Control Lab Report Experiment No. 02

8/13/2019 Control Lab Report Experiment No. 02

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Prepared by Rinku Basak

Updated by Shahriyar Masud Rizvi

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+ 15 V

- 15 V

American International University Bangladesh (AIUB)

EEE 4101: Control Systems Laboratory

Experiment # 2:Open-Loop and Closed-Loop Speed Control of a DC Motor.

Objectives:

1. To control the speed of a dc motor at no-load.2. To measure the steady-state open-loop gain of a control system.3. To observe the dead-band of a control system.4. To discover the amplifier saturation effects on system gain.5. To understand the principle of an error detector.6. To compare the behavior of a dc motor using closed-loop versus open-loop speed control.Open Loop Control of a DC Motor:

Consider the simple control system shown in Fig. 1 to vary the speed of a dc motor. It

consists of a potentiometer P, a power amplifier A, a dc motor M and a tachometer G. The

motor is geared down to drive an output shaft at low speed. The potentiometer connected to the

REFERENCE supply enables us to vary voltage V1. This is the input signal to power amplifier

A. A change in V1will produce a change in the armature voltage Va. Our knowledge of dc

motors tells us that the speed increases with an increase in armature voltage. Furthermore, the

dc tachometer acts as a transducer because it gives a voltage V2that is directly proportional to

speed. Consequently, an increase in V1produces an increase in V2. The question we now ask is

what relationship exists in the ratio V2/V1as V1 increases from zero to +10V? Does this ratio

(called system gain) depend on the value of V1, or is it constant?

Vc


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Let us first set the gain of amplifier A so that it is equal to 1. Consequently, Va will be

exactly equal to V1. On the other hand, if the gain of amplifier A is set at 5, then Va will

normally be 5 times as large as V1. However, the motor speed, and hence V2, depends on Va. As

a result, the ratio V2/V1(the system gain) will depend upon the gain (1 or 5) of the amplifier. In

the system shown in Fig. 1, V1is the control signal and also called input signal. Voltage V2 is

the output signal, which is direct measure of motor speed. It is also a direct measure of the

geared-down shaft speed. The ratio between the control signal and the output signal is called

the gain of the system. More precisely, it is called the open-loop gain G of the system. It is

called open-loop because the output signal V2 does not in any way affect the input signal V1. In

the following exercise, we shall determine the system gain when the amplifier gain equals 1.

We then determine the system gain when the amplifier gain is 5. In this latter case, the amplifier

begins to saturate for rather small values of V1. When this happens, voltage Va reaches a

maximum of about 15V and remains there, even for higher values of V 1. Because Vareaches

the saturation level of 15V, it follows that the motor speed remains the same. Consequently, V2

also remains constant. With an amplifier gain of 5, the expected relationship between V1and

V2is shown in Fig. 2.

Equipment Required:

1. Enclosure/Power Supply 88462. Connections Leads and Accessories 89443. Potentiometer 90364. Power Amplifier/Phase Shifter 90395. DC Motor/Generator 93186. Multimeter


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Procedure:

A.

1. Connect the dc motor, power amplifier and potentiometer as shown in the schematiccircuit diagram of Fig. 1.

2. Select the dc amplifier made and set the amplifier gain equal to 1.B.

When the control signal V1is very small, the resulting voltage Vamay not be enough to

cause the motor to turn, due to the friction of the bearings. In this experiment, we shall

determine the range of voltage V1over which the motor does not turn.

a) Set V1=0 so that the motor does not turn. Then gradually increase V1(positively) andrecord the critical positive voltage when motor starts turning. The approximate critical

positive voltage is V1(crit +)= ______________ V

b) Reset V1=0 and gradually increase V1 (negatively) and record the critical negativevoltage when the motor starts turning. The approximate critical negative voltage is

V1(crit -)= ______________ V

The so called dead-band of the control system ranges from the critical negative

voltage to the critical positive voltage V. This dead-band is a highly non-linear portion

of the control system.

If time permits, do parts C and D.

C.

a) Adjust the amplifier gain to 5. This will increase the overall system gain by a factor of 5.Observe what effect this has on the input-output (V2 V1) characteristics, and upon the

width of the dead-band.

D.

a) Repeat procedure 2 to measure the dead-band, but with the amplifier gain equal to 5.Determine the critical positive and negative voltages that cause the motor to start

running.

b) Critical positive voltage is V1(crit +) = ______________ Vc) Critical negative voltage is V1(crit -) = ______________ Vd) The dead-band range is between V1= ______________V and V2= ____________ VNote that the dead-band is much smaller than with a gain of 1.


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- 15 V

Error Detector:

A device that is able to add and subtract signals and give a resulting output signal is called

an error detector. The error detector in your servomechanism trainer has four terminals and one

(unseen) ground terminal. The ground is common to all four terminals. Consequently, all

signals are measured between the respective terminals and ground.

The schematic diagram of the detector is shown in Fig. 3. The input signals are e1, e2, e3and

the resulting output voltage is V1. The positive and negative symbols on the error detector tell

us whether the signals are added or subtracted by the detector. Thus, to Fig. 3, the output is

given by V1= +e1 e2 e3. We can get a better idea of how signals add and subtract by referring

to Fig. 4. The signals produced from three DC sources. According to their polarities, you can

see that e1=+10V, e2=+3V and e3=6V.

Therefore, the output voltage is

V1= +e1 e2 e3= +(+10V) (+3V) (3V) = +13V

Vc

V1

V2


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5

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- 15 V

The (+) sign in +13V means that terminal V1 is positive with respect to ground. The error

detector can simultaneously accept both ac and dc input signals. This will be discussed in AC

Servo. Fig. 5 shows the open-loop circuit to control a dc motor. The error detector uses only

one input signal VC, therefore VC = V1. The open-loop gain G=V2/V1 was previously

determined for various conditions of no-load, full-load and other disturbances. You will recall

that the gain varied from about 2.5 to 1.5 depending upon the nature of the disturbance. The

speed changed drastically even when the input control voltage V1(=VC) was held constant.

Fig. 6 is identical to Fig. 5 except that the output signal V2is fed back to the error detector.

The connections are made so that V1= V1 V2. This produces a negative feedback. In order to

obtain the same speed in Fig. 6 as in Fig. 5, we must increase the value of V1. For example, in

Fig. 5, V1= 2V and V2= 6V. Obviously, VCis equal to V1and so VC= 2V.

Because we want the motor to turn at the same speed in Fig. 6 as in Fig. 5, V 2is still 6V.

Since the amplifier gain has not been changed, V1= 2V. Therefore, VC= V1+ V2= 2 + 6 = 8V.

This means that the control signal VChas to be raised to 8V in Fig. 6, as compared to 2V in Fig.

5. In the following experiment you will discover that the closedloop gain V2/V1is much more

constant than the open-loop disturbance is introduced. The closed-loop gain is designated by

the symbol H and H = V2/VC. The closed-loop gain H tends to become even more constant (in

the face of a disturbance) if we increase the open-loop gain G. In the following experiment you

will note the effect of increased open-loop gain on closed-loop performance. A higher

open-loop gain will keep the motor speed much more constant (in closed-loop) for a given

disturbance.

Vc

V1

V2


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Equipment Required:

1. Enclosure/Power Supply 88462. Connections Leads and Accessories 89443. Potentiometer 90364. Power Amplifier/Phase Shifter 90395. DC Motor/Generator 93186. Error Detector 9037

Procedure:

A.

a) Place the friction lever in the zero position. Make sure the inertia wheel is not mountedon the shaft. Adjust the amplifier gain to 1.

b) Adjust the potentiometer so that VC= 0 V initially and then Vc increments by 1 V.Measure the value of V1. V1= _________ V. Is V1= VC?

c) Measure the value of V2. V2= __________V. Record values of Vc V1, upper and lowervalues of V2 (since V2 fluctuates), gain and a percent measure of V2 fluctuation. Take

values for both clockwise and counter-clockwise rotation of the motor. Note that, V2

fluctuation = 100* [V2 (upper) V2 (lower)] / V2 (lower). Gain = V2 (average)/ Vc.

Also, measure rpm and regulation for one set of data.

Note: If V2fluctuates, indicate its upper and lower values. The speed may vary

up and down, because the friction is not constant from one place to another

during a complete revolution of the belt.


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Complete the following table.

TABLE 1

Vc V1 V2

(upper)

V2

(lower)

V2

(average)

Gain V2

fluctuation

No-Load

Speed*

Full-Load

Speed*

Regulation*

0 0

0.5 0.5

1 1

1.5 1.5

2 2

2.5 2.5

-0.5 -0.5

-1 -1-1.5 -1.5

-2 -2

-2.5 -2.5

* Measure rpm and regulation for only one set of data.

If time is constrained, take data for only the positive values of Vc and V1 (for Table

1).

d) Measure the no-load speed of the output shaft for V1 = +2 V by counting the number ofturns in one minute. No-load speed = ____________ r/min.

e) Without changing the setting of VC, move the friction lever down to the 10 position.Record the new value of V2. If V2fluctuates, indicate the upper and lower values. V2=

__________ V.

f) Measure the full-load speed by counting turns (Friction lever in position 10). Full-loadspeed = _____________ r/min.

g) Calculate the speed regulation in percent using the following equation:

Speed Regulation = (no load speed full load speed) 100% / full load speed.Speed regulation = ___________%.

h) Plot V2 (average) versus Vc (or V1).


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B. (I)

a) Connect a wire (feedback loop) from the generator output V2to the error detectoras shown in Fig. 6. This provides the required negative feedback. Place the friction

lever in the zero position. Make sure the inertia wheel is not mounted on the shaft.

Adjust the amplifier gain to 1.

b) Adjust VCso that the speed is the same value as in step A(c). You can easily do this bymaking V1 or V2the same as in step A (c).

c) Measure VC, V1and V2.VC= _________ V, V1= ___________ V, V2= ___________ V.

Complete the following table.

TABLE 2

Vc V1 V2

(upper)

V2

(lower)

V2

(average)

Gain V2

fluctuation

No-Load

Speed*

Full-Load

Speed*

Regulation*

0 0

0.5

1

1.5

2

2.5

-0.5-1

-1.5

-2

-2.5

* Measure rpm and regulation for only one set of data.


1).

d) Measure the no-load speed of the output shaft for V1 = +2 V by counting turns.Full-load speed = __________ r/min.

e) Without making any other change, bring the friction lever down to the 10 mark. Thenmeasure VC, V1and V2. VC= _________ V, V1= _________ V, V2= __________ V.

f) Measure the speed by counting turns. Full-load speed = __________ r/min.


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g) Calculate the speed regulation in percent using the following equation:Speed Regulation = (no load speed full load speed) 100% / full load speed.

Speed regulation = ___________%.


Is speed regulation better in part B (I) as compared to part A? Also compare V2 fluctuations

and V2 (avg) vs. V1 plots for part A and part B (I). Does the transition from open-loop to

closed-loop improve these performance parameters?

B. (II)

a) Place the friction lever in the zero position. Make sure the inertia wheel is not mountedon the shaft. Keeping the feedback connection, adjust the amplifier gain to 5.

b) Complete the steps (b) through (h) just like part B (I).TABLE 3

Vc V1 V2

(upper)

V2

(lower)

V2

(average)

Gain V2

fluctuation

No-Load

Speed*

Full-Load

Speed*

Regulation*

0 0

0.5

1

1.5

2

2.5

-0.5

-1

-1.5

-2

-2.5* Measure rpm and regulation for only one set of data.


1).


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c) Measure the no-load speed of the output shaft for V1 = +2 V by counting turns.Full-load speed = __________ r/min.

d) Without making any other change, bring the friction lever down to the 10 mark. Thenmeasure VC, V1and V2. VC= _________ V, V1= _________ V, V2= __________ V.

e) Measure the speed by counting turns. Full-load speed = __________ r/min.

f) Calculate the speed regulation in percent using the following equation:Speed Regulation = (no load speed full load speed) 100% / full load speed.

Speed regulation = ___________%.


Is speed regulation better in part B (II) as compared to part B (I)? Also compare V2

fluctuations and V2 (avg) vs. V1 plots for part B (I) and part B (II). Does increasing open

loop gain improve these performance parameters?

C.

Note that the speed regulation is now much better than in the open-loop condition.

You may also have noticed that V2does not fluctuate as much as it did before. The speed is

therefore more uniform in closed-loop operation.

Report:

1. Submit the report with all graphs and questions asked during laboratoryexperiments.

2. Compare open-loop and closed-loop control systems with respect to yourexperimental data.

3. Compare closed-loop control systems with Gain = 1 and the same system withGain = 5 with respect to your experimental data.

Control Lab Report Experiment No. 02

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