transistor characteristics emt 251. outline introduction mos capacitor nmos i-v characteristics...

Transistor Transistor CharacteristicsCharacteristics

Transistor Transistor CharacteristicsCharacteristics

EMT 251EMT 251

Outline

• Introduction• MOS Capacitor• nMOS I-V Characteristics (ideal)• pMOS I-V Characteristics (ideal)

Introduction• So far, we have treated transistors as ideal

switches• An ON transistor passes a finite amount of

current– Depends on terminal voltages– Derive current-voltage (I-V) relationships

• Transistor gate, source, drain all have capacitance– I = C (V/t) -> t = (C/I) V– Capacitance and current determine speed

MOS Capacitor

polysilicon gate

(a)

silicon dioxide insulator

p-type body+-

Vg < 0

(b)

+-

0 < Vg < Vt

depletion region

(c)

+-

Vg > Vt

depletion regioninversion region

• Gate and body form MOS capacitor• Operating modes

– Accumulation– Depletion– Inversion

-V

V+

V++

Terminal Voltages• Mode of operation depends on Vg, Vd, Vs

– Vgs = Vg – Vs

– Vgd = Vg – Vd

– Vds = Vd – Vs = Vgs - Vgd

• Three regions of operation– Cutoff– Linear– Saturation

nMOS Cutoff

• No channel• Vgs < Vt

• Ids = 0 +-

Vgs = 0

n+ n+

+-

Vgd

p-type body

b

g

s d

nMOS Linear

• Channel forms– Vds = Vgs-Vgd– If Vds=0 (i.e. Vgs=Vgd)

• No electrical field tending to push current fr. d to s.

• Small positive potential Vds• Current flows from d to s

– e- from s to d

• Ids increases with Vds

• Similar to linear resistor

+-

Vgs > Vt

n+ n+

+-

Vgd = Vgs

+-

Vgs > Vt

n+ n+

+-

Vgs > Vgd > Vt

Vds = 0

0 < Vds < Vgs-Vt

p-type body

p-type body

b

g

s d

b

g

s dIds

nMOS Saturation

• Channel pinches off• Ids independent of

Vds• We say current

saturates• Similar to current

source

+-

Vgs > Vt

n+ n+

+-

Vgd < Vt

Vds > Vgs-Vt

p-type body

b

g

s d Ids

Nmos region of operation

“CUTOFF” region: VG < VT

1. Cutoff region: VGS < VT, any value of VDS

ID = 0

2. Linear (or Resistive, or Triode) region: VGS > VT, VDS < (VGS – VT)

3. Saturation region: VGS > VT, VDS > (VGS – VT)

L

WC

VVI

oxn

TGSDSAT

where

2

2

L

WC

VV

VVI

oxn

DSDS

TGSD

where

2

Example

0 1 2 3 4 50

0.5

1

1.5

2

2.5

Vds

I ds (m

A)

Vgs = 5

Vgs = 4

Vgs = 3

Vgs = 2

Vgs = 1

• We will be using a 0.6 m process for your project– From AMI Semiconductor

– tox = 100 Å

– = 350 cm2/V*s

– Vt = 0.7 V

• Plot Ids vs. Vds

– Vgs = 0, 1, 2, 3, 4, 5

– Use W/L = 4/2

14

28

3.9 8.85 10350 120 /

100 10ox

W W WC A V

L L L

pMOS I-V• All doping and voltages are inverted for pMOS• Mobility mp is determined by holes

– Typically 2-3x lower than that of electrons mn

– 120 cm2/V*s in AMI 0.6 mm process• Thus pMOS must be wider to provide same

current– In this class, assume mn / mp = 2

– *** plot I-V

Transistor Operation

• Current depends on region of transistor behavior

• For what Vin and Vout are nMOS and pMOS in– Cutoff?– Linear?– Saturation?

Schematic of CMOS Schematic of CMOS inverterinverter

nMOS Operation

Cutoff Linear Saturated

Vgsn < Vtn

Vin < Vtn

Vgsn > Vtn

Vin > Vtn

Vdsn < Vgsn – Vtn

Vout < Vin - Vtn

Vgsn > Vtn

Vin > Vtn

Vdsn > Vgsn – Vtn

Vout > Vin - Vtn

Idsn

Idsp Vout

VDD

Vin

Vgsn = Vin

Vdsn = Vout

pMOS OperationCutoff Linear Saturated

Vgsp > Vtp

Vin > VDD + Vtp

Vgsp < Vtp

Vin < VDD + Vtp

Vdsp > Vgsp – Vtp

Vout > Vin - Vtp

Vgsp < Vtp

Vin < VDD + Vtp

Vdsp < Vgsp – Vtp

Vout < Vin - Vtp

Idsn

Idsp Vout

VDD

Vin

Vgsp = Vin - VDD

Vdsp = Vout - VDD

Vtp < 0

Calculation of VIL

When the input voltage is Vin = VIL, the slope of the VTC is equal to [-1](dVout/dVin). At this point B, nMOS transistor operates in saturation while the pMOS transistor operates in the linear region. Using KCL at output node:

[Id,n(saturation)= Id,p(linear)]

At point D, input voltage is equal to VIH, the nMOS transistor operates in the linear region and pMOS transistor operates in saturation. By applying KCL at the output node:

[ Id,n(linear) = Id,p(saturation)]

Calculation of VIH

Threshold voltage is defined as Vth=Vin=Vout. For Vin=Vout, both transistor operate in saturation region, refer point C. By applying KCL at output node:

[ Idn(saturation)=Idp(saturation) ]

Calculation of Vth

Consider a CMOS inverter circuit with the following parameters, calculate the (W/L) ratios of the nMOS and pMOS transistors if the switching voltage is Vth = 1.5 V.

VDD = 3.3 V

VT,n = 0.6 V

VT,p = -0.7V

µncox = 80 µA/V2

µpcox = 30 µA/V2

Exercise 1

Exercise 2

Consider a CMOS inverter circuit with the following parameters, calculate the Switching voltage (Vth) of the circuit.

VDD = 3.3 V

VT,n = 0.6 V

VT,p = -0.7V

µncox = 80 µA/V2

µpcox = 30 µA/V2

(W/L)n = 8(W/L)p = 12

Voltage transfer characteristic

Inverter with n-Type MOSFET load

Depletion-Load nMOS Inverter

When the input voltage Vin is smaller than the driver threshold voltage VTO, the driver transistor is turned off and thus not conduct any drain current. ID,load = 0,

VOH = VDD ID,load = 0

Calculation of VOH

Assume Vin = VOH = VDD. Driver transistor operate in linear region and load transistor operate in saturation. By using KCL.

Id,driver(linear) = Id,load(saturation)

Calculation of VOL

When Vin = VIL the slope of the VTC is equal to (-1),(dVout/dVin). Driver transistor operate in saturation region and load transistor operate in linear region. KCL

Id,driver(saturation) = Id,load(linear)

Calculation of VIL

Calculation of VIH

Driver transistor operate in linear region and load transistor operate in saturation region.

Id,driver(linear) = Id,load(saturation)

Region

nMOS pMOS

A Cutoff Linear

B Saturation Linear

C Saturation Saturation

D Linear Saturation

E Linear Cutoff

Operating Regions (Summary)

Vout

Vin0.5 1 1.5 2 2 .50.

51

1.5

22.

5

NMOS resPMOS off

NMOS satPMOS sat

NMOS offPMOS res

NMOS satPMOS res

NMOS resPMOS sat

Exercise 3

VDD = 3.3 VThreshold voltage nMOS (VT,n) = 0.6 VThreshold voltage pMOS (VT,p) = -0.7VKn = 200 µA/V2

Kp = 80 µA/V2

Based on the inverter circuit and parameter in figure above, what is the value of:

i) Switching Voltage, VTH

ii) Input Voltage High, VIH

THE END