電機與電子群 專業科目(一) 共4 頁 第 1 頁
九十七學年四技二專第五次聯合模擬考試
電機與電子群 專業科目(一) 詳解
97-5-03-4 97-5-04-4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 D B D A B D B A A C B D C C B A D D D D B D D C D 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 B D C B D C B B C B B C C C C B B B A D B D B B C 第一部份:基本電學 1. (1) Po =2hp=2×746=1492W (2) 1755W 85 . 0 1492 P P o i= η = ≅ (3) 16A 110 1755 V P I= i = ≅ 2. (1) 10kΩ b 2 a A R 2 2 2 2 = lρ =ρ = (2) 20kΩ b 2 a 2 A R 2 1 1 1= lρ =ρ = (3) 0.4mA k 20 k 10 12 R R 12 I 2 1 = + = + = 3. (1) V2=E−V1=15−10=5V (2) 0.1A 50 5 R V I 2 2 = = = (3) P1=V1I=10×0.1=1W 4. (1) E=V1=V2 =12V (2) I1R1=I2R2=E=12V,因R1=2R2 則I2 =2I1=4mA,故I1=2mA (3) 3kΩ mA 4 12 I V R 2 2 2 = = = 5. (1) 因I1=0,故電橋平衡, = × =4Ω 4 8 2 R (2) Rab=(8+4)//(R+2)=12//6=4Ω (3) 0.5 1 1.5A 2 R 6 4 8 6 I = + = + + + = 6. (1) 6V 0 3 1 6 1 1 3 0 6 12 V3 = + + + + = Ω (2) V1=V3Ω+1×4=6+4=10V 7. (1) 6V 6 1 3 1 2 1 6 6 3 0 2 14 VBD = + + − + + = (2) 4A 2 6 14 IAB= − = (3) 4A 5 ) 6 ( 14 IAC= − − = (4) I1=IAB+IAC=4+4=8A (5) 2A 3 6 3 V I BD BD= = = (6) I2=I1−IBD=8−2=6A (7) I3=I1−IAB=8−4=4A (8) VAB=IAB×2=4×2=8V (9) VAC=14−(−6)=20V 8. (1) 求RN:將6Ω開路,12 V 電壓源短路,1 A 電 流源開路,RN= 12Ω (2) 求 VN : 將6Ω 短 路 , 使 用 重 疊 定 理 , 得 A 2 1 12 12 IN= + = (3) A 3 4 6 12 12 2 6 R R I I N N N + = × + = = (4) 8V 3 4 6 I 6 VAB= = × = 9. (1) RL=Rth =(8//8+2)//6=3Ω (2) 8V ) 2 6 //( 8 8 ) 2 6 //( 8 24 V1 = + + + × = (3) V2 =24−V1=24−8=16V (4) 6 2 2 V V Vth 2 1 + × + = V 18 8 2 8 16+ × = = (5) 2 L L th th max (R V R ) R P + = W 27 3 ) 3 3 18 ( 2× = + = 10. 26.55nF m 10 1 m 1 10 85 . 8 3 d A C 12 23 o r = × × × × = ε ε = − − 11. (1) CT =12//(12+12)=8μF (2) V 3 40 12 24 24 20 VAB = + × = (3) QT =Q1=CTE=8μ×20=160μC電機與電子群 專業科目(一) 共4 頁 第 2 頁 (4) 80μC 2 Q Q Q T 3 2= = = 12. (1) L1+L2+2M=120 (2) L1+L2−2M=80 (3) 將(1)式-(2)式,得40mH=4M,M=10mH (4) 將(1)式+(2)式,得2(L1+L2)=200, mH 100 L L1+ 2= (5) 因L1=4L2 ,則L1+L2=4L2+L2=100,故 mH 20 L2= ,L1=80mH (6) 0.25 40 10 20 80 10 L L M k 2 1 = = × = = 13. (1) F=BlIsinθ,當θ 90= °時,導線受力最大,即 5 . 0 5 1 1 . 0 F= × × = 牛頓 14. (1) 開關 S 接通(ON)瞬間,電容器視為短路 mA 2 k k//6 12 k 4 16 i = + = mA 3 4 3 2 mA 2 k 6 k 12 k 12 i iC = × = + × = (2) 電路穩定時,電容器視為開路 mA 1 k 12 k 4 16 i = + = ,iC=0 V 12 k 16 k 4 k 12 16 VC= × + = (3) τ=RC=(4k//12k+6k)100μF sec 9 . 0 = 經過一個時間常數後 V 584 . 7 632 . 0 12 VC= × = 15. (1) 5mS 2 m 10 R L = = = τ (2) (1 e ) 8.65A 2 20 ) e 1 ( R E i 5m t t = − = − = −τ − 故t=10mS 16. (1) 平均值Vdc =8V (2) 有效值 ) 10V 2 2 6 ( 8 V 2 2 rms= + = 17. (1) 10kΩ μ 1 . 0 10 1 C 1 XC 3 = × = ω = (2) Z=R−jXC=10k−j10k=10 2∠−45° (3) V=100∠0° (4) = ∠ ° ° − ∠ ° ∠ = = 45 2 10 45 2 10 0 100 Z V I 則i(t)=10sin(103t+45°)毫安 18. (1) 0.1 m 10 0 1 1 L 1 BL 3 = × = ω = 姆歐 (2) Y=G−jBL=0.1−j0.1=0.1 2∠−45°姆歐 (3) 5 2 2 1 . 0 1 Y 1 Z= = = 歐姆 (4) = ∠0° 2 20 V 伏特 (5) 0.1 2 2 20 VG IR = = × = 安培 (6) 0.1 2 2 20 VB IL = L = × = 安培 (7) = = ∠0°×0.1 2∠−45°=2∠−45° 2 20 VY I 安培 (8) i(t)=2 2sin(103−45°)安培 19. (1) 6A 12 120 30 120 I1= − =− (2) ) ( 6) 10A 15 120 ( I= 2+ − 2 = (3) = = =12Ω 10 120 I E Z (4) 功率因數 0.8 15 12 R Z Y G PF= = = = 20. (1) 60 V 2 100 V= ∠ ° , 30 A 2 20 I= ∠ ° (2) = ∗ = ∠ ° ∠−30°=1000∠30° 2 20 60 2 100 VI S 500 j 3 500 + = VA (3) P=500 3W,Q=500VAR (4) = =5∠30° I V Z Ω (5) 電壓v(t)相位超前電流i(t)相位30 ,負載為電° 感性 21. (1) 並聯消耗功率 R E PP= 2 (2) 串聯消耗功率 ) R X R E ( R I P 2 2 2 2 S + = = S 2 P 2 2 2 2 cos P ) X R R ( R E = θ + = (3) 500W 8 . 0 320 cos P P 2 S 2S P= θ = = 22. (1) 5kHz 20 500 1k X X f f L C o= = = (2) 500W 10 ) 2 100 ( P 2 = = 23. (1) 電容電壓相位滯後電路電流相位90° (2) 電感電壓相位超前電路電流相位90° 24. 12.2A 8 . 0 220 3 746 5 cos V 3 P I L o L = × × × = θ =
電機與電子群 專業科目(一) 共4 頁 第 3 頁 25. 第二部份:電子學 26. (1) 最大值Vm =100V (2) 180 1 t= 秒時的瞬間值 ) 60 180 1 60 2 sin( 100 ) t ( v = π× × + ° =100sin(180°)=0 (3) 60Hz 2 377 2 f = π = π ω = (4) 電壓信號v(t)=100sin(377t+60°)與電流信號 ) 60 t 377 sin( 10 ) 30 t 377 cos( 10 ) t ( i = − ° = + ° 相位差θ=60°−60°=0° 27. (A) 電洞濃度 A 822 5 1014/cm3 10 10 5 N p≅ = × = × (B) 電子濃度 3 5 14 2 10 A 2 i 4.5 10 /cm 10 5 ) 10 5 . 1 ( N n n = × × × = = (C) 本質半導體、P 型半導體及 N 型半導體中的電 子數目等於質子數目,電性為中性 (D) 多數載子為電洞,少數載子為電子 28. (1) 6mA k 1 6 12 R V E I S Z S = − = − = (2) IL(max)=IS−IZK =6−1=5mA (3) = = =1.2kΩ m 5 6 I V R (max) L Z (min) L 29. (1) 次級 10V 2 1 2 1 40 Vm= × × = (2) Vdc =0.636Vm =0.636×10=6.36V (3) PIV=2Vm=2×10=20V 30. (1) 10 48mV 100 10 8 . 4 V C R 8 . 4 V dc L ) rms ( r = = × × = (2) Vr(P−P)=2 3Vr(rms) mV 166 mV 48 732 . 1 2× × = = 31. (1) 當Vi≤−2V時,二極體導通,Vo =−2V (2) 當Vi>−2V時,二極體截止,Vo =Vi 32. (1) 射極摻雜濃度最高,但逆向耐壓最低 (2) 集極摻雜濃度最低,但逆向耐壓最高 33. NPN 電晶體操作在主動區時,VBE >0,VBC<0 0 VCE > ,則VC >VB >VE 34. (1) C C C ) sat ( CE CC ) sat ( C R 5R0.2 R4.8 V V I = − = − = (2) B B B BE BB B R 2.7R 0.7 R2 V V I = − = − = (3) 飽和條件:βIB≥IC(sat),則 C B R 8 . 4 R 2 100× ≥ 故RB≤41.6RC 35. (1) 減少IC值,可以使工作點由飽和區附近移向直 流負載線中央,因 B BE CC B V R V I = − ,IC =βIB,故 增加RB值即可減少IC 36. (1) 0.02mA k 565 ) 12 ( 7 . 0 0 IB= − − − = (2) IC=βIB=100×0.02m=2mA (3) 0−VEC−(−12)=ICRC 則VEC =12−ICRC =12−2m×2k=8V 故VCE =−VEC=−8V (4) VB=VBE =−VEB =−0.7V (5) 5.9mA k 2 0.2 12 IC(sat)= − = ,因βIB>IC(sat) 故BJT 工作於主動區 37. (1) 2mA k 1 k 5 12 R R V I E C CC ) sat ( C ≅ + = + = (2) 37.5 A k 101 k 200 0.7 12 R ) 1 ( R V V I E B BE CC B + = μ − = β + + − = 因βIB>IC(sat),電晶體已飽和,故需重新計算IB值 (3) VCC =VBE+IBRB+IERE ) R I I ( R I VBE+ B B+ B+ C(sat) E = 則 E B E ) sat ( C BE CC B R R R I V V I + − − = A 3 . 46 k 1 k 200 k 1 m 2 0.7 12 = μ + × − − = 38. (1) π π π + β − = = r R // R r R r V V V V V V C L s i o s i s o 120 k 2 k k//12 4 k 2 k 0.5 k 2 100 =− + × − = 39. (1) 1mA k 1.3 k 4.7 6 2 1 R R V V I E C CE CC E + = − = + − ≅ (2) 25.8mV 11600 ) 27 2 . 273 ( VT = + = (3) = = ≅26Ω 1 8 . 25 I V r E T e (4) 180 26 k 4.7 r R V V e C i o =− =− =− 40. (1) 0.4mA k 5 1.3 0.7 R V V R V I 2 B E BE 2 B B 2 = + = + = =
電機與電子群 專業科目(一) 共4 頁 第 4 頁 (2) 1 B B CC 2 B 2 1 I I I V R V I = + ≅ = − 故 55kΩ m 0.4 2 24 I V V R 1 B CC 1 B = − = − = (3) 因RB2開路 故 k 1.3 201 k 55 0.7 24 R ) 1 ( R V V I E 1 B BE CC B + × − = β + + − = A 6 . 73 μ = (4) 4mA k 1.3 k 4.7 0.2 24 R R V V I E C ) sat ( CE CC ) sat ( C + = − = + − = (5) 因βIB >IC(sat),故電晶體工作於飽和區 41. (1) Ai(dB) =20logAi =20log10=20dB (2) (A A ) 40 2 1 Ap(dB)= v(dB)+ i(dB) = 故Av(dB) =80−Ai(dB) =80−20=60dB (3) Av(dB) =20logAv =60,故Av =1000 (4) Av=Av1Av2,故 10 100 1000 A A A 1 v v 2 v = = − =− 42. (1) 2 GS 2 ) off ( GS GS DSS D 4) V 1 ( 16 ) V V 1 ( I I − − × = − = (2) VGS =VG−VS=−IDRS=−ID(0.5) 解(1)(2)聯立方程式,得VGS =−2V,ID=4mA (3) JFET 工作於定電流區的條件: 4 V VDG≥ GS(off) = 則VDG=VD−VG=VDD−IDRD−0≥4 故 2kΩ m 4 4 12 I 4 V R D DD D = − = − ≤ 43. (1) 對空乏型 FET 元件而言,VGS=0時, 2 ) off ( GS GS DSS D V ) V 1 ( I I = − ,因VGS可正、可負,故ID 的最大值將會大於IDSS 44. (1) 2 GS 2 ) off ( GS GS DSS D I (1 VV ) 8 (1 V4) I − − × = − = (2) VGS =VG−VS=−IDRS=−ID(0.5+0.5) 解(1)(2)聯立方程式,得VGS =−2V,ID=2mA (3) D S D DS DD D V VI I R R = − − kΩ 5 m 2 k 1 m 2 12 24− − × = = (4) ) V V 1 ( V I 2 g ) off ( GS GS ) off ( GS DSS m = − mA/V 2 ) 4 2 1 ( 4 8 2 = − − − × = (5) 5 500 m 2 1 k 5 R g 1 R V V 1 S m D i o =− + − = + − = 45. (1) 10 10 100 V V i o =− =− (2) Vo=−10Vi=10×2=−20V,因V 已超過電源o 電壓,輸出已飽和,Vo=−12V 46. (1) 12 1 12 ) 50 R 1 ( 20 100 V V i o =− × + =− =− 故R=70kΩ 47. (1) 2 1 90 sin sin = ° θ ,則θ 30 = ° (2) 工作週期 % 66 % 100 360 60 180 % 100 360 2 180 × = ° ° + ° = × ° θ + ° = 48. (1) 通帶電壓增益 k 10 k 90 1 R R 1 A 1 2 V= + = + 則AV(dB) =20log10=20dB (2) 159Hz 0.1μ k 10 2π 1 RC 2 1 fL = π = × × = ,截止點 增益為20dB−3dB=17dB (3) 截止帶電壓增益 ) R R 1 ( f f V V A 1 2 L i o V= ≅ + ) R R 1 log( 20 ) f f log( 20 A log 20 1 2 L V= + + dB 20 ) 159 f log( 20 + = (4) 當f=15.9Hz時, 20 ) 159 15.9 20log( dB 20 ) 159 f log( 20 + = + dB 0 20 20+ = − = (5) 當f =1.59kHz時,增益為20 dB 50. ) R R 2 1 ln( RC 2 T 2 1 + =