高频电子线路的.由高频功率晶体管2SC3102组成的谐振功率放大器,其工作频率ƒ = 520 MHz输出功率Po¬¬ = 60 W,VCC = 12.5 V.(1)当ηC = 60%时,试计算管耗PC和平均分量IC0值;(2)若保持Po不变
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/04 03:40:37
![高频电子线路的.由高频功率晶体管2SC3102组成的谐振功率放大器,其工作频率ƒ = 520 MHz输出功率Po¬¬ = 60 W,VCC = 12.5 V.(1)当ηC = 60%时,试计算管耗PC和平均分量IC0值;(2)若保持Po不变](/uploads/image/z/12184017-33-7.jpg?t=%E9%AB%98%E9%A2%91%E7%94%B5%E5%AD%90%E7%BA%BF%E8%B7%AF%E7%9A%84.%E7%94%B1%E9%AB%98%E9%A2%91%E5%8A%9F%E7%8E%87%E6%99%B6%E4%BD%93%E7%AE%A12SC3102%E7%BB%84%E6%88%90%E7%9A%84%E8%B0%90%E6%8C%AF%E5%8A%9F%E7%8E%87%E6%94%BE%E5%A4%A7%E5%99%A8%2C%E5%85%B6%E5%B7%A5%E4%BD%9C%E9%A2%91%E7%8E%87%26%23402%3B+%3D+520+MHz%E8%BE%93%E5%87%BA%E5%8A%9F%E7%8E%87Po%26%23172%3B%26%23172%3B+%3D+60+W%2CVCC+%3D+12.5+V.%EF%BC%881%EF%BC%89%E5%BD%93%CE%B7C+%3D+60%25%E6%97%B6%2C%E8%AF%95%E8%AE%A1%E7%AE%97%E7%AE%A1%E8%80%97PC%E5%92%8C%E5%B9%B3%E5%9D%87%E5%88%86%E9%87%8FIC0%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E4%BF%9D%E6%8C%81Po%E4%B8%8D%E5%8F%98)
xRMJ@ʀUlj"q! ƍ݈
TRQӈjmB2ɪWMRE)f|xJ2ݕ<3>ǚn*;af>t\̺,Qdrt휓@To|+nםW>҇$#$B%2?@o$-1<$OX#f\"K¢QYw[ڃvᣡSnĵ.UXnpG8!tcN NO\wϝ춖<QRV?
BӨ9*d^B- ;
S+?-=?IP%䣗7_Dučp$0($?oD g!3U[cQ> )>9wC7oH
3:]?=_1H
高频电子线路的.由高频功率晶体管2SC3102组成的谐振功率放大器,其工作频率ƒ = 520 MHz输出功率Po¬¬ = 60 W,VCC = 12.5 V.(1)当ηC = 60%时,试计算管耗PC和平均分量IC0值;(2)若保持Po不变
高频电子线路的.
由高频功率晶体管2SC3102组成的谐振功率放大器,其工作频率ƒ = 520 MHz输出功率Po¬¬ = 60 W,VCC = 12.5 V.(1)当ηC = 60%时,试计算管耗PC和平均分量IC0值;(2)若保持Po不变,将ηC提高到80%,试问PC减少多少?
高频电子线路的.由高频功率晶体管2SC3102组成的谐振功率放大器,其工作频率ƒ = 520 MHz输出功率Po¬¬ = 60 W,VCC = 12.5 V.(1)当ηC = 60%时,试计算管耗PC和平均分量IC0值;(2)若保持Po不变
(1)ηC = 60%时,PD = Po / ηC = 100 W
\x05\x05 PC = PD - Po = 40 W,IC0 = PD / VCC = 8A
(2)ηC = 80%时,PD1 = Po / ηC = 75 W
\x05\x05 PC1 = PD1 - Po = 75 W – 60 W = 15 W
\x05\x05 ΔPC = PC - PC1 = 40 W – 15 W = 25 W
\x05 可见,Po一定时,ηC提高,PD和PC将相应减小.