Icmax=gcr(Ec-Ucm)
得：Ucm=Ec-Icmax=22.56V gcrP18?52.32mW s=EcIc0=24?2.11Po=UcmIc1m=?22.56?3.91?44.10mW
22?c=Po44.10==84.3% Ps52.3230某谐振高频功率放大器，已知晶体管的饱和临界线斜率gcr = mS，Uj = V，Ec = 24 V， Eb = - V， 输
入电压Ubm = 2 V，Rc = 50 Ω。试求：Ic max，解：cos?=-Eb+UjU=-(-0.2)+0.62=0.4??=66o bmP12I2=2P2?2o=c1mRc?Ic1moRc=50=282.8mA
则 Ucm=Ic1mRc=282.8?10-3?50=14.14V
Ic1mcmax=I?(?)?282.80.421?671.7mA 1Ps=EcIc0=EcIcmax?（0?）=24?671.7?0.241=3885.1mW 集电极效率?oc=PP=2000=51.5% s3885.1cm ，ηc？?1(66o)?0.421,?0(66o)?0.241 U
