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 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

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. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

MOSFETs MOS - 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 IC’s) MESFET - metal semiconductor structure, used in high-speed GaAs devices JFET - junction FET, early type of FET ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS 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. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

If VGS > VT (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to vGS - Vt. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Symbols and conventions drain n-channel several slightly different symbols (source is often connected to the substrate which is usually grounded) + VDS - gate + VGS - source ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Symbols and conventions drain p-channel several slightly different symbols (source is often connected to VDD) + VDS - gate + VGS - source ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Output characteristics (n-channel) (linear) + VDS - An n-channel MOSFET with VGS and VDS applied and with the normal directions of current flow indicated. ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Input characteristics (n-channel) ID = K(VGS-VT)2 + VDS - ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

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) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Comparison of BJT and FET voltage controlled VGS > VT for device to be on operates in saturation region (amplifier); VDS > VGS - VT ID = K(VGS-VT)2 BJT current 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 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

MOSFET aspect ratio ID = 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 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

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 = - 2.2V ID IG = 0 + V1 VGS - n channel ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

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 = 2.2V V2 IG = 0 + VGS - ID n channel ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Prob 5.41(f) Given: VT = 2V, K = (1/2) .5 mA/V2 (f) Find VGS Equate current in load and transistor Current in transistor: ID = K(VGS-VT)2 Current in resistor: I = (5 - VGS) /100K Equate currents (5 - VGS) /100K = (1/2) .5 (VGS - 2)2 Solve for VGS VGS = 2.33V IG = 0 ID n channel + VGS - ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

5.4 MOSFETS at DC DC problem Find ID, and VGS, and VDS VGS = 5V VGS > VT, so device is on Assume device is in saturation ID = K(VGS-VT)2 ID = (0.05 mA/V2)(5-1)2 ID = 0.8 mA VDS = VDD - ID RD VDS = 10 - (0.8)6 VDS = 5.2V ID IG = 0 + VDS - + VGS - ID VT = 1V K = 0.05 mA/V2 (typical values) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

General DC problem DC problem Find ID, and VGS Assume device is in saturation ID = K(VGS-VT)2 ID = K(5 - ID RS -VT)2 18ID 2 - 25 ID + 8 = 0 Solve for ID, use quadratic formula ID = 0.89mA, 0.5mA, which is correct? For ID = 0.89mA, VGS = 5 - (0.89)6 = - 0.34V For ID = 0.5mA, VGS = 5 - (05)6 = 2V Only for ID = 0.5mA, is transistor on! IG = 0 + VDS - + VGS - ID VT = 1V, K = 0.5 mA/V2 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

DC problem: two FETs in series Find V If devices are identical IG = 0 VDD = 5V Ground device IG = 0 V V =VDD/2 = 2.5V ID device n channel ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

. 5.5 MOSFET as an amplifier ac model n channel SPICE model d g + Ro d vgs g s - ac model s g d n channel + vgs - Ro = 1/slope of the output characteristics s SPICE model ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Transconductance Transconductance = gm = dID/dVGS = 2 K(VGS-VT) = d [K(VGS-VT)2]/dVGS = 2 K(VGS-VT) Useful relation: gm = 2 K  ID ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Prob. 5.86 ac model (a) Find the resistance of an enhancement load g I + V - Rin s ac model Rin = 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/gm ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Prob. 5.86 (a) Find the resistance of an enhancement load Often, Ro >> 1/gm ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

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 9 Decrease W by a factor of 9 Increase L by a factor of 9 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

5.7 Integrated Circuit MOSFET amplifiers Resistors take up too much space on an integrated ciruit (IC) Use transistors as loads Typical amplifier DC analysis Equate current in Q1 and load I in Q1 = I in load K(VGS-VT)2 = I in load ID ID ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

ac analysis of MOSFET amplifiers g d + ID vgs - s Rin Rout ac circuit Rin =  Rout = Rload || Ro ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

ac analysis of MOSFET amplifiers iin = 0 -gmvgs g d + + vout vgs - - s Ai = iout / iin =  Av = vout/vin = -gmvgs(Ro || Rload) / vgs = -gm(Ro || Rload) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Transistor loads: depletion load + V - VGS = 0 Depletion load R = Ro || resistance of current source with 0 magnitude = Ro ||  = Ro Ro = |VA| / I Resistance is current dependent ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS amp Q2 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| = 50V p-channel mpCox = 20mA/V2 n-channel mnCox = 40mA/V2 WQ1 = Wp = 100mm WQ4 = 50mm L = 10mm Iref I ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS amp: power Given: |VT| = 1V, |VA| = 50V p-channel mpCox = 20mA/V2 Iref I Given: |VT| = 1V, |VA| = 50V p-channel mpCox = 20mA/V2 n-channel mnCox = 40mA/V2 WQ1 = Wp = 100mm WQ4 = 50mm L = 10mm Find Total power consumed Power consumed = 2IrefVDD Equate currents in Q3 and Q4 to find Iref IQ3 = IQ4 = K3(VGS-VT)2 = K4(VGS-VT)2 Note that K’s 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 = 2.5V Iref = K4(VGS-VT)2 = (1/2)(40)(50/10) (2.5 - 1)2 = 225mA Power consumed = (2) 5V 225mA = 2.25mW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS amp: DC analysis Given: |VT| = 1V, |VA| = 50V p-channel mpCox = 20mA/V2 n-channel mnCox = 40mA/V2 WQ1 = Wp = 100mm WQ4 = 50mm L = 10mm + Vout - Iref Find Vout Consider current in Q1 or Q2 Using Q1, IQ1 = K1(VGS-VT)2 where VGS = Vout 225mA = (1/2)(40)(100/10) (VGS - 1)2 Solve for VGS, VGS = Vout = 1.75V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS amp: ac analysis Given: |VT| = 1V, |VA| = 50V p-channel mpCox = 20mA/V2 n-channel mnCox = 40mA/V2 WQ1 = Wp = 100mm WQ4 = 50mm L = 10mm Iref + Vout - Find Av Av = -gm1(Ro1 || Ro2) Ro1= Ro2 = 50/ 225mA = 222KW gm = 2 K  ID = (2) [(1/2)(40)(100/10)] 1/2  225mA = 300mA/V Av = -gm1(Ro1 || Ro2) = -300(.222/2)  -33 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS multistage amp: ac analysis DC circuit ac circuit (neglects resistances of current sources) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

CMOS multistage amp: ac analysis Av of stage 1: Vout1/Vgs1 = -gm1Vgs1Ro1/Vgs1 = -gm1ro1 Av of stage 2: Vout2/Vgs2 = -gm2Vgs2Ro2/Vgs2 = -gm2ro2 Overall Av = (-gm1ro1) ( -gm2ro2) = gm1gm2ro1ro2 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Multistage CMOS amp: DC analysis Iref Q3 and Q6 form a PMOS current mirror load for Q4 Q1 and Q5 form an NMOS current mirror load for Q2 Q5 and Q6 establish the current in Q1,Q2,Q3 and Q4 The width of Q5 is adjusted to give a particular Iref ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Multistage CMOS amp: DC analysis Equate 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 Iref K3/K6 = IQ3/Iref K1/K5 = IQ1/Iref Find VD4, and VD1 = Vout from currents in those transistors Iref Given KP = 80mA/V2, KN = 100mA/V2, |VT| = 1V, VDD = 9V 100(VGS5 - 1)2 = 80((VGS5 - 9) - (- 1))2, VGS5 = 5.14V, 48.9V Find Iref, 100(5.14 - 1)2 = 1.7mA IQ3 = IQ4 = IQ2 = IQ1 because all KN’s and KP’s are equal ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002