JP2750453B2 - Brushless motor circuit for automatic door - Google Patents

Description

Description: TECHNICAL FIELD The present invention relates to a brushless motor circuit for an automatic door.

FIG. 3 is a block diagram of a conventional brushless motor circuit for an automatic door, which employs a so-called PWM inverter chopping method.

The rectifier / smoothing circuit 10 converts, for example, an AC input of 100 V into DC and supplies a high DC voltage to the chopper circuit 20. The chopper circuit 20 includes an on / off circuit (switching element such as a transistor) for chopping and a smoothing circuit, converts a DC high voltage input into a DC low voltage, and outputs it. Differential amplifier 33 calculates the deviation of the output voltage V _M of the chopper circuit 20 for the DC set voltage of, for example, + 33V, and outputs this. The pulse width modulation circuit 50 converts the output voltage φ _M of the differential amplifier 33 in proportion to the pulse width,
To set the output voltage V _M of 20. At this time, the pulse width modulation circuit
The pulse width of the chopper signal P _C1 supplied from 50 to the chopper circuit 20 is determined by comparison between the triangular wave signal of the voltage phi _M and for example 20 kHz. Rectifier / smoothing circuit 1
The switching regulator 55 includes the chopper circuit 20, the differential amplifier 33, and the pulse width modulation circuit 50. Then, the DC constant voltage V _M from the chopper circuit 20 by the switching regulator 55 is obtained.

The constant voltage V _M is applied to the stator windings of the brushless motor 70 make rotating magnetic field via the inverter circuit 60. When a three-phase brushless motor is used, the inverter circuit 60 includes, for example, three transistors forming an upper arm and three transistors forming a lower arm. However, each transistor constituting the lower arm is subjected to pulse width modulation via a Darlington connection transistor.

A position detector made up of a Hall element is arranged near a rotor made of a permanent magnet of the brushless motor 70, and a rotor position detection circuit 80 made up of a logic circuit including this detector is provided.
Detects the position of the rotor of the brushless motor 70, issues an ON / OFF command to the inverter circuit 60, sequentially supplies current to the specified stator winding, and generates a speed pulse having a repetition frequency corresponding to the motor speed. and outputs the P _R. When a Hall element is used for detecting the position of the rotor, it is difficult to obtain a speed pulse exceeding 12 pulses per rotation of the motor. Accordingly, the number of velocity pulses R _R is, for example, 1 to 12 pulses per one rotation of the motor.

F / V converter circuit 90, and outputs the repetition frequency of the velocity pulses P _R in proportion into a DC voltage. As the F / V conversion circuit 90, a circuit in which a differentiating circuit 91, a one-shot multivibrator 92, and a smoothing circuit 93 are sequentially connected in cascade is used. The differential amplifier 95 calculates a deviation of the output voltage V _R of the F / V conversion circuit 90 from the speed command voltage V _S. The pulse width modulation circuit 100, in proportion converting the output voltage phi _R of the differential amplifier 95 to the pulse width, applying a pulse width modulation in the lower arm transistors of the inverter circuit 60. The chopper signal P supplied from the pulse width modulation circuit 100 to the inverter circuit 60
The pulse width of the _C2 is determined by comparison between the triangular wave signal of the voltage phi _R and for example 20 kHz.

In the conventional automatic door brushless motor circuit described above, the DC voltage applied to the brushless motor 70 from the switching regulator 55 via the inverter circuit 60
V _M That motor voltage is constant. However, since the ON time of each transistor constituting the lower arm of the inverter circuit 60 is a pulse width modulated by the speed feedback, the brushless motor 70 is controlled to a constant speed corresponding to the magnitude of the speed command voltage V _S, the low speed The torque can be increased. At this time, the F / V converter circuit 90, differentiating circuit 91 detects the respective edges of the velocity pulses P _R, generates a pulse having a constant width one-shot multivibrator 92 starts from the edge,
A smoothing circuit 93 smoothes this pulse to obtain a DC voltage. The output pulse width of one-shot multivibrator 92 is set by a time constant circuit connected thereto.

[Problems to be Solved by the Invention] Normally, in a brushless motor that drives the opening and closing of an automatic door, for example, a high-speed mode in the range of 1000 to 3000 rpm
Only two modes, a low speed mode in the range of 50 to 500 rpm, are required.

However, the brushless motor circuit for the conventional automatic doors, including the number of velocity pulses P _R that is output from the rotor position detection circuit 80 even though less and 12 pulses per one rotation of the motor, the required range 250 The following problem arises because one speed feedback control is intended to cover the speed range of ~ 3000 rpm.

That is, the one-shot so as not to output pulses etc. Press resulting in interconnected one-shot multivibrator 92 in the motor maximum speed 3000rpm which period of the speed pulse P _R is the shortest multivibrator 92
When the output pulse width setting is small, narrow pulses are applied to the smoothing circuit 93 at long intervals during low-speed rotation. Accordingly, the ripple contained in the output voltage V _R of the output voltage, ie F / V conversion circuit 90 of the smoothing circuit 93 during low-speed rotation is very large. This pulsation is the difference amplifier 95
The pulse width of the chopper signal P _C2 that also the output voltage phi _R output from the pulse width modulation circuit 100 appear irregularly varies, the ON time of the lower arm transistors of the inverter circuit 60 varies. Therefore, conventionally, the motor speed is about 300r
If the speed is less than pm, hunting is likely to occur in the brushless motor 70, and there is a problem that vibration and torque unevenness occur particularly in the case of a lightweight door.

In order to solve the hunting problem at low speed, for example, it is conceivable to connect a rotary encoder that outputs 100 pulses per rotation to the rotating shaft of the brushless motor 70 to detect the position of the rotor, but the rotary encoder is expensive. Is a problem.

The present invention has been made in view of the special circumstances in the case of a brushless motor circuit for an automatic door in which only a two-stage mode of a high speed and a low speed is required, and is provided by switching the output voltage of a chopper circuit stepwise. It is another object of the present invention to realize smooth speed control in both modes even when the number of speed pulses per motor rotation is small.

[Means for Solving the Problems] A brushless motor circuit for an automatic door according to the present invention is a motor circuit that supplies an output of a chopper circuit to a stator winding of a brushless motor via an inverter circuit. A switching circuit for switching and outputting a high DC setting voltage when receiving a command to be performed, and a low DC setting voltage when receiving a command to perform low-speed rotation, and a chopper circuit output voltage with respect to a switching output voltage of the switching circuit. Differential amplifier that calculates the deviation of the first differential amplifier, a first pulse width modulation circuit that sets the output voltage of the chopper circuit by proportionally converting the output voltage of the first differential amplifier to a pulse width, and a brushless motor The rotor position that detects the position of the rotor and issues an ON / OFF command to the inverter circuit and outputs a speed pulse with a repetition frequency corresponding to the motor speed An F / V conversion circuit that is composed of a cascade connection of a detection circuit, a differentiation circuit, a one-shot multivibrator, and a smoothing circuit, that proportionally converts the repetition frequency of the speed pulse to a DC voltage and outputs the DC voltage, and this F / V conversion A second differential amplifier for calculating a deviation of the output voltage of the circuit from the speed command voltage, and an on-time of a switching element constituting an inverter circuit by converting the output voltage of the second differential amplifier in proportion to the pulse width. And a second pulse width modulation circuit for setting.

[Operation] The switching circuit, the first differential amplifier, the first pulse width modulation circuit and the chopper circuit constitute a switching regulator capable of obtaining a plurality of different constant voltage outputs. That is,
When a command to rotate at high speed is given to the switching circuit, a DC high voltage output is obtained from the chopper circuit, and when a command to rotate at low speed is given to the switching circuit, a DC low voltage output is obtained from the chopper circuit.

The variable output of the chopper circuit is supplied to the stator winding of the brushless motor via the inverter circuit to generate a rotating magnetic field. The speed of the rotor of this brushless motor is
The signal is fed back to the inverter circuit through the rotor position detection circuit, the F / V conversion circuit, the second differential amplifier, and the second pulse width modulation circuit. That is, the on-time of the switching element forming the inverter circuit is pulse width modulated in accordance with the magnitude of the speed command voltage, and constant speed control is performed. However, if a command to perform high-speed rotation is given to the switching circuit and a high DC voltage is applied from the chopper circuit to the inverter circuit, even if the speed command voltage is low, for example,
A high-speed mode in the range of 1000 to 3000 rpm is realized. On the other hand, when a command to perform low-speed rotation is given to the switching circuit and a DC low voltage is applied from the chopper circuit to the inverter circuit, a speed command voltage in the same range as in the high-speed mode is applied to the second circuit. , A low-speed mode in the range of, for example, 250 to 500 rpm is realized. That is, the low-speed mode can be realized without extremely narrowing the pulse width of the chopper signal provided from the pulse width modulation circuit to the inverter circuit.

In other words, the output pulse width of the one-shot multivibrator in the F / V conversion circuit for speed feedback can be set within each range of the high-speed mode or the low-speed mode, so that the output pulse width setting can be increased. . Therefore, even when the number of speed pulses output from the rotor position detection circuit is as small as 1 to 12 pulses per rotation of the motor, the output ripple of the smoothing circuit connected to the one-shot multivibrator becomes small, In both modes, smooth speed control is realized.

Embodiment FIG. 1 is a circuit diagram of a brushless motor circuit for an automatic door according to an embodiment of the present invention.

For example, a 100 V AC power supply 2 is connected to the rectifier / smoothing circuit 10. The rectifying / smoothing circuit 10 converts the AC input into DC and supplies a DC high voltage to the chopper circuit 20. In this chopper circuit 20, the output voltage of the rectifying / smoothing circuit 10 is the primary winding of the pulse transformer 21 and the main switching element.
This is applied to a series circuit with the NPN transistor 22. The base circuit of the main transistor 22 is connected to another pulse transformer 23.
Is connected to the collector circuit of another NPN transistor 24 (sub-transistor) insulated by. The secondary side of the pulse transformer 21 in which the main transistor 22 is connected to the primary winding is connected to an inverter circuit 60 via a smoothing circuit 25 composed of a diode, a coil and an output capacitor.

The voltage across the output capacitor of the chopper circuit 20 is supplied not only to the inverter circuit 60 but also to a voltage dividing circuit formed by connecting two resistors 31, 32 in series. Both resistors 31, 32
The connection point voltage V _M, are utilized to monitor the output voltage of the chopper circuit 20. That is, the voltage V _M is applied as an inverting input to the differential amplifier 33, a voltage feedback circuit 30 here
Is formed.

Reference numeral 40 denotes a switching circuit for switching the magnitude of the DC setting voltage applied to the non-inverting input terminal of the differential amplifier 33.
In this switching circuit 40, + 10VDC is the analog switch 4
1 and + 33VDC is applied to the other analog switch 42. One analog switch 41 has a speed switching signal
S _W is supplied through a NOT circuit 43, to the other analog switch 42, the speed switching signal S _W is given directly. When the speed switching signal S _W ie fast command H level is applied to the switch circuit 40, the analog switch 42 is turned on, the DC high voltage + 33V is applied to the differential amplifier 33 as a switching output voltage V _U. On the other hand, when the speed switching signal S _W ie slow command L level is applied to the switch circuit 40, the analog switch 41 is turned on, a low DC voltage + 10V is applied to the differential amplifier 33 as a switching output voltage V _U.

Output voltage phi _M of the differential amplifier 33, a pulse width modulation circuit 50
In, the chopper signal PC1 is compared magnitude output signal T ₁ of the triangular wave oscillation circuit 52 in a comparator 51. This chopper signal PC1 is used as a sub-transistor of the chopper circuit 20.
It is applied to the base of 24 and turns on / off the main transistor 22 via the pulse transformer 23. That is, the pulse width modulation circuit 50, in proportion converting the output voltage phi _M of the differential amplifier 33 to the pulse width, setting the output voltage V _M of the chopper circuit 20. The frequency of the triangular wave signal T ₁ is, for example, 20 kHz.

The above rectifying / smoothing circuit 10, chopper circuit 20, voltage feedback circuit 30, switching circuit 40, and pulse width modulation circuit 50 constitute a switching regulator 55 capable of switching the output voltage in two stages.

The output voltage of the chopper circuit 20 is applied to the stator winding of the brushless motor 70 via the inverter circuit 60 to generate a rotating magnetic field. When a three-phase brushless motor is used, the inverter circuit 60 has three PNP transistors (upper arm) 61a, 61b, 61c and three NPN transistors (lower arm)
62a, 62b, and 62c. Each transistor is connected to a flywheel diode. However, the collectors of the other PNP transistors 63a, 63b, 63c are respectively connected to the bases of the transistors 62a, 62b, 62c constituting the lower arm, and the emitters of these transistors 63a, 63b, 63c apply pulse width modulation. To the DC power supply _VBB via one common PNP transistor 64 for connection.

The brushless motor 70 has a rotor composed of, for example, two-pole permanent magnets, around which three-phase stator windings are equally arranged at 120 degrees. Further, three position detectors 81 composed of Hall elements are arranged at equal intervals of 120 degrees in the vicinity of the rotor, and these detectors supply a brushless motor 70 with a speed signal corresponding to the rotor position.
A, B and C are output respectively. These speed signals A, B, C are
It is input to the drive logic circuit 82 and is converted into ON / OFF commands U, V, W for the upper arm of the inverter circuit 60 and ON / OFF commands X, Y, Z for the lower arm. However, ON / OFF commands U, V, W for the upper arm are
a, 61b, 61c are directly driven, respectively, but the ON / OFF commands X, Y, Z for the lower arm are PNP transistors 63a,
NPN transistors 62a, 62b, 62c through bases of 63b, 63c
Are respectively driven. The drive logic circuit 82, the speed signals A, B, and generates a speed pulse P _R based on C. Rate of speed pulse P _R is, for example, 12 pulses per rotation of the brushless motor 70. However, since the circuit configuration of the drive logic circuit 82 is well known, a detailed description thereof will be omitted. Note that the position detector 81 and the drive logic circuit 82 constitute the rotor position detection circuit 80.

Speed pulse P _R is input to the F / V converter circuit 90. F / V converter circuit 90, differentiating circuit 91, be those sequentially cascaded one-shot multivibrator 92 and the smoothing circuit 93 converts the repetition frequency of the velocity pulses P _R to the direct current voltage V _R which is proportional to . The output voltage V of this F / V conversion circuit 90
_R is input to the inverting input terminal of the differential amplifier 95. This differential amplifier 95 is a DC voltage setting circuit which is a resistance voltage dividing circuit.
98 speed command voltage V _S is input to the non-inverting input terminal from,
Calculating a deviation with respect to the speed command voltage V _S of the output voltage V _R of the F / V converter circuit 90. Note that the F / V conversion circuit 90 and the differential amplifier 95
Constitutes the speed feedback circuit 96.

Output voltage phi _R of the differential amplifier 95, a pulse width modulation circuit 10
At 0, the chopper signal P _C2 is compared output signal T ₂ and the magnitude of the triangular wave oscillation circuit 102 in the comparator 101. The chopper signal P _C2, the in inverter circuit 60 and drives the base of the common PNP transistor 64, multiplying the lower arm transistors 62a of the inverter circuit 60, 62b, a pulse width modulation 62c. That is, the pulse width modulation circuit 100
In proportion converting the output voltage phi _R of the differential amplifier 95 to the pulse width is set lower arm transistors 62a, 62b, a 62c on time. The frequency of the triangular wave signal T ₂ are, for example 20kH
z.

FIG. 2 is a timing chart showing the operation of the automatic door brushless motor circuit described above.

The rectifying / smoothing circuit 10 converts, for example, a 100 V AC input supplied from the AC power supply 2 into DC and supplies a high DC voltage to the chopper circuit 20. The chopper circuit 20 converts the DC high voltage input into a DC low voltage, and supplies this to the inverter circuit 60.

On the other hand, in the position detection circuit 80, the position detector 81 outputs a speed signal A,
Output B and C respectively. These speed signals A, B, C are turned on / off by the drive logic circuit 82 to the inverter circuit 60.
It is converted into the off command U, V, W, X, Y, Z. At this time, the upper arm transistors 61a, 61b, and 61c and the lower arm transistors 62a, 62b, and 62c of the inverter circuit 60 are respectively selected to be one, and the fixed stator winding corresponding to the rotor position of the brushless motor 70 is formed. An electric current flows to generate a rotating magnetic field, and the rotor rotates. However, since the current supply to the stator windings is limited to the on-time of the common PNP transistor 64, the chopper signal P _C2 brushless motor 70 is output from the pulse width control circuit 100, the speed command voltage V _S size Is performed in accordance with the constant speed.

At this time, the speed of the H-level switching signal S _W ie fast command as shown in the left half of Figure 2 is applied to the switching circuit 40, the DC high voltage + 33V as switching output voltage V _U is the differential amplifier 33 The switching regulator 55 operates with this voltage as a reference voltage. Therefore, a high DC voltage is output from the chopper circuit 20. At this time giving a ramp input, as indicated in the figure as a speed command voltage V _S,
The speed of the brushless motor 70 changes in a high-speed mode range of, for example, 1000 to 3000 rpm.

On the other hand, the speed of the L-level switching signal S _W ie slow command as shown in the right half of the figure is applied to the switching circuit 40,
Switching regulator 55 has +10 V switching output voltage V _U
Operates as a reference voltage, and the DC output voltage of the chopper circuit 20 decreases. At this time, even if a ramp input in the same range as the high-speed mode as shown in FIG.
In the low speed mode range. In other words, it is possible to realize a low-speed mode without extremely narrowing the pulse width of the chopper signal P _C2 to provide a pulse width modulation circuit 100 to the inverter circuit 60.

In other words, the output pulse width of the one-shot multivibrator 92 in the F / V conversion circuit 90 for speed feedback can be set within each range of the high-speed mode or the low-speed mode. Can be. Therefore, the speed pulse output from the rotor position detection circuit 80
The number of P _R is at least, the output ripple of the smoothing circuit 93 becomes small, both smooth speed control can be realized in both modes.

[Effect of the Invention] As described above, the brushless motor circuit for an automatic door according to the present invention switches the output voltage of the chopper circuit stepwise between the high-speed mode and the low-speed mode in the so-called PWM inverter chopping method. According to the present invention, smooth speed control can be realized in both modes even when the number of speed pulses per motor rotation is small. That is, not only can the speed be smoothly controlled at the time of high-speed rotation, but also hunting in a low-speed region can be prevented unlike the related art.

[Brief description of the drawings]

FIG. 1 is a circuit diagram of a brushless motor circuit for an automatic door according to an embodiment of the present invention, FIG. 2 is a timing chart showing the operation of the brushless motor circuit for an automatic door of the preceding figure, and FIG. It is a block diagram of a brushless motor circuit for automatic doors. Explanation of reference numeral 10: rectifying / smoothing circuit, 20: chopper circuit, 30: voltage feedback circuit, 33: differential amplifier, 40: switching circuit, 50:
... Pulse width modulation circuit, 55 ... Switching regulator, 60 ... Inverter circuit, 70 ... Brushless motor,
80: Rotor position detection circuit, 81: Position detector, 82:
Driving logic circuit, 90 F / V conversion circuit, 91 Differentiator circuit, 9
2 One-shot multivibrator, 93 Smoothing circuit, 95 Differential amplifier, 96 Speed feedback circuit, 98 DC voltage setting circuit, 100 Pulse width modulation circuit, P _C1 , P _C2
... chopper signal, P _R ...... speed pulse, S _W ...... speed switching signal, V _S ...... speed command voltage.

Claims (1)

(57) [Claims] 1. A motor circuit for supplying an output of a chopper circuit to a stator winding of a brushless motor via an inverter circuit. A switching circuit for switching and outputting a low DC setting voltage when receiving a command to perform the operation, a first differential amplifier for calculating a deviation of a chopper circuit output voltage with respect to a switching output voltage of the switching circuit, A first pulse width modulation circuit for setting the output voltage of the chopper circuit by proportionally converting the output voltage of the differential amplifier into a pulse width, and turning on / off the inverter circuit by detecting the position of the rotor of the brushless motor A rotor position detection circuit that issues a command and outputs a speed pulse having a repetition frequency corresponding to the motor speed, a differentiation circuit, a one-shot multivibrator, and a smoothing circuit. An F / V conversion circuit configured to convert the repetition frequency of the speed pulse into a DC voltage in proportion to the DC voltage, and output the F / V conversion circuit, and calculate a deviation of the output voltage of the F / V conversion circuit from the speed command voltage. Two differential amplifiers,
A second pulse width modulation circuit for converting the output voltage of the second differential amplifier into a pulse width in proportion to a pulse width to set an on-time of a switching element constituting an inverter circuit. Motor circuit.