Chap. 5 Field-effect transistors (FET)

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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. - PowerPoint PPT Presentation

Transcript of Chap. 5 Field-effect transistors (FET)

  • 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