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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.

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Chap. 5 Field-effect transistors (FET)

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

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

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

  10. 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

  11. Comparison of BJT and FET • 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. VDD = 5V Ground DC problem: two FETs in series Find V If devices are identical IG = 0 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

  19. 5.5 MOSFET as an amplifier . 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

  20. Transconductance Transconductance = gm = dID/dVGS =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

  21. Prob. 5.86 (a) Find the resistance of an enhancement load g I d + 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

  22. 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

  23. 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

  24. 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

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

  26. 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

  27. Transistor loads: depletion load I + 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

  28. 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

  29. Iref I CMOS amp: power 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

  30. 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

  31. + Vout - Iref 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 • 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

  32. 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

  33. 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

  34. 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

  35. Iref 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 • 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 • FindIref, 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

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