modeling and control of a dc servo motorbisonacademy.com/ece761/lectures/11 servo motors.pdf ·...

Modeling and Control of a DC Servo Motor

http://www.pittman-motors.com/Brush-DC-Motors.aspx

In order to drive a robot arm, you need a motor. There are several types:

Stepper Motors: Motors with a digital input, allowing you to control the position of the motor with aresolution of 0.9 degrees typically. Stepper motors are easy to control (you specify the number of steps themotor turns at 0.9 degrees per step). However, they are very inefficient and have a limited holding torque.

Brushless DC Motors (AC Synchronous Motors): Like a stepper motor but with a 3-phase AC input. Here,speed is frequency and torque is phase lead angle. Harder to control but smaller and more efficient thanbrush-type DC motors. Typical motor used in quadcopters due to their power vs. weight ration.

Brush DC Motor: Common motor used in RC cars. Speed is determined by voltage and torque by current.

Here, we will focus on brush-type DC motors to control a robotic arm.

Brush-Type DC Motor Modeling:

A brush-type DC motor consists of a rotor and a stator.

Inner Workings of a DC Servo Motor (http://www.zgcmotor.com)

NDSU Modeling and Control of a DC Servo Motor ECE 761

JSG 1 rev March 15, 2016

The rotor is the part that spins. It consists of a shaft which contains electromagnets. To provide electricalconnection to the outside world, carbon brushes are often used. These rub against the commutator, whichswitches different portions of the stator coils in and out.

Without the commutator, the voltage produced when the motors is spun will be sinusoidal. (From duality, motorsand generators are the same thing. When you apply torque to a motor, it acts as a generator, convertingmechanical power to electrical power. When you apply current to a motor, it acts the other way, convertingelectrical power to mechanical power.) Likewise, you need a sinusoidal voltage to drive a motor without acommutator (termed an AC synchronous motor or (inaccurately) a brushless DC motor.) With the commutator,the polarity is switched, resulting in a voltage with a DC offset.

Brushless Motor Brush-Type DC Motor

Without a commutator (brushless motor), the voltage produced by a motor is sinusoidal. With a commutator (brush-type motor), different coilsare turned on and off, resulting in an output voltage with less ripple and a DC offset.

To reduce the ripple, several coils are often used, with only one energized at a given time. To further reduce theripple, the rotor is 'twisted' to even out the voltage (and torque when used as a motor).

Rotor of a DC Servo Motor showing multiple windings, twisted to reduce the ripple in the voltage (and torque) producedhttp://image.slidesharecdn.com/chapter4-dcmachineautosaved-140915220206-phpapp01/95/

Equations for a Brush-Type DC Motor (DC Servo Motor)

The equations that describe a DC servo motor are as follows:

Motors and generators are one and the same. If you spin a motor, you produce voltage proportional to the speedof the motor:



Back EMF = Kt KtdQ

dt

where Kt is a constant of the motor and Q is the motor angle. Similarly, if you apply current to a motor, youproduce torque

Torque = KtIa

where Ia is the armature current. Note that power out must equal power in

Pin Pout

VaIa T

Substituting:

Kt Ia KtIa

It isn't obvious, but the two constants are the same:

Kt Vrad/sec

NmA

The circuit model for a DC servo motor consists of an electrical side (an electromagnet) and a mechanical side(the motor).

The electrical side (the armature) has a resistance and inductance inherent with an electromagnet.

The mechanical side has inertia and friction, inherent with the rotor

The two are coupled by the torque constant, Kr:

+

-

Q

Js 2 Ds

KtIa

Kt*dQ/dt

LasRa

Va

Ia

Model for a DC Servo Motor

The equations of motion are then:

Va KtsQ Ra LasIa

KtIa Js2 DsQ

Solving gives

Q

1s

Kt

JsDLasRaKt2Va

if the output is angle (Q) or

dQ

dt

Kt

JsDLasRaKt2Va



if the output is speed.

Example: Determine the transfer function for a Pittman 14207 Motor operated at 76.4DC.

http://www.pittman-motors.com/Brush-DC-Motors.aspx

From the table:

Kt = 0.226 Nm/A

L = 8.93mH

J = 4.73E-5 kg m2

R = 5.78 Ohms

D = unknown

No-Load Speed = 3140 rpm @ 0.090 A

You can compute the resistance through a power balance

Pin 76.4V 90mA 6.88W

The electrical losses are

PR Ia2R 90mA2 5.78 46.8mW

The remaining losses must be the rotational losses

T 6.829W



The no-load speed is 3140 rpm (328.8 rad/sec)

The torque is

T 6.829W328.8rad/ sec

0.02077Nm

This then tells you the friction

T D

D T 63.17E 6 Nm

rad/sec

Plugging in numbers:

Kt

JsDLasRaKt2Va

535,050

s2648.6s121,790Va

The step response of a motor with no load is then

>> G = tf(535050, [1, 648.6, 121790]) Transfer function: 535050----------------------s^2 + 648.6 s + 121790

The 2% settling time is approximately 4 / the real part of the dominant pole >> Ts = 4 / 324

0.0123

Plot the step response for a 50V step out to 20ms

>> t = [0:0.001:1]' * 0.02;>> y = step(G,t);>> plot(t*1000,y * 50);>> xlabel('Time (ms)');>> ylabel('Speed (rad/sec)')



Step Response of a Pittman 14207 Motor with No Load and a 50V Step Input

Of course, from a robotics standpoint, what we care about is torque. If you make the input current, the torque isfrom

T KtIa

With a current source, you are specifying the torque directly (which is more useful from a control standpoint for arobotic arm).

Motors and Gears:

One problem with using a motor to drive a robotic arm is that motors are designed to spin whereas robots tend toremain at a certain spot. To couple a motor to a robot, a gear is often used.

Gear Reducer: http://motiontek.ca/gear_reducer.html



Gears reduce the speed of the shaft (or conversely, a slow speed at the output corresponds to a high speed at theinput - which is good since motors want to spin.) The relationship for a gear is

N11 N22

or

1

N2

N1

2

With a 12:1 gear reduction (N1 = 12, N2 = 1), the output shaft spins 12x slower than the input shaft.

Power has to balance, meaning

T11 T22

Substituting

T1

N1

N2

T2

With a 12:1 gear reduction, the output shaft has 12x higher torque than the input shaft. Gears act as torqueamplifiers.

One interesting feature of gears is they reduce the impedance's as seen by the motor. Using the analogy

V IR

2 T2Z2

then an impedance at the output of the gear (Z2) looks like

N1

N2

1

N2

N1

T1Z2

1 T1

N2

N1

2

Z2

The impedance as seen through a gear increase as the turn-ratio squared.

For a 12:1 gear reduction, the motor sees the outside world (the robot) as an impedance, reduced by a factor of144 (122). With a 300:1 gear reduction (not uncommon with robotics), the motor sees the outside world reducedby a factor of 90,000.

One way to make the coriolis and other forces negligible is to severely gear down the motor. This has the badside-effect of making the motor oblivious to the outside world, making things like touch difficult (upcominglecture).

Going the other way, with a 12:1 gear reduction, the motor's inertia gets amplifier by a factor of 144 (122). Thus,to model the inertial of the motor at each joint, a term of 0.0068 should be added to the diagonal of the M-matrix(which we added previously to avoid singularities in the M matrix). (slightly more if you add in the rotationalinertia of the gear itself - hence the 0.01 from before).



N2

N1

2

J 0.0068 kg; m2

The net result is the current applied to this motor is related to the torque as follows. The torque produced by themotor is:

Tm KtIa

The torque as seen by the robot with a 12:1 gear reduction is:

T1 12Tm

The current required at the motor is then

Ia Tm

Kt T1

12Kt 0.3678 T1

Ia 0.3678 T1

If you know the torque you want to apply at a given link, you also know the current.

If you use a current amplifier, you really don't care about the motor dynamics that much. All you really careabout is current (and hence torque).

Feedback Control of a DC Motor

The transfer function for a DC motor with a current input is

Gs Ia

Kt

Js2Ds Ia

To control the angle, you can use a feedback control law

Ref QIaK(s) G(s)

MotorCompensator

where G(s) is the transfer function of the motor and K(s) is a compensator to be designed. Since this motor is tobe attached to a robot with inertia of 1kg m2 (approximately), the motor dynamics on the motor-side of the gearare:

J 4.73 105 1

144 0.00699 kg m2

D 63.17 106 Nmrad/s



giving

32.33

s20.009s Ia

32.33

s2 Ia

If you let K(s) be a constant, k, then the closed-loop system will be

GKE

E R

GK1GK

R

or substituting in the transfer function for G(s)

32.33k

s232.33kR

This has roots on the jw axis, meaning the closed-loop system will not be stable.

Assume isntead that the control law is a PD compensator:

Ia w

D

PR Q

1/s32/s

Motor

Motor with PD Control

Ia PR DsThen

s2 32.33Ia

s2 32.33PR Ds

s2 32.33Ds 32.33P 32.33PR

32.33P

s232.33Ds32.33PR

With this control law, you can place the poles pretty much wherever you like. These poles then effect theresponse of the motor as

The real part of the pole tells you the 2% settling time as



t2% 4real(pole)

The overshoot for a step input comes from the angle of the pole

cos

OS exp

12

For 4% overshoot, the angle of the poles should be 45 degrees.

For a 2% settling time of 100ms and 4% overshoot,

The poles belong at -40 + j40

32.33P

s232.33Ds32.33P

32.33Ps40j40s40j40

32.33P

s280s3200

D = 2.474

P = 98.98

This has the following step response:

Closed-Loop Step Response of the DC Motor

Matlab Code:>> G = tf(3200, [1, 80, 3200]);>> t = [0:0.001:1]' * 0.2;>> y = step(G,t);>> plot(t*1000,y)>> xlabel('Time (ms)');>> ylabel('Angle (radians)');



Note that this is an error-driven feedback loop: the ouput will lag behind the set point (R) with the current drivingthe motor dictated by the error. This can be seen by looking at the frequency response of the transfer function:

3200

s280s3200Rm

The DC gain is one: the angle goes to the reference angle eventually

The high frequency gain is zero: the angle does not responsd imediately.

To speed up the transient, feedforward terms can be added:

Ia sQm

D

PRm Qm

1/s32/s

Motor

Ds

s2/32

feedforward terms

Feedback Control of a DC Motor with Feedforward Control

If you know the set point (R) and its derative and 2nd derativive, you can make the current

Ia PR Ds D R.

m 132.33R̈m

This results in the transfer funciton being

s280s3200

s280s3200Rm Rm

meaning you're tracking the reference angle of the motor (Rm) exactly. Assuming you know Rm and itsderivatives. Assuming the current commanded is possible.



Current Amplifiers

For robotics, you really want to operate a DC motor with a current source (since current is torque). One way todo this is to use an off-the-shelf amplifier such as a Advanced Motion Control ($40 on ebay, $500 retail)

Advanced Motion Control Brush-Type Servo Amplifier (www.ebay.com)

With this amplifier, you can operate it in

Voltage Mode (where the output voltage is proportional to the voltage input - making it look like a bighoggin op-amp capable of 80V and 30A, or

Current Mode (where the output current is proportional to the current input)

To set this, you need to connect the current monitor inputs (see the data manuals).

You can also turn a voltage-amplifier into a current amplfier with a resistor. Conceptually, the circuit looks likethe following:



0.1 Ohm

Vin

Motor

Vr

Vm

Using an op-amp as a current amplifier. Im = 10 Vin

The 0.1 Ohm resistor is a current sensor, which outputs 0.1V per amp. With negative feedback, the op-ampoutputs whatever voltage it takes to force V+ = V-. With the 0.1 Ohm current sensor, this controls the current tothe motor so that

Im 10Vin

The output of the Advanced Motion Control amplifier is actually a differential output (motor + and motor -outputs), resulting in a slightly different circuit to allow for a floating ground, but the idea is the same.



modeling and control of a dc servo motorbisonacademy.com/ece761/lectures/11 servo motors.pdf ·...

Documents