1 gm 5 23解:
diC23
b1 2b2vBE 3b3vBE 4b4vBE
dvBE
diCdvBE
b1 2b2V0mcos 0t 3b3V02mcos2 0t 4b4V03mcos3 0t
vBE v0
gm t
b1 2b2V0mcos 0t gm1 2b2V0m 3b4V03m
3
b3V02m 1 cos2 0 2b4V03mcos 0t b4V03m cos 0t cos3 0t 2
1
gm1 b2V0m 1.5b4V03m2
qvBEdiCaISq
2 gm vBE ekT
dvBEkTgc digm t C
dvBE
omcos 0taIq
SVomcos 0t ekT
kT
q
vBE v0
23
aISqq1 q1 q
Vomcos 0t 1 Vomcos 0t Vomcos 0t Vomcos 0t kT2 kT6 kT kT
qV I qV I qV qV
ISomcos 0t IS om cos2 0t S om cos3 0t S om cos4 0t
kT2 kT 6 kT kT gm1
qV3 IS qVom ISom
kT8 kT
3
234
IqV3 IS1
gc gm1 Som
22kT16
5 25解:i i1 i3 i2 i4
qVom kT
2
3
a0 a1 v0 vs a2 v0 vs a3 v0 vs a4 v0 vs
3
4
a0 a1 v0 vs a2 v0 vs a3 v0 vs a4 v0 vs
2
3
4
a0 a1 v0 vs a2 v0 vs a3 v0 vs a4 v0 vs
2
3
4
a0 a1 v0 vs a2 v0 vs a3 v0 vs a4 v0 vs
2
3
4
3
8a2v0vs 16a4v0vs 16a4vs3v0