Chap. 5 Fieldeffect transistors (FET)
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Transcript of Chap. 5 Fieldeffect transistors (FET)
Chap. 5 Fieldeffect transistors (FET)Widely used in VLSI
used in some analog amplifiers  output stage of power amplifers (may have good thermal characteristics if designed properly)
nchannel or pchannel structure
FET  voltage controlled device
BJT  current controlled device
Physical structure of a nchannel 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  nchannel MOSFET
PMOS  pchannel MOSFET
CMOS  complementary MOS, both nchannel and pchannel devices used in conjuction with each other (most popular in ICs)
MESFET  metal semiconductor structure, used in highspeed 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 ntype region, known as an n well. CMOS
If VGS > VT (threshold voltage), an induced, conducting nchannel forms between the drain and source. The channel conductance is proportional to vGS  Vt.
Symbols and conventionsnchannel
several slightly different symbols
(source is often connected to the substrate which is usually grounded)+VDS+ VGS drainsourcegate
Symbols and conventionspchannel
several slightly different symbols
(source is often connected to VDD)+VDS+ VGS drainsourcegate
An nchannel MOSFET with VGS and VDS applied and with the normal directions of current flow indicated. Output characteristics (nchannel)(linear)+VDS
Input characteristics (nchannel)+VDSID = K(VGSVT)2
Summary of MOSFET behavior (nchannel)VGS > VT (threshold voltage) for the device to be on
VDS > VGS  VT for device to be in saturation region
ID = K(VGSVT)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(VGSVT)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(VGSVT)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(VGSVT)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(VGSVT)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(VGSVT)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(VGSVT)2ID = (0.05 mA/V2)(51)2ID = 0.8 mA
VDS = VDD  ID RDVDS = 10  (0.8)6VDS = 5.2VVT = 1VK = 0.05 mA/V2(typical values)+ VGS +VDSIDIDIG = 0
General DC problem+ VGS +VDSIDIG = 0DC problemFind ID, and VGS
Assume device is in saturationID = K(VGSVT)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(VGSVT)2]/dVGS
= 2 K(VGSVT)
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(VGSVT)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 = 0gmvgs
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 pchannel current mirror load for Q1
Q4 and Q3 establish Iref
I = Iref due to current mirror
Given:VT = 1V, VA = 50Vpchannel mpCox = 20mA/V2nchannel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm
CMOS amp: powerGiven:VT = 1V, VA = 50Vpchannel mpCox = 20mA/V2nchannel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm
Find Total power consumed
Power consumed = 2IrefVDDEquate currents in Q3 and Q4 to find IrefIQ3 = IQ4 = K3(VGSVT)2 = K4(VGSVT)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(VGSVT)2 = (1/2)(40)(50/10) (2.5  1)2 = 225mAPower consumed = (2) 5V 225mA = 2.25mW
CMOS amp: DC analysisGiven:VT = 1V, VA = 50Vpchannel mpCox = 20mA/V2nchannel mnCox = 40mA/V2WQ1 = Wp = 100mmWQ4 = 50mmL = 10mm
Find Vout Consider current in Q1 or Q2Using Q1, IQ1 = K1(VGSVT)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 = 50Vpchannel mpCox = 20mA/V2nchannel 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(VGS5VT)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