MOSFET DC Circuits Analysis - OCW

bracebustlingElectronics - Devices

Oct 7, 2013 (3 years and 10 months ago)

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MOSFET DC Circuits Analysis
1.
Assume an operation region (usually the saturation region)
2.
Apply KVL at the gate source loop to find
V
GS
3.
Use
V
GS
from step 2 to calculate
I
D
4.
Apply KVL at the drain source loop and use
I
D
from step 3 to find
V
DS
5.
Check the validity of operation region assumptions by comparing
V
DS
to
V
DSat
6.
Change assumptions and analyze again if required.
NOTES :

An enhancement-mode device with
V
DS
=
V
GS
is always in saturation

If we have a source resistance, we need to solve the equations in steps 2
and 3 together to find ID and VGS.

If we include channel length modulation or we are in the triode region, we
will solve the equations in steps 3 and 4 together

If we include channel length modulation or we are in the triode region and
we have a source resistance, we will solve the equations in steps 2, 3, and 4
together
Bias Analysis: Example 1
Problem:Find the Q-pt (
I
D
, V
DS
)
Given:
V
TN
=1V,
K
n
=25μA/V2
Approach:Assume operation
region, find Q-point, check to see
if result is consistent with
operation region
Assumption:Transistor is
saturated,
I
G
=I
B
=0
Analysis: First, simplify circuit,
split
V
DD
into two equal-valued
sources and apply Thevenin
transformation to find
V
EQ
and
R
EQ
for gate-bias voltage
Bias Analysis: Example 1 (contd.)
S
R
D
I
GS
V
EQ
V
+
=
KVL at G-S loop,
2
2
K
n
IVV
GSTN
D
⎛⎞
⎜⎟
⎝⎠
=−
63
25103910
2
41
2
VV
GS
GS
⎛⎞⎛⎞
⎜⎟⎜⎟
⎜⎟⎜⎟
⎛⎞
⎝⎠⎝⎠
⎜⎟
⎝⎠

××
=
+−
021.705.0
2
=−+
GS
V
GS
V
V66.2,V71.2
+

=

GS
V
Since
V
GS
<V
TN
for
V
GS
= -2.71 V
and MOSFET will be cut-off,
we ignore it
V66.2
+
=
GS
V
and
I
D
= 34.4μA
DS
V
S
R
D
R
D
I
DD
V
+
+
=)(
V08.6
=

DS
V
VDS>VGS-VTN. Hence saturation
region assumption is correct.
Q-pt: (34.4μA, 6.08 V)
with
V
GS
= 2.66 V
2
2
KR
nS
VVVV
GSTN
EQGS
⎛⎞
⎜⎟
⎝⎠

=+−
KVL at D-S loop,
Bias Analysis: Example 2

Estimate value of
I
D
and use
it to find
V
GS
and
V
SB

Use
V
SB
to calculate
V
TN

Find
I
D

using above 2 steps

If
I
D

is not same as original
I
D
estimate, start again.
Find the Q-point for the shown
circuit with body effect using
2φF=0.6 V, VTO=1V, and γ=0.5V1/2:
D
I
S
R
D
I
EQ
V
GS
V
000,226

=

=
D
I
S
R
D
I
SB
V
000,22
=
=
)22(
FF
SB
V
TO
V
TN
V
φφγ
−++=
)6.06.0(5.01

+
+
=

SB
V
TN
V
2
2
6
1025
'


















×
=
TN
V
GS
V
D
I
Iterative solution can be found
by following steps:
KVL at G-S loop,
Bias Analysis: Example 2 (contd.)
The iteration sequence leads to
I
D
= 88.0 μA
V48.6000,4010)(
=

=
+

=
D
I
S
R
D
R
D
I
DD
V
DS
V
V
DS
>V
GS
-V
TN
. Hence saturation region assumption is correct.
Q-pt: (88.0μA, 6.48 V)
Bias Analysis: Example 3
Assumption:
I
G
=I
B
=0, transistor
is saturated (since
V
DS
=
V
GS
)
Analysis:
VVVIR
DDDD
DSGS
==−
2
2








−−=
TN
V
GS
V
D
R
n
K
DD
V
GS
V
2
1
2
4
10
4
106.2
3.3


























×
−=

GS
V
GS
V
V00.2,V769.0
+

=

GS
V
Since
V
GS
<V
TN
for
V
GS
=
-0.769
V and MOSFET will be cut-off,
it will be ignored.
V00.2
+
=
GS
V
and
I
D
= 130μA
V
DS
>V
GS
-V
TN
. Hence saturation
region assumption is correct.
Q-pt: (130μA, 2.00 V)
Find the Q-point for the shown circuit?
Bias Analysis: Example 4
( Biasing in Triode Region)
Assumption:
I
G
=I
B
=0,
transistor is saturated
Analysis:
V
GS
=V
DD
=
4 V
mA13.1
2
)14(
2
V
μA
2
250
=−=
D
I
VIRV
DDDD
DS
=
+
V19.2
=

DS
V
But
V
DS
<V
GS
-V
TN
. Hence, saturation
region assumption is incorrect Using
triode region equation,
DS
V
D
I
+
=

16004
DS
V
DS
V
DS
V)
2
14(
2
V
μA
250*16004−−=−
V3.2=

DS
V
and
I
D
=1.06 mA
V
DS
<V
GS
-V
TN
, transistor is in triode region
Q-pt:(1.06 mA, 2.3 V)
Find the Q-point for the shown circuit?
KVL at D-S loop,
Bias Analysis: Example 5
Assumption:
I
G
=
I
B
=0, transistor
is saturated (since
V
DS
=
V
GS
)
Analysis:
15V(220kΩ)0IV
SG
D
−−=
2
μA
50
15V(220kΩ)20
22
V
VV
SG
SG
⎛⎞
⎜⎟
⎝⎠

−−−=
V45.3,V369.0=

SG
V
Since
V
SG
= 0.369 V is less than
|VTP|= 2 V, ∴
V
SG
= 3.45 V
I
D
= 52.5μAand
V
SG
= 3.45 V
Hence saturation assumption is correct.
Q-pt: (52.5μA, 3.45 V)
TP
V
SG
V
SD
V−>
Find the Q-point for the shown circuit?
KVL at G-S loop,
MOSFET Circuits At DC

Example 6: Design the circuit of Fig. Ex6 so that the transistoroperates at
I
D
=0.3 mAand
V
D
=+1V. The NMOS transistor has
V
t
= 1V,
μ
n
C
ox
=20 μA/V2,
L
=1 μm, and
W
=30μm.

Example 7: Design the circuit in Fig. Ex7 to obtain a current
I
D
of 0.4 mA. Find
the value required for
R
and find the DC voltage
V
D
. The NMOS transistor has
V
t
= 0.5V,
μ
n
C
ox
=20 μA/V2,
L
=1 μm, and
W
=40μm.

Example 8: Design the circuit in Fig. Ex8 to establish a drain voltage of 0.1 V.
What is the effective resistance between drain and source at this operating
point? Let
V
t
= 1V and
k
n
= 1 mA/V2
Fig. Ex6
Fig. Ex8
Fig. Ex7
MOSFET Circuits At DC (contd.)

Example 9: Analyze the circuit shown in Fig. Ex9 to determine
the voltages at all nodes and the currents through all branches.
Let
V
t
= 1V and
k
n

(W/L)
= 1 mA/V2

Example 10: Design the circuit in Fig. Ex10 for the shown
currents and voltages (i.efind R, (W/L) for each transistor). Let
Vt=1 V,
μ
n
C
ox
=20 μA/V2
Fig. Ex9
Fig. Ex10
MOSFET As A Current Source

Ideal current source
gives fixed output
current regardless of
the voltage across it.

MOSFET behaves as
as an ideal current
source if biased in
the pinch-off region
(output current
depends on terminal
voltage).
NMOS Current Mirror
Assumption:
M
1
and
M
2
have identical
V
TN
,
K
n

,
λ
and
W/L
and are
in saturation.
1
'
2
1
2
M
K
W
n
IVVV
REFTN
GS1
DS1
L
λ
⎛⎞
⎛⎞
⎛⎞
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
⎜⎟
⎝⎠
=−+
2
'
2
1
2
M
K
W
n
IVVV
TN
OGS2
DS2
L
λ
⎛⎞
⎛⎞
⎛⎞
⎜⎟
⎜⎟
⎜⎟
⎜⎟
⎝⎠
⎝⎠
⎜⎟
⎝⎠
=−+
But
V
GS2
=V
GS1
22
11
1
1
MM
MM
WW
V
LL
DS2
III
O
REFREF
WW
V
DS1
LL
λ
λ
⎛⎞
⎜⎟
⎝⎠
⎛⎞
⎜⎟
⎝⎠
⎛⎞⎛⎞
⎜⎟⎜⎟
+
⎝⎠⎝⎠

=≅
⎛⎞⎛⎞
+
⎜⎟⎜⎟
⎝⎠⎝⎠
Thus, output current mirrors
reference current if
V
DS1
=V
DS2
or
λ
=
0, and both transistors have
the same (W/L)
NMOS Current Mirror: Example 11
Find the output current and the minimum output
voltage
v
o
to maintain the given current mirror in
proper operation.
Given data:
I
REF
= 50 μA
, V
O
= 12 V,
V
TN
= 1 V
, K
n

= 75 μA/V2,
λ
= 0 V-1,
(W/L)M1
= 2, (W/L)M2=10
Analysis:
2
2(50μA)
1V1.82
μA
2*75
(1)
2
1
V
I
REF
VVV
TN
GS
W
KV
n
DS
L
λ
⎛⎞

⎜⎟
⎜⎟
⎜⎟
⎝⎠
=+=+=
+
Hence,
V
omin
=VGS
–V
TN
= 0.82 V.
μA250
1
2
==

















M
L
W
M
L
W
REF
I
O
I