power consumed = 2IrefVDDEquate currents in Q3 and Q4 to find IrefIQ3 = IQ4 = K3(VGS-VT)2 = K4(VGS-VT)2Note that Ks are the same: K3 = (1/2)(20)(100/10) = K4 = (1/2)(40)(50/10)Therefore, Q3 and Q4 behave the same, so VGS3 = VGS4 = 2.5VIref = K4(VGS-VT)2 = (1/2)(40)(50/10) (2.5 - 1)2 = 225mAPower consumed = (2) 5V 225mA = 2.25mW

CMOS amp: DC analysisGiven:|VT| = 1V, |VA| = 50Vp-channel mpCox = 20mA/V2n-channel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm

Find Vout Consider current in Q1 or Q2Using Q1, IQ1 = K1(VGS-VT)2 where VGS = Vout225mA = (1/2)(40)(100/10) (VGS - 1)2 Solve for VGS, VGS = Vout = 1.75V

Iref+Vout

-

CMOS amp: ac analysisGiven:|VT| = 1V, |VA| = 50Vp-channel mpCox = 20mA/V2n-channel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm

Find Av Av = -gm1(Ro1 || Ro2)Ro1= Ro2 = 50/ 225mA = 222KWgm = 2 K ID = (2) [(1/2)(40)(100/10)] 1/2 225mA = 300mA/VAv = -gm1(Ro1 || Ro2) = -300(.222/2) -33

CMOS multistage amp: ac analysisDC circuitac circuit(neglects resistances of current sources)

CMOS multistage amp: ac analysisAv of stage 1: Vout1/Vgs1 = -gm1Vgs1Ro1/Vgs1 = -gm1ro1

Av of stage 2: Vout2/Vgs2 = -gm2Vgs2Ro2/Vgs2 = -gm2ro2

Overall Av = (-gm1ro1) ( -gm2ro2) = gm1gm2ro1ro2

Multistage CMOS amp: DC analysisQ3 and Q6 form a PMOS current mirror load for Q4Q1 and Q5 form an NMOS current mirror load for Q2 Q5 and Q6 establish the current in Q1,Q2,Q3 and Q4The width of Q5 is adjusted to give a particular Iref

Iref

Multistage CMOS amp: DC analysisEquate currents in Q5 and Q6 IQ5 = IQ6 = K5(VGS5-VT)2 = K6((VGS5 - VDD)-VT)2

Solve for VGS5, Use VGS5 to find Iref

Other current s are multiples of IrefK3/K6 = IQ3/Iref K1/K5 = IQ1/Iref

Find VD4, and VD1 = Vout from currents in those transistors

Given KP = 80mA/V2, KN = 100mA/V2, |VT| = 1V, VDD = 9V100(VGS5 - 1)2 = 80((VGS5 - 9) - (- 1))2, VGS5 = 5.14V, 48.9VFind Iref, 100(5.14 - 1)2 = 1.7mAIQ3 = IQ4 = IQ2 = IQ1 because all KNs and KPs are equal