高頻電子技術試題庫 第三章 (DOC) - 下載本文

?c=Po5904==78.5% Ps7516.826 晶體管3DG12B組成諧振功率放大器,已知電源電壓Ec = 18 V,Ucm = 16 V,Icmax = 236.5 mA。電壓利

用系數?=0.94,gcr = 0.219 S,θ = 80o,且放大器工作于臨界狀態。試求:Ps,Po,Pc ,ηc??1(80o)?0.472,?0(80o)?0.286

解:Ic0=Icmax?0(80o)?236.5?0.286?67.6mA

Ic1m=Icmax?1(80o)?236.5?0.472?111.6mA Ps=EcIc0=18?67.6?1216.8mW

11Po=UcmIc1m=?16.92?111.6?944.1mW

22Pc?Ps-Po=1216.8-944.1=272.7mW

?c=Po944.1==77.6% P1216.8s27 某諧振功率放大器工作于臨界狀態,已知Ec = 24 V,飽和臨界線斜率gcr = 0.6 S,導通角θ = 80o, Po = 2

W。試求:Ps, Pc ,ηc??1(80o)?0.472,?0(80o)?0.286 解:由于Icmax=gcr(Ec-Ucm),Ic1m=Icmax?1(?)可得

11Po=UcmIc1m=Ucmgcr(Ec-Ucm)?1(?)

2211Po=gcrEc?1(?)Ucm-gcr?1(?)Ucm2

22則Ucm2-EcUcm+即Ucm2-24Ucm+2Po=0

gcr?1(?)2?2=0

0.6?0.472解之得Ucm=23.4或0.6(舍去)

Icmax=gcr(Ec-Ucm)=0.36A=360mA

Ic0=Icmax?0(80o)?360?0.286?102.96mA Ic1m=Icmax?1(80o)?360?0.472?169.92mA Ps=EcIc0=24?102.96?2471.04mW Pc?Ps-Po=2471.04-2000=471.04mW

?c=P2000o==80.9% P2471.04s28某諧振高頻功率放大器,已知晶體管的飽和臨界線斜率gcr = 6.94 mS,Uj = 0.5 V,Ec = 24 V, Eb = - 0.5

V, 輸入電壓Ubm = 2 V,gc = 10 mS,且放大器工作于臨界狀態。試求:θ,Ic0,Ic1m,Ucm??1(60o)?0.391,?0(60o)?0.218

-Eb+UjUbm-(-0.5)+0.5=0.5??=60o 2解::cos?==Icmax=gcUbm(1-cos?)=10mA

Ic0=Icmax?0(60o)?10?0.218?2.18mA Ic1m=Icmax?1(60o)?10?0.391?3.91mA Icmax=gcr(Ec-Ucm)

得:Ucm=Ec-Icmax=22.56V gcr29某諧振高頻功率放大器,已知晶體管的飽和臨界線斜率gcr = 6.94 mS,Uj = 0.5 V,Ec = 24 V, Eb = - 0.5

V, 輸入電壓Ubm = 2 V,gc = 10 mS,且放大器工作于臨界狀態。試求:Ps, Po ,ηc??1(60o)?0.391,?0(60o)?0.218

-Eb+UjUbm-(-0.5)+0.5=0.5??=60o 2解::cos?==Icmax=gcUbm(1-cos?)=10mA

Ic0=Icmax?0(60o)?10?0.218?2.18mA Ic1m=Icmax?1(60o)?10?0.391?3.91mA Icmax=gcr(Ec-Ucm)

得:Ucm=Ec-Icmax=22.56V gcrPs=EcIc0=24?2.18?52.32mW

11Po=UcmIc1m=?22.56?3.91?44.10mW

22?c=Po44.10==84.3% Ps52.3230某諧振高頻功率放大器,已知晶體管的飽和臨界線斜率gcr = 0.5 mS,Uj = 0.6 V,Ec = 24 V, Eb = - 0.2 V,

輸入電壓Ubm = 2 V,Rc = 50 Ω。試求:Ic max, Ucm ,ηc??1(66o)?0.421,?0(66o)?0.241 解:cos?=-Eb+UjUbm=-(-0.2)+0.6=0.4??=66o 212?2Po=Ic1m2Rc?Ic1m=2PoRc==282.8mA

250則 Ucm=Ic1mRc=282.8?10-3?50=14.14V

Icmax=Ic1m282.8??671.7mA ?1(?)0.421Ps=EcIc0=EcIcmax?(=24?671.7?0.241=3885.1mW 0?)集電極效率?c=Po2000==51.5% P3885.1s





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