By James Stallcup Sr., NEC/OSHA Consultant

codeissuesMotor Design Letters andCode Letters – How TheyAre UtilizedMotor circuits must be designed to provide protectionfor motor windings and components when motors arestarting, running, and driving loads. Motor windings areprotected by overcurrent protection devices that areselected according to the type used, based upon theamount of starting current required. Overcurrent pro-tection devices are sized by percentages based on thetype motor, starting method, and design or code letter.Starting methods should be selected based on theamount of current required to start and run the motoror the amount that is reduced by utilizing a particularstarting method.

This article, one of a series, will address a variety ofmotors and their many different characteristics and why itis sometimes desirable to choose one over the other, basedon the requirements of the driven load or equipment.

Types Of MotorsTable 430.52The following are five types of motors that must be considered when sizing OCPDs to allow motors to startand run:

(1) Single-phase AC squirrel-cage,(2) Three-phase AC squirrel-cage,(3) Wound-rotor,(4) Synchronous, (5) DC.

Single-Phase Squirrel-Cage MotorsSquirrel-cage motors are how induction motors areknown in the electrical industry. An induction motoroperates on the same principles as the primary and sec-ondary windings of a transformer. When power ener-gizes the field windings they serve as the “transformer”primary by inducing voltage into the rotor, which servesas the secondary. Squirrel-cage motors have two wind-ings on the stator (the stationary windings): one is the

run winding and the other is the starting winding. Amotor with this additional starting winding on the sta-tor is called a split-phase, single-phase, induction motor;it provides the ability to both start and run. The startingwinding has a higher resistance than the running wind-ing, which creates a phase displacement between thetwo. It is this phase displacement between the twowindings that gives split-phase motors the power tostart and run.

The angular phase displacement is about 18 to 30degrees, which provides enough starting torque (twist orforce) to start the motor. The motor operates on therunning winding after the rotor starts and has reached aspeed of about 75 to 80 percent of the motor's synchro-nous speed. The starting winding is then disconnectedby a centrifugal switch. (See Figure 1)

Three-Phase Squirrel-Cage MotorsThree-phase squirrel-cage motors have three separatewindings per pole on the stator, which generate magnet-ic fields that are 120 degrees out-of-phase with eachother. Three-phase motors do not require an additionalstarting winding. A three-phase induction motor willalways have a peak phase of current. This is due to the

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POWER SOURCE• 1Ø• 120 VOLT, OR• 240 VOLT

WINDING• RUNNING

WINDING• STARTING

SWITCH• CENTRIFUGAL

POWER SUPPLY

NEC TABLE 430.52

Figure 1. The above illustrates an example of a single-phase squirrel-cage motor listed in Table 430.52 of the NEC.

Figure 1

alternating current reversing its direction of flow. Inother words, when alternating current of one phasereverses its direction of flow, a peak current will bedeveloped and as current reverses direction again, a sec-ond phase will peak, etc. Three-phase motors provide asmooth and continuous source of power once they arestarted and driving the load. (See Figure 2)

Three-phase induction motors are wound-rotormotors, and are similar in design to squirrel-cage induc-tion motors. They are three-phase motors having twosets of leads. One set consists of the main leads to thestator – the field poles – and the other set consists of thesecondary leads to the rotor. The secondary leads areconnected to the rotor through slip rings, while theother end of the leads are connected to a controller anda bank of resistors. The speed of the motor varies withthe amount of resistance added in the motor circuit. Therotor will turn slower when the resistance in the rotorcircuit is greater, and vice versa. The resistance may beincorporated in the controller or as a separate resistorbank. (See Figure 3)

Synchronous MotorsThe following are two types of synchronous motors thatare available:

(1) Non-excited, and(2) Direct-current excited.Synchronous motors are available in a wide range of

sizes and types that are designed to run at fixed speeds.A DC source is required to excite a direct-current excitedsynchronous motor. The torque required to turn therotor for a synchronous motor is produced when the DCcurrent of the rotor field locks in with the magnetic fieldof the stator's AC current. (See Figure 4)

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WINDINGS• 3Ø• DELTA MOTOR

POWERSOURCE• 3Ø DISCONNECTING

MEANSWINDINGS• 3Ø• WYE MOTOR

NOTE: THE MOTORS BELOW CAN BEEITHER NEMA DESIGN B, C, OR D.

SUPPLY

NEC TABLE 430.52

WINDINGS• 3Ø• DELTA MOTOR

POWERSOURCE• 3Ø DISCONNECTING

MEANSWINDINGS• 3Ø• WYE MOTOR

NOTE: THE MOTORS BELOW CAN BEEITHER NEMA DESIGN B, C, OR D.

SUPPLY

NEC TABLE 430.52

Figure 2. The above illustrates an example of a three-phase squirrel-cage motor listed in Table 430.52 of the NEC.

Figure 2

480 V POWER SUPPLYCONDUCTORS• 430.22(A)

SECONDARYCONDUCTORS• 430.23(A) DRUM

CONTROLLER

CONTROLLER• 430.81

DISCONNECTINGMEANS• 430.102

RESISTOR BANK• 430.23(C)

RESISTOR BANK CONDUCTORSUSED AT CONTINUOUS DUTY• 430.23(B)

NEC TABLE 430.52

Figure 3. The above illustrates an example of a three-phase wound motor listed in Table 430.52 of the NEC.

Figure 3

POWER SOURCE• DIRECT CURRENT

MOTOR• 3Ø• SYNCHRONOUS

POWER SOURCE• ALTERNATING CURRENT

ROTOR

NEC TABLE 430.52

Figure 4. The above illustrates an example of a three-phase synchronous motor listed in Table 430.52 of the NEC.

Figure 4

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codeissuesDC MotorsDirect-current only is used to operate DC motors. A DCmotor is designed with the following two main parts:

(1) The stator, and(2) The rotor (or armature).The stationary frame of the motor is called the stator.

The armature, mounted on the drive shaft, is known as therotor. The speed of a DC motor with a given load is deter-mined by the amount of current driven through the rotor.(See Figure 5)

Series DC MotorsA very high starting torque of 300 to 375 percent of thefull-load torque is provided when using series DCmotors. Loads that are required to be driven with hightorque and poorly regulated speed use this type ofmotor. The speed varies depending on the mechanicalload. Series DC motors are used in installations such astraction work, where the speed varies depending on theload on the hoist. The armature and fields are connectedin series. (See Figure 6)

Shunt DC MotorsA starting torque of 125 to 200 percent of the full-loadtorque is provided when using shunt DC motors. Loadsthat are required to be driven with constant or adjustablespeeds, but do not require high starting torque, use thistype of motor. Shunt DC motors are useful for applica-tions such as woodworking machines, printing presses,and papermaking machines. (See Figure 7)

DYNAMIC BRAKINGRESISTORS

SEPARATELYMOUNTED

POWERRESISTORS

TERMINALBOARDS

DC MOTOR

MOTORCONTROLLER

DC POWER SUPPLY

CB• 2-POLE

NEC 430.29TABLE 430.29

Figure 5. The above illustrates an example of a DC motor listed in Table 430.52 of the NEC.

Figure 5

ARMATURE

SERIESFIELD

+

SERIES MOTOR• 300% TO 375% FULL-LOAD TORQUE IS PROVIDED

TO POWERSUPPLY

SERIES DC MOTOR

Figure 6. A series DC motor provides a very high starting torque of 300 percent to 375 percent of the full-load torque.

Figure 6

ARMATURE

SERIESFIELD

+

SHUNT MOTOR• 125% TO 200% FULL-LOAD TORQUE IS PRODUCED

SHUNTFIELD

TO POWERSUPPLY

SHUNT DC MOTOR

Figure 7. A shunt DC motor provides a medium starting torque of 125 percent to 200 percent of the full-load torque.

Figure 7

ARMATURE

SERIESFIELD

+

MOTOR• 180% TO 260% FULL-LOAD TORQUE IS PROVIDED

SHUNTFIELD

TO POWERSUPPLY

COMPOUND DC MOTOR

Figure 8. A compound DC motor provides a high torque of 180 percent to 260 percent of the full-load torque.

Figure 8

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Compound DC MotorsCompound DC motors provide a starting torque of 180 to 260 percent of the full-load torque and a con-stant speed. The compound DC motor is equipped witha series and a shunt winding. The series winding is con-nected in series with the armature and the shunt windingis connected in parallel with the armature. This type ofmotor has the characteristics of both series and shuntmotors during operation. Loads such as crushers, recip-rocating compressors, and punch presses use compoundDC motors. (See Figure 8)

Calculating TorqueTo accelerate and drive a piece of equipment, the motormust be capable of producing a torque. Torque is theturning or twisting force of the motor and is measuredin foot-pounds or pound-feet.

Full-Load TorqueThe full-load torque of a motor is determined by multiplying the horsepower by 5252, and then dividingby the rpm of the motor.

Motor Tip: The value of 5252 is found by dividing33,000 foot-pounds per minute by 6.2831853 (33,000 ÷ 6.2831853 = 5,252).{6.2831853=2 x π; π=3.14159265 } (See Figure 9)

Starting TorqueThe starting torque of a motor varies with the classification of the motor. Motors are classified byNEMA as Design B, C, or D. These standardized typesare the most used motors in the electrical industry.Other types of motors classified by NEMA are Design For G motors.

Each class of motors has a different rotor design,which provides a different value of starting torque.Different values of torque, speed, current, and slip forstarting and driving the various types of loads are produced when using NEMA Design B, C, or D motors.The Design type of the motor to be selected depends onthe starting torque and running torque required to drivethe load. (See Figure 13)

Class B MotorsThe most used motor in the electrical industry is Class B. For example, the starting torque of an induction motor will increase by 150 percent of the full-load torque when using Class B design motors. Mostdesigners however, assume a starting torque of less than150 percent when using Class B induction motors tostart and run loads. (See Figure 10)

Class C MotorsThe starting torque of a squirrel-cage induction motorwill increase about 225 percent of the full-load torquewhen using Class C design motors. However, to keepfrom overloading the starting torque, designers willoften load these motors to less than 225 percent.

For example: What is the full-load torque and startingtorque of a 40 HP, Class C design induction motor operating at 1725 rpm?

Step 1: Finding full-load torqueTorque = HP x 5252 ÷ rpmTorque = 40 x 5252 ÷ 1725Torque = 210,080 ÷ 1725Torque = 121.8 ft. lbs.

Step 2: Finding starting torqueFull-load torque increased by 225%Torque = 121.8 ft. lbs. x 225%Torque = 274.05 ft. lbs.

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Figure 9

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Solution: The full-load torque is 122 ft. lbs. and the starting torque is 274 ft. lbs.

Class D MotorsThe starting torque of a squirrel-cage induction motor is increased about 275 percent of the full-load torquewhen using Class D design motors. However, to keepfrom overloading the starting torque of a motor, design-ers will often load these motors to less than 275 percent.

For example: What is the full-load torque and startingtorque of a 50 HP, Class D design induction motoroperating at 1725 rpm?

Step 1: Finding full-load torqueTorque = HP x 5252 ÷ rpmTorque = 50 x 5252 ÷ 1725Torque = 262,600 ÷ 1725Torque = 152.2 ft. lbs.

Step 2: Finding starting torqueFull-load current increased by 275%Torque = 152.2 ft. lbs. x 275%Torque = 418.6 ft. lbs.

Solution: The full-load torque is 152.2 ft. lbs. and thestarting torque is 418.6 ft. lbs.

Energy Efficient MotorsWhen designing and installing an energy efficient motor, itis most important to know the starting and running torqueof the load. The difference between the nominal and theminimum efficiency must also be determined in order forthe motor to be properly sized to start and drive the load.

Starting CurrentsMost energy efficient motors have higher starting cur-rents, which presents a real problem when one is replac-ing a standard motor. This can cause nuisance trippingof the OCPD during full-voltage start up.

There are some energy efficient motors that have startingcurrents as high as 1500 percent of the full-load current. If1700 percent per Ex. 1 to 430.52(C)(3) does not allow themotor to start and run, reduced voltage starting or a mod-ern electronic type of motor start/run technology must beutilized. Note that starting currents of energy efficientmotors vary based on manufacturer and size. Energy effi-cient motors must be selected with enough starting torqueand breakdown torque to start and run the driven loads.

The nameplate on most motors will list their startingand running kVA. It is from these values and the manufacturer data that the OCPD and conductors aresized. The motor should be loaded based upon the minimum efficiency, not its nominal efficiency. Note that an energy efficient motor lists both nominal andminimum efficiency full-load ratings on its nameplate.

Two-Speed MotorsThe full-load torque of a motor is determined by its rpm.A motor turning at 1800 rpm produces less torque than amotor turning at 1200 rpm.

For example: What is the full-load torque for a two-speed30 HP motor, operating at either 1200 rpm or 1800 rpm?

Step 1: Finding full-load torque (1200)Torque = HP x 5252 ÷ rpmTorque = 30 x 5252 ÷ 1200Torque = 157,560 ÷ 1200Torque = 131.3 ft. lbs.

Step 2: Finding full-load torque (1800)Torque = HP x 5252 ÷ rpmTorque = 30 x 5252 ÷ 1800Torque = 157,560 ÷ 1800Torque = 87.5 ft. lbs.

Figure 10

Solution: The full-load torque for 1200 rpm is 131.3 ft. lbs. and the full-load torque for 1800 rpm is 87.5 ft. lbs.

Resistor Or Reactor-Reduced StartingTo reduce the inrush starting current of a motor, a resistor or reactor-limiting starting method can be used.The starting current is reduced to 65 percent by usingeither method. The starting torque will be reduced to 42 percent (65% x 65% = 42%) if the starting current is reduced. When selecting a reduced current startingmethod, care must be taken to ensure that enough foot-pounds are provided to accelerate the load. (See Figure 11)

Code LettersTables 430.7(B) and 430.151Code letters are given to motors by manufacturers forcalculating the locked-rotor current (LRC) in ampsbased upon the kVA per horsepower per the motor'scode letter. Overcurrent protection devices shall be setabove the locked-rotor current of the motor to preventthe overcurrent protection device from opening whenthe rotor of the motor is starting. The following twomethods can be used to calculate and select the locked-rotor current of motors:

(1) Utilizing code letters to determine LRC, and(2) Utilizing horsepower to determine LRC.

Locked-Rotor Current Based Upon Code Letters430.7(B)Code letters must be marked on motor nameplates;these letters are used for determining locked-rotor current. Locked-rotor currents for particular code lettersare listed in Table 430.7(B) in kVA (kilovolt-amps) perhorsepower.

For example: What is the locked-rotor current rating fora three-phase, 208 volt, 20 horsepower motor with a codeletter B marked on its nameplate?

Step 1: Finding LRC ampsTable 430.7(B)A = kVA per HP x 1000 ÷ (V x 1.732)A = 3.54 x 20 x 1000 ÷ (208 V x 1.732)A = 70,800 ÷ 360 A = 197

Solution: The locked-rotor current is 197 amps. Note that Table 430.7(B) must be used to find the LRC of motors based on their code letters per the 1996 NECand earlier editions.

Locked-Rotor Current Utilizing HPTables 430.151(A) and (B)The locked-rotor current of a motor may be found inTables 430.151(A) and (B). The locked-rotor currentsfor single-phase and three-phase motors are selectedfrom this table based upon the number of phases, volt-age, and horsepower rating of the motor. For motorswith code letters A through G, round the nameplate cur-rent in amps up to an even number and multiply by 6 toobtain the LRC of the motor. Note that code letterscan’t be found in Tables 430.151(A) and (B) becausethey won’t be listed on the motors nameplate anymore.Motors will be marked either as Design B, C, or D toindicate which locked-rotor currents are to be selectedfrom Tables 430.151(A) and (B) based on horsepower,phases, and voltages.

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Figure 11

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codeissuesFor example: What is the locked-rotor current rating for

a three-phase, 460 volt, 50 horsepower, Design B motor?

Table method using Design letter:

Step 1: Finding LRC ampsTable 430.151(B)50 HP requires 363 A

Solution: The locked-rotor current is 363 amps.

For example: Consider a motor with a nameplate current of 63 amps and determine the LRC of the motorbased upon code letters A through G?

Rule of thumb method using code letter:

Step 1: Finding even multiple of ten numberTable 430.7(B) Round up 63 A to 70 A and multiply by 6

Step 2: Calculating LRCTable 430.7(B)70 A x 6 = 420 A

Solution: The locked-rotor current is 420 amps. Note: This method can only be used for code letters A through G.

See Figures 12(a) and (b) for calculating and selectingthe locked-rotor current of a motor.

Motor Tip: Engineers and electricians must select thelocked-rotor current rating from Tables 430.151(A) and (B) when using Design B, C, or D motors. The overcurrent protection device must be set above thelocked-rotor current of the motor so the motor can startand run. See the problem in Figure 12(b).

When code letters are used, the locked rotor currentmust either be calculated per Table 430.7(B) or by the

POWER SUPPLY• 430.22(A)

460 V, 3ØCONTROLLER• 430.81

3Ø, 460 V, 50 HPMOTOR, CODE LETTER G

MOTOR• 430.7(A); (B)• MAX. LRC• 62 A

DISCONNECTING MEANS• 430.102• 430.110(A)

GEC

GES

MBJ

USINGCODE LETTERS - 430.7(B)

TABLE 430.7(B)

APPLYING RULE OF THUMB FOR CODELETTERS A THRU G• MOTORS FLA x 6 = LRC• ROUND NAMEPLATE AMPS UP

FINDING LRC USINGMOTOR'S CODE LETTERS

ROUND UP 62 A TO 70 A70 A x 6 = 420 ALRC = 420 A

QUICK CALC

Step 1: Finding kVA multiplier430.7(B); Table 430.7(B)Code Letter G = 6.29 kVA

Step 2: Applying formulaLRC = kVA per HP x 1000 x HP ÷ V x √3 = ALRC = x 6.29 x 50 x 1000 ÷ 460 V x 1.732LRC = 395 A

Solution: The locked-rotor current is 395 amps.

FINDING LOCKED-ROTOR CURRENT USINGTHE MOTOR'S CODE LETTERS

NEC 430.7(B)NEC TABLE 430.152

Figure 12(a). For motors having code letters instead of Design Letters, the LRC must be calculated per Table 430.7(B) using the code letter of the motor.

Figure 12a

SINGLE-PHASE MOTORS• TABLE 430.151(A)THREE-PHASE MOTORS• TABLE 430.151(B)

FINDING LRC PER TABLEBASED ON DESIGN LETTER

3Ø, 460 V, 50 HPMOTOR, DESIGN B• FOR 1Ø MOTORS, SEE TABLE 430.151(A)• FOR 3Ø MOTORS SEE TABLE 430.151(B)

POWER SUPPLY• 430.22(A

460 V, 3ØCONTROLLER• 430.81

DISCONNECTINGMEANS• 430.102• 430.110(A)

Step 1: Finding LRC of 50 HP, Design B, 460 V Table 430.151(B)LRC = 363 A

Solution: The locked-rotor current is 363

FINDING LOCKED-ROTOR CURRENT PER TABLEUSING DESIGN LETTER OF MOTOR

NEC TABLES 430.151(A) AND B

GEC

GES

MBJ

Figure 12(b). Table 430.151(A) and (B) can be used to determine the LRC in amps for motors with Design Letters.

Figure 12b

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rule of thumb method, based on code letters A throughG. See the problem and Quick Calc in Figure 12(a).

See Figure 13 for a chart showing the different electri-cal characteristics for different design Type motors.

For example: What size OCPD, using a circuit breaker,is required per Figure 12(a) and (b) to start and run?

Step 1: Finding FLC of motorTable 430.15050 HP = 65 A

Step 2: Finding percentage to size OCPD (CB)Table 430.52Percentage = 250%

Step 3: Performing math65 A x 250% = 162.5 A

Step 4: Selecting OCPD (CB)430.52(C)(1), Ex. 1; 240.6(A)162.5 A = 175 A CB

Solution: A 175 amp CB will hold about 525 A (175 A x 3 = 525 A) for 4 to 9 Seconds.

Note: Inverse time circuit breakers (600 volts or less)will hold about three times their rating for different periods of time based on their frame size. It really doesnot matter if the code letter or design letter is listed todetermine LRC (starting current) to size the OCPD (circuit breaker). When sized per Table 430.52 it will belarge enough to hold the current and allow the motor tostart and run.

More information on this subject can be found in chapter16 of the book 'Stallcup's Generator, Transformer, Motorand Compressor Book,' available from the NFPA.

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Figure 13. The type of motor will determine the electrical characteristics of the design. Note that NEMA has designated the above designs for polyphase motors.

NEMADesign

StartingCurrent

BreakdownTorque

Full-LoadSlip

StartingTorque

A Normal Normal High LowB Normal Low Medium LowC High Low Normal LowD Very high Low - High

Figure 13