Chap. 5 Field-effect transistors (FET)

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


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.


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


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.

CMOS


If VGS > VT (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to vGS - Vt.


Symbols and conventions

•n-channel

•several slightly different symbols

(source is often connected to the substrate which is usually grounded)

+VDS

-+ VGS

-

drain

source

gate


Symbols and conventions

•p-channel

•several slightly different symbols

(source is often connected to VDD)

+VDS

-+ VGS

-

drain

source

gate


An n-channel MOSFET with VGS and VDS applied and with the normal directions of current flow indicated.

Output characteristics (n-channel)

(linear)

+VDS

-


Input characteristics (n-channel)

+VDS

-

ID = K(VGS-VT)2


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)


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 = IB


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


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

V1VGS -+

IDIG = 0

n channel


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

VGS -+

ID

IG = 0

n channel


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

VGS -+

ID

IG = 0

n channel


5.4 MOSFETS at DC

DC problemFind ID, and VGS, and VDS

VGS = 5VVGS > VT, so device is on

Assume device is in saturationID = K(VGS-VT)2

ID = (0.05 mA/V2)(5-1)2

ID = 0.8 mA

VDS = VDD - ID RD

VDS = 10 - (0.8)6VDS = 5.2V

VT = 1VK = 0.05 mA/V2

(typical values)

+ VGS

-

+VDS

-

ID

ID

IG = 0


General DC problem

+ VGS

-

+VDS

-

ID

IG = 0

DC problemFind ID, and VGS

Assume device is in saturationID = K(VGS-VT)2

ID = K(5 - ID RS -VT)2

18ID 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 series

Find V

If devices are identical

ID

IG = 0

n channel

IG = 0 V

VDD = 5V

Ground

device

device

V =VDD/2 = 2.5V


5.5 MOSFET as an amplifier

g

d

s

n channel

ac model

d

s

g

vgs

+

-Ro = 1/slope of the output characteristics

. d

s

g

vgs

+

-

SPICE model

Ro


Transconductance

Useful relation: gm = 2 K ID

Transconductance = gm = dID/dVGS

= d [K(VGS-VT)2]/dVGS

= 2 K(VGS-VT)


Prob. 5.86

(a) Find the resistance of an enhancement load

Rin

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

ac model

+

V

-

I gd

s


Prob. 5.86(a) Find the resistance of an enhancement load

Often,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 9•Decrease W by a factor of 9•Increase L by a factor of 9


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


ac analysis of MOSFET amplifiers

ID

ID

ac circuit

Rin =

Rout = Rload || Ro

d

s

g

vgs

+

-Rin Rout


ac analysis of MOSFET amplifiers

Ai = iout / iin =

Av = vout/vin = -gmvgs(Ro || Rload) / vgs

= -gm(Ro || Rload)

d

s

g

vgs

+

-

vout

+

-

iin = 0 -gmvgs


Transistor loads: depletion load

I

V+

-

Depletion load

VGS = 0

R = Ro || resistance of current source with 0 magnitude

= Ro ||

= Ro

Ro = |VA| / I

Resistance iscurrent dependent


CMOS amp

Iref

I

•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| = 50Vp-channel pCox = 20A/V2

n-channel nCox = 40A/V2

WQ1 = Wp = 100mWQ4 = 50mL = 10m


CMOS amp: power

Iref I



WQ1 = Wp = 100mWQ4 = 50mL = 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


CMOS amp: DC analysis



WQ1 = Wp = 100mWQ4 = 50mL = 10m

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

Iref

+Vout

-


CMOS amp: ac analysis



WQ1 = Wp = 100mWQ4 = 50mL = 10m

Iref

+Vout

-

Find Av

Av = -gm1(Ro1 || Ro2)Ro1= Ro2 = 50/ 225A = 222Kgm = 2 K ID

= (2) [(1/2)(40)(100/10)] 1/2 225A = 300A/V

Av = -gm1(Ro1 || Ro2) = -300(.222/2) -33


CMOS multistage amp: ac analysis

DC circuit

ac circuit(neglects resistances of current sources)


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


Multistage CMOS amp: DC analysis

•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

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

Iref

•Given KP = 80A/V2, KN = 100A/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 KN’s and KP’s are equal

Chap. 5 Field-effect transistors (FET)

Documents

Transcript of Chap. 5 Field-effect transistors (FET)