Transcript Document
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 0 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 1 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 2 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 3 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 4 Symbols and conventions drain •n-channel gate + VDS - + VGS - •several slightly different symbols source (source is often connected to the substrate which is usually grounded) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 5 Symbols and conventions drain •p-channel gate + VDS - + VGS - •several slightly different symbols source (source is often connected to VDD) ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 6 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 7 Input characteristics (n-channel) + VDS - ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 ID = K(VGS-VT)2 8 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 9 Comparison of BJT and FET BJT •current controlled FET •voltage controlled •VBE 0.7 V for device to be on •VGS > VT for device to be on •operates in linear region (amplifier); •operates in saturation region (amplifier); VDS > VGS - VT BE junction forward biased, BC junction reversed biased •IC = bIB •ID = K(VGS-VT)2 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 10 MOSFET aspect ratio ID = K(VGS-VT)2 K = transconductance parameter K = 1/2 K' (W/L) K' = nCox, where n 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 11 Prob 5.41(a) Given: VT = 2V, K = (1/2) .5 mA/V2 ID IG = 0 (a) Find V1 Use, + VGS - V1 ID = K(VGS-VT)2 10uA = (1/2) .5 (VGS - 2)2 Solve for VGS n channel VGS = 2.2V V1 = - 2.2V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 12 Prob 5.41(b) Given: VT = 2V, K = (1/2) .5 mA/V2 (b) Find V2 V2 IG = 0 Use, ID = K(VGS-VT)2 10uA = (1/2) .5 (VGS - 2)2 + VGS - n channel ID Solve for VGS VGS = 2.2V V2 = VGS = 2.2V ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 13 Prob 5.41(f) Given: VT = 2V, K = (1/2) .5 mA/V2 (f) Find VGS Equate current in load and transistor IG = 0 ID + VGS - n channel 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 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 14 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 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 ID IG = 0 + VDS - + VGS - ID VT = 1V K = 0.05 mA/V2 (typical values) 15 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! ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 IG = 0 + VDS - + VGS - ID VT = 1V, K = 0.5 mA/V2 16 DC problem: two FETs in series Find V IG = 0 If devices are identical VDD = 5V IG = 0 V device V =VDD/2 = 2.5V ID device Ground n channel ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 17 5.5 MOSFET as an amplifier g . d d + Ro vgs g - s ac model s g n channel d + vgs Ro = 1/slope of the output characteristics - s SPICE model ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 18 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 19 Prob. 5.86 (a) Find the resistance of an enhancement load + I g 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 20 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 21 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 22 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 ID Equate current in Q1 and load I in Q1 = I in load K(VGS-VT)2 = I in load ID ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 23 ac analysis of MOSFET amplifiers g ID d + vgs Rin s Rout ac circuit ID Rin = Rout = Rload || Ro ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 24 ac analysis of MOSFET amplifiers g iin = 0 d -gmvgs + vgs - s + vout - 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 25 Transistor loads: depletion load VGS = 0 Depletion load I + V - R = Ro || resistance of current source with 0 magnitude = Ro || Ro = |VA| / I Resistance is current dependent = Ro ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 26 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 Iref I ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 Given: |VT| = 1V, |VA| = 50V p-channel pCox = 20A/V2 n-channel nCox = 40A/V2 WQ1 = Wp = 100m WQ4 = 50m L = 10m 27 CMOS amp: power Iref I Given: |VT| = 1V, |VA| = 50V p-channel pCox = 20A/V2 n-channel nCox = 40A/V2 WQ1 = Wp = 100m WQ4 = 50m L = 10m 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 = 225A •Power consumed = (2) 5V 225A = 2.25mW ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 28 CMOS amp: DC analysis Given: |VT| = 1V, |VA| = 50V p-channel pCox = 20A/V2 n-channel nCox = 40A/V2 WQ1 = Wp = 100m WQ4 = 50m L = 10m + Vout Iref - Find Vout •Consider current in Q1 or Q2 •Using Q1, IQ1 = K1(VGS-VT)2 where VGS = Vout 225A = (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 29 CMOS amp: ac analysis Given: |VT| = 1V, |VA| = 50V p-channel pCox = 20A/V2 n-channel nCox = 40A/V2 WQ1 = Wp = 100m WQ4 = 50m L = 10m Iref + Vout Find Av Av = -gm1(Ro1 || Ro2) - Ro1= Ro2 = 50/ 225A = 222KW gm = 2 K ID = (2) [(1/2)(40)(100/10)] 1/2 225A = 300A/V Av = -gm1(Ro1 || Ro2) = -300(.222/2) -33 ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 30 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 31 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 32 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 33 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 Iref •Find VD4, and VD1 = Vout from currents in those transistors •Given KP = 80A/V2, KN = 100A/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 34