Download presentation

Presentation is loading. Please wait.

0
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

1
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

2
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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

3
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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

4
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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

5
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

6
**Symbols and conventions**

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

7
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

8
**Input characteristics (n-channel)**

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

9
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

10
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

11
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

12
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

13
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

14
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

15
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

16
**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 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 = V 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

17
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

18
**. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

19
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

20
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

21
**Prob. 5.86 (a) Find the resistance of an enhancement load Often,**

Ro >> 1/gm ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

22
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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

23
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

24
**ac analysis of MOSFET amplifiers**

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

25
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

26
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

27
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

28
**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 = 225mA Power consumed = (2) 5V 225mA = 2.25mW ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

29
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

30
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

31
**CMOS multistage amp: ac analysis**

DC circuit ac circuit (neglects resistances of current sources) ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

32
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

33
**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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

34
**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( )2 = 1.7mA IQ3 = IQ4 = IQ2 = IQ1 because all KN’s and KP’s are equal ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

Similar presentations

OK

1 Fundamentals of Microelectronics CH1 Why Microelectronics? CH2 Basic Physics of Semiconductors CH3 Diode Circuits CH4 Physics of Bipolar Transistors.

1 Fundamentals of Microelectronics CH1 Why Microelectronics? CH2 Basic Physics of Semiconductors CH3 Diode Circuits CH4 Physics of Bipolar Transistors.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on second law of thermodynamics creationism Ppt on creating brand equity Ppt on conservation of mass Ppt on tcp ip protocol architecture definition Ppt on power sharing in democracy in america Ppt on pollution of air and water Ppt on gujarati culture society Ppt on delhi metro train Ppt on working of human eye and defects of vision and their correction Ppt on review of literature in research