L20 DC MOSFET Circuits V2 - nanoHUB.org · Outline 2 1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples

1

ECE 255:

DC MOSFET Circuits

(Sedra and Smith, 7th Ed., Sec. 5.3)

Mark Lundstrom School of ECE

Purdue University West Lafayette, IN USA

Lundstrom: Fall 2019

ECE 255: Fall 2019 Purdue University

Outline

2

1)  BJT Analysis is Easy 2)  DC N-MOSFET Circuit Analysis 3)  DC N-MOSFET Circuit Design 4)  P-MOSFET Design and Analysis 5)  Examples


BJT circuit analysis

3 Lundstrom: Fall 2019

+V

CE= ? V

−

β = 100

+5 V

−5 V

RE = 4.3 kΩ

RC = 1kΩIC = ? mA Find IC and VCE

VBE on( ) = 0.7 V

Why is it so easy?


+V

CE= 4.71V

−

β = 100

+5 V

−5 V

RE = 4.3 kΩ

RC = 1kΩIC = 0.99 mA

VBE on( ) = 0.7 V

+V

BE

− Because we can guess this voltage

IC

VBE

IC = ISeVBE VT

VBE≈ 0.7 V

Why is MOSFET circuit analysis harder?


Because we cannot guess this voltage

ID

VGSV

tn

+VDS = ?−

+5 V

−5 V

RS = 4.3 kΩ

RD = 1kΩID = ?

+VGS−

ID =

kn

2VGS −Vtn( )2

Vtn > 0 V

Outline

6



IV Summary (saturation region)


VDS

IC

act VGS −Vtn

VDSsat

ID =

kn

2VGS −Vtn( )2

IG = 0 kn =

WL

′kn =WLµnCox

Vtn > 0 V( )

VDSsat =VGS −Vtn

VDS >VDSsat

VGS >Vtn

ID = IS

MOSFET circuit analysis

8

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ?

ID =

′kn

2WL

VGS −Vtn( )2

Transistor model:

′kn

2WL= 1mA/V2

Vtn = 1.0 V


ID = 1 VGS −1.0( )2 (device)

VGS = ? (circuit)

MOSFET circuit analysis (ii)

9

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ? ID = 1 VGS −1.0( )2

VGS = ?

VGS =VG −VS

VS = −5+ ID RS

VGS = 5− ID RS

ID = 5− ID RS −1.0( )2


VG

VD

VS+IDRS−

MOSFET circuit analysis (iii)

10

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ? ID = 5− ID RS −1.0( )2

ID2 − 2.06ID +1= 0

ID = 1.28 / 0.76


MOSFET circuit analysis (iv)

11

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ? ID = 1.28 mA Does not work.


Vtn = 1.0 V

MOSFET circuit analysis (iv)

12

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ? ID = 0.76 mA

VD = 5− ID RD = 2.3 V

VGS = −VS = 1.96 V >Vtn

VDS =VD −VS = 4.26 V

VDSsat =VGS −Vtn = 0.96 V

✓

✓

VS = −5+ ID RS = −1.96 V

VDS >VDSsat


Vtn = 1.0 V

MOSFET circuit analysis (summary)

13

+VDS = ?−

+5 V

−5 V

RS = 4 kΩ

RD = 3.5 kΩID = ?

Vtn = 1.0 V

ID =

′kn

2WL

VGS −Vtn( )2

1) Write the device equation:

VGS =

2) Write the circuit equation:

3) Solve the equations

4) Check solution

Outline

14



MOSFET circuit design

15

+VDS = 3.0 V−

+5 V

−5 V

RS = ? kΩ

RD = ? kΩID = 0.5 mA

ID =

′kn

2WL

VGS −Vtn( )2

Transistor model:

′kn

2WL= 1mA/V2

Vtn = 1.0 V

ID = 1 VGS −1.0( )2= 0.5

VGS = 1.71

Now design the circuit

MOSFET circuit design (ii)

16

+VDS = 3.0 V−

+5 V

−5 V

RS = ? kΩ

RD = ? kΩID = 0.5 mA VGS = 1.71

VGS =VG −VS = 1.71

VS = −1.71

RS = 6.6 kΩ

VD =VS +VDS

RD = 7.4 kΩ

✓

✓


VG

VD = 1.29

MOSFET circuit design (summary)

17

+VDS = y.yymA−

+5 V

−5 V

RS = ? kΩ

RD = ? kΩID = x.x mA

ID =

′kn

2WL

VGS −Vtn( )2

1) Write the device equation:

2) Solve for VGS:

3) Design the circuit to give that VGS

4) Design the circuit to give the desired VDS

Outline

18



IV Summary (NMOS saturation region)


VDS

IC

act VGS −Vtn

VDSsat

ID =

kn

2VGS −Vtn( )2

kn =

WL

′kn =WLµnCox

Vtn > 0 V( )

VDSsat =VGS −Vtn

VDS >VDSsat

VGS >Vtn

IV Summary (PMOS saturation region)


VSD

IC

act VSG − Vtp

VSDsat

ID =

kp

2VSG − Vtp( )2

kp =

WL

′kp =WLµ pCox

Vtp < 0 V( )

VSDsat =VSG − Vtp

VSD >VSDsat

VSG > Vtp

P-MOSFET circuit design

21

+VSD = 3.0 V−

+5 V

−5 V

RD = ? kΩ

RS = ? kΩ

ID = 0.5 mA

ID =

′kp

2WL

VSG − Vtp( )2

Transistor model:

′kp

2WL= 1mA/V2

Vtp = −1.0 V


ID = 1 VSG −1.0( )2= 0.5

VSG = 1.71

P-MOSFET circuit design (ii)

22

+VSD = 3.0 V−

+5 V

−5 V

RD = ? kΩ

RS = ? kΩ

ID = 0.5 mA

VSG = 1.71

VS = 1.71 RS = 6.6 kΩ

VD =VS −VSD

VD = −1.29

RD = 7.4 kΩ

✓

✓


P-MOSFET circuit analysis

23

+VSD = ?−

+5 V

−5 V

RD = 3.5 kΩ

RS = 4 kΩ

ID = ?

ID =

′kp

2WL

VSG − Vtp( )2

Transistor model:

′kp

2WL= 1mA/V2

Vtp = −1.0 V

ID = 1 VSG −1.0( )2

VSG = 5− ID RS


P-MOSFET circuit analysis

24

ID = 1 VSG −1.0( )2

VSG = 5− ID RS

ID = 5− ID RS −1.0( )2

ID2 − 2.06ID +1= 0

ID = 1.28 or 0.76

ID = 0.76

ID = 5−VSG( ) RS = 5−VSG( ) 4

4VSG2 − 7VSG −1= 0

VSG = 1.88 or − 0.13

VSG = 1.88

ID = 0.77

Option1: Option2: 1

2

Outline

25



Example 1: Analysis

26

VDD = +5 V

RD = 4.3kΩ

1) Operating region?

+VGS

−


ID = 0.1 VGS −1( )2

Example 1: Analysis

27

VDD = +5 V

RD = 4.3kΩ


VGS =VD

VDS =VD

VDS >VGS −Vtn

+VGS

−Saturation


ID = 0.1 VGS −1( )2

Example 1: Analysis

28

VDD = +5 V

RD = 4.3kΩ

ID = 0.1 VGS −1( )2

ID = 0.1 VGS −1( )2

+VGS

−


VD

Example 1: Analysis

29

VDD = +5 V

RD = 4.3kΩ

ID = 0.1 VGS −1( )2

ID = 0.1 VGS −1( )2

VGS = 5− ID RD = 5− 4.3ID

ID2 − 2.86ID +1.35= 0

+VGS

−

ID = 2.26 or 0.60

ID = 0.60


VD

Example 2: Design

30

VDD = +5 V

RD = ?kΩ

ID =

′kn

2WL

VGS −Vtn( )2

ID = 0.1 VGS −1( )2

Design for: ID = 0.9 mA


VD

Example 2: Design

31

VDD = +5 V

RD = ?kΩ

ID =

′kn

2WL

VGS −Vtn( )2

ID = 0.1 VGS −1( )2

ID = 0.1 VGS −1( )2= 0.9


VGS = 4.0 V

VGS =VD = 4.0 V

RD = 5− 4

0.9= 1.11kΩ


VD

Example 3: Design (i)

32

VDD = +5 V

RS = ? kΩ

ID = 0.1 VSG −1( )2




VDS >VGS −Vtn ?VS

Example 3: Design (i)

33

VDD = +5 V

RS = ? kΩ

ID = 0.1 VSG −1( )2



VSD >VSG − Vtp ?

saturation


VDS >VGS −Vtn ?VS

Example 3: Design (ii)

34

VDD = +5 V

RS = ? kΩ

ID =

′kp

2WL

VSG − Vtp( )2

ID = 0.1 VSG −1( )2



Example 3: Design (ii)

35

VDD = +5 V

RS = ? kΩ

ID =

′kp

2WL

VSG − Vtp( )2

ID = 0.1 VSG −1( )2

ID = 0.1 VSG −1( )2


0.9 = 0.1 VSG −1( )2

VSG = 4.0 V

VSG =VS = 4.0 V

RS =

5− 4.00.9

= 1.11kΩ


Example 4

36

+5 V

RD = 2 kΩ

ID = 0.1 VSG −1( )2

I = 1.0 mA

−5 V

VS = ?

VD = ?


Example 4

37

+5 V

RD = 2 kΩ

ID = 0.1 VSG −1( )2

I = 1.0 mA

−5 V

VS = ?

VD = −5+ ID RD

= −5+1× 2= −3 V

VD = ?

1= 0.1 VSG −1( )2

VSG =VS = 4.16 V

VS =VSG


Example 5

38

+5 V

ID = 0.1 VGSn −1( )2


VD = ?

ID = 0.1 VSGp −1( )2

+VGSp

−

+VSGn

− saturation


Example 5 (ii)

39

+5 V

ID = 0.1 VGSn −1( )2

VDSn =VD


VD = ?

VGSn =VD

VDSn > VGSn −Vtn( ) ID = 0.1 VSGp −1( )2

+VGSp

−

+VSGn

−

VSDp = 5−VD

VSGp = 5−VD

VSDp > VSGp − Vtp( )

saturation Lundstrom: Fall 2019

Example 5 (iii)

40

+5 V

ID = 0.1 VGSn −1( )2

VD = ?

ID = 0.1 VSGp −1( )2

+VGSn

−

+VSGp

− VGSn =VSGp Why?


Example 5 (iii)

41

+5 V

ID = 0.1 VGSn −1( )2

VD = ?

ID = 0.1 VSGp −1( )2

+VGSn

−

+VSGp

− VGSn =VSGp

VGSn +VSGp = 5

VD = 2.5

ID = 0.1 2.5−1( )2= 0.225 mA

Why?


Example 6: Analysis

42

VDD = 10 V

RD = 100 k

RG1

700 k

RG2

300 k

Transistor model:

kn = 25 µA/V2 Vtn = 1.0 V

ID =

kn

2VGS −Vtn( )2

ID = 0.025

2VGS −1( )2

mA VGS = 3

ID = 0.05 mA

VDS = 5 V >VGD -Vtn

Summary

43

1)  DC MOSFET analysis often involves solving a quadratic equation and throwing away the unphysical solution.

2)  DC MOSFET analysis is often more difficult than DC BJT analysis

3)  DC MOSFET design is usually easier than design.


DC MOSFET Circuits

Lundstrom: Fall 2019 44


Documents

L20 DC MOSFET Circuits V2 - nanoHUB.org · Outline 2 1) BJT Analysis is Easy 2) DC N-MOSFET Circuit Analysis 3) DC N-MOSFET Circuit Design 4) P-MOSFET Design and Analysis 5) Examples