高频功率放大器加习题及完整答案

1.有一谐振功率放大器，已知晶体管的gc=2000ms,Vb2=0.5V,Vcc=12V,谐振回路电阻RP=130Ω,集电极效率ηC=74.6%,输出功率P～=500mW,工作于欠压状态,试求：(1),VCM,C,C1,CO,CM(2),为了提高效率ηc，在保证VCC、RP、Po不变的条件下，将通角θC减小到600，计算对于C=600的C1，CM，ηc(3),采用什么样的措施能达到将C变为600的目的?答案：(1)①VCM=11.4V②C=900③C1=87.71mA④CM=175.42mA⑤CO=55.84mA(2)若VCC、RP、Po不变，C=600C1=100.15mA;CM=256.15mA;CO=55.84mA%3.97C(3)应改变晶体管直流偏压题解:(1)VPRVRVPPCMPCM4.11221~2~mAVPICMC71.872~1mAVPICCCC84.55~057.122~~011CMCCCCCCCMCCVVVPVPIIg查表C=900∴0()=0.3191(C)=0.5mAIICCM42.17511(2)由CCCVIP00可得当VCC、RP、Po不变的条件下mAIC84.550391.0,218.0,80.16010g时。mAIICCM15.25600mAIICMC15.10011%3.9720210~PRIPpPCC(3)改变导通角，应改变晶体管直流偏压o60cos-0.5cos-cos-bbbZbbbbbbZUUVVUVV即在Vbm=87.69mV,Vbb=0.5s时，导通角900由Vbm(600)==0.24427V==0.5∴Vbb(600)=112.135mV∴Vbm由87.69mV变为244.27mVVbb由500mV变为112.135mV可保证在输出不变时，C由900变为6002.已知谐振高频功率放大器的晶体管饱和临界线的斜率gcr=0.9s,VbZ=0.6V,电压电源VCC=18V,Vbb=-0.5V,输入电压振幅Vbm=2.5V,CM=1.8A,临界工作状态试求：(1)电源VCC提供的输入功率Po(2)输出功率P～(3)集电极损耗功率PC(4)集电极效率ηC(5)输出回路的谐振电阻RP答案:(1)WVIPAIIVVVCCCoCMCbmbbbZ5168.718418.0418.08.1232.041.0232.06444.05.25.06.0-cos00010o(2)公式见P107WIUPAIIVgIVeVUCckpCMCcrCMcckpcccckp904.5738.0162121738.0169.08.1-18--1~11min(3)%54.785168.7904.50~PPC(4)68.21738.0161CckppIUR3.某谐振高频功率放大器，晶体管饱和临界线的斜率gcr=0.5s,Vb2=0.6V,电压电源VCC=24V,Vbb=-0.2V,输入电压振幅Vbm=2V,输出回路谐振电阻RP=50Ω,输出功率Po=2W,试求(1).①CM=?(集电极电流最大值)②VCM=?(输出电压振幅)③ηC=?(集电极效率)(2).判断放大器工作于什么状态?(3).当RP变为何值时，放大器工作于临界状态，这时输出功率P～，集电极效率ηC=?答案:(1).%2625298.08.529279.71057.1452121279.7501057.14557.14578.345241.033.8333.83242421.0241.04.664.022.06.0-cos0~3-1~3-111000010PPmWVIPVRIVmAIImAIImAVPIVVVCCMCpCCMCMCCCMCCCbmbbbZo(2)当处于临界状态时CMCMCMCCcrCMIIAVVgI''361.8279.7-245.0-因此处于欠压状态(3)CMI保持不变，对吗？由bmbbbZVVV-cos可知o4.66%43.882769.1769.1308.231057.145212112.160308.230.534578.0-24--57.14578.345241.033.8333.832420~3-1~1110000PPWVIPIVRVgIVVVVgImAIImAIImAVPICCMCCCMpcrCMCCCMCMCCcrCMCMCCCMCCC