eet113-exp08-three-phase power v1.0 - pccspot.pcc.edu/~wlara/eet113/labs/eet113-exp08-three... ·...

EET-113 Experiment 8: Three-Phase Power

© 2014 Dan Kruger 1

Pre-Lab Read through this entire lab. Perform all of your calculations (calculated values) prior to making

the required circuit measurements. You may need to measure circuit component values to obtain

your calculated values. All calculations should be written on a separate piece of paper (or in

your lab notebook). They should be legible and written so that someone else can clearly

understand your thought process. This is to demonstrate your understanding of the material, as

well as aid in the troubleshooting process.

Introduction Single-Phase Power: A single voltage source powering a circuit is an example of a single phase power system. It

supplies the circuit with a single sinusoidal voltage waveform. Any current flowing out of its

positive terminal must be returned to its negative terminal. This means we have a single voltage

source and two conductors that are piping the source’s current--one to the load, and one back

from the load.

Remember Kirchhoff’s current law (KCL). Basically, it says that any current flowing into a node

must be flowing out of that node via some other path(s). What goes in must come out. This not

only applies to any circuit node, but to any closed region of a circuit. During this introduction,

think of our voltage source as the closed region to which KCL applies. If our voltage source is

sourcing a current, we must also have a route in which current returns.

In higher-current systems, such as power transmission lines, the size of the conductors (lines)

contributes directly to the cost of building and maintaining the lines.

With a single-phase system, you would need to build two lines and you would have all of your

current flowing through both wires. This means we are supplying our load(s) with only a single

voltage source and we have to buy and maintain two wires that can handle 100% of the current,

meaning they will have to be physically large.

Three-Phase Power: A three-phase power system is a circuit that has three AC voltage sources. Each of the sources

provides a voltage sinusoid that is exactly 120° (one third of a cycle), apart from the other two

sources.

S

V SI

SI Figure 1: Single-Phase System



With the three voltage sources out of phase from each other, current flowing out of any one

source must be returned via the current flowing through the other two sources.

Figure 2 depicts a three-wire three-phase voltage source. With three energized lines, a load can

be connected between any two of the three lines. This type of connection is called a line-to-line

connection.

Applying KCL to the entire three-phase source, we can see that 0

A B CI I I+ + = . This means

that all three energized lines are carrying current. There is no dedicated return path for the

current. At any given instant in time, each line is both sourcing its load, and acting as a return

path for the other lines. They all must work together because of KCL. In other words, we get

three voltage sources for the price of three lines, as opposed to one voltage source for the price of

two lines.

In the case of the four-wire system in Figure 3, there is a neutral line that can return current to the

power source. In this case, KCL says that

A B C NI I I I+ + = . When the system’s load is

nearly balanced, the neutral wire will only be required to carry little or no current, meaning it

may be allowed to be much smaller than the conductors for the lines A, B, and C.

The overall objective for an efficient power system is to deliver the power needed by the load as

cheaply and efficiently as possible. To deliver more power to a load at a given voltage, more

current must be carried by the transmission lines. If you wanted to double the amount of power

Figure 2: Three-Wire Three-Phase

System

A

V

A

I

B

V

BI

C

V

C

I

neutral

A

V

AI

BV

BI

CV

CI

neutral

N

I

Figure 3: Four-Wire Three-Phase

System



transmitted by a set of 3kV transmission lines, you would want to double the current carried by

the lines. So, simply double the thickness (more precisely, double the cross-sectional area) of

your transmission lines, and this will double the current capacity.

There is a problem with this reasoning. Due to the magnetic field caused by the current, a

conductor’s ability to efficiently carry current diminishes as you try to push more and more

current through it. There is something called the skin effect, where, for higher current situations,

current is actually pushed towards the outside of the conductor. This means that the center

portions of the wires are not being used, and that doubling the cross-sectional area of a wire does

not necessarily double its current capacity.

So, because of the skin effect, it is more efficient to split the current between 3 different

conductors, rather than have all the current go through two conductors.

Another advantage of three-phase power is that it works more naturally with the functioning of

both generators and motors. And for a balanced load condition, power delivered/consumed is

constant.

In summary, the advantages of a three-phase power system over a single-phase system are given

below.

• More efficient current flow, because more of each line is carrying current.

• In many cases, a neutral line can either be very small or not needed at all.

• It works more naturally with generators and motors.

• The flow of power is constant, as opposed to being sinusoidal.

Required Equipment: • Oscilloscope (Tektronix TDS 2002C)

• Function Generator

• Phase Tripler circuit board

• R1 = R2 = R3 = 1kΩ, R4 = 220Ω



Part A: Wye-Connected Load With and

Without a Neutral Connection 1) Setup your phase tripler per the instructions in Appendix A.

2) Set your function generator to generate a 20Vp-p 60Hz sinusoid with no DC offset, and

attach it to the phase tripler’s input. You will need to use a cable that has a BNC

connector on both ends.

3) Use your oscilloscope to verify that each of the phase tripler’s output voltages are 20Vp-p

and that they are all 120° apart. If you see any significant imperfections in voltage

amplitudes or phases, revisit step A1.

4) On your circuit board, assemble the wye-configured load depicted in Figure 4. For now,

we will be using a balanced load, so use a 1kΩ resistor for 1Z , 2Z , and 3Z . Connect

your three-phase voltage source to points “a”, “b”, and “c” as labeled in Figure 4.

Connect the neutral point “n” to your phase tripler’s ground terminal.

5) Use your DMM to measure the rms value of the AC current flowing between the load’s

neutral point “n” and the phase tripler’s ground. Your load should be close to a balanced

load (all of the load’s phase impedances are roughly equal) so you should measure very

little current.

6) Remove the neutral node’s connection to ground, and measure the rms voltage at point

“n”.

Figure 4

1Z

2Z

3Z

020p p

V∠ °

−

12020p p

V∠ °

−

12020 p pV∠− °

−



7) Redraw this circuit with the phase tripler represented as three different voltage sources.

Keep 2Z and 3Z the same as 1kΩ resistors, but change 1Z to a 220Ω resistor and make

sure the neutral node’s connection to ground has been removed. This means we will have

an unbalanced load and no neutral line.

8) Using nodal analysis and your circuit components’ nominal values, solve for the neutral

node’s voltage in the circuit you’ve drawn in part A7.

9) Build the circuit from part A7 (the unbalanced load with no neutral line) and measure the

voltage at point “n”. Use percent error to compare it to your result from part A8.

10) With no neutral line in place, does a drop in Z1’s resistance have an effect on the power

delivered to Z2 and Z3?

11) With a neutral line in place (as it was in part A4), what effect would a drop in Z1’s

resistance have on the power delivered to Z2 and Z3?



Part B: LTspice Simulations

This part will involve simulations only. “Measure” will be used to mean “obtain simulated a

value for”.

1) Simulate the wye-connected load from Part A in LTspice with for the balanced and

unbalanced loads used previously. For the situations without a neutral connection,

measure the load’s neutral point’s voltage (peak value). For the situations with a neutral

connection, figure out a way to measure the current (peak value) flowing between the

load’s neutral point and ground.

For the voltage sources, use a the same configuration used in Part A, except change the

voltage source magnitudes to 3kV. We will use this to simulate an electric generator.

• balanced load with neutral connection

• balanced load without neutral connection

• unbalanced load with neutral connection

• unbalanced load without neutral connection

2) From your results, under what condition(s) does the drop in Z1’s resistance effect the

power delivered to the other two phases?

3) Under what conditions is the neutral line actually used (when does it actually carry

current)?

4) Figure 5 is a simplified depiction of a power plant’s 3-phase generator delivering power

to a residential customer. Power is first generated, then its voltage is stepped up through

a transformer so it can be efficiently transmitted over high-voltage power lines. Nearby a

consumer’s home, the voltage is then stepped down, and a center-tapped secondary coil is

used to provide the home with a +120Vrms line, a -120Vrms line, and a neutral line.

Research how to simulate a transformer in LTspice. Research how to simulate a

transformer with a center tap in LTspice. Simulate one phase of the circuit in figure 5.

You will have to figure out your transformation ratios based on the voltages given. For

the residential load, connect a 50Ω resistor between the +120V and ground terminals, and

one between the -120V and ground terminals. Use a 65Ω resistor to represent the

power line losses.



Appendix A: Setting up the Phase Tripler

013.8

AV

kV∠ °

0230 kV∠ °

12013.8

BV

kV∠ °

120230 kV∠ °

12013.8

CV

kV∠− °

120230 kV∠− °

240V

120V

-120V

(Transformer and load not shown)

(Transformer and load not shown)

Figure 5: Simplified Power

Distribution



1) To power the phase tripler, attach your DC power supply to give 12V to the “+12V”

terminal and -12V to the “-12V” terminal. Connect the “GND” terminal to your

power supply’s “COM” terminal.

2) Use a cable with BNC connectors on both ends to attach your function generator to

the phase tripler’s “IN” terminal. Set your function generator to produce a 20Vp-p,

60Hz sinusoid with no DC offset.

3) Attach your circuit’s ground to the phase tripler’s remaining “GND” terminal.

4) Put the outputs “P000” and “P120” on CH1 and CH2 of your oscilloscope,

respectively. Fine tune your function generator amplitude so that voltage on the

output “P000” reads as close as possible to 20Vp-p.

5) Set your oscilloscopes Measure function to measure CH1’s peak-to-peak amplitude,

CH2’s peak-to-peak amplitude, and CH2’s phase (source 1) with respect to CH1’s

phase (source 2).

6) Adjust pot (potentiometer) “P1” so that P120’s phase is 120° ahead of P00.

7) Adjust pot “A1” so that P120’s amplitude is the same as P000’s amplitude.

8) Connect CH2 of the oscilloscope to P240.

9) Adjust pot (potentiometer) “P2” so that P240’s phase is 120° behind P00.

10) Adjust pot “A2” so that P240’s amplitude is the same as P000’s amplitude.

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