第一章 高频小信号谐振放大器习题解答
1-1解:根据品质因数的公式得:
Qo?1R?oCo?1?212
5?2??1.5?106?100?10?12Lo?11?H?113?H 22226?124?foC4??1.5?10?100?10??谐振时,回路电流 Io?Ums1mV??0.2mA R5?UCom?ULom?QoUms?212Ums?212mV
1-2解:1-1端短路时, ?o?1LC
所以:L?111Q??100 又因为 : ??253?H226?12?orC?oC2??10?100?10??所以 r?11??15.9? 6?12?oQC2??10?100?10?1001-1端接Zx时, Zx为Rx和Cx的串联,
CCx200Cx?100,?100;解得:Cx=200pF,C?Cx200?Cx?L?L由于:Q0=0,QL=0rRLQr501所以:L===,QL?50,RL=(r+Rx)=2r
Q0RL10022??106?253?10-6解得:RX=r==?=15.9?Q0100?0L因而,未知阻抗由15.9?的电阻与200pF得电容串连组成。1-4 解:C'?C2?CL?22pF
所以接入系数
C'221p?'??
C?C120?222把RL折合到回路两端,得:
R'o?Ro9?5?k??11.25k?
4p2C1C'20?22??又 C??CS??5???pF?18.3pF
20?22?C1?C'?谐振频率为fP?12?LC??12?0.8?10?18.3?10?6?12Hz?41.6MHz
谐振阻抗为RP?Qo?PL??100?2??41.6?106?0.8?10?6???20.9k? 总电导为
g??111?111??'?????S?236.7?10?6S 333?RSRLRP?10?1011.25?1020.9?10?因而 R??1?4.2k? g?1?1?20.2
236.7?10?6?2??41.6?106?0.8?10?6最后得QL?g??PLB?2?f0.7?fP?2.06MHz QL1-6 (1)证明:因为回路总电导g??R1111??? RRSRPRLfO QL并且QL??OLfO 而B?所以B?R??OL???OLfOR
g??1B? R?OLfO又因为 ?O?2?fo?1LC
?ofo?所以 g??证明结束
1LC?12?LC?1 B?2?f0.7 2?LC2?f0.7?4??f0.7C
L2?LC(2)g??4??f0.7C?6.28?6?106?20?1012S?7.536?10?4S
11111???4?GL????7.536?10?4??S?5.536?10S?33RLRPRS?10?1010?10?
1??1.8k?
5.536?10?41-10解:(1)Y参数等效电路图如下:
RL?
uce yfeube
yie
yreuce
gp
yoe
C
Yie2
ube