Bai Tap Ky Thuat Do Luong_2

8/3/2019 Bai Tap Ky Thuat Do Luong_2

1/21

DO LL/ONG dien bAitap

CHLTONG I: DO DiEN AP VA DONG DiEN

1.1 Mot ampe-ke dung cO cau do tCf dien c6 dien trd cO cau do R(m) =99Q va dong

lam lech toi da Imax = 0,lniA. Dien trd shunt Rs = IQ. Tinh dong dien tOng cQng

di qua ampe-ke trong cac trLfdng hOp:

a) kim lech tOi dab) 0,5Din; (FSD = Imax, full scale deviation)

c) 0,25Dm

a) kim lech toi da

Dien ap hai dau cO cau do:

Vn,=Im.Rm=0,lmA.99Q=99mV

IsRs=Vn,=> Is

Vm _ 9,9mV

Rs ~ 129,9mA

Dong tOng c6ng:I = Is + I = 9,9 + 0,1 = 10mA

b)0,5D,\

Im= 0,5 . 1mA = 0,05mA

Vm = Im.Rm = 0,05mA.99Q = 4,95mV

Vm 4.9SmVIs == = 4.95mA

Rs in

I = Is+ Im= 4.95mA + 0,05mA=5mA

c)0,25mA:

Im= 0,25.0,1mA = 0,025mA

Vn,= In,Rm= 0,025mA.99Q = 2,475mV

Trang 1


2/21

Vm lOOmV

DO ll/Ong dien bAi tap

Vm 2,475Io==- = 2A75V

Rs 1

1.2 Mot cO cau do tLf dien c6 1= 100]jA, dien trd nOi khung quay R= IKQ. Tinh dien

trd shunt mac vao cO cau do de trd thanh mot ampe-ke tLTOng Lfng v6i hinh 1.1.

a) Dm = 100mA = tam do 1

b) Dn, = lA = tam do 2

Giai:

a) d tam do 100mA

Vn,= ImRm= 100.1 = lOOmV

It = Is+ Im => Is = It -Im = lOOmA - lOOpA = 9,9mA

Rs ==-= 1,001QIs 99,9mA

b) O tam do lA:V,= I,Rn,= 100mV

Is= It - Im = lA- 100)_iA= 999,9mA

R= = i0 = 0,10001flIs 999,9mA

1.3 Mot cO cau do tCf dien c6 ba dien trd shunt dl/Oc mac theo kieu shunt ayrton sCf

dung lam ampe-ke. Ba dien trd c6 trj so Ri=0,05Q, R2=0,45Q, R3=4,5Q, Rm= IkQ,

Imax= SOpA, CO mach do nhu' hinh sau, tinh cac trj s6 tam do cCla ampe-ke

Hinh B.1.3

Giai:

Trang 2


3/21

Vs 50mV

Is =" = 777 = lA.I = 50pA+lA=l,00005A = lA

0i?n trd t m Im = Imax =100]jA

looy100

=999KQ

Trang 3


4/21

DO LWONG DIEN bAitAp

Tai do lech 0,75 Dm

In,= 0,75.100pA = 75pA

V= URs+ Rm) 75pA(999kQ +lkQ)=75V

Tai do lech 0,5 Dm

Im = 50 ]lA

V= 50 viA(999 kQ+lkQ)=50V

Tai do lech 0,25 Dm

V= 25pA(999 kQ+lkQ)=25V

1.5 Mot cO cau do tLf dien c6 Imax=50 pA; Rm =1700 Q dLTOc sLf dUng lam von ke DC

CO tarn do lOV, 50V, lOOV. tmh cac dien trd tarn do theo hinh sau:

a)

Hinh B.1.5

Giai

Theo hinh a:

R+R. =V

V loy= >R, =--Rm =--1700n = 198,3/cn

Inax 50M

R, = -17002 = 998,3kQSOfjA

R = - 1700a = l,9983Mn' SOjuA

Theo hinh b:

Trang 4


5/21

m

RJ/R,


50M

+1 +2 = 7

R, = -R,-Rm = - 198,3ki-1700= SOOkQImax 50//A

y3 y,R+R+R+R- -- >i?3 - ---R-R-R

max

= - SOOkQ. -198,Ska -1700 = IMQ.SOfjA

1.6 Mot vonke c6 tarn do 5V, dLTOc mac vao mach, do dien ap hai dau dien trdR2 nhLf hinh sau:

a) Tmh dien ap Vr2 khi chLTa mac Vonke.

b) Tmh Vr2 khi mac von ke, c6 dp nhay 20kQ/V.

c) Tinh Vr2 khi mac von ke, c6 dp nhay 200kQ/V

+

e

2V J

70kn

. > T50.n

Hinh B.1.6

Giai:

a) Vr2 khi chLfa mac Vonke.R2

y2 =-E-= i2y-R1+R2

SOkO.

70kQ. + SOkQ.= 5y

b)V6i von ke c6 dp nhay 20kQ/V.Rv=5V.20kQ/V=100kQ

Rv//R2=100kQ//50kQ=33,3kQ

= E= i2y-

Vr2= "2

333kQ.

70kQ. + 33,3kQ.= 3,87V

Trang 5


6/21

DO LUONG dien bAi tap

c)V6i von ke c6 dp nhay 200kQ/V

Rv=5V.200kQ/V=lkQ

Rv//R2=lMQ//50kQ= 47,62kQ

47,62/(f2y2 = 12V-=4,86V70kQ. + 47,62m4,86V '

1.7 Mot cO cau do tLf dien c6 Ifs= lOOpA va dien tr73 cO cau do Rm =lkQ

dLfpc sLf dung lam vonke AC c6 V tarn do = lOOV. Mach chlnh llAi c6

dang cau sCr dung diode silicon nhU hinh ve, diode c6 VF(dinh) =0,7V

a) tfnh dien trd nOi tiep Rs

b) Tinh dp lech cua vonke khi dien ap dUa vao vonke la 75V va 50V (trj hieu

dung-RMS).

c) Tinh dp nhay cua von ke. Tin hieu do la tin hieu xoay chieu dang sin.

-SVv-HiD.

Hinh B.1.7

Giai:

a) Tinh Rs:

Day la mdch chlnh lulltoan ki nen ta c6 quan he:

Ip(tri dlnh)= Itb/0,637

Vm (trj dlnh)= yf2V

CO cau do c6:

Trang 6

D,y


7/21

DO nhay= ' = 9,009kQ./V

DO LL/ONG DIEN

= I,, = 100/uA = 157/uA0,637

taco:

bAitAp

1A14V,,-2V, 1A14V,-2V,

Rs + Rm Ip

-Rm

_ (l,414.100y)-(2.0,7V)

157

b)KhiV = 75y

-lkD. = 890,7kD.

I,, = 0,637/ = 0,6371,414V-2Vp-- = 0,637

Rs +R.

(I,414x75y)-(2x0,7y)

890,7kQ. + lkQ.

I,, = 75M

KhiV = 50y

* 890,7/cf2 + l/cf2

c)Z = 157fjA => I(RMS) = 0,707/P = 0,707xl57;iA = 111

R =looy

= 900,9ka.lllfjA

Qnn Qko

lOOy

1.8 Mot cO cau do tCf dien c6 Ifs = SOpA; Rn, = 1700Q ket hOp v6i mach chlnh llAi

ban ki nhLT hinh sau. Diod silicon Di c6 gia trj dong dien thuan If (dinh) toi thieu

la 100 pA. Khi dien ap do bang 20% Vtimdo, diode c6 Vf = 0,7V, von ke c6 Vtam

do = 50V.

a) Tfnh Rs va Rsh

b) Tinh dp nhay cua Vonke trong hai trl/dng hpp: c6 D2 va khong c6 D2

Hinh B.1.8

Giai:

a)Tmh Rs va Rs

Trang 7


8/21


0 day sLf dung chlnh llAi ban ki nen ta c6:

Ip=Itb/(0,5.0,673): trj dinh trong trl/dng hOp chlnh llAi ban ki

CO cau do CO Ifs = Itb = 50 ]jA=> Im= 50 )_iA/(0,5.0,673) = 157 )-iA(tridinh)

Khi V= 20% Vtd, IF(dlnh) c6 gia trj 100 pA. Vay khi V= Vtd, IF(dlnh) c6

gia trj:

20%

hn => hn 500 -157 = 343juA

= 157/JAX1700Q. = 266,9mV

266,9. V,,78

343M

_ I,4i4y,, -y, -y,

Rs

l,414y,-y,-y, l,414x50y-266,9my-0,7y

Ip 500//A

h)Tmh do nhdy:

Co D2 trong ban ki dLTOng, dong qua D1 c6 gia trj dinh: If=500 pA

Trong ban ki am, dong qua vonke c6 gia trj dinh:

J1 1.414.50VRs 139,5/(n

hieucuns = 0,707.500 = 353,5fjAiRMR)c

SOViRMR)

353,5fjAiRMR)

Donhay = = 2,8m/V50y

Khong CO D2:

Trong ban ki dU'Ong:lF(dinh) = 500 pA. Trong ban ki am: 1 = 0

Trong chu ki cua tin hieu:

Ihieu dung ~0,5I F(dinh)

V6i Ila dong dien mach chinh chay qua Rs trong ban ki dLfOng.

Trang 8


9/21

ii-,. j (I,smoxrd, =

DO LL/ONG DIEN bAitAp

2

ZI g 4

I = 0,5.500juA = 250juAsov

R =-

= ZOOkQ250/zA

Do nhay:== 4ka/V~ 5oy

1.9 Mot ampe ke sLf dung cO cau do tCf dien c6 cau chlnh iLTu va bien dong nhLT

hinh ve. Biet rang cO cau do c6 Ifs = 1mA va Rm = 1700Q. Biet dong c6 Nth =

500; Nso= 4. Diode c6VF(dlnh) = 0,7V; Rs=20kQ. Ampe ke lech toi da khi dong

sOf cap Ip = 250mA. Tmh gia trj Rl.

Hinh B.1.9

Giai:

Chlnh iLTu toan ki nen ta c6:

Itb _ 1mA

Im(trjdinh)- ~ 0,673 " '

Dien ap Em d hai dau cu6n thiJ bien d6ng(tri dinh):

Em = (Rm+Rs) + 2Vf = l,57mA(20kQ + 1700Q) + 1,4V= 35,5V

Es(tr/hieu dung) = (0,707.35,5V) = 25,1V

Dong lam lech tOi da cO cau do c6 trj hieu dung I:

I = ll,lltb = ll,l.lmA=ll,lmA

Ta c6:

Trang 9


10/21

DO LUONG DIEN bAi TAP

N 4Ith. =Iso= 250mA-= 2mA

500

ItKu = -f, +-1 => = 2mA-11,1mA = 0,89mA; (v6i Iq=Iquacacauao)

0,89mA

CHirONG II: DO DiEN TRO

2.1 Cho Eb = 1,5; Ri= 15kQ; Rm =lkQ; R2 = IkQ; Imax = 50pA. Xac djnh trj s6 dpc

CUd Rx khi lbImax; ImV2 Imax; Im3/4 Imax

Gi'ai:

Tai Im =Iinax = 50]jA; Vm = Imax >< Rm = 50]jA X ikQ = 50mA.

y, SOmV Do do: - ~ - 50/zA. NhLT vay dong dien: lb = lOOpA.

R.2 iKii

E,Vay R+Ry#Neu R, + R R2 URm 500n.

1 sv# . ' . = 15/(n. +15kQ = 15kQ; Rx = OQ.

100//A

Khi Im =1/2 Imax = 25]iA; Vm = 25mV => I2 = 25pA.

i,5ySuy ra lb = 50uA. Vay Rx + Ri # ~r7 ; Rx # 15kQ.

50;zA

TLTOng tli nhLT cach tmh tren. Im = 3/4 Imax = 37,5pA.

lb = Im + 12 = 37,5)aA + 37,5)aA = 75pA.

1,5VRx + Ri = ' = 20kQ, Rx = 5kQ.

/ Id

2.2 Mot ohm-ke loai noi tiep c6 mach do (Hinh dLfdi day). Nguon Eb = 1,5V,cOf cau do CO Its = lOOpA. Dien trd Ri + Rm = 15kQ.

a)Tinh dong dien chay qua cO cau do khi Rx= 0.

b)Tinh trj gia Rx de cho kim chl thi c6 dO lech bang 1/2 FSD, 1/4 FSD, 3/4 FSD

(FSD: dp lech toi da thang do.)

Trang 10


11/21

DO LL/ONG dien

E

Hinh B.2.2

Giai:

bAitap

a. =

1,5V

R+R,+R 0 + lSkQ.:100/zA(fsD).

b. Dp lech bang 1/2 FSD:

Im =lOOjuA

- SOpA (vi cO cau do tuyen tmh.)

1,5V-15kQ.15kQ

I I 50juA

Dp lech bang 1/4 FSD:

I == 25fjA- R,=-15kQ. = 45kQ.4 25fjA

Dp lech bang 3/4 FSD:

L = 0,75 X 100]jA = 75]iA; R, =1,5V

75//A

-15ka = 5ka.

2.3 Mot ohm-ke c6 mach do nhu' hinh sau. Biet Eb =1,5V, Ri = 15kQ; Rm =

50Q; R2 = 50Q; cO cau do c6 Its = SOpA.

Tmh trj gia Rxkhi kim chl thj c6 dp lech toi da: (FSD); 1/2 FSD va 3/4 FSD.

Trang 11


12/21

T Chmd '0* f

DO LL/ONG DIEN

r*

Hinh B.2.3

Giai:

Khi kim lech toi da (FSD):

Im= 50)_iA; Vm = Im.Rm = 50)_iAx50Q = 2,5mV.

i?2 son

Dong dien mach chmh: lb = I2 + Im = 50]jA + 50]jA = lOOpA.

15kQ.100

Rx= ( Rx+ Ri) - Ri = 15kQ - 15kQ = 0Kim lech 1/2 FSD:

, l,25mV ,In, = 25viA; Vm = 25]jA x 5OQ = l,25mV; 12 = = 25juA

lb = 25)aA + 25)aA = 50]jA.

bAitAp

K+1=15V

50= 30kQ.; Rx = 30kQ - 15kQ = 15kQ.

Kim lech 3/4 FSD:

In, = 0,75 X SOpA = 37,5]jA; Vn,= 37,5]jAx50Q = l,875mV.

I2-l,875mV

502

1,5V

37,5/zA; lb = 37,5viA + 37,5]jA = 75viA.

= 20ka => i? = 20ka - iska = ska75//A

Trang 12


13/21

" ' ' R,+R, 0 + 15kQ.


2.4 Mot ohm-ke co mach do nhiu hinh bai 3. c6 nguOn Eb giam xuOng chl con

1,3V. Tmh trj gia m6i cua R2 ?.?lai cac gia trj Rx tLTOng LTng v6i dp lech

cua kirn: 1/2 FSD, 3/4 FSD.

Giai:

Rx = 0; = 86,67/zA

L = SOpA (FSD); 12 = lb - L = 86,67pA - SOpA = 36,67)aA.

y 2 SmVY= URn, = 50]jA X 50Q = 2,5mV; Rz = 68,18

I2 30,0/jUA

Khi kirn lech 1/2 FSD:

In, = 25)aA; Vn,= 25)aA x 5OQ = 12,5mV

i?2 68,12Ib=Im + I2 = 25)aA + 18,3)aA =43,33]-iA

_ 1,3V

I, 43,33MKhi kirn lech 3/4 FSD:

L = 0,75 X SOpA = 37,5)aA; Vn, = 37,5)aA x 5OQ = l,875mV.

l,875my 12 = = 27,5M: Ib=37,5]jA + 27,5viA = SSpA.

V i3y

i? +R =-20/(f2=> R20kQ-15kQ.SkQ

I, GSjuA

2.5 Tmh dong dien chay qua cO cau do va dp lech cUa kim chl thj cUa ohm-ke

CO mach do nhLT hinh ve khi ta sLf dung tam do Rxl trong hai trLTdng hpp:

a)Rx = 0

b)Rx=24Q

Trang 13


14/21

DO LUONG DIEN

StVHtl&lnl

3.e2ki-

bAitap

EhfnJi2. aygjcfi

SerrjMo IQH

:s3eKfi eOkl il470t:n JtflDkrt ukn

R*10k Hilk R.i.W R=10RkI

Khda _lm do

+ 1ISV

'Cdng EC' di CKuyen

aJ

t.&v

R.-Wi

b)

Hinh B.2.5

Giai:

Mach tLTOng dLTOng cua ohm- ke khi ta sLfdung tarn do Rxl trong hai

truwowg\ngf hOp Rx = 0 va Rx = 24Q nhLT sau:

J_W_

Rx-0; b 14Q. + [l0a//{9,99kD. + 2,87SkD. + 3,82kQ.)]

15V

142+ (102//16,685/(f2= 62,516mA

Dong Im chay qua cO cau do:

ion/ = 62,516mA-

10a + 16,685kQ.

In, = 37,5pA = Ifsi Khi kim lech toi da.

Rx = 24Q:

h =l,5y

, , ,,= 31,254mA24n + 14n(lOn //(l6,685/(n))

T = 31,254mA102

10f2 + 16,685/cn18,72/zA: kim lech 1/2 FSD.

2.6 Tfnh dong dien chay qua cO cau do va dp lech cCla kim chl thj cUa ohm-ke

CO mach nhLT hai 5, khi sLf dung tam do RxlOO va RxlOk trong tru'5ng hpp

Rx = 0.

Trang 14


15/21

In, = cncim 37,5/(n = If,: Kim chl thj lech tOi da.

DO LUONG DIEN bAitap

70ii ea6-(i E,a75lcfl

1.5V 3.a2i


16/21


2.7 Ta do dien trd bang cach dung phLTOng phap V va A dLTOc mac re dM.

Ampe-ke chl 0,5A,von ke chl 500V.Ampe ke c6 Ra = 10Q,10kQ/V. Tmh gia

tri R.

Hinh B.2.7

Giai:

E + Ea = 500V; I = 0,5A

R = = = 10002I 0,5A

R = lOOOQ - Ra = lOOOQ - ICQ =990Q.

2.8 Cac ampe-ke, von ke va dien trd R d bM 2.7 dl/Oc mac re ngan. Hay tmh

dp chl cua von ke va ampe-ke (nguon cung cap van la 500V).

Hinh B.2.8

Giai:

Noi trd cCia von ke :

Rv= lOOOV X lOkQ/V =10MQ

Rv // R = lOMQ 11 990Q = 989,9Q

Trang 16


17/21


, SOOVx{R//R) 5000yx 989,92 Do chl cUa vonkG ~-1-\~~ 495V

R+{R+R) loa + 989,9n

_ E _ 495V. DO Chi cLIa ampe-ke: -/ + /.-

CHLTONG III:

DO DIEN DUNG, DiEN CAM VA HO CAM

3.1.Cho cau do nhLT hinh ve , biet Ci =0.1|jF va tl s6 R3/R4 c6 the chlnh dLTOc

thay doi trong khoang : 100/1 va 1/100 . Hay tmh Cx ma cau c6 the do dLTOc.

Giai:

Ta c6: Cx = C1R3/R4. Vdi: R3/R4 =100/1

=>Cx = 0,1ijF(100/1) =10|jF

V6i : R3/R4 =1/100 => 0,1ijF(1/100) =0,001ijF

Vay cau CO tarn do : tCf 0,001|jF lOpF

3.2. Cho cau dien dung nhLT hinh sau, thanh phan mau Ci =0,lpF ;R3 =10kQ. Biet rang

cau can bang khi nguOn cung cap co f = lOOHz; Ri =125Q va R4 = 14,7Q . Hay tinh gia

trj Rs, Cs va he so ton hao D cua tu?

Trang 17


18/21

DO LirONG DIEN

Ta CO : Cs =CiR3/R4;

Hinh B.3.2

Giai:

bAitAp

0,ljuFxl0kQ.Cs=-

14,7kQ.

= 0.068|aF;

i?iXJ?4 125nx 14,7mRs= = --=183.3Q

R. lOkQ.

D =CsRs = 2n . lOOHz xo,068iiFX 183,8Q = 0,008

3.3. Cho cau dien dung nhLThinh sau, thanh phan mau Ci =0,1]jF ;R3 =10kQ. Biet rang

cau can bang khi nguOn cung cap co f = lOOHz; Ri =125Q va R4 = 14,7Q . Hay tinh gia

trj Rs, Cs va he s6 tOn hao D cCia tu?


19/21

Rs = ' = --=183.3Q

DO LL/ONG DIEN bAitAp

Hinh B.3.3

Giai:

Ta CO : Cs =CiR3/R4;

0,luFxl0m

ujka

125axl4,7ka

i?3 lO/cH

D =CsRs = 2n . lOOHz xo,068|jFX183,8Q = 0,008

3.4.Cau Maxwell do dien cam dung thanh phan mau C3 = 0,l|iF, nguOn cung cap c6

tan so f=100Hz. Cau can bang khi Ri =l,26kQ; R3= 470Q va R4 =500Q .Tmh trj gia

dien cam Ls, dien trd Rs va he s6 pham chat Q cUa cuQn day.

Hinh B.3.4

Giai:

Ta CO :Ls =C3RiR4 =0,l|iFXl,26/(a x 500 = 63mH

R:R.

l,26kQx500Q.

470a l.,34kD.

coLS Ittx 100Hz x63mH Q= -=-= 0,03

R. l,34kQ.

Trang 19


20/21

DO LL/ONG dien bAitAp

3.5. Cau CO nguOn cung cap f= lOOHz can bang khi C3 =0,1|jF, Ri =l,26kQ , R3 =75Q

va R4 =500Q. Tmh dien cam Lp ,dien trd Rp va he s6 pham chat Q cUa cuQn day?

Hinh B.3.5

Giai:

Lp- - 0,lpFXl,26/(nx SOOQ = 63mH

R,R. l,26kax500Q.

75Q

Q= =-= 212(oLp 2n X lOOHz x 63mH

3.6. Hay tmh thanh phan tLfOng dLfOng Ls,Rs cUa cuQn day c6 :Lp =63Mh ; Rp = 8,4kQ (

R X

Rs = / \;the: Rp = 8,4kQ ; Rp = 7,056x 10'; Xp = coLP

>Xp =2 TT X 100HzX63mH =39,6Q

Xp=l,57xlO';Xp +i?p=7,056 XIQ

8,4/(nx 1,57x10

7,056x10'= 0,187 ;

Trang 20


21/21

DO ll/Ong dien bAi tap

7,056x10' X 39,6Xs = 39,6Q

7,056x10'

X, 39,6nLs ==--= 63mH

a IttxIOOHz

3.7. Hay tinh thanh phan tLTOng dLTOng Cp ,Rp cua tu dien c6 Rs =183,8Q va Cs

=0,068Mf (f=100Hz).

Gi'ai:

Ta c6: Rp =( Rs'+Xs' )/Rs; Rs' = (183,8)' =33,782x10'

Xs =l/2TTfCs = 1/(2tt.100Hz.0,68]jF) =23,405. lO'Q

X 5=5,478.10

Rp =( 33,78.10' +5,478xl0')/183 = 2,99MQ

Rl+X 33,78x10'+5,478x10Xp =- -=23,41.lO'Q

Xs 23,405x10'

Cp = l/(2TT.100Hz.23,41kQ)= 0,068|jF

Trang 21

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Bai Tap Ky Thuat Do Luong_2