习题2-11 利用微变等效电路法估算图P2-11(a)电路的电压放大倍数、输入电阻及输出电阻。已知：Rb1=2.5k?,Rb2=10k?, Rc=2k?,Re=750?,RL=1.5k?,Rs=0,VCC=15V，三极管的输出特性曲线如图(b)所示。解：首先估算Q点，可求得：U?UBEQRb1ICQ?IEQ?BQ?3.07mAUBQ?VCC?3VRb1?Rb2Re?i??C?150由图可知，IBQ?20?A?iB26rbe?300?(1??)?1.6k?IEQUCEQ?VCC?ICQRc?6.56V&???(Rc//RL)??80.6AurbeRi?Rb//rbe?0.89k?Ro?Rc?2k?习题2-12 上题中，如Rs=10k?，则电压放大倍数&U&Au?o??&Ui解：&URiAus?o?Au??6.6&UiRi?Ro习题2-13 在图P2-13的放大电路中，设三极管的β=100，UBEQ=-0.2V，rbb’=200?。①估算静态时的IBQ，ICQ和UCEQ；②计算三极管的rbe值；③求出中频时的电压放大倍数；④若输入正弦电压，输出电压波形为，试问三极管产生了截止失真还是饱和失真？应该调整电路中的哪个参数（增大还是减少）？-VCCRV?UBEQb-10V解：①IBQ?CC?20?A490k?RcC2Rb3k?C1IEQ?ICQ??IBQ?2mA++&URLo&UCEQ?VCC?(?ICQRc)??4VUi_3k?_②rbe?rbb??(1??)③2626?200?101??1.5k?IEQ2&???(Rc//RL)??100④是截止失真，应减少RAubrbe习题2-14 在图P2-14的电路中，设β=50，UBEQ=0.6V。①求静态工作点；②画出放大电路的微变等效电路；③求电压放大倍数、输入电阻和输出电阻。+VCCV?IR?U?(1??)IR解：①CC+12VBQbBEQBQeRbRc470k?C2VCC?UBEQ12?0.73.9k??IBQ???20?AC1Rb?(1??)Re470?51?2+RL&ReIEQ?ICQ??IBQ?1mAUi2k?3.9k?_UCEQ?VCC?ICQRc?IEQRe?6.1V②微变等效电路26r?300?(1??)?1.63k?③beIEQ&???(Rc//RL)??0.94Aurbe?(1??)Re+&Uo_?UirbeRb??IbRLReRc?UoRi?Rb//??rbe??1???Re???84.9k?Ro?Rc?3.9k?习题2-15 设图P2-15中三极管的β=100, UBEQ=0.6V,rbb’=100?, VCC=10V, Rc=3k?, Re=1.8k?, RF=200?, Rb1=33k?, Rb2=100k?, 负载电阻RL=3k?，电容C1、C2、C3均足够大。①求静态工作点；②画出微变等效电路；③求电压放大倍数Au；④设RS=4k?，求AuS；+VCC⑤求Ri和Ro。Rb2Rb1Rc解：①UBQ?VCC?2.48VC2Rb1?Rb2C1ICQ?IEQ?UBQ?UBEQRF?Re?0.94mARs+Re+IBQ?ICQ??9.4?A&+Ui&Rb1Us--RFRLCe&Uo-UCEQ?VCC?ICQRc?IEQ?RF?Re??5.3V②微变等效电路Rb?Rb1//Rb2Us+U?i&--Rs+Rbrbe?R?IbLRFRc?Uo③rbe?rbb??(1??)26?2.89k?IEQAu???(Rc//RL)??6.5rbe?(1??)RF④Ri?Rb1//Rb2//??rbe??1???RF???11.96k?Aus?RiAu??4.87Ri?RsRo?Rc?3k?-③rbe?rbb??(1??)26?2.89k?IEQAu???(Rc//RL)??6.5rbe?(1??)RF④Ri?Rb1//Rb2//??rbe??1???RF???11.96k?Aus?RiAu??4.87Ri?RsRo?Rc?3k?习题2-16 P106解：①VCC?IBQRb?UBEQ?(1??)IBQRe?IBQ?VCC?UBEQRb?(1??)Re?10?ARbC1+VCCC2ReRL+IEQ?ICQ??IBQ?1mAUCEQ?VCC?IEQRe?6.4V&Ui-+&-Uo②rbe?300?(1??)26?2.93k?IEQ&?RL=∞时，Au?1???Rerbe?(1??)Re?0.99?0.97&?RL=1.2k?时，Au?1???Re//RLrbe?(1??)Re////RL③RL=∞时，Ri?Rb//??rbe??1???Re???283k?RL=1.2k?时，Ri?Rb//??rbe??1????Re//RL????87.6k?④?r?Rs??rbeRo??be//R??29??e1???1???P107RbC1+Rc+VCCC2C3ReRL习题2-17 &U?Rco1??&Urbe?(1??)Rei&U(1??)Reo2?&r?(1??)RUibee&Ui-&Uo1&Uo2&??U&当Rc=Re时，Uo1o2uituo1tuo2t+&UiRbrbeRe??Ib+&Uo1-+R&Uo2c--VCC=15VRb1=2.5k?RCRb2=10k?Rb2?&UiUoReRLRc=3k?RL=3k?Rb1VCC解：①β=50CbRb1U?UBEQUBEQ=0.6VUBQ?VCC?3VRe?BQ?1.2k?Rb1?Rb2IEQrbb’=300?IBQ?ICQ??0.04mA?40?AI=2mA②ICQ?IEQ?2mAEQUCEQ?VCC?ICQRc?IEQRe?6.6V26&??(Rc//RL)?77.8r?r?(1??)?963?③bebb?AuIEQrberRi?be?18.8?Ro?Rc?3k?1??习题2-18 P107C1C2习题2-19 P107解：(a)共基电路(b)共射电路(c)共集电路(d)共射电路(e)共射－共基电路习题2-20 P107Rd4iD/mA3VDD2105101520RguI+uO?iDuGS=3V2V1.5V1VuDS/VQ+-VGG-解：①直流负载线方程：uDS?VDD?iDRd?20?5.1iDUGSQ?2V,UDSQ?8.3V,IDQ?2.3mA?iDg??1.7mS②m?uGSRo?Rd?5.1k?③Au??gmRd??8.7VDD=20VVGG=2VRd=5.1k?Rg=10M?习题2-21 P108R2①Au= -gm(RD//RL) ?-22.5R?C1Ri= RG+ (R1//R2) ?20 MgR+o= RD=15 k?U&②不接Ci-R1S 时，微变等效电路为GDRgU?gsRgmU?gs1R2SRdRLRsGDRgU?gsR1R2SgmU?gsRdRLRsU&i?U&gs?I&dRs?(1?gmRs)U&gsU&o??gmU&gs(Rd//RL)?AU&ogm(Rd//RL)u?U&???i1?g?9mRsR+VdDDC2+RLU&oRsCs-VDD=30VRd=15k?Rs=1k?Rg=20M?R1=30k?R2=200k?RL=1M?gm=1.5mSVDD=30VRd=15k?Rs=1k?Rg=20M?R1=30k?R2=200k?RL=1M?gm=1.5mS
模拟电子技术基础简明教程 第三版 习题答案
