INTERFACING PROGRAMS FOR 8085

PROGRAM TO INTERFACE 16 LEDS TO 8255 PORTS AND DISPLAY THEM AS PER THE SPECIFIED PATTERN.01234567LA10101010AAHLB0101010155HTHE PROGRAM BELOW IS USED TO DISPLAY PATTERN AS ABOVE AND CERTAIN SEQUENCES.8000 MVI A 80 3E ;8001 80 ;8002 OUT CWD3 ;8003 43 ;8004 MVI A AA 3E ;8005 AA ;8006 OUT PAD3 ;8007 40 ;8008 CALL DLYCD ;8009 00 ;800A 85 ;800B CMA 2F ;800C OUT PBD3 ;800D 41 ;800E CALL DLYCD ;800F 00 ;8010 85 ;8011 MVI C 08 0E ;8012 08 ;8013 RRC 0F ;8014 OUTPBD3 ;8015 41 ;8016 CALLDLYCD ;8017 00 ;8018 85 ;8019 DCR C 0D ;801A JNZ 8013 C2 ;801B 13 ;801C 80 ;801D MVI A AA 3E ;801E AA ;801F MVI C 08 0E ;8020 08 ;8021 RRC 0F ;8022 OUT PAD3 ;8023 40 ;8024 CALL DLYCD ;8025 00 ;8026 85 ;8027 DCR C 0D ;8028 JNZ 8021 C2 ;8029 21 ;802A 80 ;802B JMP 8004 C3 ;802C 04 ;802D 80 ;

INTERFACING 8085 TO DAC

TO GENERATE SQUARE WAVEFORM

TO GENERATE SQUARE WAVEFORM USING DAC JUST WE HAVE TO SEND TWO DIFFERENT VALUES TO DAC THAT IS FOR LOW LEVEL AND HIGH LEVELAND A PROPER DELAY.THE DELAY DEFINES THE FREQUENCY OF SQUARE WAVE.THE PROGRAM IS AS FOLLOWS :

0000 LXI SP8500 31 ;0001 00 ;0002 85 ;http://people.sinclair.edu/nickreeder/eet114/PageArt/wavesquare.gif0003 MVI A 80 3E ;0004 80 ;0005 OUT CWD3 ;0006 40 ;0007 MVI A 00 3E ;0008 00 ;0009 OUT PAD3 ;000A 41 ;000B CALL DLYCD ;000C 00 ;000D 85 ;000E CMA 2F ;000F JMP 0009 C3 ;0010 09 ;0011 00 ;

PROGRAM TO GENERATE TRIANGULAR WAVE USING DAC

0000 LXI SP8500 31 ;0001 00 ;0002 85 ;0003 MVI A 80 3E ;0004 80 ;0005 OUT CWD3 ;http://www.epuniversity.org/tech/image/triangular-sine-wave.jpg0006 40 ;0007 MVI A 00 3E ;0008 00 ;0009 OUT PAD3 ;000A 41 ;000B CALL DLYCD ;000C 00 ;000D 85 ;000E INR A 3C ;000F CPI FF FE ;0010 FF ;0011 JNZ 0009 C2 ;0012 09 ;001300 ;0014 OUTPAD3 ;0015 41 ;0016 DCR A 3D ;0017 CALL DLYCD ;0018 00 ;0019 85 ;001A JNZ 8014 C2 ;001B 14 ;001C00 ;001D JMP 8009 C3 ;001E 09 ;001F00 ;PROGRAM TO GENERATE SAWTOOTH WAVEFORM USING DAC

http://www.till.com/articles/QuadTrapVCO/images/disc-plot-saw.gif0000 LXI SP8500 31 ;0001 00 ;0002 85 ;0003 MVI A 80 3E ;0004 80 ;0005 OUT CWD3 ;0006 40 ;0007 MVI A 00 3E ;0008 00 ;0009 OUT PAD3 ;000A 41 ;000B CALLDLYCD ;000C 00 ;000D 85 ;000E INR A 3C ;000F JMP 0009 C3 ;0010 09 ;0011 00 ;

PROGRAM TO GENERATE STAIRCASE WAVE USING DAC

TO GENERATE STAIRCASE WAVEFORM IT IS REQUIRED TOCALUCLATE THE STEP SIZE. THE STEPSIZE IS CALUCLATED AS FOLLOWS :IF NUMBER OF STEPS = NTHEN STEP SIZE = (MAX. AMPLITUDE / N)THE MAXIMMUMVALUE OFAN 8-BIT SYSTEM ISFFH. AND THE MAXIMUM AMPLITUDE IS 5V. SO FOR 5 STEPS THE STEPSIZE IS

FF/N = 255/5 = 51 = 33H0000 MVI A 80 3E ;0001 80 ;0002 OUT CWD3 ;0003 40 ;0004 MVI A 00 3E ;0005 00 ;0006 OUT PAD3 ;0007 41 ;0008 CALLDLYCD ;0009 00 ;000A 85 ;000B ADI 33 C6 ;000C 33 ;000D JNC 0006 D2 ;000E 06 ;000F 00 ;0010 JMP 0004 C3 ;0011 04 ;0012 00 ;

PROGRAM TO GENERATE EXPONENTIAL WAVEFORM USING DAC.

0000 LXI SP8500 31 ;0001 00 ;0002 85 ;0003 MVI A 80 3E ;0004 80 ;0005 OUT CWD3 ;0006 40 ;0007 MVI A 00 3E ;0008 00 ;0009 STC 37 ;000A RAL 17 ;000B OUT PAD3 ;000C 41 ;000D CALLDLYCD ;000E 00 ;000F 85 ;0010 CPI 40 FE ;0011 40 ;0012 JNZ 000A C2 ;0013 0A ;001400 ;0015 RAR 1F ;0016 OUT PAD3 ;0017 41 ;0018 CALL DLYCD ;0019 00 ;001A 85 ;001B CPI 01 FE ;001C 01 ;001D JNZ 0015 C2 ;001E 15 ;001F 00 ;0020 JMP 000A C3 ;0021 0A ;0022 00 ;

http://www.ece.uci.edu/docs/hspice/hspice_2001_2-43.jpg

PROGRAM TO GENERRATE SINUSOIDAL WAVE USING DAC.

TO GENERATE SINUSOIDAL WAVEFORM WE USE LOOKUP TABLE METHOD.LET V= VPSIN + VOIF WE SELECT THE AVERRAGE VALUE OF THE SINE WAVE (THAT IS = 0O) AT THE MIDDLE OF THE AVAILABLE RANGE THE WAVEFORM CAN BE OF MAXIMUM POSSIBLE AMPLITUDE. THE FULL SWING IS LIMITED TO FFH IN AN 8-BIT SYSTEM. IF AVERAGE LEVEL V0IS SELECTED TO BE 80H THE VP(PEAK EXCURSION) CAN BE FFH 80H = 7FH. THUS THE WAVEFORM WILL TAKE FFH AT ITS POSITIVE PEAK(AT 90O)AND 01H(80H 7FH) AT ITS NEGATIVE PEAK(AT 270O). PUTTING THE VALUES IN ABOVE EQUATION,V= 7FH SIN + 80HIF WE CALUCLATE SAMPLES WITH 5OINTERVALS VWILL TAKE VALUES AS FOLLOWS :8000 MVI A 80 3E ;8001 80 ;http://www.sweethaven02.com/ModElec/DcAc/acee/fig0109.gif8002 OUT CWD3 ;8003 40 ;8004 LXI H 8500 21 ;8005 00 ;8006 85 ;8007 MVI C 72 0E ;8008 72 ;8009 MOV A M 7E ;800A OUT PAD3 ;800B 41 ;800C CALLDLYCD ;800D 00 ;800E 85 ;800F INX H 23 ;8010 DCR C 0D ;8011 JNZ 8009 C2 ;8012 09 ;8013 80 ;8014 JMP 8004 C3 ;8015 04 ;8016 80 ;

(DEG)V(DEG)V(DEG)V(DEG)V

58BH95FFH18575H27501H

1096H100FDH1906AH28003H

15A1H105FAH1955FH28505H

20ABH110F7H20055H29009H

25B6H115F3H2051AH2950DH

30C0H120EEH21041H30012H

35C9H125E8H21537H30518H

40D2H130E1H2202EH3101FH

45DAH135DAH22526H31526H

50E1H140D2H2301FH3202EH

55E8H145C9H23518H32537H

60EEH150C0H24012H33041H

65F3H155B6H2450DH3354AH

70F7H160ABH25009H34055H

75FAH165A1H25505H3455FH

80FDH17096H26003H3506AH

85FFH1758BH26501H35575H

90FFH18080H27001H36080H

THE SAMPLED VALUES ARE KEPT IN THE MEMORY STARTING FROM 8500 AND CALLED ONE BY ONE TO OBTAIN A SINUSOIDAL WAVEFORM THE FREQUENCY OF SINUSOIDAL WAVEFORM DEPENDS ON THE DELAY CALUCLATION.8255 PROGRAMMABLE PERIPHERAL INTERFACE

D7D6D5D4D3D2D1D0

PORT C PC3 PC01 = INPUT0 = OUTPUT

PORT B PB7 PB01 = INPUT0 = OUTPUT

MODE0 MODE 01 MODE 1

1 I/O MODE0 BSR MODE

PORT C PC7 PC41 = INPUT0 = OUTPUT

PORT A PA7 PA01 INPUT0 OUTPUT

MODE00 MODE 001 MODE 11X MODE

WE KNOW THAT 8085 MICROPROCESSOR DOES NOT CONTAIN ANY PORTS TO IT TO INTERFACE PERIPHERAL DEVICES HENCE IT REQUIRES AN ADDITIONAL CHIP SO AS TO INTERFACE PERIPHERAL DEVICES FOR THIS TO BE ACHIEVED WE USE A PROGRAMMABLE PERIPERAL INTERFACE 8255 IC.THIS IC CONSISTS OF 3 PORTS NAMELY PORT A, PORT B, PORT C EACH OF WIDTH 8-BIT . HERE IN THIS CHIP TWO PORTS(PORT A AND PORT B) WORK AS8-BIT PORTS BUT PORT C CAN WORK AS 8-BIT PORT OR AS TWO 4-BIT PORTS.HERE WE CAN CONCLUDE THAT OUR 8255 PROVIDES US THE TOTAL OF 24 LINES (3 PORTS) WHICH CAN BE USED EITHER AS INPUT OR OUTPUT PORTS.BUT TO DEFINE ALL THIS PORTS AND MODES WE REQUIRE A WORD TO DEFINE ALL THIS WHETHER TO SELECT PORTS AS INPUT OR OUTPUT OTHERWISE FOR MODE SELECTION ETC., THIS WORD IS KNOWN AS CONTROL WORD.THE 8085 CAN OPERATE IN THREE MODES OF OPERRATIONSIMPLE I/O MODE.

STROBED I/O MODE.

BIDIRECTIONAL DATA TRANSFER.

DURING INTERFACING WE USE SIMPLE I/O MODE FOR INTERFACING DACS, ADCS, LED DISPLAYS ETC.,STEPS TO INTERFACE EXTERNAL DEVICES WITH 8255WRITE THE PROPER CONTROL WORD TO CONTROL WORD REGISTER

IDENTIFY THE ADDRESSES OF THE PORTS (PORT A, PORT B, PORT C)

FROM THE MANUAL.THE ADDRESSES FOR THE PORTS AND CONTROL WORD REGISTER FOR THE KIT WE ARE USING ARE AS FOLLOWS :

CONTROL WORD-PORT A-PORT B-PORT C-BY USING THIS ADDRESSES WE CAN EITHER ACCEPT DATA AS INPUT OR WE CAN SEND DATA OUTPUT TO CONTROL THE DEVICE.THE LAST STEP IS TO PREPARE THE ALGORITHM AND WRITE THE PROGRAM AS PER THE REQUIREMENT.

INTERFACING ADC TO 8085

TO INTERFACE ADC TO MICROPROCESSOR WE REQUIRE TO FOLLOWTHE FOLLOWING STEPS:The analog part of the circuit consists of a high input impedance buffer a1, precision integrator a2 and a voltage comparator.

The converter first integrates the analog input signal vafor a fixed duration of 2nclock periods.

Then it integrates an internal reference voltage vrof opposite polarity until the integrator output is zero.

The number n of clock cycles required to return the integrator to zero is proportional to the value of vaaveraged over the integration period.

Hence, n represents the desired output code.

The operation of the circuit is as follows :Before the start command arrives, the switch sw1 is connected to ground and sw2 is closed. Any offset voltage present in the a1, a2, comparator loop after integration, appears across the capacitor caztill the threshold of the comparator is achieved.

The capacitor cazthus provides automatic compensation for the input offset voltage of all the three amplifiers.

When sw2 opens, cazacts as a memory to hold the voltage required to keep the offset nulled.

At the arrival of start command at t=t1, the control logic opens sw2 and connects sw1 to vaand enables the counter starting from zero.

The circuit uses an n-stage ripple counter and therefore the counter resets to zero after counting 2npulses.

The analog voltage vais integrated for a fixed number 2ncounts of clock pulses after which the counter resets to zero.

If the clock period is t, the integration takes place for a time t1= 2nx t and the output is ramp going downwards as in fig 2.

The counter resets itself to zero at the end of the interval t1and the switch sw1 is connected to the reference voltage (-vr). The output voltage vowill now have a positive slope. As long as vois negative , the output of the comparator is positive and control logic allows the clock pulse to be counted. However, when vobecomes just zero at time t=t3, the control logic issues an end of conversion(eoc) command and no further clock pulses enter the counter.

It can be shown that the reading of the counter at tsis proportional to the analog input voltage va

In fig 2,t1= t2 t1= (2ncounts / clock rate) Andt3 t2= (digital count n / clock rate)For an integrator,v0= (-1/rc)v(t)The voltage vowill be equal to v1at the instant t2and can be written asv1= (-1/rc ) va(t2 t1)So,va(t2 t1) = vr(t3t2)Putting, the values of (t2 t1) = 2nand (t3 t2) = n, we get,va= (vr) (n/2n).The main disadvantage of the dual slope ADC is long conversion time.

Program to InterfaceADC to 8085 Microprocessor :8000 MVI A 82 3E ;8001 82 ;8002 OUT 43 D3 ;8003 43 ;8004 MVI A 02 3E ;8005 02 ;8006 OUT 40 D3 ;8007 40 ;8008 NOP 00 ;8009 NOP 00 ;800A NOP 00 ;800B NOP 00 ;800C NOP 00 ;800D NOP 00 ;800E MVI A 01 3E ;800F 01 ;8010 OUT 40 D3 ;8011 40 ;8012 LXI H 02C0 21 ;8013 C0 ;8014 02 ;8015 DCX H 2B ;8016 MOV A H 7C ;8017 ORA L B5 ;8018 JNZ 8015 C2 ;8019 15 ;801A 80 ;801B MVI A 04 3E ;801C 04 ;801D OUT 40 D3 ;801E 40 ;801F MVI B 00 06 ;8020 00 ;8021 INR B 04 ;8022 IN 41 DB ;8023 41 ;8024 ANI 01 E6 ;8025 01 ;8026 JNZ 8021 C2 ;8027 21 ;8028 80 ;8029 MVI A 02 3E ;802A 02 ;802B OUT 40 D3 ;802C 40 ;802D MOV A B 78 ;802E STA 8501 32 ;802F 01 ;8030 85 ;8031 HLT 76 ;

Integrator o/p voltage Vo

N-Cycles

T1= 2nT

t1

t2

AUTOZERO

Time T

t3

Integrate (-VR)

Integrate Va

V1

Fig :- 2 :- integrated o/p waveform of dual slopeADC

d0

dn

N STAGE

COUNTER

START

Control logic

G:\aa.jpgFig : 1 Functional diagram of a dual slope ADC

EOC

INTERFACING STEPPER MOTOR TO 8085:MOTOR ELECTRICAL BASICSA step motor is a constant output power transducer, where power is defined as torque multipliedby speed. This means motor torque is the inverse of motor speed. To help understand why a stepmotors power is independent of speed, we need to construct (figuratively) an ideal step motor.

An ideal step motor would have zero mechanical friction, its torque would be proportional toampere-turns and its only electrical characteristic would be inductance. Ampere-turns simplymean that torque is proportional to the number of turns of wire in the motors stator multiplied bythe current passing through those turns of wire.

Anytime there are turns of wire surrounding a magnetic material such as the iron in the motorsstator, it will have an electrical property called inductance. Inductance describes the energystored in a magnetic field anytime current passes through this coil of wire.

Inductance (L) has a property called inductive reactance, which for the purposes of thisdiscussion may be thought of as a resistance proportional to frequency and therefore motorspeed.

According to ohms law, current is equal to voltage divided by resistance. In this case wesubstitute inductive reactance for resistance in ohms law and conclude motor current is theinverse of motor speed.

Since torque is proportional to ampere-turns (current times the number of turns of wire in thewinding), and current is the inverse of speed, torque also has to be the inverse of speed.

In an ideal step motor, as speed approaches zero, its torque would approach infinity while atinfinite speed torque would be zero. Because current is proportional to torque, motor currentwould be infinite at zero speed as well.

Electrically, a real motor differs from an ideal one primarily by having a non-zero windingresistance. Also the iron in the motor is subject to magnetic saturation, as well as having eddycurrent and hysteresis losses. Magnetic saturation sets a limit on current to torque proportionalitywhile eddy current and hysteresis (iron losses) along with winding resistance (copper losses)cause motor heating.

Motors convert electrical energy into mechanical energy. A stepper motor converts electrical pulses into specific rotational movements. The movement created by each pulse is precise and repeatable, which is why stepper motors are so effective for positioning applications.

Permanent Magnet stepper motors incorporate a permanent magnet rotor, coil windings and magnetically conductive stators. Energizing a coil winding creates anelectromagnetic field with a north and south pole as shown in figure 1. The stator carries the magnetic field which causes the rotor to align itself with the magnetic field. The magnetic field can be altered by sequentially energizing or stepping the stator coils which generates rotary motion.

Figure 2 illustrates a typical step sequence for a two phase motor. In Step 1 phase A of a two phase stator is energized. This magnetically locks the rotor in the position shown, since unlike poles attract. When phase A is turned off and phase B is turned on, the rotor rotates 90clockwise. In Step 3, phase B is turned off and phase A is turned on but with the polarity reversed from Step 1. This causes another 90rotation. In Step 4, phase A is turned off and phase B is turned on, with polarity reversed from Step 2. Repeating this sequence causes the rotor torotate clockwise in 90steps.

The stepping sequence illustrated in figure 2 is called one phase on stepping.A more common method of stepping is two phase on where both phases of themotor are always energized. However, only the polarity of one phase is switchedat a time, as shown in figure 3. With two phase on stepping the rotor aligns itselfbetween the average north and average south magnetic poles. Since bothphases are always on, this method gives 41.4% more torque than one phase onstepping, but with twice the power input.

Half Stepping

The motor can also be half stepped by inserting an off state betweentransitioning phases. This cuts a steppers full step angle in half. For example, a90stepping motor would move 45on each half step, figure 4. However, halfstepping typically results in a 15% - 30% loss of torque depending on step ratewhen compared to the two phase on stepping sequence. Since one of the windingsis not energized during each alternating half step there is less electromagneticforce exerted on the rotor resulting in a net loss of torque.

Bipolar WindingThe two phase stepping sequence described utilizes a bipolar coil winding.Each phase consists of a single winding. By reversing the current in the windings,electromagnetic polarity is reversed. The output stage of a typical two phase bipolar drive is further illustrated in the electrical schematic diagram and stepping sequence in figure 5. As illustrated, switching simply reverses the current flowthrough the winding thereby changing the polarity of that phase.

Unipolar WindingAnother common winding is the unipolar winding. This consists of twowindings on a pole connected in such a way that when one winding is energized amagnetic north pole is created, when the other winding is energized a south pole iscreated. This is referred to as a unipolar winding because the electrical polarity,i.e. current flow, from the drive to the coils is never reversed. The stepping sequence is illustrated in figure 6. This design allows for a simpler electronic drive.However, there is approximately 30% less torque available compared to a bipolarwinding. Torque is lower because the energized coil only utilizes half as muchcopper as compared to a bipolar coil.

AccuracyThe accuracy for can-stack style steppers is 6 - 7% per step, non-cumulative.A 7.5stepper will be within 0.5of theoretical position for every step, regardlessof how many steps are taken. The incremental errors are non-cumulative becausethe mechanical design of the motor dictates a 360movement for each fullrevolution. The physical position of the pole plates and magnetic pattern of therotor result in a repeatable pattern through every 360rotation (under no loadconditions).

ResonanceStepper motors have a natural resonant frequency as a result of the motorbeing a spring-mass system. When the step rate equals the motors natural frequency,there may be an audible change in noise made by the motor, as well as anincrease in vibration. The resonant point will vary with the application and load,but typically occurs somewhere between 70 and 120 steps per second. In severecases the motor may lose steps at the resonant frequency. Changing the step rateis the simplest means of avoiding many problems related to resonance in a system.Also, half stepping or micro stepping usually reduces resonance problems. Whenaccelerating to speed, the resonance zone should be passed through as quickly aspossible.

TorqueThe torque produced by a specific rotary stepper motor is a function of: The step rate The current through the windings The type of drive used(The force generated by a linear motor is also dependent upon these factors.)Torque is the sum of the friction torque (Tf) and inertial torque (Ti).T= Tf+ TiThe frictional torque (ounce-inches or gram-cm) is the force (F), in ounces orgrams, required to move a load multiplied by the length, in inches or cm, of thelever arm used to drive the load (r) as shown in figure 8.

Tf= F r

Figure 8.Frictional torque is the force (F)required to move a load multiplied by the lengthof the lever arm (r).

The inertial torque (Ti) is the torque required to accelerate the load (gram-cm2).

It should be noted that as the step rate of a motor is increased, the back electromotive force (EMF) (i.e. the generated voltage) of the motor also increases. This restricts current flow and results in a decrease in useable output torque.

AC Synchronous MotorsStepping motors can also be run on AC (Alternating Current). However,one phase must be energized through a properly selected capacitor. In this casethe motor is limited to only one synchronous speed. For instance, if 60 hertz isbeing supplied, there are 120 reversals or alterations of the power source. Thephase being energized by a capacitor is also producing the same number of alterations at an offset time sequence. The motor is really being energized at theequivalent of 240 steps per second. For a 15rotary motor, 24 steps are requiredto make one revolution (24 SPR). This becomes a 600 RPM synchronous motor.

In the case of a linear actuator the linear speed produced is dependent on theresolution per step of the motor. For example if 60 hertz is supplied to a .001/step motor the resulting speed is .240 per second (240 steps per second times.001/step). Many of HSIs stepping motors are available as 300 or 600 RPMAC synchronous motors.

DrivesStepper motors require some external electrical components in order to run.These components typically include a power supply, logic sequencer, switchingcomponents and a clock pulse source to determine the step rate. Many commercially available drives have integrated these components into a complete package. Some basic drive units have only the final power stage without the controller electronics to generate the proper step sequencing.

Bipolar DriveThis is a very popular drive for a two phase bipolar motor having four leads.In a complete driver/controller the electronics alternately reverse the current ineach phase. The stepping sequence is shown in figure 5.

Unipolar DriveThis drive requires a motor with a center-tap at each phase (6 leads). Instead of reversing the current in each phase, the drive only has to switch current from one coil to the other in each phase (figure 6). The windings are such that this switching reverses the magnetic fields within the motor. This option makes for a simpler drive but only half of the copper winding is used at any one time. This results in approximately 30% less available torque in a rotary motor or force in alinear actuator as compared to an equivalent bipolar motor.

L/R DrivesThis type of drive is also referred to as a constant voltage drive. Many ofthese drives can be configured to run bipolar or unipolar stepper motors. L/Rstands for the electrical relationship of inductance (L) to resistance (R). Motor coilimpedance vs. step rate is determined by these parameters. The L/R drive shouldmatch the power supply output voltage to the motor coil voltage rating for continuous duty operation. Most published motor performance curves are based onfull rated voltage applied at the motor leads. Power supply output voltage levelmust be set high enough to account for electrical drops within the drive circuitry for optimum continuous operation.Performance levels of most steppers can be improved by increasing the appliedvoltage for shortened duty cycles. This is typically referred to as over-drivingthe motor. When over-driving a motor, the operating cycle must have sufficientperiodic off time (no power applied) to prevent the motor temperature risefrom exceeding the published specification.

Chopper DrivesA chopper drive allows a stepper motor to maintain greater torque or forceat higher speeds than with an L/R drive. The chopper drive is a constant currentdrive and is almost always the bipolar type. The chopper gets its name from thetechnique of rapidly turning the output power on and off (chopping) to controlmotor current. For this setup, low impedance motor coils and the maximum voltage power supply that can be used with the drive will deliver the best performance. As a general rule, to achieve optimum performance, the recommended ratio between power supply and rated motor voltage is eight to one. An eight to one ratio was used for the performance curves in this catalog.

Microstepping DrivesMany bipolar drives offer a feature called microstepping. Microsteppingelectronically divides a full step into smaller steps. For instance, if one step of alinear actuator is 0.001 inch, this can be driven to have 10 microsteps per step. Inthis case, one microstep would normally be 0.0001 inch. Microstepping effectivelyreduces the step increment of a motor. However, the accuracy of each microstephas a larger percentage of error as compared to the accuracy of a full step. Aswith full steps, the incremental errors of microsteps are non-cumulative.

Fatigue / LifeWith proper application, HSIs linear actuators deliver up to 20 million cyclesand HSIs rotary motors provide up to 25,000 hours of service. Ultimately motorfatigue and resultant life are determined by each customers unique application.The following definitions are important for understanding motor lifeand fatigue.

Continuous Duty:Running a motor at its rated voltage.

25% Duty Cycle:Running a motor at double its rated voltage on an L/Rdrive. The motor is on approximately 25% of the time. The motorgenerates about 60% more output than at rated voltage. Note, dutycycle is not related to the load placed on the motor.

Life:A linear actuators life is the number of cycles that the motor is able tomove at a prescribed load and maintain step accuracy. Rotary motor lifeis the number of hours of operation.

One Cycle:A linear actuators cycle consists of extending and retractingback to the original position.

Terminology :

Detent or residual torque:The torque required to rotate the motors outputshaft with no current applied to the windings.

Drives:A term depicting the external electrical components to run a StepperMotor System. This will include power supplies, logic sequencers, switching components and usually a variable frequency pulse source to determine the step rate.

Dynamic torque:The torque generated by the motor at a given step rate. Dynamictorque can be represented by PULL IN torque or PULL OUT torque.

Holding torque:The torque required to rotate the motors output shaft while thewindings are energized with a steady state D.C. current.

Inertia:The measure of a bodys resistance to acceleration or deceleration.Typically used in reference to the inertia of the load to be moved by a motor or theinertia of a motors rotor.

Linear step increment:The linear travel movement generated by the leadscrewwith each single step of the rotor.

Maximum temperature rise:Determined by the resistance rise method, motorunmounted in free air and energized with a steady state D.C. current.

Pull in torque:The load a motor can move without missing steps when startedat a constant pulse rate.

Pull out torque:The load a motor can move when at operating speed. This isnormally substantially greater than the Pull in torque.

Pulse rate:The number of pulses per second (pps) applied to the windings of themotor. The pulse rate is equivalent to the motor step rate.

Pulses per second (PPS):The number of steps that the motor takes in onesecond (sometimes called steps per second). This is determined by the frequencyof pulses produced by the motor drive.

Ramping:A drive technique to accelerate a given load from a low step rate, to agiven maximum step rate and then to decelerate to the initial step rate without theloss of steps.

Single step response:The time required for the motor to make onecomplete step.

Step:The angular rotation produced by the rotor each time the motor receivesa pulse. For linear actuators a step translates to a specific linear distance.

Step angle:The rotation of the rotor caused by each step, measured in degrees.Steps per revolution:The total number of steps required for the rotor torotate 360.

Torque to inertia ratio:Holding torque divided by rotor inertia.

0000 MVI A 80 3E ;0001 80 ;0002 OUT 43 D3 ;0003 43 ;0004 MVI A 88 3E ;0005 88 ;0006 OUT 40 D3 ;0007 40 ;0008 CALL8050 CD ;0009 50 ;000A 80 ;000B RLC 07 ;000C JMP 0006 C3 ;000D 06 ;000E 00 ;

8050 PUSH PSW F5 ;8051 LXI H 0001 21 ;8052 01 ;8053 00 ;8054 LXI D 0FFF 11 ;8055 FF ;8056 0F ;8057 DCX D 1B ;8058 MOV A D 7A ;8059 ORA E B3 ;805A JNZ 8057 C2 ;805B 57 ;805C 80 ;805D DCX H 2B ;805E MOV A H 7C ;805F ORA L B5 ;8060 JNZ 8054 C2 ;8061 54 ;8062 80 ;8063 POP PSW F1 ;8064 RET C9 ;