MOSFET As An Amp & Switch Package.pdf
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10/31/2004
section 4_4 The MOSFET as an Amp and Switch blank
1/1
4.4 The MOSFET as
an Amp and Switch
Reading Assignment:
pp. 270-280
Now we know how an enhancement MOSFET works!
Q:
A:
1.
2.
HO: The MOSFET as an Amp and Switch
Jim Stiles
The Univ. of Kansas
Dept. of EECS
10/31/2004
The MOSFET as an Amp and Switch
1/14
The MOSFET as an
Amp and Switch
Consider this
simple
MOSFET circuit:
5.0V
Q:
Oh, goody—you’re
going to
waste
my time
with another of these
pointless
academic
problems. Why can’t you
discuss a circuit that
actually
does
something?
R
D
=1K
v
O
v
I
K=0.75
mA/V
2
A:
Actually, this circuit
is
a fundamental electronic
device! To see what this circuit does, we need to
determine its
transfer function
( )
v fv
=
.
I
Q:
Transfer function
! How can
we determine the transfer
function of a MOSFET circuit!?
A:
Same
as with junction diodes—we determine the output v
O
for each device mode, and then determine
when
(i.e., for what
values of v
I
) the device is in that mode!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
O
10/31/2004
The MOSFET as an Amp and Switch
2/14
First, note that
regardless
of the
MOSFET mode:
5.0V
i
D
v v . v
GS
=− =
00
I
R
D
=1K
and:
K=0.75
mA/V
2
v
O
D
v v . v
=− =
00
+
v
DS
-
O
O
v
I
+
v
GS
-
From KVL, we can likewise conclude that:
= =−
Now let’s ASSUME that the MOSFET is in
cutoff
, thus
ENFORCING i
D
=0.
v v
50
. iR
DS
O
D D
Therefore:
v
O
= −
50
50 0 1
50V
. iR
.
.
DD
=−
()
=
Now, we know that MOSFET is in cutoff
when
:
v vV.
GS
= <=
I
t
10
Thus, we conclude that:
v
=
50V when
.
v .
I
<
10 V
Jim Stiles
The Univ. of Kansas
Dept. of EECS
I
O
10/31/2004
The MOSFET as an Amp and Switch
3/14
Now, let’s ASSUME that the MOSFET is in
saturation
, thus
ENFORCE:
i KvV
= −
( )
( )
( )
2
D
GS
t
= −
Kv V
2
I
t
=
075
. v.
−
10
2
I
And thus the output voltage is:
v
O
=−
. iR
DD
=− −
. . v .
( )
()
( )
10 1
2
I
=− −
. . v .
10
2
I
Now, we know that MOSFET is in saturation
when
:
v vV.
GS
= >=
I
t
10
and when:
D
v vvVv.
=> −=−
OGS
t
I
10
This second inequality means:
vv.
OI
>−
10
50 075
. . v .
−
( )
− > −
10
2
v .
10
I
I
0 075
>
. v.
( ) ( )
− + − −
10
2
v.
10 50
.
I
I
Solving this quadratic, we find that the
only
consistent solution
is:
v . .
v .
I
− <
<
10 20
30
I
Jim Stiles
The Univ. of Kansas
Dept. of EECS
50
50 075
50 075
10/31/2004
The MOSFET as an Amp and Switch
4/14
Meaning that the MOSFET is in saturation when v
I
> 1.0 and v
I
<
3.0. Logically, this is same thing as saying the MOSFET is in
saturation when 1.0 < v
I
< 3.0.
Thus we conclude:
v
=− −
( )
2
I
10 when 10
. v .
<<
I
30 V
Finally, let’s ASSUME that the MOSFET is in
triode
mode, thus
we ENFORCE:
i KvVvv
=
⎡
2
( )
GS
− −
t
DS
DS
⎤
=
075 2
.
⎡
( )
v . v v
− −
10
2
⎤
I
O
O
And thus the output voltage is:
v
O
=−
. iR
DD
=− − −
50 075 2
⎡
( )
()
( )
v . v v
I
10
O
O
2
⎤
1
=− − −
50 075 2
⎡
v . v v
I
10
O
O
2
⎤
Rearranging this equation, we get the quadratic form:
075
. v . v . v .
O
2
− −
(
15
I
05
)
O
+ =
50 0
The solutions of which are:
v
=
(
15
.v .
I
− ±
05
) (
15
.v .
I
− −
05 150
)
2
.
O
15
.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
50 075
. . v .
O
⎣
⎦
D
2
⎣
⎦
50
. .
⎣
⎦
. .
⎣
⎦
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