电路与电子技术计算题

1、电路与电子技术计算题1在图所示电路中，求各支路电流。(2)画出微变等效电路，求电压放大倍数A：、输入电阻ri和输出电阻ro1 = 20 -H=l2=312 =£=!12 = 4AI3 = I1 + I21311+ 4R1 * I1 + I3*R3 = 10联立求解，得I1 = 2AI3 = 6A2.电路如图所示，已知2k Q, Rl = 6k Q晶体管(1)计算静态值Ib，Ic，UceU cc = 12V, RB1 = 68k Q , Rb 2 = 22k Q , RC = 3k Q, RE = (其中be止200 + P空进行计算)。I C(Ube = 0.7V);IBQ = (V

2、CC - UBEQ) / RbIBQ=(12 - 0.7)/ 68 * 103=17小ICQ =60 * 0.17 = 10 mAUCEQ = VCC - ICQ * CR=12- 10* 3=-18VUCEQ << UBEQ,即VC< VB<VE ,晶体管工作在作在饱和区。rbe200A = Uo/UI= 200-10026/10= (M5kQ(RC/RL) / (Rb+rbc)=-100*0.45/ (68+0.45)=-0.66ir = Ui / Ii = Rb+rbe = 68.45k Qor = Uo / Io = RC = 68k3. 在图所示电路中，开关S

3、在t=0时刻合上，试求:(1)S闭合瞬间各支路的电流和各元件上的电压；解：ul=U; ur=0; uc=0 (2)电路达到新的稳定状态后各支路电流和各元件上的电压。k L+陀二C解：UL=0； UR=U ； UC = U4. 电路如图所示，求各支路电流。1|bA<CD C)12V4£2lil1解：11 = 8/2 = 4A12 = 12/4 = 3A11 + I2 + I3 = 013 = - 7A5.电路如图所示,已知 Ucc = 12V, Rb1 = 30kQ , Rb2 = 10kQ , Rc = 3kQ, Re=1.5k Q Rl = 6k Q 晶体管P = 40。(其

4、中 - 200 + P 进行计算)。I C(1)计算静态值Ib , IU CE ( U BE 0.7V)；IBQ = (VCC - UBEQ) / Rb=(12 - 0.7)/ 30 * 103=38 AICQ = IBQ=60 * 0.38 = 22.8 mACRUCEQ = VCC - ICQ *=12 - 22.8* 3=-56.4VUCEQ << UBEQ,即VC< VB<VE ,晶体管工作在作在饱和区。rbe200+ p 26 = 200+ 100* 26 / 22.8 = 0.11k Q IcAu = Uo/Uip (Rc/Rl) / (Rb+rbe)=-1

5、00*0.11/ (30+0.11)=-0.37ri = Ui / I i = Rb+rbe = 30.11k Qr0 = Uo / Io = RC = 30kQ3U =12V, t=0时将开6.电路如图所示，已知， R1=R2=R3=3kQ , C=10pF, 关S断开。试求电压uc和Uo的变化规律。解：求解一阶电路的三要素。1)初始值：由Cu (0+) = Cu (0-) = 0V 画出t=0+的等效电路，如下图所示得 Cu (0+)= U*R2/ (R1+R2)=12* 1/2 = 6V电路处于新的稳定状态，如下图所示(换路后的电路中2)稳态值：当时，C开路)U+ 3 = U* R1/

6、( R1+R2+R3) = U * 1/3= 12 * 1/3 = 4VU+(祠=U* R1/ ( R1+R2+R3)=U * 1/3= 12 * 1/3 = 4V3)时间常数：将电路中的电源置零，如下图所示R0 = R3/(R1+R2) = 3*6/(3+6) = 2k匸 R0*C = 2*103*103*10-12 =2*10-6S4)将三要素代入UC( t)=uc (叼 + uc (0+)- UC (0-) * e -t/=4 + 0-4* e-t/2* 10-6 = 4(1- e -5*105t)V* e -t/Uo (t) =Uo d + OU ( 0+) - Uo ( 0-)=4

7、+6 -4 * e-t/2* 10-6 = 4+ 2e-5* 105tV/Q.(t)=件(辺)十°-)=4 0- 4* r A 皿= 4(1- e 亠皿(T)=(丈)-心 CO-)M +6 显*y4亠 21沪 1C?V7.在图所示电路中，已知 Ej = 90V, E2 = 60V, R1 = R2 = 15Q, R3 = 30Q, R4=Rs = 15 Q,求各条支路的电流。I 2E2解：Uab = E1/(R1 +R4) + E2/(R2+R5) / (1 / (R1+R2) + 1/R3 +1/(R2+R5)Uab = (90/30 + 60/30) / (1/30 + 1/30

8、 + 1/30)=5 / (1 / 10)=50V13 = Uab/ R3 = 50/ 30 = 5/3A12 = E2 - Uab / R2+R5 = 1/3A11 = I3 - I2 = 4/3A8.电路如图所示，晶体管的P=60, Ube = 0.7V。(其中止200 + P计算)。I C(1)试用直流通路估算静态值I B， IC，UCE ；解：IBQ = Vcc-UBEQ/Rb = 12 - 0.7/480 = 0.02mAUCEQ = VCC- ICQRc = 12 - 1.2 * 3 = 8.4VUCEQ >> UBEQ,即VC< VB<VE ,晶体管工作在

9、作在放大区Au，输入电阻Ri和输出电阻Ro(2)画出微变等效电路，求电压放大倍数+VCC购knRcCUT解：C beQ 200= 200 +nr 26 42 二2367kQAu = Uo/UI(RC/RL) / (Rb+rbe)什 12V)=-60*23.67/(3+23.67)=-53ri = Ui / Ii =Rb+rbe= 503.67k Qro = Uo / Io = Rc = 3k Q9. 电路如图示，已知R1=8Q , R2=12Q, L = 0.6H，U =220V。S闭合后试求ijt)iL(0+)= iL(0-) = 220/ 20 = 11iL( - ) = 220/12 =

10、 1833T = L/R2 = 0.6 = 0.05iL(t) = 18.33 + (11 - 18.33) e " ”0'0510. 电路如图所示，求支路电流I1, I2 , 13。10QL0+10Q20Q解：用节点分析法，弥尔曼定理Uab = ( 4/10 + 2/10 - 0.1)/ ( 1/10 + 1/20 + 1/10)= 2VI2 = Uab / 20 = 0.1AUab = -10 * 11 + 4 11 = (4 - Uab) / 10 = 0.2A13 = 11 - I2 - 0.1 = 0A11. 如图所示电路中，开关S原来在“1位置，在t=0瞬间换接到

11、“2位置, 试求 uC (t)及 ic(t)。100 Q=】0 V叱宁10 pF 士 50 V -解 :开关换接瞬间，根据换路定律#c（oo）»50 卩r= AC = 100x10x10' s根扌居三要素法求，得叫（0 = %（8）+吩 0）=-50+10-（-50）严=-50 + 6站1叫=-5血=50 + 50-60严”=6宀100 10012.晶体管放大电路如图所示，已知Ucc = 12V, Rc = 3kQ , Rb = 240kQ晶体管P=50, Ube = 0.7V。（其中 -200 + P进行计算）I c(1)试用直流通路估算静态值IB, Ic , UcE ；Ibq = Vcc-UBEQ/Rb = 12 - 0.7/240 = 0.05mAIcq = P*I BQ = 50 * 0.05 = 2.5 mAUcEQ = VCC- I cqRc = 12 - 2.5 * 3 = 4.5V