FET Summary Lesson (1)

7/25/2019 FET Summary Lesson (1)

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FET Lesson:

BJT is a current-controlled deviceFET is a voltage-controlled device

BJT: bipolar deviceFET: unipolar device

FET: high input impedanceBJT: higher sensitivity to changes in the applied signal

FET: more temperature stable than BJTFET: usually smaller in construction than BJT

ID = ISIG = 0 (gate current)

VDS > VP, JFET is a current-source device

IDSS means Drain-to-Source current with a Short-circuit connection from Gate to Source.- maximum drain current for JFET @ VGS = 0 V & VDS > |VP|

VGS(off) = -VP

Ohmic region: voltage-controlled resistance region- resistance is controlled by the applied gate-to-source voltage.

2

1

P

GS

OD

V

V

rr

ro = resistance with VGS = 0Vrd = resistance at a particular level of VGS

N-channel JFET: gate is p-type material, drain-to-source is n-typeP-channel JFET: gate is n-type material; drain-to-source is p-type

ID = IDSS( 1 VGS/VP)2

2 types of FET1. JFET (Junction FET)2. MOSFET (Metal Oxide Semiconductor FET)

- Depletion type- Enhancement type (source & drain are totally separated by a diff channel)


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The greater the applied reverse bias, the wider the depletion region.

IG (gate current) is zero..

Seen in datasheet: VGS(off) = VP

As voltage VDS is increased from 0 to few volts, the current will increase.

The relative straightness of the plot reveals that for the region of low values of VDS, theresistance is essentially constant.

As VDS increases & approaches a level referred to as VP (pinch-off voltage), a depletion regionwill widen causing a reduction in the channel width.

As VDS is increased beyond VP, the region of close encounter between the two depletion regionswill increase in length along the channel, but the level of ID remains essentially the same.

Therefore, once VDS > VP, JFET has the characteristics of a current source.

p

n

p

+

+

VGS

VDD

VDS = VP

D

G

S

Saturation region VGS = 0 V

ID

IDSS

VP

VDS

n-channel resistance

Increasing resistance due to narrowing channel

Ohmic region

VGS =2 V

VGS = VP


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IDSS- drain-to-source current with a short circuit connection from gate to source- maximum drain current for a JFET & is defined by the condition,

VGS = 0 V & VDS > VP

Voltage-controlled Resistor

Ohmic region:

- variable resistor as controlled by applied gate-to-source voltage

2

P

GS

Od

V

V1

rr

rO resistance with VGS = 0 Vrd resistance of a particular level of VGS

For an n-channel JFET with rO equal to 10k (VGS=0 V, VP= 6 V), then

rd = 40k @ VGS = 3 V

JFET Symbol

D

G

S

N-channel

D

G

S

P-channel

ID = IDSS VDS

+

LOAD


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Transfer Characteristics

For BJT,

IC

= f(IB)

IC = IB is the constant while IB is the controlled variable

For FET,

2

P

GSDSSD

V

V1II

- IDSS & VP are the constants- Raise to 2nd power means non-linear relation between ID & VGS.

Fixed Bias

Since IG 0 AVRG = IGRG

= (0 A)(RG)= 0 Volt

ID

IDSS

VGS VDS

ID

VGS = 0 V

VGS =1 V

VGS =2 V


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The circuit becomes

Using KVL,

VGG VGS = 0

VS = 0

VDS = VD VS

VD = VDS + VS

VGS = VG VS

VG = VGS + VS

Power Dissipation

DDSD IVP

Derating factor:

The dissipation rating decreases by constant value for every increase in temperature of

certain value above 25C. See datasheet for each FET.

VGS =VGG

VDS = VDD IDRD

VD = VDS

VG = VGS


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SMALL SIGNAL ANALYSIS

For BJT, amplification factor is beta ()

For FET, transconductance factor is gm

ID = gmVGS

GS

Dm

V

Ig

P

GS

P

DSSm

V

V1

V

2Ig

When VGS = 0 (max transconductance curve)

P

DSSmo

V

2Ig

P

GSmom

V

V1gg

Example:

For JFET having IDSS of 10 mA and VP of 5V,

a. Find gmob. Find gm @ VGS = 0.5V

Solution:

a.

4mS5

10mA2

V

2Ig

P

DSSmo

b. @VGS = 0.5V

5V

0.5V14mS

V

V1gg

P

GSmom

= ___________


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2

P

GSDSSD

V

V1II

P

GS

DSS

D

VV1

II

P

GSmom

VV1gg

DSS

Dmom

I

Igg

FET Input Impedance

Zi =

Typical values:

JFET: 1000 MMOSFET: 1015

FET Output Impedance

ZO = rD

OS

dy

r1

D

DSd

I

Vr

VGS is constant

IDSS

VDS

ID

VGS = 0 V

VGS =1 V

VGS =2 V

VDS

ID


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FET Equivalent Circuit

Note: Vgs code used @AC levelVGS code used @DC level

JFET FIXED BIASED

Equivalent circuit:

rd

gm

Vgs

G

S

D

VgsRG RDZ

i ZO

Vin Vout

Input Impedance

Gi RZ

dDO rRZ


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dDgsmO rRVgV

igs VV

dDimO rRVgV

i

dDim

i

OV

V

rRVg

V

VA

dDmV rRgA

Example:

Find gm, rd, Zi, ZO, AV.

a.

mmhos.V

V

V

mA

V

V

V

Ig

P

GS

P

DSSm 8751

8

21

8

1021

2

b. kmhosy

rOS

d 2540

11

c. MRZ Gi 1

d. 8521851252 .kkrRZ dDO

e. 472385185218518751 ..m.ZgrRgA OmdDmV


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JFET SELF-BIAS

A. Bypassed

RG

RS

RD

C1

C3

C2

Vout

+VDD

Vin

Equivalent circuit:

rd

gm

Vgs

G

S

D

VgsRG RD

Zi ZO

Vin Vout

Note: same with Fixed Bias equivalent circuit.

Input Impedance

Gi RZ

dDO rRZ

dDgsmO rRVgV

igs VV

dDimO rRVgV

i

dDim

i

OV

V

rRVg

V

VA

dDmV rRgA


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2

P

GSDSSDq

V

V1II

SDGS RIV

2

P

SDDSSDq

V

RI-1II

2

222

1

P

SDq

P

SDq

DSS

Dq

V

RI

V

RI

I

I

01122

2

2

Dq

DSSP

SDq

P

S I

IV

RI

V

R

In quadratic equation,

2

2

P

S

V

Ra

DSSP

S

IV

Rb

12 1c

roots

a

acbbxIDq

2

42

Possible root values:

Notes: q-pt must be within the transfer curve.


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Example:

Answers:Idq = 12.783mA, 2.816mA

Since IDSS is only 10 mA, IDq = 2.816 mA.

VGSq = -2.816 V

gm = 0.00176887 mhos

rd = 50 k

Zi = 1 M

ZO = 1155.91766

AV = 2.04467342


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B. Unbypassed

Input Impedance

Gi RZ

d

SDSm

DO

r

RRRg

RZ

1

d

SDSm

DmV

rRRRg

RgA

1


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Example:

Answers:Idq = 11.4187mA, 3.1527mA

IDq = 3.1527mA

VGSq = -3.1527 V

gm = 0.00221455 mhos

rd = 50 k

Zi = 1 M

ZO = 999.833284

AV = 2.2141811


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JFET Voltage Divider Bias

R2 RS

RD

C1

C2 Vout

+VDD

Vin

R1

C3

Zi ZO

AC equivalent circuit:

dDgsmO rRVgV

igs VV

dDimO rRVgV

i

dDim

i

OV

V

rRVg

V

VA

dDmV rRgA

Zi = R1 || R2

ZO = rd || RD


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JFET Source Follower (Common-Drain) Configuration

AC Equivalent Circuit:

Gi RZ

mSdO1/gRrZ

Ogsi VVV

Oigs VVV

SdOimO RrVVgV SdOmSdimO RrVgRrVgV

SdimSdmO RrVgRrg1V

Sdm

Sdm

i

OV

Rrg1

Rrg

V

VA


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MOSFET

Channel formation in the N-Channel enhancement-type MOSFET

VDG = VDS - VGS

As VGS is increased beyond the threshold level, the density of free carriers in the inducedchannel will increase, resulting in an increased level of drain current.

However, if we hold VGS to a constant value while increasing the VDS, the drain current will reacha saturation level.

VDS sat = VGS VT


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For values of VGS less than the threshold level the drain current of an enhancement-typeMOSFET is 0 mA.

For levels of VGS > VT,

ID = k(VGS VT)2

2

TonGS

onD

VV

Ik


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P-Channel Enhancement-type MOSFET

Symbols:


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Sample Data Sheet

Using the datasheet, determine k.

22

310

3

VV

mA

VV

Ik

TonGS

onD

k = 0.061 x 10-3A/V2


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MOSFET also used in analog circuits.

MOSFET can be used as precision resistors w/c can have higher controlled resistance than BJT.

Main advantage of BJT over MOSFET is the ability to handle larger current in a smaller space.

In logic circuits, BJT has an advantage over MOSFET. BJTs are able to drive more gates (largerfanout) because they can output more current than MOSFET. Many chips use MOSFET asinput while BiCMOS (BJT-FET mixed) as outputs.

In high speed switching, BJT doesnt have larger capacitance from the gate which whenmultiplied by the resistance of the channel gives the intrinsic time constant of the process. Athigher frequency, MOSFET operates slower.

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FET Summary Lesson (1)