电子模拟技术基础,几个试题求会的人来解答

发布网友发布时间：2023-02-12 14:21

共2个回答

热心网友时间：2023-04-26 14:09

1. Icq=(Ucc-Uceq)/Rc=(12V-6V)/3kΩ=6V/3kΩ=2mA
Ibq=Icq/β=2mA/100=0.02mA
Rb=Ucc/Ibq=12V/0.02mA=600 kΩ
βRc//RL/rbe=100Rc//RL/rbe=100mV/1mV=100
Rc//RL/rbe=1
Rc//RL=rbe=1.4kΩ
3RL/(3+RL)=1.4 kΩ
3RL=(3+RL)1.4 kΩ
(3-1.4)RL=3x1.4 kΩ
1.6RL=4.2 kΩ
RL=2.625 kΩ
2. RL=∞时，Uom=Uceq-Uces=6V-0.6V=5.4V
RL=3 kΩ时，Uom=Rc//RLIcq=3 kΩ//3 kΩ×2mA=3V
3.(1)Rb=25k//5k=4.2k
α=5/(25+5)=1/6
Ib=(αUcc-Ube)(Rb+βRe)=(12/6-0.7V)/(4.2k+100×1.3k)=0.01mA
Ic=βIb=100×0.01mA=1mA
Uceq=Ucc-(Rc+Re)Icq=12V-(5k+1.3k)×1mA=5.7V
rbe=100+26mV/0.01mA=2700Ω=2.7kΩ
rbe+βRe1=2.7 kΩ+100×0.3kΩ=32.7kΩ
Av=-βRc//RL/( rbe+βRe1)=-100×5k//5k/32.7kΩ=-76.5
Ri=Rb//( rbe+βRe1)=4.2k//32.7kΩ=3.7kΩ
Ro=Rc=5kΩ
(2) Ib=0.01mA
Ic=βIb=200×0.01mA=2mA
Uceq=Ucc-(Rc+Re)Icq=12V-(5k+1.3k)×2mA=0V
(3) Av变小，Ri变大
根据《模拟电子技术简明教程》（元增民，清华大学出版社）解答

热心网友时间：2023-04-26 14:10