ELEC 5565 - Electric Drives and Ctrl Lab

8/2/2019 ELEC 5565 - Electric Drives and Ctrl Lab

1/13

ELEC 5565 - Electric Drive Design and Control Laboratory

LABORATORY REPORT

Student name Student ID Mark for Content Mark for

Presentation

Mark for

Conciseness

Achievement Total

PartA

PartB

PartC

/10 /20 /20 /15 /15 /20 /80

It is the students responsibility to get his/her report signed.

ACHIEVEMENT Stage 1 Stage 2 Stage 3 Stage 4 Stage 5

Full marks will only be given if your calculations, derivation of formula and results have been explained clearly. Ifadditional space is needed, please attach separate sheet(s). Note that a high quality report which deserves a good mark is

the one that has legible writing, well-sorted equations and well-presented diagrams. This report should be typed out and

diagrams can be computer generated if that helps you to produce a more presentable report. The due date for the lab

report will be announced later.

Part A: Open Loop Steady-State Response

1) Comment on the operation of the chopper when applied to the speed control of a d.c. motor.

From the circuit, when the switch is switched on, the input voltage Vs is applied directly to the

motor and during this period, the path of the current is indicated by the red line in figure below.

When the switched is turning off, the current will freewheel through the diode. When the current isfreewheeling through the diode, the motor voltage is clamped at (almost) zero. The speed of the

M

Vs

+

_

V

A

B

L

S

D


2/13

motor is determined by the average of motor voltage, Vdc. Which in turn depends on the total

cycle time (T) for which the switch is turning on. If the on period and off period are defined as:-

Ton = kT (1)

Toff= (1-k)T (2)

Where 0< k


3/13

70% 6280

Table 2: Duty ratio and speed of open loop system

Fig 1: Graph of Speed Vs Duty Ratio for Vs =12V

As we changed the value of Vs to 6V, the graph and output obtained is as below:-

Fig 2: Graph of Speed Vs Duty Ratio for Vs =6V

As we can see, the speed versus duty ratio has a linear relationship. As Vs is decreasing to half of

the voltage, it is obviously seen that the speed is also linearly decreasing to half of the speed when

Vs is equal to 12V (3000 rpm).


4/13

4) Based on Table 1 and the graph derived in 3), describe how the chopper can be used (in open-

loop scheme) to regulate the speed of a d.c. motor. Are there any drawbacks in applying this

approach for speed regulation? If yes, what are they? Compare this scheme with the closed-loopfeedback system shown in Fig. 2. Note that in this closed-loop scheme, the error between the

measured speed (from the tacho) Vspeedand the desired speed value, Vspeed*

is used to regulate VREF

and hence the duty ratio of the chopper. Explain how this scheme outperforms that of the openloop.

The chopper can be used (in open loop scheme) to regulate the speed of a d.c motor as the speed could be

varied by switching the input voltage periodically on and off for varying intervals. Which it is understood

that the speed control is affected by the duty cycle, k. As the duty ratio increases, the speed of a d.c motor

also keeps increases and vice versa. However this is only referring to the analogue control and as far as

possible it is limited to some aspects. In practice, there would be only few potentiometer adjustments

once the drive has been implemented.

In other words, there are drawbacks by applying this approach for speed regulation. We can consider on

what can be done if we were controlling the motor manually. If the speed was found to be below targetand we want to provide more current in order to make acceleration, we would have raised the input

voltage. We have to ensure to do this carefully, however to be mindful of the danger of the creation of an

excessive current due to delicate balance exists between the motor back e.m.f and applied voltage, Vs. We

will end up by keeping an eye on the ammeter all the times to avoid circuit to be blown up as the speed

approached targets. We also would have to adjust back the currents in order to avoid set of speed from

overshooting. This is clearly noted that the system needs to be manually calibrated in order to reach the

desired speed. This action should be overcome automatically by the drive system illustrated in figure 2.

With the closed loop system, the overall performance is dependent on the feedback signal quality as the

speed- proportional voltage is provided by the tacho-generator. With this new approach, we noticed that

as the reference speed is greater than the actual speed, there will be a speed error signal. A speed errorindicates that the acceleration is required. So when Vref increases, so does the motor speed, thereby

provides a motor acceleration. As the speed rises and the speed error reduced, the Vref therefore reduce to

obtain a smooth targeted speed. Thus the control scheme in figure 2 clearly performs a better performance

of required speed as compared to open loop control system.

Part B: Design of Interface-Control Circuit

1) Show why and how the three inverting amplifier circuits shown in Figs. 3 to 4 are used to

implement the scheme highlighted by the grey box as well as the P+I controller in Fig. 2.

Figure 3 shows that an inverting amplifier circuit is used to sum up the negative feedback signal (-

Vspeed) and the reference speed (Vspeed*) to produce an error signal (Verror). This Verror then is

passed through Proportional and Integrator controller (PI controller), in which case the actual anddemanded speed will be under the steady state conditions.

Verror can be derived as follows:-

10 iR

V

i

peeds=


5/13

Which this is obtained from nodal analysis

Then the job of comparing the speed reference and actual speed from tacho and amplifying theerror signal is carried out by speed error amplifier.

A good speed controller will result in zero steady state error and have well-damped response to

step changes in the demand speed. The integral term in PI controller caters for the requirement ofzero steady state error.

As long as speed control loop functions accordingly, the motor speed can never exceed the

reference signal. When there is a small error in speed, the speed references increases in

proportional to speed and thus ensure the linear system behavior with a smooth approach to thetarget speed. However when the speed error exceeds the limits, the output of the speed error

amplifier will start to saturate and there would be no further increase in the speed reference. It

means there is no possibility of exceeding the rated value no matter how large the speed error

becomes.

In other words, we aimed that while setting up the circuit, the steady state motor speed should

correspond exactly with the speed reference.

2) ExpressKP andKI in terms ofRPCandRIC. How much can KP andKI be varied for this setup?

Summarize your observation for the effect ofRPC andRIConKPandKI. What other parameters canbe changed ifKPandKIare to be further adjusted?

Let us consider the proportional circuit. The Transfer function of the proportional amplifier is

given by,

(1)

Where ,

Substituting above in equation (1) we obtain (2)

Therefore output of the proportional control in time domain is (3)

Zin

Zf

sVe

sUp=

)(

)(

PC

f

e

p

R

R

sV

sU=

)(

)(

e

PC

f

p VR

RtU =)(

20*

iR

V

i

speed=

i

speed

i

peed

f

error

R

V

R

Vs

R

Vii

==

=+

*021

*speedspeederror VVV +=


6/13

Where the proportional gain is,

Similarly

The transfer function of the Integrator controller is given by,

(1)

Where ,

Substituting above in equation (1) we get (4)

Therefore output of the proportional control in time domain is (5)

Where

As we understood by now, by only a simple proportional amplifier is in the loop, it is appreciatedthat in order for there to be a steady state value of Vref, there must be a finite speed error which a P

controller would not allow the targeted speed to be achieved. Therefore by adding the Integral

amplifier the steady state error should be eliminated. Kp should be varied such that the transientresponse to the step changes in the Vref should be fast and well damped. The potentiometer,RPC is

adjusted as to optimize the transient speed response. While KI should be varied till the steady state

motor speed corresponds exactly with the speed reference. Extra care should be taken whileperforming adjustments for Kp and KI as both are related and dependent on each other, by varying

one of these may change of the effect of the other as well.

From the observations, we realized that when the value of Rpc is decreasing, the gain of Kp will be

increased. Similarly happened when the value of RIc is decreasing, the gain of KI will be increased.

Thus these characteristics allow the Vref increases in proportion to the speed, thereby ensuring alinear system behavior with the smooth approach to the speed as targeted.

Other parameters that possibly changed to further adjustment forKPandKI are by adjusting the

value of Rffor Kp (Proportional amplifier) and changing the C value forKI(Integral Amplifier).

PC

f

pR

RK =

Zin

Zf

sV

sU

e

p=

)(

)(

sRCsVsU

ICe

p 11)()(

=

dtVCR

tU eIC

I =1

)(

IC

IRC

K

=1


7/13

3) Explain the function of integral controller. Relate the explanation to the operations of the

amplifiers in Fig. 4. Include the explanation on how you verify the rate of change of UIwith the

components you used.

Integral controller will ensure that there is a zero steady state error in the control system. Hence in

order to eliminate the steady state error speed, the arrangement of speed controller to have integral(I) term as well as a proportional (P) term as in Fig 4. A PI controller can have a finite output even

when there is a zero input to the system. By means that the rate of change UI can be achieved tozero steady state error together with this arrangement of PI controller. Thus to verify the rate of

change of UI by the components used is given as follows:-

4) Using the op-amps TL072, describe the limits imposed on the outputs, UPand UI. Discuss theireffects on the system performance.

By using op-amps TL072, the circuit only can be constructed to a limited circuitry. So it does sometimes

causes problem from the instrumentation connected to the controller. One of the effects to the system

performance is that it may take some time for the output to ramp up to the desired speed when the loop

started. Another problem that may arise is that the small amount of noise can cause a large change of

amount in the output. Thus this may be lead to some disturbance to system performance. We also might

face a problem of not getting the steady state error closest to the zero as well. As the proportional term

can also produce a very undesirable output subjected to instantaneous step changes as when the set point

of Vspeed varies.

)()20(

1)(

1

)20(

1

)(

)(

)20()(

)(

/1

0

20

0

tdVKRiCF

tU

sKRisCFsV

sU

KRisCFsU

sV

C

U

KRi

VI

errorI

error

I

I

error

FS

Ierror

+

=

+

=

+=

=

+

=


8/13

Part C: Tuning of P+I Controllers and Evaluation Controller Design

1) Describe your controller tuning approach in detail.

As our goal in the controller design is to eliminate the response (overshoot) and at the same time

minimize the rise time response, so the controller need to be tuned in order to achieve thisobjective. By using the mechanical switch with the switch operates repetitively and the ratio ofon/off time is varied, the average load voltage can be varied continuously. So this arrangement is

often referred as choppers due to the input supply is chopped on and off. When we used a

constant repetition frequency, the width of the on pulse can be varied in order to control the outputvoltage. Provided that the chopping frequency is high enough, the speed remains almost constant at

a value controlled by d.c level of the chopped waveform. So by the process of some trial and error

with some aggressive tuning ofRPCandRIC,we tried to eliminate the response (speed) overshoot aswell as minimizing the rise time response.

As in any difference between the actual and desired speed is been amplified, the output will serve

as the input to the PI controller. Hence for the example if the actual speed is less than the requiredspeed, the speed amplifier will demand Vref proportional to the speed error, and the motor finallyaccelerate in order to minimize the speed error. As we keep tuning the controller in order to get the

best response of speed and response time, we have found that the motor keep accelerate and

decelerate as we tried our best to get to the finest tuning. The tuning method that we used is firstwe set the Integral term to zero. Then we increased the P until the output oscillates. After that we

increased the Integral term until we found that the oscillation has stopped. Finally we increased P

until the loop reached to its reference. We have successfully tuned the controller by balancing the

Proportional and Integral term by adjusting RPC and RIC to the step response at the level ofsatisfactory speed based on the guidelines tabulated below:-

Effects of Increasing parameters

Parameter Response Rise

Time

Response

Overshoot

Response

Settling Time

Steady state

Error

KP Decrease Increase Small Change Decrease

KI Decrease Increase Increase Eliminate

Table 3: General rule of tuning P+I controller

2) Based on the results obtained, evaluate your control design.

From the results as shown below, we can evaluate our results by tuning the controller and we have

achieved a good speed controller. A good speed controller will result in zero steady state error.

This can be achieved by adjusting the potentiometer of RIC or KI as the integral term caters for the

requirement of zero steady state error and eliminating the speed overshoot. While KP is responsibleto decrease the rise time as the transient response depends on the setting of the proportional gain

and time constant. These speed stability potentiometer allowed us to optimize the transient speed


9/13

response as obtained in the figure below. The results of the simulation shows that it almost reached

to overdamped response with minimum rise time as we required.

Fig 3: Speed response with a good control design

3) Discuss on why a bad control system is not suitable for the speed control of a d.c. motor.

A good control system will ensure that the closed loop control system should react quickly to

eliminate the steady state error and perform an error correction without a delay, while a badcontrol system with high electrical or mechanical inertia will have a slow response because of the

delay. When there is a time lag between the error sensing and the corrective action completion, the

system may overshoot. The delay may due to the armature current response or the inertia

associated with lower acceleration and also from the stress on the mechanical parts of the motor.This will make the system to oscillate and will possibly become unstable. This will lead to an

instantaneous loss of torque and finally will reduce the efficiency of the motor. The Figure below

shows an example of undesirable speed with overshoot and unstable response.


10/13

Fig 4: Results of speed with a bad control design

4) Setting TL = 0, derive the closed loop transfer function, (Vspeed(s)/Vspeed*(s)) in terms of the

controller resistances (i.e. RPCandRIC). Use MATLAB to perform the step response to find themost suitable sets ofRPCandRIC. You have to obtain some values for the variables in the list above

from the given datasheet. You may have to consider the external inductance when entering the

values forLa. You may also need to include the scalar gain for the tacho as well as the PWM

generation process. Show your derivation and comment on the choice ofRPCandRICyou made.

The closed loop transfer function obtained from the system below:-

Fig. 5. Control scheme block for the speed regulation of the d.c. motor

DC Motor

Tacho-

Generator

P+I

ControllerSwitching

Circuit

Vspeed

*

+

Vspeed

Verror

Varm

N

VREF


11/13

From the closed loop we obtain:-

(1)

Re-arrange the equation above and we get:-

(2)

Then we obtained Vspeed again from the closed loop in fig 5 yield the equation as below:-

(3)

Rearrange the equation (3) and we get:-

Then Substitute (3) into (1), we obtained the equation as below:-

Finally, the closed loop transfer function can be derived as below:-

(4)

As and

Substituting above into equation (4), Thus the closed loop transfer function in terms of control

resistance is given by:-

To obtain the value of RPC and RIC, we use the formula below and by adjusting the gain of

proportional and integral term, the resistance value can be determined.

speedspeederror VVV = *

*speedspeederror VVV =+

)(s

KsKVV IPerrorspeed

+=

sKsK

VV

IP

speed

error

+

=

*)1( speedIP

speed VKsK

sV =+

+

)1(

1

*+

+

=

IP

speed

speed

KsK

sV

V

PC

f

p

R

RK =

IC

IRC

K

=1

PCfPCIC

ICfPC

RsRRCR

sCRRRsT

+

=

)()(

Verror - 0

1 k + RPC=

0 - UP

10 k

UP = -10 k Verror

1 k + RPC

UP = KP Verror =10k Verror

1k + RPC

KP =10k

1k + RPC


12/13

So by determining the right value of Kp and KI in MATLAB simulation using the trial and errormethod, we have chosen the value of RIC = 0.09 ohm and RPC = 100k ohm as these values could

provide the best response of control action according to our specific design requirement. As these

values are been obtained, which it described the controller responsiveness to the error, finally makethe overshoot to minimum as well as the rise time. If we choose the proportional gain to be very

high, the system will be very unstable and as well as if we try to adjust the integral gain to be too

high it can cause the system to overshoot. Thus by balancing these two terms, and by choosing the

right value of RIC and RPC or KI and Kp are the only way that will ensure the optimal control can be

achieved and finally reached to system stability. Figure below shows the step response of the speedwith PI controller in the system.

Fig 6: Step Response with Proportional Integral Controller

UI = KI Verror.dt =RIC + 20 k

Verror.dt

KI =1

RIC + 20


13/13

5) In this lab, the motor is continuously unloaded. Based on the control scheme you designed

above, describe the circuit operation (including the motor) when a load is suddenly added.

If the load is suddenly added to the continuously unloaded motor we will suddenly increase thespeed reference from zero value to full value. The speed error would be 100%, so the output of

Vref from the speed amplifier immediately will be saturated to the maximum value as a reaction tocorrespond to the maximum (rated) value of the motor. Therefore the motor speed will be at the

rated value and it will accelerate at a full torque. The back emf and the motor speed rises at theconstant rate, the input voltage increases steadily so that the difference (V-E) is sufficient to drive

rated current (I) through the armature resistance and cause the motor to run at the rated speed.

6) What further improvements are there for the speed control?

As we discussed on further improvements for the speed control the digital implementation should have

been replaced the analogue circuitry in order to get the better performance and results. As many drives

use digital feedback as the digital controllers as it offers tremendously a freedom with an added

flexibility. For an example programmable ramp-down, ramp-up, minimum and maximum speed and etc.In addition they also offered an ease of interfacing while linking to other devices and host computers, to

the controllers and also self tuning. Other benefits represented by a digital speed feedback is that it offers

a user friendly diagnostic while providing to the local and remote user with data of historical on every

state of all drives key variables.

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ELEC 5565 - Electric Drives and Ctrl Lab

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