# MOSFET DC BIASING

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```					Analogue Electronics I
EAB 2014

Jan 2009

Lecture 8 - MOSFETs
DC Biasing

1
Topics To Cover

• DC Biasing For Depletion-Type and
Enhancement-Type MOSFETs

2
MOSFETs Biasing… why?
We are interested in finding the quiescent
point (Q-point) of the response in the
circuit (current and/or voltage)
Steps to obtain the Q-point
1. Plot the transfer characteristics curve
(ID vs VGS)
2. Draw the DC load line (network characteristics)
superimposed on the curve
3. Obtain the Q-point by finding the intersection
point of the two plots.

3
Depletion type MOSFETs
Use Shockley’s Equation for simpler
calculation @ shorthand method
VGS versus ID                                      2
⎛    V GS   ⎞
VGS           ID      I D = I DSS   ⎜1 −
⎜           ⎟
⎟
0           IDSS                   ⎝    VP     ⎠
0.3 VP      IDSS / 2
0.5 VP      IDSS / 4
VP           0

4
Example 1 (voltage-divider)
For the circuit given,
determine:
a)   IDQ and VGSQ
b)   VDS

5
Example 1 …
ID(mA)
11
Step 1: Plot the Transfer Characteristics Curve.                 10
9
8
Using Shockley’s Equation:                                       7
IDSS   6
When VGS = 0, ID = IDSS                                          5

ID = 0, VGS = VP                                            4
3
When VGS = VP/2 = -1.5 V,                                        2
ID = IDSS/4 = 1.5 mA         VGS(V)                         1

-3   -2   -1               1
VP

6
Example 1 …
ID(mA)
11
Consider when VGS = +1 V,                                                    10
9
2
⎛    V GS   ⎞                                          8
I   D   = I   DSS
⎜1 −
⎜           ⎟
⎟                                          7
⎝    VP     ⎠                                   IDSS   6
2
⎛      1 ⎞
= (6 mA ) ⎜ 1 −
5
⎟
⎝     − 3 ⎠                                              4
3
= 10 . 67 mA
2

VGS(V)                         1

-3   -2   -1               1
VP

7
Example 1 …
Step 2: Draw the DC load line.
Find equation for VGS using
voltage divider relationship:   Characteristics curve
⎛     10 M     ⎞
VG = ⎜              ⎟ ×18V = 1.5V
⎝ 10 M + 110 M ⎠                     DC Load Line
VGS = VG − I D RS = 1.5V − I D (750 Ω )
Intersection
Set I D = 0 mA                                      (Q-point)

VGS = VG = 1.5V
Set VGS = 0V
VG
ID =    = 2 mA
RS

8
Example 1 …
Step 3: Obtain the Q-point values
a) From the plot:
IDQ = 3.1 mA and VGSQ = – 0.8 V

b) VDS = VDD – ID (RD + RS)
= 18 – (3.1m)(1.8k + 750)
= 10.1 V

9
Example 2 (self-bias)
For the network given,
determine:
a)   IDQ and VGSQ
b)   VD

10
Example 2 (self-bias)
Simplify the self-bias network:       20 V
20 V

6.2kΩ                         6.2kΩ
Vo

IDSS=8mA                     IDSS=8mA
Vi                                                        V P= -8V
VP= -8V

1M Ω               2.4kΩ                         2.4kΩ

11
Example 2 (self-bias)
Step 1: Plot the Transfer Characteristics Curve.
Using Shockley’s Equation:                                  ID(mA)
When VGS = 0, ID = IDSS                                              11
10
ID = 0, VGS = VP
9
IDSS
When VGS = VP/2 = -4 V,                                              8
ID = IDSS/4 = 2 mA                                             7
6
Consider when VGS = +1 V,                                            5
2
⎛    V   ⎞                                          4
I D = I DSS ⎜ 1 − GS ⎟
⎜
⎝     VP ⎟
⎠                                          3

2                                       2
⎛     1 ⎞
= (6 m ) ⎜ 1 −   ⎟                                           1
⎝     −8⎠       VGS(V)

= 10 . 13 mA                      -8 -7 -6 -5 -4 -3 -2 -1
VP
1   2

12
Example 2 (self-bias)
Step 2: Draw the DC load line.
From the input loop:
VGS = − I D RS = − I D (2.4 kΩ )   Characteristics curve

Set I D = 0 mA
VGS = 0V
(Q-point)

Set       VGS = −6V
VG
ID = −      = 2.5 mA
RS

13
Example 2 (self-bias)
Step 3: Obtain the Q-point values

a) From the plot:
IDQ = 1.7 mA and VGSQ = – 4.3 V

b) VD = VDD – IDRD
= 20 – (1.7m)(6.2k)
= 9.46 V

14
Enhancement-Type MOSFETs
Different equation is used:
I D = k (V GS − V T               )2
I D ( on )
k =
(V   GS ( on )   − VT   )2

VGS versus ID
VGS                              ID
VGS(on)                         ID(on)
VT                              0

15
Example 3 (feedback-biasing)

For the circuit given,
determine:
a)   IDQ and VDSQ

16
Example 3 …
Simplify the self-bias network:
12 V                         12 V

2kΩ                           2kΩ

Vo
10M Ω               I D(on)= 6mA                 I D(on)= 6mA

VGS(on)= 8V                  VGS(on)= 8V
Vi
VGS(th)= 3V                  VGS(th)= 3V

17
Example 3 …
Step 1: Plot the Transfer Characteristics Curve.
I D (on )
Find value of k:        k =
(V   GS (on ) − V T   )
2

6m
=
(8 − 3 )2
= 0 . 24 × 10 − 3 A / V        2

Select a few points:
For VGS = 6 V,                     For VGS = 10 V,
ID = k(VGS – VGS(T))2              ID = (0.24m) x (10 – 3)2
= (0.24m) x (6 – 3)2                 = 11.76 mA
= 2.61 mA

18
Example 3 …

19
Example 3 …
Step 2: Draw the DC load line.

Characteristics curve
From output circuit:
VDS = VGS
= VDD – IDRD
Intersection
ID = 0                                                     (Q-point)
VGS = VDD = 12 V

VGS = 0
ID = VDD / RD = 6 mA

20
Example 3 …
Step 3: Obtain the Q-point values

a) From the plot:
IDQ = 2.75 mA and VDSQ = 6.4 V

21
Example 4 (voltage-divider)
For the circuit given,
determine:
a)   IDQ
b)   VGSQ
c)   VDSQ

22
Example 4 (voltage divider)
Step 1: Plot the Transfer Characteristics Curve.
I D (on )
Find value of k:        k =
(V GS (on ) − V GS (T ) )2
3m
=
(10   − 5 )2
= 0 . 12 × 10 − 3 A / V     2
Select a few points:
For VGS = 15 V,                 For VGS = 20 V,
ID = k(VGS – VGS(T))2           ID = (0.12m) x (20 – 5)2
= (0.12m) x (15 – 5)2              = 27 mA
= 12 mA

23
Example 4 (voltage divider)
ID(mA)

40

30

20

10
ID(on)

V GS(V)
5        10       15   20   25   30
VGS(th)   VGS(on)

24
Example 4 (voltage divider)
Step 2: Draw the DC load line.
⎛ R2      ⎞
VG   =⎜⎜R +R     ⎟ V DD
⎟
⎝ 1   2   ⎠
= 18 V
From the input circuit:            At ID = 0,
VGS = VG – IDRS                    VGS = 18 V
= 18 – ID(0.82k)
At VGS = 0,
ID = 21.95 mA

25
Example 4 (voltage divider)

Characteristics curve

Intersection
(Q-point)

26
Example 4 (voltage divider)
Step 3: Obtain the Q-point values
a) From the plot:
IDQ = 6.7 mA and VGSQ = 12.5 V
b) From the output circuit:
VDS = VDD – ID(RS + RD)
= 40 – (6.7m)(0.82k + 3k)
= 14.4 V

27

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Description: MOSFET DC Biasing for Depletion and Enhancement. Cover Voltage Divider, Self-Bias, and Feedback Bias.