1、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=130,集电极效率C=74.6%,输出功率P=500mW,工作于欠压状态,试求：(1),VCM ,C,C1 ,C0 ,CM(2),为了提高效率c，在保证VCC、RP、Po不变的条件下，将通角C减小到600，计算对于C =600的C1，CM，c(3),采用什么样的措施能达到将C变为600的目的?题解:(1) P= = VCM=11.4V C=g1(C)g1()=VCC=1.568 查表C =9000()=0.319 1()=0.5C1m=87.69mACM=175.38mAC0= ICM0
2、()=55.95mA(2) P和不变，则C1M和不变(P= =2C1MCM=224.27mACO= CM0(600)=48.89mAc=1(600)=85.5%(3)改变导通角，应改变偏量和Vbm由 gc=C1M不变，CM(900)=175.38mA CM(600)=224.27mAVbm(900)=87.69mV= =0.5V即在Vbm=87.69mV, Vbb=0.5s时，导通角900由Vbm(600)=0.24427V=0.5Vbb(600)=112.135mVVbm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600答案：
3、(1) VCM=11.4VC =900 C1=87.69mA CM=175.38mA CO=55.75mA(2) 若VCC、RP、Po不变，C =600C1=87.69mA ; CM=224.27mA ;CO=48.89mA(3) Vbm由87.89mV变为224.27mVVbb由500mV变为112.135mV 2.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,Vb2=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C (5)输出回路的谐振电阻RP答案:Po=7516.8mVP=5904mWPC=1612.8mWC=78.5%RP=21.68题解:临界时:CM=gcr(VCC-VCM)可得VCM由= 得=63.90查表0(63.90)=0.2321(63.90)=0.4104 / 4
浙ICP备2021020529号-1 | 浙B2-20240490 
