Rotating DC Motors Part II


 Lora Watkins
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1 Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors usully consist of mny winings of reltively thin wire. Recll tht thin wires hve lrger resistnce thn thick wires. The equivlent circuit then must inclue resistor R which ccounts for the totl resistnce of the rmture wining. Figure 1 shows the equivlent circuit for motor. V represents the pplie voltge which cuses the rmture current to flow, R is the resistnce of the rmture, n E is the generte or inuce bck EMF in the rmture. I R V E = K υ  Figure 1: ermnent Mgnet Motor Equivlent Circuit Figure 1 shows tht rmture current, I, flows from the positive terminl of the voltge source, V, n the vlue of I is etermine by I = ( V E )/ R. The bck EMF cts to limit the flow of rmture current, but E g must lwys be less thn V in orer for I to be positive. The equivlent circuit for the motor cn be nlyze using ny circuit technique lerne erlier in this course. II. ower in Motors The power input to motor is simply the input current multiplie by the pplie voltge: = I V. Due to electricl n mechnicl losses in the motor, the mechnicl power out of the motor must be less thn the electricl power in. The most obvious electricl loss is ue to the rmture resistnce, elec loss = I R. As iscusse in previous sections, power losses lso occur ue to:  friction between prts of the mchine
2  mgnetic inefficiencies of the mteril use  ir resistnce (winge) of the rotting rmture  smll resistnce cross the brushes use to couple current onto the commuttor These losses cn be pproximte collectively by specifying n overll vlue for mechnicl power lost,. As with ny other system which oes not store energy, the electricl power input to motor must equl the sum of the mechnicl power output n the vrious loss mechnisms: = + + The output power cn be clculte from the power blnce eqution bove, or from the prouct of output torque n ngulr velocity: = T lo Mechnicl power is often expresse in units of horsepower, hp. This is nonsi unit of power equl to 746 wtts. If torque is expresse in N m n ngulr velocity is expresse in units of rins per secon, then the units out of = T lo will be expresse in wtts. To convert to horsepower, simply ivie the nswer in wtts by 746. A power conversion igrm is useful tool to ocument the power flow in motor. It clerly shows tht the input power is electricl n tht the output power is mechnicl. The power evelope,, enotes the chnge from electricl to mechnicl power. = V I = EI = T = T lo = I R = T Figure : ower Conversion Digrm II.3 Efficiency Efficiency of system cn be generlly efine s the rtio of power output reltive to power input. In motors this becomes the mechnicl power out reltive to the electricl power in: η = /
3 The expression for power out, = T lo, cn be recst to provie simple eqution for efficiency s follows. Erlier in this chpter, evelope torque ws given s T = K. Multiplying both sies of this eqution by the ngulr velocity,, gives: I T = K v I But T equls the power evelope,. Thus: = K v I Bck EMF ws lso efine erlier s E = K. Hence, we cn rrive t new eqution for mechnicl power evelope s shown t the she line in Figure bove: = T = KI = EI In mny cses the friction losses will be smll compre to the rmture loss. If we ssume tht mechnicl losses cn be ignore, the efficiency of motor cn be expresse simply s the rtio of bck EMF to pplie voltge becuse: T = T + T lo n if: T = 0 then: T = T lo hence: = T = T = KI = EI lo n EI E η = (100%) = (100%) = (100%) V I V Agin, this finl eqution ssumes ll mechnicl losses re ignore.
4 Here is summry of the rotting motor equtions: Rotting Motor Equtions in = mechnicl If no lo, then out n T out = 0, n =, T = T electricl = out ower eqution ( out = lo ) T T = T out Torque eqution (T out = T lo ) energy conversion here (T = /, or, = T ) ( subscript mens evelope ) in = V I = I R = E * I = T = K v I E = K v T = K v I η = out / in
5 Exmple II1: A permnent mgnet motor is rte for 5 V, A n 1300 rpm. If the mchine is 90% efficient t rte conitions fin R n K if T = N m. SOLUTION: We use the given rte electricl informtion to etermine the power into the motor. = V I = ( 5V )( A) = 50 W The efficiency vlue then enbles us to clculte the output power t rte conitions. = ( 0.90)(50 W ) = 45 W We cn clculte the power loss t rte conitions by The power evelope must be = + Now, we cn solve for To fin R : = T 1300 rev 1min π r = ( N m) min 60 s 1rev = 4.55 W = 4.55 W + 45 W = W K. = T = K I π W = K ( A) 1300rpm 30 K = 0.18 V s V = RI + E V = RI + K π 5 = R ( A) + (0.18 V s)(1300) 30 R = 0.11 Ω
6 A power conversion igrm shows nother pproch to this problem if we consier electricl power loss in the rmture. = V I = EI = T = T lo = I R = T From the power conversion igrm. 50 W = 0.45 W R = = 0.45 W = I R = (A) = 0.11 Ω W + 45 W R
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