bai_tap_ky_thaut_dien_tu_1476

8/6/2019 bai_tap_ky_thaut_dien_tu_1476

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Trang 1

TRNG I HC SPHM K THUT TP.HCM

KHOA: IN T

B MN: C S K THUT IN T

Tn hc phn: K THUT IN T M hc phn:1162010

SVHT:3

Trnh o to:i hc

A - NGN HNG CU HI KIM TRA NH GI KIU T LUN.

Chng 1: DIODE V MCH NG DNG

1. Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 11.1

PIV : n p phn cc ngc.

ID : dng in qua Diode.

V, V

D: in p ngng dn ca Diode.

Is : dng in bo ha.

VT : in p nhit.

: hng s ph thuc vo vt liu. 12

Tk : nhi t kelvin Tk = Tc +273 q : i n tch q = 1,6 x 10-19 C k : h ng s Boltzman. k = 1,38 x 10-23 J/0K

1.2

1 TD VVSD eII

q

kTV kT

1.3

Bi ton 1: Cho ng vo Vi, xc nh v v dng sng ng ra Vo

Bi ton 2: Cho mch n p dng zener, cho ng vo, tm ng ra.

Bi ton 3: Cho mch n p dng zener, cho ng ra, tm ng vo.

Bi ton 4: Tm ID, Vo, xc nh cng logic2. Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 1

Mc tiu kim tra nh gi Ni dung

Mc Nh Cc kin thc cn nh:

Phng trnh ca diode

Biu mu 3a


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Trang 2

Ngng dn ca diode Si v Ge, in p PIV ca diode trong ccmch chnh lu.

Mc Hiuc cc kinthc hc

Cc thng s gii hn ca diode

Hiu c hot ng ca cc mch chnh lu bn k, ton k, cngthc tnh in p ra trung bnh, dng in ra trung bnh trn ti

Cc loi diode khcKh nng vn dng cc kinthc hc

Cc kin thc m sinh vin phi bit vn dng :

Xc nh dc trong tng bi ton c th ngng dn ca diode

Kh nng tng hp: Bi ton 1: Cho ng vo Vi, xc nh v v dng sng ng ra Vo

Bi ton 2: Cho mch n p dng zener, cho ng vo, tm ng ra.

Bi ton 3: Cho mch n p dng zener, cho ng ra, tm ng vo.

Bi ton 4: Tm ID,Vo, xc nh cng logic

3. Ngn hng cu hi v p n chi tit chng 1tt Loi Ni dung im

1 Cu hiCho Vi. V dng sng ng ra VoVi Diode l Si.Vi = Vmsin(wt)

1.5

p n

D

R

VI

VO

Si


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Trang 3

2 Cu hi 1.5

p n

3 Cu hi 1.5

p n

Vi

-

Si

++

-

Vo3.3k

V

-

+

-

VoVi

+

Ideal

8.2k

Si

VT =-0.7V

-Vm +VT

t0

t

Vo

VT = V+0.7V

-Vm - VT

0

-VT


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4 Cu hi 1.5

p n

5 Cu hi 1.5

p n

6 Cu hi 1.5

VT = V+ 0.7v

-

Vi

-

++

R

v

Vo

Si

-

Vi

+

-

+

VoR

vSi

+

Vi Vo

-

R

-

v+ Si

t

0

Vo

VT = -V-0.7V

VT

-Vm - VT


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Trang 5

p n

7 Cu hi 2

p n

Si

-

Si

Si

-

+

Vo

Si

+

Vo

VT1 = 1.4V

VT2 = -1.4V

t

-Vm - VT2

0

t0

Vo

VT = V-0.7V

VT = V-0.7V

Vm + VT


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Trang 6

8 Cu hi 2

p n

9 Cu hi 2

p n

VT1 = 1.4V

VT2 = -1.4V

t

Vm - VT1

0

-

VoIdeal Diodes

-

+

+

5.6k 5.6k 5.6k

Si

-

Si

Si

-

+

Vo

Si

+

VT1 = (Vm -0.7v)/2

VT2 = (-Vm +0.7v)/2

t

Vm - VT

0


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Trang 7

10 Cu hi 1.5

p n

11 Cu hi 2

p n

Si

Si

+

Vo

-5.6k

5.6k

5.6k

VT1 = (Vm -0.7v)/2

VT2 = (-Vm +0.7v)/2

t

Vm - VT1

0


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Trang 8

12 Cu hi 2

p n

13 Cu hi 1

p n

14 Cu hi 1

VoVi

-

+R+

-

-Vm

t0

V0

T2

T

- 0.7Vt0

V0

T

2

T

- 0.7V

- -

Vi= 110V (rms) Ideal

++

2.2K

Vo (Vdc)

Si

Si


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Trang 9

p n

15 Cu hi 2

p n

16 Cu hi 2

R

Si Vo

+

-

10K

i

-

+1K

Vo

+

-

+

10K

R

-

1K

Si

0.7V

t0

V0

T

2

T

VT = 0.7V

0.7V

t0

V0

T

2

T

Vm11

10

VT =0.77V

Vm


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Trang 10

p n

17 Cu hi 1.5

p n

18 Cu hi 1.5+ +

-v

Vi Vo

R

-

+

VoVi

+Si

--

5.6k

v

0

VT

t

V0

T

2

T

VT = v+ 0.7V

Vm

-0.7V

t0

V0

T

2

T

-0.77V

Vm11

10


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Trang 11

p n

19 Cu hi 1.5

p n

20 Cu hi 1.5

p n

VT t

V0

VT = -v+ 0.7VVm

0

TVT

t0

V0

2

T

VT = -v - 0.7V

VT

t0

V0

VT = v + 0.7V

VoVi

+

-v-

R+

Si

Vo

--

R+

Vi

v

+

Si


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Trang 12

21 Cu hi 1.5

p n

22 Cu hi 1.5

p n

23 Cu hiCho bit ng ra ca mch Vo l baonhiu?

1.5

-

Vo

Si+

-

Vi

+

V1 V2

Si

15k

-10V

Vo0V

2.2K

Si

Si

-10V

-10V

Si

R

v

+

Vi

+

1v

-

Vo

-2

Si Si

VT1

t0

V0

2

T

VT1= -v1 + 0.7V

VT2= v2 - 0.7V VT2

T

VT2

t0

V0

2

T

VT2= -v1 - 0.7V

VT1= v2 + 0.7V VT1

T


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Trang 13

Si

Si

Vo

1K

0V

-10V

Si

-10V

IZ

Vi

-

+

RS

RL

IL

IZM

VZ

p n Vo = 9.3V24 Cu hi

Cho bit ng ra ca mch Vo l baonhiu?

1.5

p n Vo = -9.3V25 Cu hi

a. Xc nh RL v IL VRL = 10V.b. Xc nh cng sut cc iVi IZM = 32mA, Vi = 50V, Vz = 10V, RS = 1k.

2

p n a.

kR

VV

VRR

kI

VR

mAIII

mAR

VVI

VVVV

L

Zi

ZS

L

L

Z

L

ZML

S

i

25.1250

2501050

101

25.18

10

83240

401

1050

10

min

min

max

min

0

0

b. mAIVP ZMZZM 3203210

26 Cu hi a. Hy xc nh VL, IL, IR vi RL = 180.b. Xc nh gi tr ca RL c c cng sut cc I PZmax =

400mW.c. Xc nh gi tr nh nht ca RL zener diode c th hot

ng c.Cho VZ = 10V, RS=110, Vi = 50V

3


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Trang 14

p n a.

ARR

VI

mAR

VI

VVVV

LS

i

L

L

LL

LL

17.0180110

50

55180

10

100

b.

9.7613.0

10

13.004.017.0

4010

400maxmax

L

L

L

ZmazLL

Z

Z

Z

I

VR

AIII

mAV

PI

c.

5.27

110

50

1110

1

1

0

0

V

VRR

RR

RVV

i

SL

SL

L

i

27 Cu hi Hy xc nh gi tr ca Vi sao cho VL = 9V v zener diode hotng khng qu cng sut.Cho RL = 1k, PZM = 300mW, R = 100.

3

p n

mAV

PI

VVVV

Z

ZM

ZM

ZL

3.339

300

90

Chn: mAII ZMZ 33.33.3310

1

10

1min

-

+

Vi RL

RS

IZ IL

Vi

RS

RL

IZ IL


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Trang 15

VVV

V

IIRVVIIRV

VVV

mAR

VI

i

i

LZMSZiLZSZ

iii

L

L

L

1310

93.331.09933.31.09

91

9

min

maxmin

28 Cu hi Cho mt s mch diode nh hnh v. Cc s liu khc cho sntrn s . V mch tng v tnh in p ng ra.

1

p n in p ng ra :VO = -(24 - 0,3 -0,7 ) 6,8K / 6,8K + 2,2K = - 17,37V

1

29 Cu hi

Tnh in p ng ra Vout vdng in I chy trong mi s . Bit in tr thun ca diodekhng ng k.

1

p n

S (a) : V OUT = + 0.7VI = ( 12 -0,3 - 0,7 ) / 2,2K = 5mA

S (b) : I = ( 22 -0,7 ) / ( 2,2K + 2,2K ) = 4,84 mA

6,8K

Ge Si

24V +

24V

0,3V 0,7V

6,8K

2,2K Vout

SiVout

Ge

(a)

(b)

2,2K

22V

Vout

0,7V

+

+

(b)


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Trang 16

VOUT

= I . 2,2K = 4,84 mA . 2,2K = 10,65V

29 Cu hiTnh in p ng ra VOUTv dng in I chy trong s . Bit intr thun trn diode khng ng k.

1

p n

I = 11 / ( 2,2K + 3,3K ) = 2 mAVOUT = 2mA . 3,3K = 6,6 V

30 Cu hi Cho mt s mch diode nh hnh v. Tnh dng in qua minhnh R1, R2, R3 . Cho mt s mch diode nh hnh v.Tnh dng in qua mi nhnh R1, R2, R3. B qua in tr thunca diode. Cc s liu khccho trn s .

1

p n Hai nhnh R 1 v R 2 khng c dng i qua .

Dng i qua nhnh R3 l I = ( 24 - 0,7 - 0,7 ) / 3,3K = 6,85 mA31 Cu hi Xc nh in p ng ra Vo v dng in qua mi diode trong mchsau. B qua in tr thun ca diode khi dn.

1

p n Dng in qua nhnh diode Ge l I1=2,47 mA

-

- + +

-24V

Si Si Si

Si

R1 R2 R3

2K

2K

1K


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Trang 17

Dng in qua nhnh diode Si l I 2 = 2,27 mAin p ng ra VO = 1K .( 2,47 mA + 2,27mA ) = 4,74 mA

32 Cu hi Xc nh in p ng ra Vo v dng in qua diode trong mch sau.B qua in tr thun ca diode khi dn.

1

p n RT

= 150 . 100 / 150 + 100 =

60 E

T= 24 . 100 / 100 + 150 =

9,6 V

VO = 0,7 VI = 9,6 - 0,7 / 60 = 0,15 mA

33 Cu hi Cho 1 s mch diode sau, tnh dng in qua mi diode v inp ng ra V0, cc s liu khc cho trn s .

1

p n Diode Ge ON , diode Si OFFDng qua Si bng 0Dng qua Ge I = ( 20 + 5 - 0,3 ) / 6,8K = 3,63 mAV O = 20 - 6,8 K . 3,63 mA = - 4,7 V

34 Cu hi Cho 1 s diode sau , v mch tng ng, b qua in tr thunca diode, xc nh dng in qua diode v in p ng ra VO , ccs liu khc cho trn s.

1.5

+

+

Vo

-


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Trang 18

p n S tng ng:

Dng in qua diodev in p ng ra:

VVo

mAI

65,105.2,2

84,42,22,2

7,022

1

35 Cu hi Cho 1 s o diode sau , ve mach tng ng, bo qua ien trthuan cua diode ,tnh dong ien qua moi diode va ien apngo ra VO , cac so lieu khac cho tren s o.

42V3,3KSi

Ge

Vout

1

p n Mch in tng ng :Dng in qua diode v in p ng ra:

)(3,0

)(4,123,3

3,07,042

VVo

mAI

36 Cu hi Cho mt mch n p c in p ng ra 10V., in tr ti RLbinthin trong phm vi 250


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Trang 19

PZMAX

= 10 . 32 mA = 320 mW

37 Cu hi Cho m t m ch n p c i n p ng ra 10V (hnh 2), zener c cngsu t tiu tn cc tn cc ti u l 20mW v cc i l 1W. Tnh i ntr t i nh nh t v ln nh t cho php Zener c tc d ng n p?Cc s li u khc cho trn s .

3

p n IS

= (22 - 10 ) /0,1K = 120 mA

I ZMIN = 20mW / 10 = 2 mA

I ZMAX = 1000 mW / 10 = 100 mA

I LMAX = 120 -2 = 118 mA R LMIN = 10 /118 mA = 84

I LMIN = 120 - 100 = 20 mA R LMAX = 10 /20 mA = 500

38 Cu hi Cho mt mch n p sau. Bit Zener c VZ=10V, PZMAX =400mW.a) Xc nh RL cc i Zener tiu tn cng sut ln nht chophp.b) Xc nh RLcc ti Zener trng thi ON.

3

p n IS

= ( 20 - 10 ) / 0,22K = 45,45 mAIZMAX = 400mW / 10 = 40 mAILMIN = 45,45 - 40 = 5,45 mARLMAX = 10 / 5,45mA = 1,83KILMAX = 45,45mARLMIN = 10 / 45,45 mA = 220

39 Cu hi Cho mt mch n p sau c in p ng ra Vo khng i 20V cungcp cho ti RL=1K v in p ng vo VIN thay i t 30V n50V. Xc nh ga tr ca in tr ni tip Rs v dng cc i qua

3

RL

(+)

(-)

100

UI=22V UO=10VZ

(+)

(-)

ILIR

RL

(+)

(-)

220

UI=20V ULZ

(+)

(-)


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Trang 20

Zener.

p n IL = 20 / 1K = 20 mA

ISMIN

= 20mA

RS

= ( 30 - 20 ) / 20 mA = 500

ISMAX

= ( 50 - 20 ) / 0,5K = 60 mA

IZMAX

= 60 - 20 = 40 mA

40 Cu hi Cho mt mch n p sau, zener n p 12 V c cng sut tiu tn cci 150mW v BJT c li dng =100 v b qua VBE. Tnh intr ti ln nht cho php Zener khng b qa ti? Cc s liu khccho trn s .

3

p n IZmax=150mW/12=12,5mAIS=(36-12)/1,5K=16mAIBmin=16-12,5=3,5mAICmin=100.3,5mA=0,35ARLmax=12/0,35=34,3

Chng 2: PHN CC TRANSISTOR1. Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 21.1Cu to BJT, FET

Hat ng BJT, FET

Cc s ni dy, c tuyn V-A ca BJT, FET

Mi quan h gia h s alpha v beta ca BJT

RL

(+)

(-)

RS

UI ULZ

(+)

(-)

RL

NPN

+


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Trang 21

B

C

I

I

BCE III

11

1

JFET , D-MOSFET thphng trnh Shockley:2

1 GS D DSS

P

VI I

V

Cc mch phn cc cho BJT v FET1.2

Bi ton 1:Tm im tnh Q, xc nh VB , VE, VC

Bi ton 2:Thit k mch phn cc

2. Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 2Mc tiu kim tra nh gi Ni dung

Mc Nh Cc kin thc cn nh:

BJT

B

C

I

I

BCEIII

1

CBOCBOCEO III )1(

BBBBCE IIIIII )1(

JFET , D-MOSFET thphng trnh Shockley:2

1 GS D DSS

P

VI I

V

SDII

AIG

0

SDII

AIG

0


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Trang 22

Cc thng s gii hn ca BJT v FET


Hiu c :

c tuyn V-A

Cc dng mch phn cc

Kh nng vn dng cc kinthc hc

cc kin thc m sinh vin phi bit vn dng :

So snh cc mch mc kiu EC,BC,CC

So snh cc mch mc kiu SC, GC,DC

So snh BJT v FET

Kh nng tng hp: Bi ton 1:Tm im tnh Q, xc nh VB , VE, VC

Bi ton 2:Thit k mch phn cc

3. Ngn hng cu hi v p n chi tit chng 2tt Loi Ni dung im

1 Cu hi Tnh ton in p phn cc v IC cho mch in hnh sau: 2

p n

VkmAVRIVV

mAAII

Ak

V

R

VVI

CCCCCE

BC

B

BECCB

83,62,235,212

35,2)08,40(50

08,47680

7,012

+

VCE10F

C2

= 85

C1 IC

RC3,3K

Ng vo ac

VCC = + 12V

RB240K

Ng ra acIB

= 50


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Trang 23

2 Cu hi Tnh VC v IC cho mch in hnh sau: 2

p n

VkmAVRIVV

mAAII

Ak

V

R

VVI

CCCCC

BC

B

BECCB

6,93,376,322

76,332,31120

32,31680

7,022

3 Cu hi Tnh ton in p phn cc VCE v dng in IC trong mch in hnhsau:

2

p n

VkmAkmAVRIRIVVmAAII

Ak

V

kk

V

RR

VVI

EECCCCCE

BC

EB

BECC

B

1,91635,32635,320

635,3)35,36(100

35,36531

3,19

1101430

7,020

1

VCC = 22V

VC

VB

20F

C1

ICRC3,3K

Vi

RB680K

Vo

IB

= 120

IECE40F

VoC2

10F

+

C1

10F

ICRC2 k

Vi

VCC = + 20V

RB430K

IB

VCE = 100

RE1 k


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Trang 24

4 Cu hi Tnh ton gi trin trRC nu c VC = 10V trong mch in hnh sau:

:

3

p n

mAAII

A

k

V

kk

V

RR

VVI

BC

EB

BECCB

635,3)35,36(100

35,36

531

3,19

1101430

7,020

1

C

CCCCC

R

RIVV

310635,32010

Suy ra :

kRC 75,2

10635,3

10203

Chn RC =2,7k.5 Cu hi Tnh ton gi tr RB transistor hat ng trng thi dn bo ha 3

p n Ta c:

EC

CC

EC

CECCC

CECECCC

RR

V

RR

VVI

VIRRV

2.0

(Transistor dn bo ha => VCE = 0.2V)

IECE

40F

VoC2

10F

+

C1

10F

ICRC1k

Vi

VCC = + 15V

RB

IB

VCE = 50

RE

0.5 k

IECE40F

VoC2

10F

+

C1

10F

ICRC

Vi

VCC = + 20V

RB

430K

IB

VCE = 100

RE1 k


25/54

Trang 25

/

7.0

C

ECCC

B

EEBECCB

EEBEBBCC

I

RIV

I

RIVVR

RIVRIV

6 Cu hi Tnh ton in p phn cc VCE v dng in IC trong mch in hnhsau (p dng phng php tnh gn ng)

3

p n

VVVVVV

VkmAVRIVV

mAk

VI

R

VI

VVVVVV

VVRR

RV

ECCE

CCCCC

C

E

EE

BEBE

CC

BB

BB

03,123,133,13

33,1310867,022

867,05,1

3,1

3,17,02

2229,339

9,3

21

2

7 Cu hi Cho s phn cc mt BJT NPN Si nh sau. Bit VBE=0,7V, lidng =100. Tnh:a) dng in tnh IC, IE, IBb) in p tnh VCEv in p b nhit VE

Cc s liu khc cho trn s .

2

IC

VCC = + 22V

RB1390k

RC10k

RB23,9k

CE10F

IE RE1,5k

VB 10F

10F

VC

VE

VCEVi

Vo

= 140

C1

C2

10K

100+


26/54

Trang 26

p n IC I

E= 100 I

B

6 = 10K . IB

+ 0,7 + 0,1K .100 IB

I B = 0,265 mA

I C I E = 26,5mA

VCE = 36 - 1K.26,5 - 0,1K.26,5 = 6,85V

V E = 0,1K.26,5 = 2,65V8 Cu hi Tnh dng in phn cc IE v in p VCE cho mch in hi tip in

p hnhsau:

3

p n in tr hi tip RBl tng ca hai in tr mc gia cc C v cc B:

VVV

kkmAVRRIVV

mAAII

Akkk

VRRR

VVI

ECECCCE

BE

ECB

BECCB

72,528,410

2,1302,110

02,103,20511

03,202,1351250

7,0101

Vi

C3

RC3 k

VCC = 10V

Vo

RE1,2 k

C310F

C2

10F

10F

R1 R2

IC

CI

IB

100k 150k

IE

+

= 50VCE


27/54

Trang 27

9 Cu hi Tnh dng in cc thu ICv in p VCcho mch in phn cc phn hnh sau:

3

p n

VkmAVRIVV

mAAII

Akk

V

RRR

VVI

CCCCC

BC

ECB

BECCB

02,124,249,218

49,22,3375

2,335104,276300

7,018

1

10 Cu hi Cho s phn cc mt BJT NPN Si nh sau. Bit VBE=0,7V, lidng =100. Tnh:a) dng in tnh IC, IE, IBb) in p tnh VCECc s liu khc cho trn s .

2

p n I C I E = 100 I B

12 = 220K . I B + 0,7I B = 0,051 mA

IC I E = 5,1 mA

VCE

= 12 - 1,2K .5,1mA = 5,88V

11 Cu hi Cho mt s khuch i Darlington gm 2 BJT T1 v T2. li dngca T1 v T2. li dng ca T1 v T2 ln lt l 1 = 80 v 2=20. Bit dng ng

3

Vi

C3

RC2,4 k

VCC = 18V

Vo

RE510

C310F

C2

10F

10F

R1 R2

IC

CI

IB

150k 150k

IE

+

= 75VCE

CE50F

VC

220K


28/54


29/54


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Trang 30

15 Cu hi Tnh gi tr in tr RE, RC v RCcho mch khuch i transistor viin tr n nh RE hnh sau. H s khuch i dng tiu biu catransistor l 90 ti im c IC = 5mA.

2.5

p n im lm vic c chn t cc thng s ca ngun v transistor l ICQ= 5mA v VCEQ = 10V.

VVVV CCEQ 220101

10

1

in tr cc pht:

4005

2

mA

V

I

VR

QC

EE

in tr cc thu c tnh bng:

k

mA

V

mA

V

I

VVVR

Q

QQ

C

ECECC

C

6,15

8

5

21020

Tnh dng in cc nn bng:

AmAI

IQ

Q

C

B

56,5590

5

Ta thy, in tr cc nn c tnh bng:

A

V

A

V

I

VVVR

Q

Q

B

EBECC

B 56,55

3,17

56,55

27,020

16 Cu hi Thit k mch in phn cc cho mt mch khuch i nh hnh sau.Vi = 150, IC = 1mA, VCQ = VCC/2.

2.5

IC = 2 mA

VCC = + 20V

RB RC

2N4401

CE50F

IB

RE

VB

10F

10F

VC

VE

VCE = 10VVi

Vo

= 140

C1

C2

+


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Trang 31

p nChn VVVV CCEQ 6,11610

1

10

1 v tnh RE:

kmA

VI

VR

Q

Q

C

EE 6,1

16,1

: VVV

V CCCQ 82

16

2

VVVVVVQQQ ECCE

4,66,18

Sau tnh RC:

k

mA

V

I

VVVR

Q

QQ

C

ECECC

C8

1

6,14,616

(s dng in tr 8,2k)

Tnh VBQ:VVVVVV BEEB QQ 3,27,06,1

Cui cng, tnh RB1 v RB2:

kk

RR EB 2410

6,1150

10

11

V t: VVVRR

RQBCC

BB

B 3,221

2

kRB 1431 (s dng 150k)

IC = 10 mA

VCC = 16V

RB1 RC

RB2 CE100F

IE RE

VB

10F

10F

VC

VE

VCE = 8VVi

Vo

(min) = 80

C1

C2

IBI1

I2


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17 Cu hi Xc nh dng cc mng IDv in p cc mng ngun VDScho mchin phn cc c nh hnh sau:

2

p n VVV GGGS 5,1

VV

VmA

V

VII

P

GS

DSSD 69,44

5,11121

2

VkmAVRIVV DDDDD 4,62,169,412 VVVVVV SDDS 4,604,6

18 Cu hi Xc nh in p phn cc VDS v dng in IDcho mch in hnhsau:

2

p n Ta c: do dng in IG = 0

2

1

0

p

GSDSSD

SDSDSGGS

V

VII

RIRIVVV

Gii h phng trnh trn ta c: ID, VGS VVIRRV DSDSDDD 20 => VDS = VDD(RD + RS)ID

+ 12V

G

S

D

ID

VDS

1M

1,5V

1,2k

IDSS = 12mAVP = 4V

Vo

Vi

1M

+ 20V

IDSS = 8 mAVP = 6V

750

1,5k

VDD


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19 Cu hi Tnh in p phn cc VDSv dng in ID 2

p n Ta c:

2

21

1

1p

GSDSSD

SD

BB

BDDSGGS

V

VII

RIRR

RVVVV

Gii h phng trnh trn ta c: ID v VGS VVIRRV DSDSDDD 12 => VDS = VDD(RD + RS)ID

20 Cu hi Tm in p phn cc v dng in ID ca mch: 2

p n Ta c:

)2(

)1(1

21

2

2

SDGSGGS

GG

GDDG

P

GSDSSD

RIVVVV

RR

RVV

V

VII

Gii h phng trnh (1) v (2) ta c: ID v VGS

+

100uF

CG

12K

R8

+

10uF

C3

12 V

2.2K

R10

D

K30A

2.2KR9

12K

R7RB1

RB2

RD

VDD

Rg1

Rs

Vi C1

Rg2

RdC2

Vdd

Vo


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VDD = (RD + RS)ID + VDS=>VDS = VDD(RD + RS)ID

Chng 4: OPAMP V MCH NG DNG

1Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 4

1.1

1.2Khuch i o :1

20

R

R

V

VA

i

V

1.3Khuch i khng o: 20

1

1i

RV V

R

1.4

Bi ton 1:Xc nh mch o hay khng o, cho v iv vo

Bi ton 2: Mch khuch i ghp 2 Opamp , xc nh vo

2Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 4Mc tiu kim tra nh gi Ni dung

Mc Nh cc kin thc cn nh:

Khuch i o :1

20

R

R

V

VA

i

V

Khuch i khng o: 20

1

1i

RV V

R


Hiu c hot ng ca cc mch khuch o v khng o, ngdng.

+VCC

-VCC

+VS Vd = Vi+ - Vi-

AVf

+VSf

-VSf -VS

V0

AV0

c tuy n truy n t khi c h i ti p m


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Trang 35


Cc kin thc m sinh vin phi bit vn dng :

Mch cng o du

Mch cng khng o du

Mch khuch i vi sai

Kh nng tng hp: Bi ton 1:Xc nh mch o hay khng o, cho v iv vo

Bi ton 2: Mch khuch i ghp 2 Opamp , xc nh vo

3Ngn hng cu hi v p n chi tit chng 4

tt Loi Ni dung im1 Cu hi Cho mt mch khuch o OPAMP c cc s liu cho trc nh

hnh v. Bit ng vo sin bin nh 0,5V, ng ra sin bin nhnh 20V.

15V

-15V

Vo

R2

4,7K

ViR1

a) Tnh li p v in tr hi tip. b) Vi li p tm c cu a v gi s ng sin bin nh 1V,v dng sng in p ng ra.

2

p n a)in p nh ng ra l 10V, ng vo l 0,5V. li p l10/0,5=20. in tr hi tip R2 =20.4,7K=94K

b)VIN=sint[V]Vout=-20sintBin nh dng +20V>ngun dng +15V nn b xn bi+15VBin 3nh m -20V


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a) Tnh in p ca mch.b) V dng sng in p ng ra VOUT so vi in p ng vo V IN. Cnhn xt g v nh dng v m ca dng sng in p ng ra?

p n A V = - 10K / 1K = -10

VOUT = ( - 10 ) 2 sint = - 20 sint [V]

2.5

3 Cu hi Cho m t m ch khu ch iOPAMP sau, bi t i n png vo VIN=12 sint[V].

a) Tnh l i i n p c am ch.b) V d ng sng i n png ra VOUT so v i i n png vo VIN. Tnh bin nh nh c a i n p ng ra.

2

p n A V = 1

VOUT = VIN = 12 sint [V]Bin nhnh ng ra l 18V

T/2 T

15V

-15V

Vout

t

-

+

T/2 T

9V

-9V

Vout

t


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Trang 37

4 Cu hi Cho mch khuch i o OPAMPa) Tnh li pb) in vo ct in p ng ra khi bit in p ng vo cho

trc ca bng sau

UI(V) UO(V)

-3-2

-1

+1

+2

+3

2

p n l i p = - ( 100K // 100 K ) / 10K = -5

UI(V) UO(V)

-3 +15

-2 +10

-1 +5

+1 -5

+2 -10

+3 -155 Cu hi Cho mt mch khuch a o sau:

a) Tnh li in pb) V dng sng in p ng ra theo in p ng vo V IN (t) = 2 sin

2

UO

+

-

UI


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Trang 38

100t (V)p n li p = - 100 K / 5 K = -20

V OUT = (-20) ( 2 sin 100 t ) = - 40 sin 100 t [V}

6 Cu hiCho mt s mch OPAMP nh hnh v. Bit V I1 =+3V v VI2=-4V. Tnh in p ng ra. Cc s liu khc cho trn s .

2

p n VO = ( - 22K / 2,2K ) ( +3 ) + ( 1 + 22K / 2,2K ) ( -4 ) = -74V boha mVO = - 15V

7 Cu hi Cho mt mch khuch o OPAMP c cc s liu cho trc nhhnh v. Bit ng vo sin bin nh 0,5V, ng ra sin bin nhnh 20V.

a)Tnh li p v in tr hi tipb)Vi li p tm c cu a v gi s ng vo sin bin

nh 1V, v dng sng in p ng ra.

15V

-15V

Vo

R2

4,7K

ViR1

2

p n li p = - 10 /0,5 = -20= 20 . 4,7K = 94K

30V

T/2 T

Vout

t

15V

-15V

Vo

22K?

2,2K?Vi1

Vi2

T/2 T

15V

-15V

Vout

t


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Trang 40

10 Cu hi Cho mt mch khuch i OPAMP sau. Bit in p ng voVIN=0,5sint [V]

a)V dng sng in ng ra Vo1 v Vo2 theo VIN bit in pVo2 c bin nh nh ln nht cho php m khng b xndo bo ha

b)T tnh li p ton mch v in tr hi tip R thamn iu kin

(Cc s liu khc cho sn trn s )

2

-

T/2 T

15V

-15V

Vo2

t

4 7K

R10K

15V

-15V

15V

-15V

++

2K


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Trang 41

p n OPAMP 1 c AV1 = + 6

VO 1 = 3 sin t [V]

VO 2 = - 15 sin t [V]OPAMP 2 c AV2 = - 15 / 3 = -5 li in p ton mch AV = (+6) . (-5) = -30R = 4,7K.5 = 23,5 K

11 Cu hi Cho mt mch khuych i o sau, tnh phm vi gii hn ca inp ng vo OPAMP khng b bo ha. Cc s liu khc cho trns .

1

p n

)(5,110

15

1010

100

maxVV

Av

i

Chng 6 : I S BOOLE V MCH LOGIC1Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng 6

1.1 Cac cng logic co bn

1.2 Cac quy tc c bni s Boole

a. 1 = a

(Vi b = 1)

a + 1 = 1

(Vi b = 1)

-

-

-


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Trang 42

a. 0 = 0

(Vi i b = 0)

a.a = a

a. a = 0

a(a + b) = a

a + 0 = a

(Vi b = 0)

a + a = a

a + a = 1

(Vi = 1)

a + a.b = a

1.3

nh l De Morgan

nh l 1: BABA .

o li : A + B = BA.

nh l 2: BABA .

o li: A.B = BA

1.4

Bi ton 1: T bng s tht xy dng biu thc Boole , v s logic.Bi ton 2:Cho biu thc Boole ,lp bng s tht,n gin biu thc biu din thnh s logic

2Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 6

Mc tiu kim tra nh gi Ni dung

Mc Nh cc kin thc cn nh:

Cac quy tc c bni s Boole

nh l De Morgan


Hiu c hot ng ca cc cong logic

Kh nng vn dng cc kinthc hc cc kin thc m sinh vin phi bit vn dng :Xy dng biu thc BooleV s logicLp bng s thtn gin biu thc v ly kt qa biu din thnh s logic

Kh nng tng hp: Bi ton 1: T bng s tht xy dng biu thc Boole , v s logic.


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Trang 43

Bi ton 2:Cho biu thc Boole ,lp bng s tht,n gin biuthc biu din thnh s logic


tt Loi Ni dung im

1 Cu hi Cho mt bng s tht sau ya) Xy dng biu thc Boole dng tng ca tch.b) V s logic

Input Out put

A B C Y

0 0 0 1

0 0 1 1

0 1 0 00 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

2

p n a) Y = A B C + A B C + A B Cb) S logic

2 Cu hi Cho mt s logic sau:

A

B

C

Y

2


44/54

Trang 44

a) Vit hm Boole ng ra.b) Lp bng s tht.

p na) Biu thc Boole Y = A B + B C

= 000 + 001 + 000 + 100

b) Bng s tht

Input Out put

A B C Y

0 0 0 1

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

3 Cu hi Cho mt biu thc Boole sau:

1) Lp bng s tht

2) n gin biu thc v ly kt qa biu din thnh s logic

2

p n )(011101111 BACACBCABCCBABCAY Input Out put

A B C Y

0 0 0 0

ABCCBABCAY

AB

C

Y


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Trang 45

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 01 1 1 1

4 Cu hi Cho mt biu thc Boole:

a) Lp bng s tht.b) n gin biu thc bng phng php i s, ri biu din thnh s logic.

2

p n ABACBCCABCBABCAABCY 110101011111 A B C

Y

A B C Y

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

5 Cu hi Cho mt cng logic sau y:

A

B

Y

High

Low

High

LowOutput

Inputs

1 2 3 4 5 6 7 8

V dng xung ng ra vi cng cc xung ng vo A v B.

1

CABCBABCAABCY


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Trang 46

p n

6 Cu hi Cho mt s logic sau:A

B

C

Y

a) Vit hm Boole ng ra.b) Lp bng s tht

2

p nY = A B + B C = 000 + 001 + 100

A B C Y

0 0 0 1

0 0 1 10

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 0

7 Cu hi Dng ton cng NAND xy dng 3 cng c bn AND, OR, NOT 1.5p n

A A

HLHLH

A

B

Y

BABA

A A

B B

AB ABAB


47/54

Trang 47

8 Cu hi Dng ton cng NOR biu din 3 cng logic c bn AND, OR, NOT. 1.5p n

9 Cu hi Dng ton cng NOR, v s logic biu thc Boole sau:Y=(A+B)(C+D)

1.5

p nA

B

A

C

D

A+B

C+D

A+B+C+D =(A+B)(C+D)

10 Cu hi Cho m bng s tht sau y:

a) Xy dng biu thc Boole dngtng ca tch b) n gin biu thc v biu dinbng s logic

Input Out put

A B C Y

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 01 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

2

p nY = A B C + A B C + A B C

BA

A

B

A

BA = AB

B

A

B

A

A

A A


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Trang 48

= A B C + A C

11 Cu hi Cho mt s logic sau:

A

B

C

Y

2

p n a) CBABY b)

Input Out put

A B C Y

0 0 0 0

0 0 1 1

0 1 0 00 1 1 0

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

12 Cu hi Cho mt bng s tht sau y

Input Out put

A B C Y0 0 0 1

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

2


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Trang 49

a) Xy dng biu thc Boole dngtng ca tchb) V s logic

1 1 0 0

1 1 1 1

p n a) ABCCBAY b) S logic:

Y

ABC

13 Cu hi Cho mt s logic sau

AB

C

Y

a) Vit hm Boole ng ra.b) Lp bng s tht

2

p n a) ))(( CBBAY

b) Input Out put

A B C Y

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1


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Trang 50

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 1

14 Cu hi Cho biu thc:

a) V s logicb) Lp bng s tht

2

p n a) CBABY A

B

C

Y

b) Input Out put

A B C Y

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 11 1 1 1

Chng 7: MCH M

1Cc ni dung kin thc ti thiu m sinh vin phi nm vng sau khi hc xong chng7

1.1Bng trng thi ca JKFF, TFF

CBABY

FF tc ng c nh xu ng

K

JCK

Q

QK

JCK

Q

Q

Flip-flop tc ng cnh ln

Flip-flop JK.

0


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Trang 51

Bi ton 1:Thit k mch m ln bt ng b MOD M

Bi ton 2: Xc nh s Flip Flop JK cndng cho mch m MODM?V s logic, gin xungv tn s ng ra

2Cc mc tiu kim tra nh gi v dng cu hi kim tra nh gi gi chng 7

Mc tiu kim tra nh gi Ni dungMc Nh cc kin thc cn nh: Bng trng thi ca JKFF, TFF


mch m ln bt ng b MOD M


Thit k mch m ln bt ng b

Kh nng tng hp: Bi ton 1:Thit k mch m ln bt ng b MOD MBi ton 2: Xc nh s Flip Flop JK cn dng cho mch mMODM?V s logic, gin xung v tn s ng ra


tt Loi Ni dung im1 Cu hi Kho st mch m MOD 7 dng Flip Flop JK qua cc bc phn tch, v

s logic, vDnh xung m, lp bng m.

3

p n a) Phn tch : MOD 7 : m t 0 n t 0 n 6 v s cn xa 7 , dng 3FF JKb) V s logic:

Flip-flop T.

Ck


52/54

Trang 52

JQ

Q K

JQ

Q K

JQ

Q K

CK CK CK

1

1

1

1

1

1

RESET

FFC FFB FFA

CLOCKf/7

7

f

c) V dng xung m

Clock

K

J

QA

QB

QC

1

1

d) Bng m MOD 7

Xungm C B A S m

1 0 0 0 0

2 0 0 1 1

3 0 1 0 2

4 0 1 1 35 1 0 0 4

6 1 0 1 5

7 1 1 0 6

2 Cu hi a) C bao nhiu Flip Flop JK cn dng cho mch m MOD 8?b) V s logic v tn s xung clock nu ng ra bit MSB c tn

s l 15 Hz.

2

p n

F = 8.15HZ = 120 HZ3 Cu hi a) C bao nhiu FlipFlop JK cn dng cho mch m MOD 15? 2

JQ

Q K

JQ

Q K

JQ

Q K

CK CK CK

1

1

1

1

1

1

RESET

FFC FFB FFA

CLOCKf/8


53/54

Trang 53

b) V s logic v tnh tn s xung clock n ng ra bit MSB c tn sl 25Hz.

p n a. 4 FFb.

Tn s xung clock=15.25=375Hz

4 Cu hi a) C bao nhiu Flip Flop JK cn dng cho mch m MOD8?b) V s logic v tnh tn s xung clock nu ng ra ca bit 22c tn s200 Hz.

2

p n

a. 3 FFb. Tn s xung clock = 8 . 200 Hz = 1600Hz

Ngn hng cu hi thi ny c thng qua b mn v nhm cn b ging dy hc phn.

Tp.HCM, ngy . 29. . thng . . 6. . nm . .2007 . . . .

Ngi bin son(K v ghi r h tn, hc hm, hc v)

ThS L Hong Minh

T trng b mn:GVC Vi nh Phng . . . . . . . . . . . (K v ghi r h tn, hc hm, hc v)

Cn b ging dy 1: THSL Hong Minh. . . . . . . . . . . (K v ghi r h tn, hc hm, hc v)

Cn b ging dy 2: THS Nguyn Th Lng . . . . . . . . . . (K v ghi r h tn, hc hm, hc v)

1f

f/15

1

1 1 1

1 1

JQ

Q K

SET

CLR

JQ

Q K

SETJQ

Q K

CK CK CK

1

1

1

1

1

1

RESET

FFC FFB FFA

CLOCKf/8


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