A.) B.) = 0.995 0.995 = 212.63 1 212.63 20 = 0.906 = 90.6% 20
ο»ΏSOLUTIONS:
Problem 1:
A.)
=
=
1.00 1.050
=
20 21
= 1 - = 20
B.)
=
1.001.01 1.0250(1-.01)
=
0.995
0.995 = 1 - 0.995 = 212.63
212.63 - 20
=
20
= 0.906 = 90.6%
Problem 2:
=
=
1.00 20
=
50
=
=
1.00 1.0025 50 0.9975
=
20.1
20.1 - 20 = 20 = 0.00501 = 0.501%
Problem 3:
Assume BJT works in forward active region and the capacitors are very large so they can be treated as open circuits.
8 - 0.6 = 400 = 18.5 = = 100 18.5 = 1.85 = 10 - 4000 0.00185 = 0.6 = 0 (there is a capacitor creating an open circuit in DC. )
Problem 4:
For the MOSFET to be in saturation -
+ 2 2 - 0.5 -0.5
=
2
(
-
)2
=
4
- 1
300
10-6
8 4
(0
-
(-2)
-
0.5)2
=
4
- 1
= 4 - 0.00135 1 -0.5
1 3.33
Problem 5:
Assuming that 1 and 2 are in saturation
1
=
2
2
(
-
)2
=
2
(
- )2
300
10-6 22
10
(0
-
(-2)
-
0.5)2
=
75
10-6 21
50
-
2
-
(-0.5)2
= 0.55 or 0.816. Since output has to be VDD Vout VSS we will choose 0.55 V
Problem 6:
For quiescent values that capacitors act as open circuits, so the voltage is simply,
=
28 - 90K
-
10K
=
28 - 10 90K
= ( + 1)
=
(101)
28
-
10
( 90K
+
0.6)
=
2K
0.4 V
=
1 V
1 = IE = 0.99 2000 = 495 ? = 28 - 4000 = 26.02 V = 0
Problem 7:
a)for the same voltage drop, the ratio of currents between two BJT is a the same as the ratio of
their areas (you can verify that to yourself) 1 = 1 = 1
2 2 5
= 1 + 2 = 6 1
= 1 + = 1 + 61
1
5
1 = + 6 -> = 2 = 51 = + 6
Assuming that is large = 5 = 7.5
b) Similarly to part (a), for two transistors in saturation with the same VGS, their currents will be
a ratio of their W/L
1
1 2
=
1 2
=
5 20
=
1 4
2
= 4 = 6
Problem 8:
:
=
2 1
2
:
=
2 1
1
These are very useful structures that are called "Current mirrors". They allow current to be mirrored over to other branches with a gain that is dependent on the geometric ratios of MOSFETS/BJT's used.
Problem 9: Code:
TestBench:
Waveform Output:
................
................
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