Activity4 Chan
Activity4 Chan
Activity4 Chan
Exercise 6
The Autotransformer
EXERCISE OBJECTIVE When you have completed this exercise, you will know what an autotransformer
is. You will be able to connect an autotransformer so that it operates as either a
step-up or a step-down unit. You will also be able to determine the primary and
secondary voltages and currents of an autotransformer. You will know how the
power rating of an autotransformer compares with the power rating of a similarly
sized conventional power transformer.
DISCUSSION OUTLINE The Discussion of this exercise covers the following points:
▪ Autotransformer operation
▪ Autotransformer circuit analysis
▪ Power rating of conventional transformers and autotransformers
Autotransformer Autotransformer
Load .
. Load
In step-down operation, on the other hand, the relationship between the voltage
ratio, the current ratio, and the turns ratio of an autotransformer is equal to:
𝐸𝑃𝑟𝑃𝑃. 𝐼𝑆𝑒𝑐. 𝑁𝐴−𝐶 𝑁𝑃𝑟𝑃𝑃.
= = =
𝐸𝑆𝑒𝑐. 𝐼𝑃𝑟𝑃𝑃. 𝑁𝐵−𝐶 𝑁𝑆𝑒𝑐.
These relationships are true only when voltages 𝐸𝐵−𝐶 and 𝐸𝐴−𝐶 of the
autotransformer are in phase (which is usually the case in any autotransformer),
and thus, add together. As for conventional power transformers, the relationship
between the current ratio and the turns ratio of an autotransformer is the reciprocal
(inverse) of the relationship between the voltage ratio and the turns ratio of the
autotransformer. Furthermore, the apparent power at the primary winding of an
autotransformer is virtually equal to the apparent power at the secondary winding
of the autotransformer, as for any conventional power transformer.
The rules that apply to the operation of conventional power transformers also
apply to autotransformers. This means that:
1. The voltage ratio (i.e., the ratio of the primary voltage to the secondary
voltage) of an autotransformer is equal to its turns ratio 𝑁𝑃𝑟𝑃𝑃. ⁄𝑁𝑆𝑒𝑐. .
Autotransformer
𝐸𝑆 .
100 V
𝐸𝑃𝑟𝑃𝑃. 𝑁𝑃𝑟𝑃𝑃.
=
𝐸𝑆𝑒𝑐. 𝑁𝑆𝑒𝑐.
𝐸𝑆𝑒𝑐. 300 V
𝐼 = = = 1.00 A
𝑆𝑒𝑐. 𝑅𝐿𝑜𝑎𝑑 300 Ω
As the calculations show, the current (i.e., the primary current 𝐼𝑃𝑟𝑃𝑃. ) flowing in the
autotransformer winding tap is equal to the sum (𝐼𝑆𝑒𝑐. + 𝐼𝐶𝑜𝑚𝑚𝑜𝑛) of the currents
flowing in the two sections of the autotransformer winding. Consequently, the value
of the current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 flowing in the turns of the winding that is common to the
primary and secondary is equal to the difference (𝐼𝑃𝑟𝑃𝑃. − 𝐼𝑆𝑒𝑐. ) between the
primary and secondary currents. This must be taken into account when selecting
the size of the wire used for this section of the autotransformer winding. The
lower the turns ratio, the greater the difference between the primary and secondary
current values, and thus, the higher the value of current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛.
Autotransformer
.
.
𝐸𝑆 2000 turns
.
400 V 500 turns
𝑅𝐿𝑜𝑎𝑑
300 Ω
𝐸𝑃𝑟𝑃𝑃. 𝑁𝑃𝑟𝑃𝑃.
=
𝐸𝑆𝑒𝑐. 𝑁𝑆𝑒𝑐.
𝐸𝑆𝑒𝑐. 100 V
𝐼 = = = 4.00 A
𝑆𝑒𝑐. 𝑅𝐿𝑜𝑎𝑑 25 Ω
. 100 V 𝐸𝑆𝑒𝑐.
1A 1A
Figure 38. Conventional power transformer with a power rating of 100 VA.
Autotransformer
100 V
1A
100 V
1A
Autotransformer
100 V
1A
𝐼𝑆𝑒𝑐. (max. = 2 A)
100 V
1A
Figure 39. Step-up and step-down autotransformers built using the conventional power
transformer in Figure 38.
. 200 V 𝐸𝑆𝑒𝑐.
6A 3A
Autotransformer
200 V
3A
100 V
𝐸 . = 100 V 6A
Figure 40. Autotransformer with a power rating of 900 VA built using a conventional power
transformer with a power rating of 600 VA.
PROCEDURE
High voltages are present in this laboratory exercise. Do not make or modify any
banana jack connections with the power on unless otherwise specified.
4. Connect the Power Input of the Data Acquisition and Control Interface to the
Power Output of the 24 V AC Power Supply module. Turn the 24 V AC
Power Supply module on.
6. Connect the USB port of the Data Acquisition and Control Interface to a
USB port of the host computer.
In the LVDAC-EMS Start-Up window, make sure that the Data Acquisition and
Control Interface and the Four-Quadrant Dynamometer/Power Supply are
detected. Make sure that the Computer-Based Instrumentation function for the
Data Acquisition and Control Interface is available. Select the network voltage
and frequency that correspond to the voltage and frequency of your local ac
power network, then click the OK button to close the LVDAC-EMS Start-Up
window.
8. Connect the equipment as shown in Figure 41. In this circuit, the two
120 V - 1 A windings are connected in series-aiding configuration to formthe
single winding of the autotransformer.
Transformer
120 V
1A
E1
120 V
1A
E2 𝑅𝐿𝑜𝑎𝑑
9. Make the necessary switch settings on the Resistive Load so that the
resistance of the resistive load is infinite.
10. Determine the turns ratio of the autotransformer just set up, as well as the
voltage at the secondary winding when a voltage of 100 V is applied to the
primary.
Turns ratio = 2
❑ Yes ❑ No
− Do not start the AC Power Source yet. This will be done in the next
section of this exercise.
12. In LVDAC-EMS, open the Metering window. Make the required settings to
measure the rms (ac) values of the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃. and
current 𝐼𝑃𝑟𝑃𝑃. (inputs E1 and I1, respectively), secondary (load) voltage 𝐸𝑆𝑒𝑐.
and current 𝐼𝑆𝑒𝑐. (inputs E2 and I2, respectively), andcurrent
𝐼𝐶𝑜𝑚𝑚𝑜𝑛 (input I3) flowing in the lower portion of the autotransformer winding.
In this section, you will start the ac power source. You will set the resistance of the
resistive load to 57 Ω. You will measure the autotransformer voltages and currents,
and analyze the results. You will determine the autotransformer maximum
secondary (load) voltage and current, as well as its power rating.
14. On the Resistive Load, decrease the resistance to 57 Ω. In the Four- Quadrant
Dynamometer/Power Supply window, readjust the Voltage (V at no load)
parameter so that the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃. indicated in the
Metering window is as close as possible to 100 V.
15. In the Metering window, measure the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃.
and current 𝐼𝑃𝑟𝑃𝑃. , secondary (load) voltage 𝐸𝑆𝑒𝑐. and current 𝐼𝑆𝑒𝑐. , and
current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 flowing in the lower portion of the autotransformer winding.
Record the values below.
16. Is the autotransformer secondary (load) voltage 𝐸𝑆𝑒𝑐. measured in the previous
step approximately equal to the voltage calculated in step 10?
❑ Yes ❑ No
Is the secondary (load) current 𝐼𝑆𝑒𝑐. measured in the previous step virtually
equal to the sum of the currents (𝐼𝑃𝑟𝑃𝑃. and 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 ) flowing in the two portions
of the autotransformer winding?
❑ Yes ❑ No
Apparently, the KCL direction is different from the expected formula
for Step-Down which is IPri = ISec + Icommon.
❑ Yes ❑ No
Transformer
120 V
1A
E2 𝑅𝐿𝑜𝑎𝑑 1
120 V
1A
E1
20. Make the necessary switch settings on the Resistive Load so that the
resistance of the resistive load is infinite.
21. Determine the turns ratio of the autotransformer just set up, as well as the
voltage at the secondary winding when a voltage of 50 V is applied to the
primary.
❑ Yes ❑ No
23. On the Resistive Load, decrease the resistance to 200 Ω. In the Four-Quadrant
Dynamometer/Power Supply window, adjust the Voltage (V at no load)
parameter so that the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃. indicated in the
Metering window is as close as possible to 50 V.
24. In the Metering window, measure the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃.
and current 𝐼𝑃𝑟𝑃𝑃. , secondary (load) voltage 𝐸𝑆𝑒𝑐. and current 𝐼𝑆𝑒𝑐. , and
current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 flowing in the lower portion of the autotransformer winding.
Record the values below.
25. Is the autotransformer secondary (load) voltage 𝐸𝑆𝑒𝑐. measured in the previous
step approximately equal to the voltage calculated in step 21?
❑ Yes ❑ No
*There is only a small difference with the value of ESec. between steps
21 and 24 by 4.21.
Is the primary current 𝐼𝑃𝑟𝑃𝑃. measured in the previous step virtually equal to the
sum of the currents (𝐼𝑆𝑒𝑐. and 𝐼𝐶𝑜𝑚𝑚𝑜𝑛) flowing in the two portions of the
autotransformer winding?
❑ Yes ❑ No
*Although the result of IPri. may not be exact, it is still considered valid
with a small difference of 0.004.
❑ Yes ❑ No
© Festo Didactic 579437 101
Exercise 6 – The Autotransformer ⬥ Procedure
In this section, you will calculate the power rating of a conventional transformer
built using the same windings of the transformer module used to implement the
step-down and step-up autotransformers studied in the previous sections. You
will compare the power rating of these autotransformers with the power rating of
the conventional transformer built using the same windings.
28. Determine what would be the power rating of a conventional 120 V:120 V
power transformer implemented with the same two 120 V - 1 A windings used
to build the step-down and step-up autotransformers in Figure 41 and Figure
42, respectively.
29. Compare the power ratings of the step-down and step-up autotransformers
calculated in steps 17 and 26, respectively, to the power rating of the
conventional power transformer calculated in the previous step. What
conclusions can you draw?
Comparing the power ratings from steps 17 and 26, if the voltage and
current in the secondary side decrease (step-down), the power rating will
decrease. However, is the voltage and the current increases (step-up), the
power rating will increase. We can say from here that the power rating can
be adjustable depending on the voltage and current delivered to the load
(ESec. & ISec.)
What size would the 120 V:120 V conventional transformer need to be to have
a power rating similar to the power rating of the autotransformers used in the
previous sections of this exercise? Briefly explain why.
Usually, it would just need two the same winding to build a 120V:120V.
Despite that, it is neither a step-down nor step-up transformer, the
secondary voltage and current would just be the same as the primary side.
30. Connect the equipment as shown in Figure 43. Notice that the resistive load
consists of two resistors (𝑅𝐿𝑜𝑎𝑑 1 and 𝑅𝐿𝑜𝑎𝑑 2) connected in series. Connect
two resistor banks of the Resistive Load in parallel to implement resistor
𝑅𝐿𝑜𝑎𝑑 2.
Transformer
120 V
1A
𝑅𝐿𝑜𝑎𝑑 1
40 A input
E2
24 V
5A
E1
𝑅𝐿𝑜𝑎𝑑 2
40 A input
31. Make the necessary switch settings on the Resistive Load so that the
resistance of the resistive load is infinite.
32. Determine the turns ratio of the autotransformer just set up, as well as the
voltage at the secondary winding when the nominal voltage of 24 V is applied
to the primary.
34. On the Resistive Load, set the resistance values of 𝑅𝐿𝑜𝑎𝑑 1 and 𝑅𝐿𝑜𝑎𝑑 2to
171 Ω and 86 Ω, respectively. In the Four-Quadrant Dynamometer/Power
Supply window, readjust the (V at no load) parameter so that the
autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃. indicated in the Metering window is as
close as possible to 24 V.
35. In the Metering window, measure the autotransformer primary voltage 𝐸𝑃𝑟𝑃𝑃.
and current 𝐼𝑃𝑟𝑃𝑃. , secondary (load) voltage 𝐸𝑆𝑒𝑐. and current 𝐼𝑆𝑒𝑐. , and
current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 flowing in the lower portion of the autotransformer winding.
Current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛 = A
36. Is the autotransformer secondary (load) voltage 𝐸𝑆𝑒𝑐. measured in the previous
step approximately equal to the voltage calculated in step 32?
❑ Yes ❑ No
Is the primary current 𝐼𝑃𝑟𝑃𝑃. measured in the previous step virtually equal to the
sum of the currents (𝐼𝑆𝑒𝑐. and 𝐼𝐶𝑜𝑚𝑚𝑜𝑛) flowing in the two portions of the
autotransformer winding?
❑ Yes ❑ No
37. Determine the maximum secondary (load) current 𝐼𝑆𝑒𝑐. which the
autotransformer can provide, as well as the resulting power rating of the
autotransformer.
a Use the secondary voltage 𝐸𝑆𝑒𝑐. determined in step 32 to determine the power
rating of the autotransformer.
Power rating = VA
Power rating = VA
40. Compare the power rating of the step-up autotransformer calculated in step
37 to the power rating of the conventional transformer calculated in the
previous step.
Is the power rating of the autotransformer still higher than the power rating of
the conventional transformer built with the same windings?
❑ Yes ❑ No
41. Compare the increase in power rating obtained using the autotransformer with
a 1:6 turns ratio to that obtained using the autotransformers with 2:1 and 1:2
turns ratios. What conclusions can you draw?
42. Close LVDAC-EMS, then turn off all the equipment. Disconnect all leads and
return them to their storage location.
Notice: "Step 33 - 42 was not conducted and intentionally left blank due to the insufficient amount
of time."
CONCLUSION In this exercise, you learned what an autotransformer is, and how to connect an
autotransformer so that it operates as either a step-up or a step-down unit. You
saw how to determine the primary and secondary voltages and currents of an
autotransformer. You also saw how the power rating of an autotransformer
compares with the power rating of a similarly sized conventional power
transformer.
REVIEW QUESTIONS 1. What is the main difference between conventional power transformers and
autotransformers?
4. Consider the step-up autotransformer shown in Figure 44. Assuming that this
autotransformer has 300 turns of wire in the primary winding and 900 turns of
wire in the secondary winding, calculate the value of the current 𝐼𝐶𝑜𝑚𝑚𝑜𝑛flowing
in the lower part of the autotransformer winding.
The value of the current Icommon is 4 A.
Transformer
𝐸𝑆
.
80 V
5. What are the two turns ratios which maximize the power rating increase of an
autotransformer in comparison to a conventional power transformer built using
the same windings? Explain briefly.
The two turns ratios are 3:1 (N1:N2) and k=33.33%. This shows that
the autotransformer is operational to supply a load of 120 Ω.
Notice: Proof of Connection is unavailable, but the experiment was conducted on the 16 th of
November 2022.
Data Gathered
For Step-Down:
For Step-Up:
Conclusions
Autotransformers are transformers that rely on only one winding that both the primary and the
secondary sides share. It also works as either a step-down or step-up like a conventional power
transformer that uses two windings. Since the point of this experiment is to analyze the functions of an
autotransformer, the measured output may not be as exact as expected but its values were able to prove
its operation throughout the simulation despite the discrepancies. We should consider that we are running
a practical autotransformer.
Considering that the one winding in an autotransformer is two of the same size windings connected
in series, the primary and the secondary sides are connected where one side is connected to one winding
and the other connected to two of the windings connected in series. In a step-down, the primary side is
connected to both windings, in series, while the secondary side is connected to one winding, in which the
connection takes place where one end is connected in between the two windings. In a step-up, the
primary side is connected to one of the two windings, where one end of the connection takes place in
between the two windings, while the secondary side is connected to the two windings, in series.
The primary and secondary currents can be determined through a KCL equation. Although the
windings may serve the number of turns in an autotransformer, they can be treated temporarily as a short
circuit. The KCL equation is depicted differently between the step-up and step-down. During the
simulation, the measured values of the step-down autotransformer did not associate with the standard
KCL formula for step-down, ISec = IPri + Icommon. However, it follows the step-up KCL formula, IPri = ISec +
Icommon. This would probably mean that the polarity voltage supply (+) affects the direction of the primary
current IPri since it moves toward the node. What also matters is the direction of the common current
Icommon. In the step-down KCL formula, the common current is depicted that it moves toward the node. In
the simulation, the measured value indicates that the common current moves away from the node. The
LabVolt may not have follow the KCL formula specifically for step-down but rather that it follows one KCL
formula for both step-down and step-up, which is IPri = ISec + Icommon.
Usually, the power on both sides in autotransformers is equal. However, in this simulation, we can
determine whether it is step-down or step-up by simply multiplying the voltage and current delivered to
the load on the secondary side. Based on computations of the measured values, if the delivered power
computed (secondary) is greater than 1 VA, then it is a step-up autotransformer. But, if the delivered
power (secondary) is less than 1 VA, then it is a step-down autotransformer. We could concur that the
output can be variable depending on input voltage and the connection of the series winding in an
autotransformer.
2.
3.
4.
5.