Solution 3:
(a) The current Id,
Id = (0 - 0.75 - 0.75 + 5.2) / (0.907 + 4.98) = 0.628 mA
Voltage at the base of the transistor, Qb
= -0.628 * 0.907 = -0.57 v
Hence, VBB = -0.57 - 0.75 = -1.32 v
(b) The current IRB,
IRB = (-1.32 - 0.75 - 0.75 + 5.2) / (3.06 + 0.72)
= 0.63 mA
Voltage VBB' = -1.32 - 0.75 - 0.75 - 0.72 * 0.63 = -3.27 v
(c) The output voltages,
V1 = VAND
&
V2 = VNAND
The truth tables are shown.
|
A |
B |
VAND |
|
0 |
0 |
0 |
|
0 |
1 |
0 |
|
1 |
0 |
0 |
|
1 |
1 |
1 |
|
A |
B |
VNAND |
|
0 |
0 |
1 |
|
0 |
1 |
1 |
|
1 |
0 |
1 |
|
1 |
1 |
0 |