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Stepper Motor Power Loss-Reduction Circuit

IP.com Disclosure Number: IPCOM000044498D
Original Publication Date: 1984-Dec-01
Included in the Prior Art Database: 2005-Feb-06
Document File: 3 page(s) / 66K

Publishing Venue

IBM

Related People

Barcomb, JG: AUTHOR [+4]

Abstract

The circuit described reduces power losses in a stepper motor driving circuit by approximately 50% during times when the motor is detented or stopped. Fig. 1 illustrates the circuitry employed, and Fig. 2 shows a typical timing chart and voltage waveform diagram that results from the operation of Fig. 1. With the A phase winding on and the A phase winding off and the voltage + PED clamped at 5 volts positive, the stepper motor, comprising the motor winding L1 and L2 in Fig. 1, will be detented. Turning to Fig. 2 in the timing chart, clock pulses are shown in the top line for a 20 kilohertz +5-volt clock with a 10% duty cycle. Beginning at the left side of the timing chart in Fig.

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Stepper Motor Power Loss-Reduction Circuit

The circuit described reduces power losses in a stepper motor driving circuit by approximately 50% during times when the motor is detented or stopped. Fig. 1 illustrates the circuitry employed, and Fig. 2 shows a typical timing chart and voltage waveform diagram that results from the operation of Fig. 1. With the A phase winding on and the A phase winding off and the voltage + PED clamped at 5 volts positive, the stepper motor, comprising the motor winding L1 and L2 in Fig. 1, will be detented. Turning to Fig. 2 in the timing chart, clock pulses are shown in the top line for a 20 kilohertz +5-volt clock with a 10% duty cycle. Beginning at the left side of the timing chart in Fig. 2 in the third line from the top and comparing it with the clock line in the first line, it will be observed that the negative-going clock pulse has set the trigger on. The trigger 1 in Fig. 1 is triggered by incoming clock pulses on line 2 and reset by a reset on line 3. The trigger drives transistor 4 which, in turn, will operate transistor 5 to apply current to windings L1 and L2. However, only winding L1 will conduct due to the fact that the phase A signal on line 6 is on and the not phase A on line 7 is off. Thus, transistor 8 will be on and transistor 9 will be off. The voltage in the timing chart bottom line will rise to the +32-volt supply level and the current I2 in winding L2 will be 0. Current I1 will reach approximately 2.1 amps and will keep the total flux in the coils L1 and L2 constant. I1 will flow through transistor 5, winding L1 and transistor 8 to ground through the sense resistor R1. The current will increase until I1 reaches about 0.7 amp. At this time the current will produce a voltage V1 of about .07 volt in the resistor R1. This voltage, amplified by the amplifier 10, which has a gain produced by the ratios of R1, R2 and R3 of about 9.1 to 1, will result in an output voltage from the amplifier of about 0.64 volt. This is applied to a comparator circuit 11 which is the level-sensing comparator that supplies a reset to the trigger circuit 1 over line 3. This will turn off the transistor 5. When transistor 5 turns off, current cannot flow from the 32-volt power supply through the transistor 5 to the node N1. Because flux cannot instantaneously change in the bifilar windings of the motor L1 and L2 and must remain constant, the current I1 will drop to 1/2 value and a current I2 shown in the opposite direction through winding L2 will exist...