In all servomechanisms, error signal is fed to an error-correcting device. This device then positions the output (valve, antenna, recorder pen etc.) so that it corresponds to the input. In order to do so, error-correcting device must be able to respond very quickly and have the ability to reverse direction. The most common device used in an error-correcting device is a servomotor. Servomotors can be either a.c. or d.c. While a.c. servomotors are used frequently, some applications require the use of d.c. servomotor.
D.C. Servomotor
A d.c. servomotor is usually a split-field motor. The field may consist of two separate windings or one center-tapped field winding which functions as two separate winding. One winding will cause motor rotation in opposite direction than other winding. These motors may be of shunt-type or series-type as shown in figure below:
In servomechanism, switch shown above would be replaced by an active circuit, perhaps a transistor circuit, and the field input would be amplified error signal. The magnitude, phase, and polarity of error signal would determine how fast and in what direction the motor will rotate.
A.C. Servomotor
An a.c. servomotor is normally a two-phase induction motor. In fact major application of a two-phase induction motor is in control systems. In two phase induction motor, two field windings electricaly at right angles are used.
For control applications one stator phase winding is used as reference field and is supplied from a constant voltage source at rated frequency, while other stator winding is excited with a variable voltage usually proportional to the error in the system. Schematic of two-phase induction motor is given below:
In figure (a), the series capacitor, C₂, is added to insure exciting currents in the two windings to be 90° apart in phase. If error voltage applied across control winding is in phase with the reference voltage, there will be clockwise rotation, whereas is error voltage is negative, motor will rotate conterclockwise as in this case phase shift will be 270°. Capacitor C₁ is added to increase the impedance of the load on error amplifier and thus reduce current. The value of capacitor, which form a resonant circuit with the field windings is usualy given in the specification by manufacturers.
Figure (b) shows schematic in which electronic amplifier provides 90° phase shift between the input voltage and amplified output voltage.
Analysis of Two Phase Induction Motor
For analysing the performance, the motor may be considered to be a continuous two-phase balanced induction motor, operating from an unbalanced two-phase sinusoidal power supply.
The unbalanced two-phase supply is resolved into two balanced two-phase systems-one of positive sequence and other negative sequence. The actual motor with unbalanced power supply may therefore condidered to be replaced by two balanced two-phase motors mechanically coupled to each other-one operating with a positive sequence and other negative sequence, at slip (s) and (2-s) respectively. The net torque of the motor at any speed then may be found by computing positive sequence motor torque and negative sequence motor torque at that speed and taking algebraic sum of these two torques.
Equivalent circuit of these two motors is shown in the figure below:
Vr = Vrp + Vrn
Vc = Vcp + Vcn … (1)
Also, Vcp = jVrp, and Vcn = - jVrn
Vc = j(Vrp - Vrn), or
jVc = (Vrn - Vrp) … (2)
From (1) and (2), we get,
Vrp = ½(Vr - jVc), and
Vrn = ½(Vr + jVc)
Taking axis of Vr as reference, we have
Vr = Vr + j0 and
Vc = 0 + jVc + j0; so
Vrp = ½(Vr + Vc) + j0; and
Vrn = ½(Vr - Vc) + j0 … (3)
Similiarly, analogous expressions for currents can be obtained.
Note - The subscript p stands for positive sequence and subscript n stands for negative sequence.
Example: A 400 V, 50 Hz, two-phase, four-pole servo-motor with a cage rotor has following parameters:
R1 = 2.1 Ω/phase; R2' = 7.25 Ω/phase; X1 = 2.0 Ω/phase; X2' = 2.0 Ω/phase; Xm = 47.34 Ω/phase.
The motor is operated with 115 V across the reference field winding and 69 V across the control winding, the latter leads the former by 90°.
Determine the stator currents, output power factor and torque of the motor with s = 0.5. Assume windage and iron losses at above slip and voltage to be 61.5 W. Stator windings have equal turns.
Solution:Taking voltage across the reference winding as reference, from equation (3), we have:
Vrp = ½(115+69) + j0 = (92 + j0) V
Vrn = ½(115-69) + j0 = (23 + j0) V
Total impedance of positive and negavive sequence circuits are:
Zp = Rp + jXp = (R₁ + jX₁) + ((R₂'/s + jX₂') || jXm) = (2.1 + j2) + (12.29 + j 5.53) = (14.39 + j7.53) Ω;
Zn = Rn + jXn = (R₁ + jX₁) + ((R₂'/(2-s) + jX₂') || jXm) = (2.1 + j2) + (4.41 + j 2.35) = (6.51 + j4.35) Ω;
Irp = Vrp/Zp = (92 + j0) / (14.39 + j7.53) = (5.02 - j2.63) A = 5.67 A
Irn = Vrn/Zn = (23 + j0) / (6.51 + j4.35) = (2.44 - j1.63) A = 2.93 A
Phase currents are:
Ir = Irp + Irn = (5.02 - i2.63) + (2.44 - j1.63) = (7.46 - j4.26) A
Ic = Icp + Icn = jIrp - jIrn = (2.63 + j5.02) - (1.63 + j2.44) = (1.00 + j2.58) A
Power delivered across the air gap by positive sequence motor (neglecting stator losses) is:
Pgap,p = 2·Irp²·R₂'/s = 932.32 W
Power delivered across the air gap by negative sequence motor (neglecting stator losses) is:
Pgap,n = 2·Irn²·R₂'/(2-s) = 82.99 W
Internal mechanical power developed is (1-s) times the total air-gap power or
Pmech = (1-s) (Pgap,p - Pgap,n) = (1 - 0.5) (932.32 - 82.99) = 424.67 W
Net output power
P = Pmech - windage and iron losses = 424.67 W - 61.5 W = 363.17 W.
Torque developed
τ = P/((1-s)(2πn/60)) = 4.62 N·m
Constructional Aspects
Rotor of a two-phase servo motor can be ordinary squirrel cage type (used for comparatively larger ratings) or in general it is drag-cup type rotor as shown in the figure below. This reduces rotor weight and therefore the moment of inertia and also increases rotor resistance, which is essential to avoid single-phase operation, when Vc becomes zero at some point.
Effect of rotor resistance on torque-speed characteristics of a two-phase servomotor and the necessity of high rotor resistance motor to prevent single-phase operation (when Vc = 0) is shown in figure below—
Cooling
Servo motor normally operates at very low speeds, therefore fans mounted on rotor shaft are ineffective in a servo motor.Generally blowers driven by a small single-phase induction motor (operated from the same source as that of reference field) are used for cooling servo motor above 10 W. The output for a particular frame size can be increased by two to three times by such cooling arrangement.
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