2018年度上半年华工平时作业电路与电子技术计算题

1、H电路与电子技术计算题1在图所示电路中，求各支路电流。町=2门2 3 12 -ch-尸 IQ解：|2 = 4A|3 = |1+|2|3 =|1 + 4R1 * 11 + |3*R3 = 10联立求解，得I1 = 2AI3 = 6A2.电路如题图所示，已知U cc2k Q, Rl = 6k Q晶体管=60。=12V, RB1 = 68kQ , RB2 = 22kQ , RC = 3kQ, RE =26进行计算)。1 C(其中 rbe 200o(Ube = 0.7V);(1)计算静态值Ib , Ic , Uce解：(1)|BQ = (Vcc - UBEQ) / Rb=(12 - 0.7)/ 68

2、* 103=17 AICQ = IbQ=60 * 0.17 = 10 mAUcEQ = Vcc - Icq * Rc=12- 10* 3=-18VUCEQ << Ubeq,即Vc< VbvVe，晶体管工作在作在饱和区。(2)rbe20026 = 200 + 100* 26 / 10 = 0.45k QI CAu = Uo/Ui(Rc/Rl) / (Rb+rbe)=-100*0.45/ (68+0.45)=-0.66 ri = Ui / li = Rb+rbe = 68.45k Q ro = Uo / Io = Rc = 68k Q3在图所示电路中，开关S在t=0时刻合上，试求

3、：(1) S闭合瞬间各支路的电流和各元件上的电压；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。UL=0； UR=U ； uc=Ube4 电路如图所示，求各支路电流。12V解:Ii= 8/2 = 4A|2 = 12/4 = 3AIi + I2 + I3 = 0|3 = - 7A5.电路如题图所示，已知 Ucc = 12V, RB1 = 30k Q , RB2 = 10k Q , RC = 3k Q, RE =1.5k Q Rl = 6k Q晶体管 =40。(其中rbe 200进行计算)。I c(1) 计算静态值 Ib , Ic , UcE ( Ub

4、e = 0.7V)；Ibq = (Vcc - Ubeq) / Rb=(12 - 0.7)/ 30 * 103=38 AIcq = Ibq=60 * 0.38 = 22.8 mAUcEQ = Vcc - Icq * RC=12 - 22.8* 3=-56.4VUCEQ << Ubeq,即Vc< Vb<Ve，晶体管工作在作在饱和区。(2)画出微变等效电路，求电压放大倍数Au、输入电阻ri和输出电阻rorbe 20026 = 200 + 100* 26 / 22.8 = 0.11k QI cAu = Uo/Ui(RC/RL) / (Rb+rbe)=-100*0.11/ (30

5、+0.11)=-0.37 ri = Ui / Ii = Rb+rbe = 30.11k Qr。= Uo / Io = Rc = 30k QRi叫-4-6 电路如图所示，已知, 断开。试求电压R2 R3 =3kQ , C=103pF, U =12V, t=0 时将开关 SUc和Uo的变化规律。CTkL>解：求解一阶电路的三要素。1)初始值：由 Uc (0+) = Uc (0-)画出t=0+的等效电路，如下图所示得uc (0+) = U*R2/ (R1+R2) = 12* 1/2 = 6V=0V/-Oc开路)2)稳态值：当t= 时，电路处于新的稳定状态，如下图所示(换路后的电路中uC()=

6、U* R1/( R1+R2+R3) = U * 1/3= 12 * 1/3 = 4VUo () = 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) = 2 k Q= R0*C = 2*10 3*103*10-12 =2*10 -6S4)将三要素代入uc (t) =uc ()+ uc (0+)- uc (0-) * e-t/-t/2* 10-6 = 4(1- e -5*105t)V=4 + 0-4* euo (t) =uo ()+ Uo (0+)- Uo (0-

7、) * e-t/=4+ 2e-5* 105tV=4 +6 -4 * e-t/2* 10-67.在图所示电路中，已知 Ei = 90V, E2 = 60V, Ri = R2 = 15Q, R3 = 30Q, R4 = R5=15Q,求各条支路的电流。+EiRiF2I3|F3Ft解:/ (R1+R2) + 1/R3 +Uab = E1/(R1 +R4) + E2/(R2+R5) / (1 1/(R2+R5)Uab = (90/30 + 60/30) / (1/30 + 1/30 + 1/30) =5 / (1 / 10)=50VI3 = Uab/ R3 = 50/ 30 = 5/3AI2= E2

8、- Uab / R2+R5 = 1/3A26计算)。I CI1 = I3 - I2 = 4/3A8.电路如图所示，晶体管的=60, Ube = 0.7V。(其中rbe 200(1 )试用直流通路估算静态值IB , IC , UCE ；IBQ = Vcc-U BEQ/Rb = 12 - 0.7/480 = 0.02mAIcq = *l BQ = 60 * 0.02 = 1.2 mAUCEQ = VCC- IcqRc = 12 - 1.2 * 3 = 8.4VUCEQ >> Ubeq,即Vc< VbvVe，晶体管工作在作在放大区。(2) 画出微变等效电路，求电压放大倍数 Au ,

9、输入电阻Ri和输出电阻 凡rbe 200= 200 + 100* 26 / 1.2 = 23.67k QI CAu = Uo/Ui(Rc/Rl) / (Rb+rbe)=-60*23.67/ (3+23.67)=-53 ri = Ui / li = Rb+rbe = 503.67k Qro = Uo / Io = Rc = 3k QRhCl :II-z3IrcRe rt宁乜I9.电路如图示，已知 R1=8Q , R2=12Q , L = 0.6H, U =220V。S闭合后试求ijt)解：iL(0+)= i L(0-) = 220/ 20 = 11iL( ) = 220/ 12 = 18.33=

10、L/R2 = 0.6 = 0.05iL(t) = 18.33 + (11 - 18.33) e-也0510.电路如图所示，求支路电流|1, I2 ,13。LlOQlOQbl I+广2 Qo.1A OV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)= 2V12 = Uab / 20 = 0.1AUab = -10 * 11 + 411 = (4 - Uab) / 10 = 0.2A13 = 11 - 12 - 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“2位置, 试求 Uc (t)

11、及 ic(t)。ZZI100+%千10吋50 V 一答：开关E换接瞬间，根据换路定律他(CT)二:U 机(T) = 10厂= -50Fr=；cr = 100xl0xl0-s，得二©)十叫0)=-50 + 10- (a = -50 -+ 驱叫 .小 50-先 C) -50 + 50- g =0 100根据三要素法求10012.晶体管放大电路如图所示，已知U CC = 12V, RC = 3k Q , RB = 240k Q 晶体管50，Ube = 0.7V。(其中 rbe 20026进行计算)I C(1)试用直流通路估算静态值Ib，I C ， U CE ；IBQ = Vcc-U BEQ/Rb = 12 - 0.7/240 = 0.05mAICQ = *I bq = 50 * 0.05 = 2.5 mAUceq = VCC- I cqRc = 12 - 2.5 * 3 = 4.5VUceq >> Ubeq,即Vc< VbvVe，晶体管工作在作在放大区。(2)静态时(Ui = 0), Ci、C2上的电压各为多少？并标出极性。