additional notes and examples.pdf

ME2143/ME2143E – Sensors and Actuators  

Part II ‐ Additional Notes and Examples 

CHUI Chee Kong, PhD 

Associate Professor 

Department of Mechanical Engineering 

National University of Singapore 

Review of Magnetic Field 

d"'l' p* A* "t@. E t a.t e, o,<? o"]

b = t b, [, b3 b"]

q.b 2- iatbtt-t

= a.,If t a.L b, + '+4*L^

[^ € *.('.'A''o* T-*\ t

@.. b = \^\\ttcc.:S o

[^1, \[ t *,* +(^- *fo"h,.^' t'*St[ (**5^,-["-t".)

+ eu *,^X b

*^gu

c*:,( 4 ou.<..a'- A n,'Lt:ce. vt

e.-.h^ ci t^"....d*t"^" $*"-t"0

0=?"CaO= I

Flux = AB

Figure 1

+L"-- o *ll

-t{^- ^T(".-

vr,ua't-a-.,F-c [lox # tt^*+ {(^'

f = Ats cta e

ACoil of l- turn

,**!-"a a-pl

"l lLlo J"*e\ B

"*t ( L

Coil of l- turn

B=1 T

e.:tt^€-- -a'L..- 'sSre-T'eL +

c-t^--yJ , ff

a.,\ t* .t"'"^3e-

{t ,*-"}L tL*- c,-)J

(7t< 6o degrees

\B cos 60

z Flux = AB cos 60

Figure 2

.H--- $..-\J B

; 'ti*^,,....j f o

9, e^*- fi. t

(,[

;

pb'f

6 r.*r^^gt.

C^) (F.5..,*- \) A Ac,.r-rr.ri.r "o.'( 4 (o **,ns

*.,^A a^{,E-o. + ,\ lo-1 *t ; p(o<e{ ; 4

t r,ti [."^ ntalae-ttc l.'-(d * [\'tx A"^s\ B

+ (0-z T so t1^.* {*Te. }t.^ t,",h +l^- +\^'a^S vtorrno'-\'

# = AB\v

F(** [;hk*;"",.\=Nf =NAB

r'f ub "T^lt t uaxzrl.t l(,.^ 'f{^.S

prar-Qur.cesil4oirc*'f4[t*'rrl

a.r\ .,r*[ 4 I vt(* , -L*^ tt^t lLx

; ,e)*.e-.4 t ?4rr.o wr'(h&t I ce,,,' "d

= to y e1 x[o-L*L x (o-tT

= + x (6-3 {Ub fit,.,l'-s.

( D ( F.1,.*.- z) Tt*- satqe- .o,{ A '"of^*c"t *G.^#

a,^ ot,'s *Q*,L )b '*,;Adte eo t'{""}

if t.ra s flyL ( o o 'in f''' d'e("t B '

F(** (;^(-o5as, A = VAB cay (o'

: 4 r (o-3 r o.S

.= L r (o{ 2<19/bt,.ry.

Rnr .\es *t^,,\L *t^L eot'l ,

AA = zr lo+r&r6t-.a.s - +,c (a-ifi4if,^-,.,

= -z ^"L.lb -h.rcr

M^n ",

*-t"( r , e-rft Scerer*: l-,-^te-

crr-v=L €Jiarlt\

A-lra--,k

o* -J

I-t".^,5

++4 VTl

C*o* P^ 4.,^- .l

C..= @ xb =

i,jK

^^ 4y qL

b.. by b"

€(^._ "^5"--F*J.-

= (atb,-6=u.y)i --

(a*1. -azbr)j fCa"!) *ayb/) h

= eit s,-l e

-gz l{T.rLok,.-^

F.*-l

Bf = xY':> I 1' .,) ."o^ I2-

\ :-t#' t+'ro= o I

\, o o\

: O' -'(+*(a'uL r t')

= -4xto-a ( Nl

all

I

,,ItII{,/*eu

d zutY- { }"Bt^

4., qr

[,i,k &

,,r-** u'41 C,"[-,t'lS ^ltL *1,.,e gl &-^(.

axql .

,-L* Xau"^,yle-

A" et.c-t",-a (t= -t.Gcr2 )<

[o E ,^ 1r ; {L p.osu,Fr\^<.

Tt* nra.trr,ta-tac $(o* -u\

pcs "4.--* y d.-*dd^

F,x -t{*"^^^A

en\ tt*- e-(eet{,r* .

L

A

(o-ttC ) ta^,ejs ".-*-

X dw*d..Lt-;r[;tr*

;f.*,*c$..,^-r "[ +L.. S^-.q

ltsr )r ts(*t{.6o 2 rto

C r r)uy-

t( - t.6o2 r

t -. (.,."'S "-l-r^(

L-= 7urcB

Ja

-"L)(tut ^ls)M, x

I U. a^y t*(e_ If-r.6ue'rd' fi a iln i

,")q/z '* a I

*L(r

tu,,_

Uy . M-' a*e- u*,-'*

\,*-4f-ec t.;e- a<e4t

tn-ff'W :

D.;.- cs; :

t'"-f, o r r

*e-cnt**n'-<-d

:l?(C>

Z

i-l

4."r., C*G.

Eo*qrt.

A urrie- "t Gte ! = ( * .ouvt\e^ q. cc!?c^er4*-

,t to A ?unY**s'Jo- 'ta c( J.'*t"q "[ B E' o' 5 T

"

C.="^^n^l- 'tt= *-'$"-*(* + +t* d--=* "''^-

*tt€- c-Jre

G- .{ ,$'o't(^{- t-':"'')"q + (*t* -!' '^^J a- ck&s&"'i:^ts

I = t/Bg;@)e .), 'eL€- *-SG tt4'-t,^ +t'*- u-':<- **J I'"-tJ "

I z (to A) C I u) ( u.s T) si;^ (1"'),\ -&=to'

=(to A)( r,,^) (o.sr)(t) *.?*r,-^,.J,u

F.5A/t*

?o"n)\" Do*rT

g \ l*"-^5.L lul'^-t*Y\X

+

Tolv*o*e*e/ i^,tsVr ^1c.:iI

!.t''c<- ol--n+-'J_ e1**\SYt

^^-\ '*"'isVa'^-*- t * <'

J"-l *lu^ wtr "

tr** EMF

E l - ct,r*ro['i'e- t-'^e C eH F')

Ll a LA*^/ d'{ \ #-=

7el

F*r-Aoo', t",^t .

t

)-

Example: Magnetic Field around a Long Straight Wire

Consider a long straight wire carrying current I out of the page.

Find expressions for the magnetic field intensity and magnetic flux density in

the space around the wire. Assume that the material surrounding the wire has

permeability µ.

By symmetry and the right-hand rule, B and H fall in a plane perpendicular to

the wire (i.e., in the plane of the paper) and are tangent to circles having their

centers at the wire.

The magnitude of H is constant for a given radius r. Applying Ampere’s Law,

The magnetic flux density is

Example: Flux Density in a Toroidal Core

Find an expression for the magnetic flux density on the center line of the core in

terms of the number of coil turns, the permeability of the core, and the physical

dimensions.

We assume that the coil is wound in a symmetrical manner all the way around

the toroidal core. By symmetry, the field intensity is constant in magnitude

along the dash circular center line.

Applying Ampere’s Law to the dashed path, we obtain

Hl = H2πR = NI

Solving for H,

Cof€

the total flux and the flux linkages.

The flux is equal to the product of the flux density and the cross sectional area:

L.NL uNIrZ0 : BA - Sptrr',: - ZR .

Note that all the flux links all of the turns, we have

uNzlrZ)": N4: --TR-

f ; lt^t' 'o^B**tl"^ *;l .

2-? Til

/

B ( ^5*** hJJ *^;* )

/f C* S,,*'{'z l"'"W

*1** )

t,,..-,,- f.*l. e € @ , tt** ,% ,r'n-o"&'*-J Jt* &'^-\ E

f.e-r.^a.;'s .jr,. esr^e_

H 1^2,&o.ee.-s .ta {

GA -$^---.,r s*-*.,ra-{"J. 4w

E".-,,"^-y (e

C*) h ,*T d-'*:*t

4;C

LsA

lr -V t)

S.r- "".r-^&--l \

q.+*--1.-a.$r1 fL-x

| .r,^ d."r* tL.C*'t"^-1',^t€-

^* a-

,-- .,i^e-

)a-arrf ea

+{4-

oJ-+I-U"-,^"-+.,

W\r/2

B

H

B=

=i.Hf

5

-,

>ff

,i", oi ,

7:

M\5, .-s.u tt*" &^iq ,

B=

* r r lo-1 eL /A-(o

( 4,/* to-r L"L/ f\'^ \ (Jd A)

,1/CtK(o'z *)

--lrrtD'T+

DC Motors 

Example

A 50 hp DC motor operates from 220 V DC source with losses 3350 W under

rated full load conditions.

The full load speed is 1150 rpm.

Under no-load conditions, the speed is 1200 rpm.

Find source current and efficiency with full load.

The input power to the dc motor is

Pin = Vsource Isource = Pout + Ploss

Multiply the number of horsepower by 746 to convert from horsepower to

watts.

Substituting values and solving for the source current we have

220 Isource = 50 x 746 + 3350

Isource = 184.8 A

Also we have

n= (Pout/Pin) x 100% = [(50 x 746)/(50 x 746 + 3350)] x 100% = 91.76%

Example

a. Which type of motor would have the most difficulty in starting a high-

inertia load from standing start?

The synchronous motor has zero starting torque and would not be able to

start a high-inertia load.

b. Which should not be operated without a load?

The series-field dc motor should not be operated without a load because

its speed becomes excessive.

Example: Idealized Linear Machine

(a) Assuming that the bar is stationary at t=0, compute the initial current and

the initial force on the bar. Also, determine the final (i.e. steady-state)

speed assuming that no mechanical load is applied to the bar.

(b) Suppose that a mechanical load of 4 N directed to the left is applied to the

bar. In steady state, determine the speed, the power delivered by VT, the

power delivered to the mechanical load, the power lost to heat in the

resistance RA, and the efficiency.

From the circuit,

(c) Suppose that a mechanical pulling force of 2 N directed to the right is

applied to the moving bar. In steady state, determine the speed, the power

taken from the mechanical source, the power delivered to the battery, the

power lost to heat in the resistance RA, and the efficiency.

When a pulling force applied to the bar to the right, the bar speeds up, the

induced voltage exceeds VT, and current circulates counterclockwise.

Thus, the machine operates as a generator.

In steady state, the force induced by the field is directed to the left and

equals the pulling force.

e,Jk^+ ttf* + t^- $l.Q-.[ $t.*^td^ u .,t*.tUJ tg*Z"f

Example: Idealized Linear Machine

?2T

(a) ins that

ia1 fl fiuml lied

Initially,lbr r* : 0, we have f4 : 0, aild the initiai current is given by

Vr ')iA(o+) : fr: 6ft :40 A

The resulting initlal f*rce on the bar is

;1tl+) :8lix(0t) *+S#4't}#+- - $*n. *._..o

=LCa"3){to} = 2t }-1 . =-)In ste*dy state with no Xoad, the inducrd voltage equal* the battery valtage. Thus, ia ''c-a-e-e"4

we havee7 =Bht-VT.

Solving f*r tlte vek:city and substitutiag value$ rve get

vr ?, 9.273tt: ---i-:

-

-6|:66,:m/sBt 40.3,t2

tamec directed totstate the the

resistance,Ra" and the efficiency.

Because the rn*chnni*al force oppars$ thc rnotion of the bar, we have rnolor

acti*n" In steeidy state, the net fol'ce on the bal i* zEra*the folce created by the

r::agnetic fteld eqlnls the l*ad force. Thus, we otltai*

f : Blia :.fioa<l

S*lving for the cun'cnt and substituting values, we find that

, -l'io"t- 4

-r*-tA: EJr =2<oi) :)-.rr A

L

From the circuit,r*0

d-eel | 661cA : Y7 * Rai* * 2" - A.*S(W) *WY

Neiw, we can lind the stcady-s(ate spsed:

,, - "* =ry-):'flr-* z.at s ^/,' Bl lto.T ' '

The mechanical power delivered 6 ,n-l:ilf ( r- r1

Pm : ftoaau - at|ffi{l :PffiW

The porver taken fii:m the battely is (.6L1 t7 .3jP1=. VTin *2t ,ffi :/b'trf W

The power dissipated in the lesistance is

As a chcck, w'c notc that pl * Pm * 716 to rvithin rounding elrol'. Finally'

the efflciency ol converting electrical power ftom the battery into mechanical

power is

pR:i2uR:ffi 2'>>L \'J

n * yx 1oo06 =ffi :'ili* :eW"i"r j.? 3

t.l ,

I{r"e1-g

i,-y * *$

(c) Suppose that a mechanical pulling force of 2 N directed to the right is

applied to the moving bar. In steady state. determine the speed" the power

taken from the mechanical source. the power delivered to the batterv. the

power lost to heat in the resistance Ra. and the efficiency.

When a putling force applied to the bar to the right, the bar speeds up, the

induced voltage exceeds Vr, dlud current circulates counterclockwise.

Thus, the machine operates as a generator.

In steady state, the force induced by the field is directed to the left and

equals the pulling force.

J',-Blis=/irtrttSolvi::g for: i; ;::rd suh*titu{i*g vi*1ues, ri'e fincl thal

*f,.6e 7'33 3

3-313 z. t( 7e,1 x V7* R;ix - 2 * $.05y)-rt *p#3 Y

' 'fPutt 2,r--!'/4 il x$.3\I

L

+

vr

Now. we can firtd the steady-s,ot. tryif,,

ea 'H 7'(r(o:fi:ffi:J,-Trtw/s

vThe mechanical power delivered by the pulling force is

pnt :.fpuilu:z&f#t : y,*{tv1 'zz2.

The porver absorbecl by the lrattery ,t ,. , ,, C C e 1

pt : Vria : Zy*{tl :12.-r{w

f=0

The power dissipated in the resistance isa- rss

;,a : {R :9"6#fu{r15

,t: #x roo% :Wr!r{*:u**}^st 7 -*? 'Y**,1-La2* i^7**f

L " rp*- - 7-.-"-\ o-1^^,,."-o{e--'s't.u 4- sL*^t D c ,**L^,-

ft.^ w;$-r-Xe- 1.,^-, ,

Va D ReE,q f FA g

9.^ = T&=., : hF I* C t*-aa"^'^ilal efl)

-?-_ I A".,\ \

-

,-ft Ra(I

hJC;e

vr=R-ftr-+Eq

sd!e4_ €^= KpL,"^ CF €1*)

F;"* O V1 : P-*+Td*., ,fftEq

R,1

.:trY6-'' lA",= tr (vr 3(rr"")

o

Example: Shunt-Connected DC Motor

A 50-hp DC shunt motor has the

magnetization curve shown below.

The DC supply voltage is 240 V, the

armature resistance is 0.065 ohms,

the field resistance is 10 ohms, and

the adjustable resistance is at 14

ohms. At a speed of 1200 rpm, the

rotational loss is 1450 W. If this

motor drives a hoist that demands a

torque of 250 Nm independent of

speed, determine the motor speed

and efficiency.

Equivalent circuit:

The field current is given by:

Next, we use the magnetization curve to find the machine constant KФ.

From the curve, the induced armature voltage EA = 280 V at IF = 10 A, and

nm=1200 rpm.

Since mA KE and the machine speed in radians per second is 60

2 mm n ,

the machine constant is:

V228.2)60/2(1200

280

m

AEK .

Solving for speed,

To find efficiency, we compute the output power and input power,

- sf-J o:La*aele'-ddc-S kuctet - ftvr,trz"fuJ

J) g t'{-s+d""

S**-

Sr,r...^.

S*.*- Eft = tr/ *^.,-.$ $ = Kr IF /

Va r R*Ir + R*T* + E4 s=rJ_p, I

V1 = ReI+t RoTe-re+_@

tlt?

T5-rrA

€e = K Kr To *.^ : k q u^^ T,q ---@S",b O .,). I

RrI* +R(IB + KKpL..'*E6

VaAF tR*+ + KKp*^

S ,r."- TJ.", = 96 Tn = k' k v TrT+ = K Kp q"

3r @; c ,

^r-^. 1J.., =

2_

K Kp Vr( Rr r RA +rKi=il1

Example: Series-Connected DC Motor

the resistances. rotatiothe power output" and then the new speed and output power if the load torque

increases to 24 Nm.

Since we are neglecting losses, the output torque and power are equal to the

2xda61 :!'t1s1 X

*:125.7rad/s

Pclevl : Poutl : r,in1Ioutl : 150tt W

V l,r.,z \ 24

t't1p77:848.5 rPm

Finally, the output power with the heavier loacl is

Pout2 : Trlev?.@na - 2133 W

developed torque and power, respectively.

The angular speed is

and the output power is

Q)nt7: 0)nt'1

which corresponds 1o

Settingftl:ltp:O,ffi

KKF-vi Y7rd"u:m:KKfr,Thus. for a fixed supply volt ageVT-,torque is inversely proportional to speed squared,and r /e can write

I.l*,'l - 4,,,

Ta""z fir,1

Solving far a4,2 and substituting values, we have

t" "1..*- ^ t1'r="-\ .X,^4il +' o 9-t.."et*c.^r..-or[eA DL

wrofar. r

AC Signals and Power 

F4Ezl +\lMEzt+3e Syu-^-"+-1 Nat- € AC n.3--{n .-i (""'-",

C"G(UT CfiEG FdlSG

A 3*-it^so.zfJ u'AFae-: vCt2= Vonc^C*+e)

Ar*.**'( :tLJ O ,i .I .[-2*r,-)

t** = -* k T.

v&): V,^ oa Cuv +e)

Vr.*^ c,,t (wt + e ) - V,^

^* tn^-n* , * c*+ e) a t

oo1 t'*o* t e s a.

+ =-e- t tra.F an

'r- LtT5,'... <r. cJ = f t

-p- E - g'7.

\- k4a.x >Tl

Rta-l\ r-^-l -) A.-5. ..a , {.tr-* = ,.-J,o.ag " +

'L F clTL ta,tzr( = ffi

C9

-rT66C,

Vn-. ), J* F--^.r, d. S.^- -r4.h; v--L* . U.{^ g^. (wv^!i,.,*- b.^n.^r. , + t -."1 t" cJcrJ*J. St*-

+t-F 4 v"l*.-g* t Yl"*I t. a\ }^e4 *-'+a.^-44

VCt ) = V* cz's C uL + g )

(-) 2-

V n^,

vi*o(*? +e) ,{evn^^s:J+ I!_

t Oea

ME2143/]VIE2143E AC Sienals and Power: Examples

Example: Power Delivered to a Resistance by a Sinusoidal Source

Suppose thnt a voltage given by v(/) - 100 ecs(100nr) V is applied to a 50-Q resis-tance. Sketch v(t) to scale versus time. Find the rms value of the voltage and theaverage power delivered to the resisJance. Find the power as a function of time and

T:llf:20 ms.

p(r) (V)

100

The peak value of the volt*ge is Vn * 1{ifi Y. Thus. the rrns value is Y'*'Vr,r/&:7A.77V. Then, the average pcwer is

p-.* : vil' - {7a'7D2 : 1oo w,avs R 50

The power as a lunction oJ time is given by

sketch to scale.

From the given v(t) equation,

ar : 1002u, andf: atl2n : 50 }Jz,

P.,,,- = 100

v(t): Vu,.t'E(*t+e)

t(rns)

-r 00

7t(t1 :+ :IIb#Sg, = 2oocos2(1oozr) w

(La- f*-fua-*'*. k-t{t"-*'*'

CI ***\ luo V

Itk

,;(ms)

A.G UC.ll- Ttuw'a-tt'

2t) i,tg '---7

/lm

p(r) (w)

{oo Q,

Ll..t*1 f L*"-= +, A-SJ S.;t'*Y" c&s

Vr :- V. -t Ut

"+ (t> {-}": zo Lqs"

= zo co.c-vs*)'r j Lo *;^('-ts "J

Lo Z-Y5 "

X Lozs (rs" ) ?-

Vu- ( c'4 o

z (+" tct

",o (-Y 5')

"

V* cea (tc 'oi

U,r.^

- (q"1yJ

Vv*c-t e -J Uu'^ s^l $

r@<.va)4 e

(1.r?zz- go.

: o . s 31+

2 *o"ol-^ -a/r l6-(}

(

-) lz 22.h

: 2q "*11

L q "1 *l c''4' ( '-+

Vu-r

s-i- O) =

(}t"-,^ e

U,,* C,./r e

F "t-" t'-'*^* *-"!. t* l*fl t

!

-)

SL*

I

Example: Analysis of a Wye-Wye system with 0.2H inductance and 50 ohms

resistance

A balanced positive-sequence wye-connected 60 -Hz three-phase source has

line-to-line voltage of Vr : 1000 V. This source is connected to a balanced

wye-connected load. Each phase of the load consists of a0.2-H inductance inseries with a 100-ohms resistance. Find the line-to-neural voltages, the line

currents and the power delivered to the load. Assume that the phase of Vun is

zero' .--"VL = fT Vy

ntt'?u.t''& d! r'"-- {-l* \ - ,.,.!..-t, " !

rr*r{-$e- ' -- !a"r i*.]t" -,*;,.J q 1 t{aqd

?&s line-t*-n*xfr*l v*l**g* is 1S00 / r,6 = 577,4 V, trll* phuxo arg*e w*s*px*1{iu,d in *hs p:.*b.**n'r sf*lemsrt" $* r** wiJl *ssxxs th*? the pk*se ofVsruisa&r*. ?hexwe**v*, 'ld**u Vu,n ,ve* i \i^*-to*'u-"-"-[ *t'"g*J .^*.f*t--rc.,*^ce t

Y**SIT,Sc0' U*,*3??"4x.-1t0" {n*S7?.'{El?il" L["^"b. L )?ha niruuit {ur th* s phuse is Eh*wn hel*ry, {}ilo *xT **nsidur * x**tnsl*p*rx***i*N ** axisl i:t * **lqnqqd Y-Y *qnnuuti'sn &ve* if *xu is **tplryui *ul ly pru*e,nt.)

fun * j#S--n*

+iuL= +i7f.faJLgd

M-.-v, , I

I : t*r .-q;. {

.f??.sg

Yh* **phas*, 1ix* +xrroxt {*? Von 577.4t".Qre.{ *

1* 100;J?5AO

r!

* 4.610/* -37.A?' !u"cr a f :q z.(

j?qry*rL

{- z-tst'Tli:e *urr*::?x f ur ph*sar * snd s dra. 1f'rs s*mo e.us€p? fur pi'r*se.

fr* * 4,*rfid" #jS' 3*r * 4.$1"0dq4j$-ar-:J.r)--r2oo t-)-i-1 .b)o+12.ro

n - *1&.*${#} - ,f,8#@{33t$r.#4"} * 3.Iss k}rY

AC Motors 

Example: DC to AC power conversion for 3-phase induction motor

A 4-pole induction motor drives a load at 2500 rpm. This is accomplished by using an electronic converter to convert a 400-V DC source into a set of 3-phase AC voltages.

Given that the frequency of the AC voltages is 86.8 Hz assuming that the slip is 4%. The load is 2 hp. If the DC-to-AC converter has a power efficiency of 88% and the motor has a power efficiency of 80%, estimate the current taken from the DC source.

Solution:

Pout,motor = 2 hp x 746 = 1492 W The input power to the motor is: Pin,motor = Pout,motor/ηmotor = 1492/0.80 = 1865 W Pout,converter = Pin,motor The input power to the converter is: Pin,converter = Pout,converter/ηconvertor = 1865/0.88 = 2119.32 W

Finally, the current taken from the 400-V source is:

I = Pin,converter/400 = 2119.32/400 = 5.298 A

Example: Slip and Frequency of rotor current in a 3-phase induction motor

A 5-hp four pole 60-Hz 3-phase induction motor runs at 1750 rpm under full-load conditions. Determine the slip and frequency of the rotor current at full load.

Estimate the speed if the load torque drops in half.

Slip frequency (in Hz), fslip = s f = 0.02778 x 60 = 1.667 Hz.

In the normal range of operation, slip is approximately proportional to output power and torque.

At half power, we estimate that s = 0.02778/2 = 1.389%.

This corresponds to a speed of 1775 rpm.

Torq ue-Speed Characteristlcs(continue)

Example:

The torque-speed characteristics ofa 60-Hz induction motor and a loadare shown in the figure. How manypoles does the motor have? lnsteady-state operation, find thespeed, the slip and the outputpower.

(Answers: 4-pole motor; 1400 rpm,22.22%,3665 W)

Copyf Oht il r0l 1, kaen t&i$on, inc

F*"" s**- kn,,'**- , &-(r^d spe4-\ * lScrg rprl,.r

4, = tE ot> rpra+ I + -po(e- il,to+b,/

S (l.r*[y - s*olc- Ope..a#t.na ; .* th- u*e-sec*r^a['

4 tt- tr?*-tp^-.t c-t.-a..ac*<.r$+.?" "[ tLc,^..t#.^. o./-.1 tt^a+ + (rtr-ol

.

+ T;t = 2-S N,^,. -J 4'^^ =- \to6 .p,^,.

s(rp, s tr $- = zL.zz1u)i^ = # An s- (+6. 6 n ,A /s

c:*+ft.-} pdwq,/, P"*+ = T"+ ur,. = 3(6 S U

Stepper Motor – Operation 

Off

Off

On

On

   N 

S 

On

On

On

Off Off

Off

Off

Off

Off

Off

Off

Off

Off

Off

Off

N

N

N

N

t= t0 = t4 t= t1

t= t2 t= t3

Half stepping to double the resolution.

On 

Comparison of Electric Motors  

Stepper motor – Small power device; Accurate, repeatable positioning application, for example, 

moving the head in an ink‐jet printer. 

Brushless DC motor – permanent‐magnet stepping motor with position sensor and enhanced control 

unit. Low‐power application in difficult environment.    

T 

nm

Holding torque  Cutoff speed Stepper motor

AC 3‐phase induction motor

AC single‐phase induction motor 

with auxiliary winding for self‐

starting 

= Split phase motor 

The synchronous motor has zero 

starting torque and would not 

be able to start a high‐inertia 

load. 

The series‐field dc 

motor should not be 

operated without a 

load because its speed 

becomes excessive. 

ns