# Laboratory manual for Electronics: Transformers and Motors: Transformer Basics

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Objectives:

• Discuss the construction of an isolation transformer.

• Determine the winding configuration with an ohmmeter.

• Connect a transformer and make voltage measurements.

• Compute the turns-ratio of the windings.

A transformer is a magnetically operated machine that can change values of voltage, cur rent, and impedance without a change of frequency. Transformers are the most efficient machines known. Their efficiencies commonly range from 90% to 99% at full load.

Transformers can be divided into several classifications such as:

Isolation Auto Current

A basic law concerning transformers is that all values of a transformer are proportional to its turns-ratio. This does not mean that the exact number of turns of wire on each winding must be known to determine different values of voltage and current for a transformer.

What must be known is the ratio of turns. E.g., assume a transformer has two windings. One winding, the primary, has 1,000 turns of wire and the other, the secondary, has 250 turns of wire, as shown in Ill. I1. The turns-ratio of this transformer is 4 to 1 or 4:1 (1,000/250 = 4). This indicates there are four turns of wire on the primary for every one turn of wire on the secondary.

Helpful Hint: A basic law concerning transformers is that all values of a transformer are proportional to its turns-ratio.

Ill. I1 All values of a transformer are proportional to its turns-ratio.

Transformer Formulas There are different formulas that can be used to find the values of voltage and current for a transformer. The following is a list of standard formulas: where: NP = Number of turns in the primary NS = Number of turns in the secondary EP = Voltage of the primary ES = Voltage of the secondary IP = Current in the primary IS = Current in the secondary

The primary winding of a transformer is the power input winding. It is the winding that's connected to the incoming power supply. The secondary winding is the load winding or output winding. It is the side of the transformer that's connected to the driven load, seen in Ill. I2. Any winding of a transformer can be used as a primary or secondary winding provided its voltage or current rating isn't exceeded. Transformers can also be operated at a lower voltage than their rating indicates, but they cannot be connected to a higher volt age. Assume the transformer shown in Ill. I2, for example, has a primary voltage rating of 480 volts and the secondary has a voltage rating of 240 volts. Now assume that the primary winding is connected to a 120 volt source. No damage would occur to the transformer, but the secondary winding would produce only 60 volts.

Isolation Transformers

Ill. I2 Isolation transformer.

Ill. I3 The current through an inductor rises at an exponential rate.

Ill. I4 Voltage spikes are generally of very short duration.

Ill. I5 The isolation transformer greatly reduces the voltage spike.

The transformers shown in Ill.s I1 and I2 are isolation transformers. This means that the secondary winding is physically and electrically isolated from the primary winding. There is no electrical connection between the primary and secondary winding. This transformer is magnetically coupled, not electrically coupled. This "line isolation" is often a very desirable characteristic. Since there is no electrical connection between the load and power supply, the transformer becomes a filter between the two. The isolation transformer will attenuate any voltage spikes that originate on the supply side before they are transferred to the load side. Some isolation transformers are built with a turns-ratio of 1:1. A transformer of this type will have the same input and output voltage and is used for the purpose of isolation only.

The reason that the transformer can greatly reduce any voltage spikes before they reach the secondary is because of the rise time of current through an inductor. The current in an inductor rises at an exponential rate, as shown in Ill. I3. As the current increases in value, the expanding magnetic field cuts through the conductors of the coil and induces a voltage that's opposed to the applied voltage. The amount of induced voltage is proportional to the rate of change of current. This simply means that the faster current attempts to increase, the greater the opposition to that increase will be. Spike voltages and currents are generally of very short duration, which means that they increase in value very rapidly ( Ill. I4). This rapid change of value causes the opposition to the change to increase just as rapidly. By the time the spike has been transferred to the secondary winding of the transformer, it has been eliminated or greatly reduced ( Ill. I5). Another purpose of isolation transformers is to remove or isolate some piece of electrical equipment from circuit ground. It is sometimes desirable that a piece of electrical equipment not be connected directly to circuit ground. This is often done as a safety precaution to eliminate the hazard of an accidental contact between a person at ground potential and the ungrounded conductor. If the case of the equipment should come in contact with the ungrounded conductor, the isolation transformer would prevent a circuit being completed to ground through a person touching the case of the equipment. Many alternating current circuits have one side connected to ground. A familiar example of this is the common 120 volt circuit with a grounded neutral conductor, as seen in Ill. I6. An isolation transformer can be used to remove or isolate a piece of equipment from circuit ground.

Ill. I6 Isolation transformer used to remove a piece of electrical equipment from ground. Equipment has no connection to ground.

Excitation Current

There will always be some amount of current flow in the primary of a transformer even if there is no load connected to the secondary. This is called the excitation current of the transformer. The excitation current is the amount of current required to magnetize the core of the transformer. The excitation current remains constant from no load to full load. As a general rule, the excitation current is such a small part of the full load current it's often omit ted when making calculations.

(cont...)

Electric Motors + Motor-Control Systems
(Nested-content Series)

## Motor Control Electronics

Electronic systems and controls have gained wide acceptance in the motor control industry; consequently it has become essential to be familiar with power electronic devices. This section presents a broad overview of diodes, transistors, thyristors, and integrated circuits (ICs) along with their application in motor control circuits.

GOALS:

• 1. Understand the operation and applications of different types of diodes.
• 2. Understand the operation and applications of different types of transistors.
• 3. Understand the operation and applications of different types of thyristors.
• 4. Understand the operation and circuit function of different types of integrated circuits.

### Semiconductor Diodes

Diode Operation

The PN-junction diode, shown in Ill. 1, is the most basic of semiconductor devices.

This diode is formed by a doping process, which creates P-type and N-type semiconductor materials on the same component. An N-type semiconductor material has electrons (represented as negative charges) as the current carriers while the P-type has holes (represented as positive charges) as the current carriers. N-type and P-type materials exchange charges at the junction of the two materials, creating a thin depletion region that acts as an insulator. Diode leads are identified as the anode lead (connected to the P-type material) and the cathode lead (connected to the N-type material).

The most important operating characteristic of a diode is that it allows current in one direction and blocks current in the opposite direction. When placed in a DC circuit, the diode will either allow or prevent current flow, depending on the polarity of the applied voltage. Ill. 2 illustrates two basic operating modes of a diode: forward bias and reverse bias. A forward-bias voltage forces the positive and negative current carriers to the junction and collapses the depletion region to allow current flow. A reverse-bias voltage widens the depletion region so the diode does not conduct. In other words, the diode conducts current when the anode is positive with respect to the cathode (a state called forward-biased) and blocks current when the anode is negative with respect to the cathode (a state called reverse-biased).

Ill. 1 PN-junction diode. Depletion region; Cathode; Junction N-type material P-type material; Anode PN-junction diode Cathode Typical diode Anode Cathode Diode symbol Anode PN

Ill. 2 Diode forward and reverse biasing. Depletion decreases Forward-bias voltage (Current flows) ; Depletion increases Reverse-bias voltage (Current flow blocked)

Ill. 3 Single-phase, half-wave rectifier circuit. Transformer

Rectifier Diode

Rectification is the process of changing AC to DC. Because diodes allow current to flow in only one direction, they are used as rectifiers. There are several ways of connecting diodes to make a rectifier to convert AC to DC. Ill. 3 shows the schematic for a single-phase, half-wave rectifier circuit. The operation of the circuit can be summarized as follows:

• The AC input is applied to the primary of the transformer; the secondary voltage supplies the rectifier and load resistor.

• During the positive half-cycle of the AC input wave, the anode side of the diode is positive.

• The diode is then forward-biased, allowing it to conduct a current to the load. Because the diode acts as a closed switch during this time, the positive half-cycle is developed across the load.

• During the negative half-cycle of the AC input wave, the anode side of the diode is negative.

• The diode is now reverse-biased; as a result, no cur rent can flow through it. The diode acts as an open switch during this time, so no voltage is produced across the load.

• Thus, applying an AC voltage to the circuit produces a pulsating DC voltage across the load.

Diodes can be tested for short-circuit or open-circuit faults with an ohmmeter. It should show continuity when the ohmmeter leads are connected to the diode in one direction but not in the other. If it does not show continuity in either direction, the diode is open. If it shows continuity in both directions the diode is short-circuited.

Inductive loads, such as the coils of relays and solenoids, produce a high transient voltage at turnoff. This inductive voltage can be particularly damaging to sensitive circuit components such as transistors and integrated circuits. A diode-clamping, or de-spiking, circuit connected in parallel across the inductive load can be used to limit the amount of transient voltage present in the circuit. The diode-clamping circuit of Ill. 4 illustrates how a diode can be used to suppress the inductive voltage of a relay coil. The operation of the circuit can be summarized as follows:

• The diode acts a one-way valve for current flow.

• When the limit switch is closed, the diode is reverse-biased.

• Electric current can't flow through the diode so it flows through the relay coil.

• When the limit switch is opened, a voltage opposite to the original applied voltage is generated by the collapsing magnetic field of the coil.

• The diode is now forward-biased and current flows through the diode rather than through the limit switch contacts, bleeding off the high-voltage spike.

• The faster the current is switched off, the greater is the induced voltage. Without the diode the induced voltage could reach several hundreds or even thou sands of volts.

• It’s important to note that the diode must be connected in reverse bias relative to the DC supply voltage. Operating the circuit with the diode incorrectly connected in forward bias will create a short circuit across the relay coil that could damage both the diode and the switch.

Ill. 4 Diode connected to suppress inductive voltage. Relay coil; Limit switch; CR; Relay coil; Limit switch; Normal operating current; CR; Current at turnoff

Ill. 5 Single-phase, full-wave bridge rectifier circuit. Fairchild Semiconductor, fairchildsemi.com. Second half-cycle; First half-cycle

The half-wave rectifier makes use of only half of the AC input wave. A less-pulsating and greater average direct cur rent can be produced by rectifying both half-cycles of the AC input wave. Such a rectifier circuit is known as a full-wave rectifier. A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave rectification. This is a widely used configuration, both with individual diodes and with single-component bridges where the diode bridge is wired internally. Bridge rectifiers are used on DC injection braking of AC motors to change the AC line voltage to DC, which is then applied to the stator for braking purposes. The schematic for a single-phase, full-wave bridge rectifier circuit is shown in Ill. 5. The operation of the circuit can be summarized as follows:

• During the positive half-cycle, the anodes of Dl and D2 are positive (forward-biased), whereas the anodes of D3 and D4 are negative (reverse-biased). Electron flow is from the negative side of the line, through D1, to the load, then through D2, and back to the other side of the line.

• During the next half-cycle, the polarity of the AC line voltage reverses. As a result, diodes D3 and D4 become forward-biased. Electron flow is now from the negative side of the line through D3, to the load, then through D4, and back to the other side of the line. Note that during this half-cycle the current flows through the load in the same direction, producing a full-wave pulsating direct current.

Some types of direct current loads such as motors, relays, and solenoids will operate without problems on pulsating DC, but other electronic loads will not. The pulsations, or ripple, of the DC voltage can be removed by a filter circuit. Filter circuits may consist of capacitors, inductors, and resistors connected in different configurations. The schematic for a simple half-wave capacitor filter circuit is shown in Ill. 6. Filtering is accomplished by alternate charging and discharging of the capacitor. The operation of the circuit can be summarized as follows:

• The capacitor is connected in parallel with the DC output of the rectifier.

• With no capacitor, the voltage output is normal half-wave pulsating DC.

• With the capacitor installed, on every positive half cycle of the AC supply, the voltage across the filter capacitor and load resistor rises to the peak value of the AC voltage.

• On the negative half-cycle, the charged capacitor provides the current for the load to provide a more constant DC output voltage.

• The variation in the load voltage, or ripple, is dependent upon the value of the capacitor and load.

A larger capacitor will have less voltage ripple.

For heavier load demands, such as those required for industrial applications, the DC output is supplied from a three-phase source. Using three-phase power, it’s possible to obtain a low-ripple DC output with very little filtering. Ill. 7 shows a typical three-phase full-wave bridge rectifier circuit. The operation of the circuit can be summarized as follows:

• The six diodes are connected in a bridge configuration, similar to the single-phase rectifier bridge to produce DC.

• The cathodes of the upper diode bank connect to the positive DC output bus.

• The anodes of the lower diode bank connect to the negative DC bus.

• Each diode conducts in succession while the remaining two are blocking.

• Each DC output pulse is 60° in duration.

• The output voltage never drops below a certain voltage level.

Ill. 6 Capacitor filter. No capacitor; Load voltage waveforms; Small capacitor; Large capacitor

Ill. 7 Three-phase, full-wave bridge rectifier. DC output voltage waveform; Three-phase supply; Load

Zener Diode

A zener diode is a type of diode that permits current to flow in the forward direction like a standard diode, but also in the reverse direction if the voltage is larger than the breakdown voltage, known as the zener voltage. This reverse-bias current would destroy a standard diode, but the zener diode is designed to handle it. The specified zener voltage rating of a zener diode indicates the volt age at which the diode begins to conduct when reverse biased.

Zener diodes are used to provide a fixed reference volt age from a supply voltage that varies. They are commonly found in motor control feedback systems to provide a fixed level of reference voltage in regulated power supply circuits. Ill. 8 shows a typical zener diode regulator circuit. The operation of the circuit can be summarized as follows:

• Input voltage must be higher than the specified zener voltage.

• The zener diode is connected in series with the resistor to allow enough reverse-biased current to flow for the zener to operate.

• Voltage drop across the zener diode will be equal to the zener diode's voltage rating.

• Voltage drop across the series resistor is equal to the difference between the zener voltage and the input voltage.

• The zener diode's voltage remains constant as the input voltage changes within a specified range.

• The change in input voltage appears across the series resistor.

Two back-to-back zener diodes can suppress damaging voltage transients on an AC line. A metal oxide varistor (MOV) surge suppressor functions in the same manner as back-to-back zener diodes. The circuit of Ill. 9 is used to suppress AC voltage transients. The varistor module shown is made to be easily mounted directly across the coil terminals of contactors and starters with 120 V or 240 V AC coils. The operation of the circuit can be summarized as follows:

• Each zener diode acts as an open circuit until the reverse zener voltage across it exceeds its rated value.

• Any greater voltage peak instantly makes the zener diode act like a short circuit that bypasses this volt age away from the rest of the circuit.

• It’s recommended that you locate the suppression device as close as possible to the load device.

Light-Emitting Diode

The light-emitting diode (LED) is another important diode device. An LED contains a PN junction that emits light when conducting current. When forward-biased, the energy of the electrons flowing through the resistance of the junction is converted directly to light energy.

Because the LED is a diode, current will flow only when the LED is connected in forward-bias. The LED must be operated within its specified voltage and current ratings to prevent irreversible damage. Ill. 10 illustrates a simple LED circuit. The LED is connected in series with a resistor that limits the voltage and current to the desired value.

Ill. 8 Typical zener diode regulator circuit. AC input; Series resistor; Unregulated; DC supply; Zener diode symbol; Regulated; DC voltage

Ill. 9 Suppressing AC voltage transients. Varistor; AC supply; Load

Ill. 10 Light-emitting diode (LED). Cathode Anode; Long lead anode; Series resistor; Light Anode PN; Cathode

The main advantages of using an LED as a light source rather than an ordinary light bulb are a much lower power consumption, a much higher life expectancy, and high speed of operation. Conventional silicon diodes convert energy to heat. Gallium arsenide diodes convert energy to heat and infrared light. This type of diode is called an infrared-emitting diode (IRED). Infrared light is not visible to the human eye. By doping gallium arsenide with various materials, LEDs with visible outputs of red, green, yellow, or blue light can be manufactured. Light-emitting diodes are used for pilot

lights and digital displays. Ill. 11 shows a typical seven-segment LED numerical display. By energizing the correct segments, the numbers 0 through 9 can be displayed.

Photodiodes

Photodiodes are PN-junction diodes specifically designed for light detection. They produce current flow when they absorb light. Light energy passes through the lens that exposes the junction. The photodiode is designed to operate in the reverse-bias mode. In this device, the reverse-bias leakage current increases with an increase in the light level. Thus, a photodiode exhibits a very high resistance with no light input and less resistance with light input.

There are many situations where signals and data need to be transferred from one piece of equipment to another, without making a direct electrical connection. Often this is because the source and destination are at very different voltage levels, like a microprocessor that is operating from 5 V DC but is being used to control a circuit that is switching 240 V AC. In such situations the link between the two must be an isolated one, to protect the microprocessor from overvoltage damage.

The circuit of Ill. 12 uses a photodiode as part of an optocoupler (also known as an optoisolator) pack age containing an LED and a photodiode. Optocouplers are used to electrically isolate one circuit from another. The only thing connecting the two circuits is light, so they are electrically isolated from each other.

Optocouplers typically come in a small integrated circuit package and are essentially a combination of an optical transmitter LED and an optical receiver such as a photodiode. The operation of the circuit can be summarized as follows:

• The LED is forward-biased, while the photodiode is reverse-biased.

• With the push button open, the LED is off. No light enters the photodiode and no current flows in the input circuit.

• The resistance of the photodiode is high, so little or no current flows through the output circuit.

• When the input push button is closed, the LED is forward-biased and turns on.

• Light enters the photodiode so its resistance drops switching on current to the output load.

QUIZ

1. Compare the type of current carriers associated with N-type and P-type semiconductor materials.

2. State the basic operating characteristic of a diode.

3. How is a diode tested using an ohmmeter?

4. What determines whether a diode is forward- or reverse-biased?

5. Under what condition is a diode considered to be connected in forward bias?

6. What is the function of a rectifier diode?

7. Explain the process by which a single-phase half wave rectifier changes AC to DC.

8. What is the purpose of a clamping, or despiking, diode?

9. A single-phase half-wave rectifier is replaced with a full-wave bridge type. In what ways will the DC output change?

10. How does a capacitor filter operate to smooth the amount of pulsation (ripple) associated with rectifier circuits?

11. What advantages are gained by using three-phase rectifiers over single-phase types?

12. In what way is the operation of a zener diode different from that of a standard diode?

13. Zener diodes are commonly used in voltage regulation circuits. What operating characteristic of the zener diode makes it useful for this type of application?

14. State the basic operating principle of an LED.

15. How many LEDs are integrated into a single LED numerical display?

16. Explain how a photodiode is designed to detect light.

<<Prev. | Next>>

Transformer Calculations

In the following examples, values of voltage, current, and turns for different transformers will be computed.

Example #1: Assume the isolation transformer shown in Ill. I2 has 240 turns of wire on the primary and 60 turns of wire on the secondary. This is a ratio of 4:1 (240/60 = 4).

Now assume that 120 volts is connected to the primary winding. What is the voltage of the secondary winding?

The transformer in this example is known as a step-down transformer because it has a lower secondary voltage than primary voltage.

Now assume that the load connected to the secondary winding has an impedance of 5 ?. The next problem is to calculate the current flow in the secondary and primary windings. The current flow of the secondary can be computed using Ohm's law since the voltage and impedance are known.

Now that the amount of current flow in the secondary is known, the primary current can be computed using the formula:

Notice that the primary voltage is higher than the secondary voltage, but the primary cur rent is much less than the secondary current. A good rule for transformers is that power in must equal power out. If the primary voltage and current are multiplied together, the result should equal the product of the voltage and current of the secondary.

Primary 120 _ 1.5 = 180 volt-amps Secondary 30 _ 6 = 180 volt-amps

Helpful Hint: A good rule for transformers is that power in must equal power out.

Example #2: In the next example, assume that the primary winding contains 240 turns of wire and the secondary contains 1,200 turns of wire. This is a turns-ratio of 1:5 (1,200/240 = 5). Now assume that 120 volts is connected to the primary winding. Compute the voltage output of the secondary winding.

Notice that the secondary voltage of this transformer is higher than the primary voltage.

This type of transformer is known as a step-up transformer.

Now assume that the load connected to the secondary has an impedance of 2,400 ohm. Find the amount of current flow in the primary and secondary windings. The current flow in the secondary winding can be computed using Ohm's law.

Now that the amount of current flow in the secondary is known, the primary current can be computed using the formula:

Notice that the amount of power input equals the amount of power output.

Primary 120 _ 1.25 = 150 volt-amps

Secondary 600 _ 0.25 = 150 volt-amps

Calculating Transformer Values Using the Turns-Ratio

As illustrated in the previous examples, transformer values of voltage, current, and turns can be computed using formulas. It is also possible to compute these same values using the turns-ratio. There are several ways in which turns-ratios can be expressed. One method is to use a whole number value such as 13:5 or 6:21. The first ratio indicates that one winding has 13 turns of wire for every 5 turns of wire in the other winding. The second ratio indicates that there are 6 turns of wire in one winding for every 21 turns in the other.

A second method is to use the number 1 as a base. When using this method, the number 1 is always assigned to the winding with the lowest voltage rating. The ratio is found by dividing the higher voltage by the lower voltage. The number on the left side of the ratio rep resents the primary winding and the number on the right of the ratio represents the secondary winding. E.g., assume a transformer has a primary rated at 240 volts and a secondary rated at 96 volts, as shown in Ill. I7. The turns-ratio can be computed by dividing the higher voltage by the lower voltage.

RATIO: 240/96 RATIO: 2.5:1 Notice in this example that the primary winding has the higher voltage rating and the secondary has the lower. Therefore, the 2.5 is placed on the left and the base unit, 1, is placed on the right. This ratio indicates that there are 2.5 turns of wire in the primary winding for every 1 turn of wire in the secondary.

Now assume that there is a resistance of 24 ohm connected to the secondary winding. The amount of secondary current can be found using Ohm's law.

Ill. I7 Computing transformer values using the turns-ratio.

The primary current can be found using the turns-ratio. Recall that the volt-amps of the primary must equal the volt-amps of the secondary. Since the primary voltage is greater, the primary current will have to be less than the secondary current. Therefore, the secondary current will be divided by the turns-ratio.

To check the answer, find the volt-amps of the primary and secondary.

Primary = 240 _ 1.6 = 384

Secondary = 96 _ 4 = 384

Now assume that the secondary winding contains 150 turns of wire. The primary turns can be found by using the turns-ratio, also. Since the primary voltage is higher than the secondary voltage, the primary must have more turns of wire. Since the primary must contain more turns of wire, the secondary turns will be multiplied by the turns-ratio.

P S × =

P 150 2.5 × =

P 375 turns =

In the next example, assume a transformer has a primary voltage of 120 volts and a secondary voltage of 500 volts. The secondary has a load impedance of 1,200 ohm. The secondary contains 800 turns of wire ( Ill. I8). The turns-ratio can be found by dividing the higher voltage by the lower voltage.

Ill. I8 Calculating transformer values.

The secondary current can be found using Ohm's law.

In this example the primary voltage is lower than the secondary voltage. Therefore, the primary current must be higher. To find the primary current, multiply the secondary current by the turns-ratio.

IP = IS _ Turns-ratio IP = 0.417 _ 4.17 IP = 1.74 amps

To check this answer, compute the volt-amps of both windings.

Primary 120 _ 1.74 = 208.8

Secondary 500 _ 0.417 = 208.5

The slight difference in answers is caused by rounding off of values.

Since the primary voltage is less than the secondary voltage, the turns of wire in the primary will also be less. The primary turns will be found by dividing the turns of wire in the secondary by the turns-ratio.

LABORATORY EXERCISE

Name _______ Date __________

Materials Required:

480-240/1volt, 0.5-kVA control transformer Ohmmeter AC voltmeter, in-line or clamp-on. (If a clamp-on type is used, a 10:1 scale divider is recommended.) These experiments are intended to provide the electrician with hands-on experience dealing with transformers. The transformers used in these experiments are standard control transformers with two high-voltage windings rated at 240 volts each generally used to pro vide primary voltages of 480/240, and one low-voltage winding rated at 120 volts. The transformers have a rating of 0.5 kVA. The loads are standard 100 watt lamps that may be connected in parallel or series. It is assumed that the power supply is 208/120 volt three phase four wire. It is also possible used with a 240/120 volt three-phase high leg system, pro vided adjustments are made in the calculations.

As in industry, these transformers will be operated with full voltage applied to the windings.

The utmost caution must be exercised when dealing with these transformers. These transformers can provide enough voltage and current to seriously injure or kill anyone. The power should be disconnected before attempting to make or change any connections.

Caution: These transformers can provide enough voltage and current to seriously injure or kill anyone.

The transformer used in this experiment contains two high-voltage windings and one low voltage winding. The high-voltage windings are labeled H1 - H2 and H3 - H4. The low-voltage winding is labeled X1 - X2.

1. Set the ohmmeter to the Rx1 range and measure the resistance between the following terminals:

H1 - H2 ____________ ? H1 - H3 ____________ ? H1 - H4 ____________ ? H1 - X1 ____________ ? H1 - X2 ____________ ? H2 - H3 ____________ ? H2 - H4 ____________ ? H2 - X1 ____________ ? H2 - X2 ____________ ? H3 - H4 ____________ ? H3 - X1 ____________ ? H3 - X2 ____________ ? H4 - X1 ____________ ? H4 - X2 ____________ ? X1 - X2 ____________ ?

2. Using the information provided by the measurements from step 1, which sets or terminals form complete circuits within the transformer?

These circuits represent the connections to the three separate windings within the transformer.

3. Which of the windings exhibits the lowest resistance and why?

4. The H1 - H2 terminals are connected to one of the high-voltage windings and the H3 - H4 terminals are connected to the second high-voltage winding. Each of these windings is rated at 240 volts. When this transformer will be connected for 240 volt operation, the two high-voltage windings are connected in parallel to form one winding by connecting H1 to H3 and H2 to H4, as shown in Ill. I9. This will provide a 2:1 turns ratio with the low-voltage winding.

When this transformer is operated with 480 volts connected to the primary, the high voltage windings are connected in series by connecting H2 to H3 and connecting power to H1 and H4, as shown in Ill. I10. This effectively doubles the primary turns, pro viding a 4:1 turns-ratio with the low-voltage winding.

5. Connect the two high-voltage windings for parallel operation as shown in Ill. I9.

Assume a voltage of 208 volts is applied to the high-voltage windings. Compute the volt age that should be seen on the low-voltage winding between terminals X1 and X2. _ volts.

6. Make certain that the incoming power leads are connected to terminals H1 and H4 as shown in Ill. I9. Apply a voltage of 208 volts to the transformer and measure the voltage across terminals X1 and X2. ____ volts.

7. The measured voltage may be slightly higher than the computed voltage. The rated voltage of a transformer is based on full load. It is normal for the secondary voltage to be slightly higher when no load is connected to the transformer. Transformers are generally wound with a few extra turns of wire in the winding that's intended to be used as the load side. This helps overcome the voltage drop when load is added. The slight change in turns-ratio does not affect the operation of the transformer to a great extent.

Ill. I9 High-voltage windings connected in parallel.

Ill. I10 High-voltage windings connected in series.

8. Turn off the power to the transformer.

9. Disconnect the wires connected to the transformer and reconnect the transformer as shown in Ill. I10. The two high-voltage windings are connected in series by connecting H2 and H3 together. This connection changes the turns-ratio of the transformer from 2:1 to 4:1. Make certain that the incoming power is connected to terminals H1 and H4.

10. Assume that a voltage of 208 volts is applied to the high-voltage windings. Compute the voltage across the low-voltage winding. ___ volts

11. Turn on the power and apply a voltage of 208 volts to the transformer. Measure the volt age across terminals X1 and X2. ___ volts

12. Turn off the power. Disconnect the power lines that are connected to terminals H1 and H4. Don't disconnect the wire between terminals H2 and H3.

13. In the next part of the exercise, the low-voltage winding will be used as the primary and the high-voltage windings will be used as the secondary. If the high-voltage windings are connected in series, the turns-ratio will be 1:4, which means that the secondary voltage will be four times greater than the primary voltage. The transformer has now become a step-up transformer instead of a step-down transformer. Assume that a voltage of 120 volts is connected to terminals X1 and X2. If the high-voltage windings are connected in series, compute the voltage across terminals H1 and H4. ___ volts

14. Connect the transformer as shown in Ill. I11. Make certain that the voltage applied to terminals X1 and X2 is 120 volts and not 208 volts. Also make certain that the AC voltmeter is set for a higher range than the computed value of voltage in step 13.

Caution: The secondary voltage in this step will be 480 volts or higher. Use extreme caution when making this measurement. Be sure to wear safety glasses at all times.

Ill. I11 The incoming power is connected to terminals X1 and X2.

Ill. I12 The transformer has a turns-ratio of 1:2.

15. Turn on the power and measure the voltage across terminals H1 and H4. ____ volts

16. Turn off the power supply.

17. Disconnect the lead between terminals H2 and H3. Reconnect the transformer so that the high-voltage windings are connected in parallel by connecting H1 and H3 together and H2 and H4 together as shown in Ill. I12. Don't disconnect the power leads to terminals X1 and X2. The transformer now has a turns-ratio of 1:2.

18. Assume that a voltage of 120 volts is connected to the low-voltage winding. Compute the voltage across the high-voltage winding.____ volts

19. Make certain the power leads are still connected to terminals X1 and X2. Turn on the power and apply 120 volts to terminals X1 and X2. Measure the voltage across terminals H1 and H4.___ volts

20. Turn off the power supply and disconnect all leads to the transformer. Return the components to their proper place.

QUIZ:

1. What is a transformer?

2. What are common efficiencies for transformers?

3. What is an isolation transformer?

4. All values of a transformer are proportional to its:

5. A transformer has a primary voltage of 480 volts and a secondary voltage of 20 volts. What is the turns-ratio of the transformer?

6. If the secondary of the transformer in question 5 supplies a current of 9.6 amperes to a load, what is the primary current (disregard excitation current)?

7. Explain the difference between a step-up and a step-down transformer.

8. A transformer has a primary voltage of 240 volts and a secondary voltage of 48 volts.

What is the turns-ratio of this transformer?

9. A transformer has an output of 750 volt-amps. The primary voltage is 120 volts. What is the primary current?

10. A transformer has a turns-ratio of 1:6. The primary current is 18 amperes. What is the secondary current?

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