Autotransformers

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Objectives:

• Discuss the operation of an autotransformer.

• Connect a control transformer as an autotransformer.

• Calculate the turns-ratio from measured voltage values.

• Calculate primary current using the secondary current and the turns-ratio.

• Connect an autotransformer as a step-down transformer.

• Connect an autotransformer as a step-up transformer.

Ill. 1 Autotransformer used as a step-down transformer.

Ill. 2 Autotransformer used as a step-up transformer.

Ill. 3 Autotransformer with multiple taps.

The word auto means self. An autotransformer is literally a self-transformer. It uses the same winding as both the primary and secondary. Recall that the definition of a primary winding is a winding that's connected to the source of power and the definition of a secondary winding is a winding that's connected to a load. Autotransformers have very high efficiencies, most in the range of 95% to 98%.

In Ill. 1 the entire winding is connected to the power source, and part of the winding is connected to the load. In this illustration all the turns of wire form the primary and part of the turns form the secondary. Since the secondary part of the winding contains fewer turns than the primary section, the secondary will produce less voltage. This autotransformer is a step-down transformer.

In Ill. 2 the primary section is connected across part of a winding and the secondary is connected across the entire winding. In this illustration the secondary section contains more windings than the primary. This autotransformer is a step-up transformer. Notice that autotransformers, like isolation transformers, can be used as step-up or step-down transformers.

Determining Voltage Values

Autotransformers are not limited to a single secondary winding. Many autotransformers have multiple taps to provide different voltages as shown in Ill. 3. In this example there are 40 turns of wire between taps A and B, 80 turns of wire between taps B and C, 100 turns of wire between taps C and D, and 60 turns of wire between taps D and E. The primary section of the windings is connected between taps B and E. It will be assumed that the primary is connected to a source of 120 volts. The voltage across each set of taps will be determined.

There is generally more than one method that can be employed to determine values of a transformer. Since the number of turns between each tap is known, the volts-per-turn method will be used in this example. The volts-per-turn for any transformer is determined by the primary winding. In this illustration the primary winding is connected across taps B and E.

The primary turns are, therefore, the sum of the turns between taps B and E (80 + 100 + 60 = 240 turns). Since 120 volts is connected across 240 turns, this transformer will have a volts-per-turn ratio of 0.5 (240 turns/120 volts = 0.5 volt-per-turn). To determine the amount of voltage between each set of taps, it becomes a simple matter of multiplying the number of turns by the volts-per-turn.

A-B (40 turns _ 0.5 = 20 volts) A-C (120 turns _ 0.5 = 60 volts) A-D (220 turns _ 0.5 = 110 volts) A-E (280 turns _ 0.5 = 140 volts) B-C (80 turns _ 0.5 = 40 volts) B-D (180 turns _ 0.5 = 90 volts) B-E (240 turns _ 0.5 = 120 volts) C-D (100 turns _ 0.5 = 50 volts) C-E (160 turns _ 0.5 = 80 volts) D-E (60 turns _ 0.5 = 30 volts)

Ill. 4 Determining voltage and current values.


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

Motor Control Electronics

Integrated Circuits (ICs)

Fabrication

An integrated circuit (IC), sometimes called a chip, is a semiconductor wafer on which thousands or millions of tiny resistors, capacitors, and transistors are fabricated.

IC chips provide a complete circuit function in one small semiconductor package with input and output pin connections, as illustrated in Ill. 37. Most integrated circuits provide the same functionality as "discrete" semiconductor circuits at higher levels of reliability and at a fraction of the cost. Usually, discrete-component circuit construction is favored only when voltage and power dissipation levels are too high for integrated circuits to handle.

Integrated circuits may be categorized as either digital or analog, depending on their intended application. Digital ICs operate with on/off switch-type signals that have only two different states, called low (logic 0) and high (logic 1). Analog ICs contain amplifying-type circuitry and signals capable of an unlimited number of states.

The analog and digital processes can be seen in a simple comparison between a light dimmer and light switch.

A light dimmer involves an analog process, which varies the intensity of light from off to fully on. The operation of a standard light switch, on the other hand, involves a digital process; the switch can be operated only to turn the light off or on.

Ill. 37 Integrated circuit (IC). IC motor controller module; Top; Chip; Input/output pin connections

Ill. 38 Operational amplifier (op-amp). Standard power supply; Output; Negative supply; Positive supply; Inverting input; Noninverting input

Ill. 39 741 op-amp voltage amplifier circuit.

Ill. 40 Operational amplifier circuit con figured as a voltage comparator.

Operational Amplifier ICs

Operational amplifier ICs (often called op-amps) take the place of amplifiers that formerly required many discrete components. These amplifiers are often used in con junction with signals from sensors connected in control circuits. An op-amp is basically a high-gain amplifier that can be used to amplify low level AC or DC signals. The schematic symbol for an op-amp is a triangle, shown in Ill. 38. The triangle symbolizes direction and points from input to output. The connections associated with an op-amp can be summarized as follows:

• The op-amp has two inputs and a single output. The inverting input (-) produces an output that is 180° out of phase with the input. The second input, called the noninverting input (+), produces an output that is in phase with the input.

• The DC power supply terminals are identified as +V and -V. All op-amps need some type of power supply, but some diagrams won’t show the power supply terminals, as it’s assumed that they are always there. The supply power will be determined by the type of output the op-amp is required to produce. E.g., if the output signal needs to produce both positive and negative voltages, then the power supply will need to be a split, or differential, type with both positive and negative voltages and a common ground. If the op-amp needs to produce only positive voltages, then the power supply be the standard, or traditional, DC type.

The op-amp is connected in a number of ways to perform different functions. Ill. 39 shows the schematic diagram of a 741 op-amp circuit con figured as an AC inverting amplifier . A split-type power sup ply, consisting of a positive supply and an equal and opposite negative supply, is used to operate the circuit. The operation of the circuit can be summarized as follows:

• Two resistors, R1 and R2, set the value of the volt age gain of the amplifier.

• Resistor R2 is called the input resistor and resistor R1 is called the feedback resistor.

• The ratio of the resistance value of R2 to that of R1 sets the voltage gain of the amplifier.

• The op-amp amplifies the AC input voltage it receives and inverts its polarity.

• The output signal is 180° out of phase with the input signal.

• The op-amp gain for the circuit is calculated as follows:

Op-amp gain = R1 / R2

Ill. 40 shows the schematic diagram of an operational amplifier circuit con figured as a differential amplifier or voltage comparator with its output signal being the difference between the two input signals or voltages, V2 and V1. A light-dependent resistor (LDR) is used to sense the light level. When LDR is not illuminated, its resistance is very high, but once illuminated with light, its resistance drops dramatically. The circuit is operated with a standard DC power supply and without a feedback circuit. The operation of the circuit can be summarized as follows:

• The resistor combination R1 and R2 form a fixed reference voltage input V2, set by the ratio of the two resistors.

• The resistor combination LDR and R3 form the variable voltage input V1.

• When the light level sensed by the LDR drops and the variable output voltage V1 falls below the reference voltage at V2, the output from the op-amp changes, activating the relay and switching the connected load.

• Likewise, as the light level increases the output will switch back, turning the relay off.

• The preset resistor R3 value can be adjusted up or down to increase or decrease resistance; in this way it can make the circuit more or less sensitive.

Ill. 41 The 555 IC timer. Ground, Trigger, Output Threshold, Discharge, Reset Control voltage Vcc 8

555 Timer IC

Ill. 42 A 555 IC interval timer. (a) Light switch interval timer; (b) The 555 interval timer circuit; Switch trigger

The 555 timer IC is used as a timer in circuits requiring precision timing as well as an oscillator to provide pulses needed to operate digital circuits. Ill. 41 shows the pinout functional block diagram of the 555 chip. The internal circuitry of the chip is made up of a complex maze of transistors, diodes, and resistors.

The light switch interval timer shown in Ill. 42a is an ideal application for lights that are needlessly left on or forgotten. The dial is set for the amount of time the lights are to be left on, from 1 minute to 18 hours. Time intervals are initiated on the closure of the switch. After the preset time has elapsed the lights will go off automatically regardless of the state (open or closed) of the switch.

Ill. 43 The 555 CD motor speed controller.

Ill. 44 Microcontroller used in motor control applications.

Ill. 45 Embedded microcontroller.

Ill. 42b shows a 555 timer version of an interval timer. The operation of the circuit can be summarized as follows:

• The time period is determined by the value of the two external timing components R and C.

• When the switch opens the external capacitor is held discharged (short-circuited) by a transistor inside the timer.

• When the switch is closed, it releases the short circuit across the capacitor and triggers the LED into conduction. At this point the timing period starts.

• Capacitor C starts to charge through resistor R.

• When the charge on the capacitor reaches two-thirds of the source voltage, the timing period ends and the LED is automatically switched off.

• At the same time the capacitor discharges to ready for the next triggering sequence.

A common method of DC motor speed control is pulse-width modulation (PWM). Pulse-width modulation is the process of switching the power to a device on and off at a given frequency, with varying on and off times.

This relation between on and off times is referred to as the duty cycle . Ill. 43 shows the 555 timer used as a pulse width modulator oscillator that controls the speed of a small permanent-magnet DC motor. While most DC motor drives use a microcontroller to generate the required PWM signals, the 555 PWM circuit shown will help you to understand how this type of motor control operates. The operation of the circuit can be summarized as follows:

• The voltage applied across the armature is the aver age value determined by the length of time that transistor Q2 is turned on compared to the length of time it’s turned off (duty cycle).

• Potentiometer R1 controls the length of time the output of the timer will be turned on, which in turn controls the speed of the motor.

• If the wiper of R1 is adjusted to a higher positive voltage, the output will be turned on for a longer period of time than it will be turned off.

Microcontroller

An electronic motor controller can include means for starting and stopping the motor, selecting forward or reverse rotation, selecting and regulating the speed, regulating or limiting the torque, and protecting against over loads and faults. As integrated circuitry evolved, all of the components needed for a controller were built right onto one microcontroller chip. A microcontroller (also called a digital-signal controller or DSC) is an ultra-large-scale integrated circuit that functions as a complete computer on a chip, containing a processor, memory, and input/ output functions.

Ill. 44 shows a typical microcontroller IC that can be used in a variety control applications.

Microcontrollers are most often "embedded," or physically built into, the devices they control, as shown in Ill. 45. An embedded microcontroller is designed to do some specific task, rather than to be a general-purpose computer for multiple tasks. The software written for embedded systems is often called firmware, and is stored in read-only memory or flash memory chips rather than a disk drive. Microcontrollers often run with limited computer hardware resources: small or no keyboard and screen, and little memory.

The block diagram of Ill. 46 shows a micro controller used to control the operation of the inverter section of a variable-speed AC motor drive. The rotating speed of AC induction motors is determined by the frequency of the AC applied to the stator, not the applied voltage. However, stator voltage must also drop to prevent excessive current flow through the stator at low frequencies. The microcontroller controls the volt age and frequency and sets the proper stator voltage for any given input frequency. Two phase currents are measured and returned to the microcontroller along with rotor speed and angular-position information from the encoder/tachometer.

Ill. 46 Microcontroller used to control the operation of the inverter section of a variable-speed AC motor drive. Current feedback signals; Mechanical feedback signals (speed and position); Three-phase motor; Encoder/tachometer

Power inverter; Microcontroller;

Electrical Discharge (ESD)

Static charge is an unbalanced electrical charge at rest.

Typically, it’s created by insulator surfaces rubbing together or pulling apart. One surface gains electrons, while the other surface loses electrons. This results in an unbalanced electrical condition known as static charge.

When a static charge moves from one surface to another, it becomes electrostatic discharge (ESD). Ill. 47 shows a common example of ESD. When a person (negatively charged) contacts a positively charged or a grounded object, electrons will move from one to the other. ESD, which is a bit annoying but certainly harmless to humans, can be lethal to sensitive electronic devices.

All integrated circuits are sensitive to electrostatic discharge to some degree. If static discharge occurs at a sufficient magnitude, some damage or degradation (the IC is weakened and will often fail later) will usually occur.

Damage is mainly due to the current flowing through ICs during discharge. Basically what happens is that a relatively large amount of heat is generated in a localized volume significantly faster than it can be removed, leading to a temperature in excess of the material's safe operating limits.

Ill. 48 shows an antistatic wrist strap band used to prevent a static charge from building up on the body by safely grounding a person working on sensitive electronic equipment. The wrist strap is connected to ground through a coiled retractable cable and resistor. An approved grounding wrist band has a resistance built into it, so it discharges static electricity but prevents a shock hazard when the wearer is working with lower circuit voltages.

Other precautions that should be taken when working on integrated circuits include:

• Never handle sensitive ICs by their leads.

• Keep your work area clean, especially of common plastics.

• Handle printed circuit boards by their outside corners.

• Always transport and store sensitive ICs and control boards in antistatic packaging.

Digital Logic

Logic circuits perform operations on digital signals. In digital or binary logic circuits there are only two values,

0 and 1. Logically, we can use these two numbers or we can specify that:

0 = false = no = off = open = low

1 = true = yes = on = closed = high

Using the binary two-value logic system, every condition must be either true or false; it cannot be partly true or partly false. While this approach may seem limited, it can be expanded to express very complex relationships and interactions among any number of individual conditions.

One reason for the popularity of digital logic circuits is that they provide stable electronic circuits that can switch back and forth between two clearly defined states, with no ambiguity attached. Integrated circuits are the least expensive way to make logic gates in large volumes. They are used in programmable controllers to solve complex logic.

Logic is the ability to make decisions when one or more different factors must be taken into account before an action is taken. Digital circuits are constructed from small electronic circuits called logic gates. The and gate is a logic circuit that has two or more inputs and a single output. Ill. 49 shows the traditional and IEC symbols used for a two-input and gate. The operation of the and gate is summarized by the table. Such a table, called a truth table, shows the output for each possible input. The basic logic that applies is that if all inputs are 1, the output will be 1. If any input is 0, the output will be 0.

Ill. 49 Two-input and gate.

Ill. 50 Two-input OR gate.

Ill. 51 The NOT function.

Ill. 52 Two-input NAND gate.

Ill. 53 Two-input NOR gate.

Ill. 55 Hard-wired circuit and the equivalent PLC logic program for an OR logic control function. Hard-wired OR function; Programmed OR function

An OR gate produces a 1 output if any of its inputs are 1s. The output is 0 if all the inputs are 0s. An OR gate can have two or more inputs; its output is true if at least one input is true. Ill. 50 shows the traditional and IEC symbols used for a two-input OR gate along with its truth table.

The simplest logic circuit is the NOT circuit . It per forms the function called inversion, or complementation, and is commonly referred to as an inverter. The purpose of the inverter is to make the output state the opposite of the input state. Ill. 51 shows the traditional and IEC symbols used for a NOT function along with its truth table. Unlike the and / OR gate functions, the NOT function can have only one input. If a 1 is applied to the input of an inverter, a 0 appears on its output. The input to an inverter is labeled A and the output is labeled (read "NOT A"). The bar over the letter indicates the complement of A. Because the inverter has only one input, only two input combinations are possible.

A NAND gate is a combination of an inverter and an and gate. It’s called a NAND gate after the NOT-AND function it performs. Ill. 52 shows the traditional and IEC symbols used for a two-input NAND gate along with its truth table. The bubble on the output end of the symbol means to invert the and function. Notice that the output of the NAND gate is the complement of the output of an and gate. A NAND gate can have two or more inputs. Any 0 in the inputs yields a 1 output. The NAND gate is the most commonly used logic function. This is because it can be used to construct an and gate, OR gate, inverter, or any combination of these functions.

A NOR gate is a combination of an inverter and an OR gate. Its name is derived from its NOT-OR function.

Ill. 53 shows the traditional and IEC symbols used to represent a two-input NOR gate along with its truth table. The output of the NOR gate is the complement of the output of the OR-function output. The output Q is 1 if NOR inputs A or B are 1. A NOR gate can have two or more inputs and its output is 1 only if no inputs are 1.

The term hard-wired logic refers to logic control functions that are determined by the way devices are interconnected. Hard-wired logic is fixed in that it’s changeable only by changing the way the devices are connected. In contrast, programmable logic, such as that used in programmable logic controllers (PLCs), is based on basic logic functions, which are easily changed by modifying the program. PLCs use logic functions either singly or in combination to form instructions that will determine if a device is to be switched on or off.

Ill. 54 shows a hard-wired circuit and the equivalent PLC logic program for an and logic control function. Both normally open input limit switches (LS1 and LS2) must be closed to energize the output solenoid valve. This control logic is implemented in the hard-wired circuit by connecting the two limit switches and the solenoid valve all in series.

The PLC program uses the same inputs (LS1 and LS2) and output (SOL) devices connected to the PLC, but implements the logic via the program rather than the wiring.

Ill. 55 shows a hard-wired circuit and the equivalent PLC logic program for an OR logic control function. Either one of the two normally open push buttons (PB1 or PB2) is closed to energize the contactor coil (C). This control logic is implemented in the hard-wired circuit by connecting the two input push buttons in parallel with each other to control the output coil. The PLC pro gram uses the same inputs (PB1 and PB2) and output (C) devices connected to the PLC, but implements the logic via the program rather than the wiring.

AND / OR logic use normally open input devices that must be closed to supply the signal that energizes the loads.

NOT logic energizes the load when the control signal is off.

Ill. 56 shows a hard-wired circuit and the equivalent PLC logic program for a NOT logic control function. This example is that of the NOT function used to control the interior lamp of a refrigerator. When the door is opened, the lamp automatically switches on. The switch controlling the lamp is a normally closed type that is held open by the shut door. When the door is opened, the switch returns to its normal closed state and the load (lamp) is energized.

For the load to remain energized there must not be a signal from the switch input. To keep the lamp on, the normally closed contact must not change its normally closed state.

Ill. 54 Hard-wired circuit and the equivalent PLC logic program for an and logic control function. Hard-wired and function; Programmed and function

Ill. 56 Hard-wired circuit and the equivalent PLC logic program for a NOT logic control function. Hard-wired NOT function; Programmed NOT function; Normally closed held open

QUIZ:

1. Describe the makeup of an integrated circuit.

2. What advantages does an IC have over discrete component circuit construction?

3. What type of circuit is not suited to integrate onto a chip?

4. Compare the functioning of digital and analog

integrated circuits.

5. What is an operational amplifier?

6. An operational amplifier is connected as a voltage amplifier. How is the gain of the circuit determined?

7. Explain the operation of an op-amp when con figured as a voltage comparator.

8. What are the two major applications for the 555 timer IC?

9. Give a brief explanation of how a 555 interval timer circuit works.

10. A 555 timer is con figured as a pulse-width modulator to vary the speed of a DC motor. How exactly does it operate to change the speed of the motor?

11. List some of the control tasks that a microcontroller designed for control of an electric motor may be required to perform.

12. What does the term embedded microcontroller refer to?

13. Explain the role of a microcontroller when used to control the inverter section of a variable-speed AC motor drive.

14. In what way can electrostatic discharge damage an integrated circuit?

15. List some of the precautions that should be taken when handling sensitive IC chips.

16. What can the terms logic 0 and logic 1 represent?

17. What makes digital circuitry so popular?

18. It’s desired to have a lamp turn on when one of three switches is closed. What control logic function would be used?

19. What control logic function energizes the load when the control signal is off?

20. What control logic function is used to implement five inputs connected in series with the requirement that all five must be closed to energize the load?

21. Compare the way in which hard-wired logic differs from programmable logic.

REPAIR/ TROUBLESHOOTING SCENARIOS:

1. One of the diodes of a single-phase full-wave rectifier is mistakenly connected backward in the bridge configuration. What effect will this have on the resultant DC output voltage and current flow through the diodes?

2. The insulation resistance of a motor operated by an electronic drive is to be tested using a megger. What precaution should you take? Why?

3. If the triac in the internal circuit of a lamp dimmer were to fail shorted, what effect would this most likely have on the operation of the circuit?

4. In what ways would the approach to trouble shooting for integrated circuits be different from that for a circuit constructed with discrete components?

MORE QUESTIONS:

1. What circuitry would have to be incorporated into an LED pilot light module rated to be operated directly by a 240-V AC source?

2. The best source to verify correct operation of modular electronic components is the operating manual. Why?

3. What are the advantages of using a triac rather than a rheostat for lamp dimming applications?

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Using Transformer Formulas

The values of voltage and current for autotransformers can also be determined by using standard transformer formulas. The primary winding of the transformer shown in Ill. 4 is between points B and N and has a voltage of 120 volts applied to it. If the turns of wire are counted between points B and N, it can be seen there are 120 turns of wire. Now assume that the selector switch is set to point D. The load is now connected between points D and N. The secondary of this transformer contains 40 turns of wire. If the amount of voltage applied to the load is to be computed, the following formula can be used:

Assume that the load connected to the secondary has an impedance of 10 ohm. The amount of current flow in the secondary circuit can be computed using the formula:

The primary current can be computed by using the same formula that was used to compute primary current for an isolation type of transformer.

The amount of power input and output for the autotransformer must also be the same.

Primary 120 _ 1.333 = 160 volt-amp

Secondary 40 _ 4 = 160 volt-amps

Now assume that the rotary switch is connected to point A. The load is now connected to 160 turns of wire. The voltage applied to the load can be computed by:

The amount of secondary current can be computed using the formula:

The primary current can be computed using the formula:

The answers can be checked by determining if the power in and power out are the same.

Primary: 120 _ 21.333 = 2,560 volt-amps

Secondary: 160 _ 16 = 2,560 volt-amps

Ill. 5 Current divides between primary and secondary.

Current Relationships

An autotransformer with a 2:1 turns-ratio is shown in Ill. 5. It is assumed that a voltage of 480 volts is connected across the entire winding. Since the transformer has a turns-ratio of 2:1, a voltage of 240 volts will be supplied to the load. Ammeters connected in series with each winding indicate the current flow in the circuit. It is assumed that the load produces a current flow of 4 amperes on the secondary. Note that a current flow of 2 amperes is supplied to the primary.

I PRIMARY = I SECONDARY/ Ratio

If the rotary switch shown in Ill. 4 were to be removed and replaced with a sliding tap that made contact directly to the transformer winding, the turns-ratio could be adjusted continuously. This type of transformer is commonly referred to as a Variac or Powerstat depending on the manufacturer. The windings are wrapped around a tape-wound torroid core inside a plastic case. The tops of the windings have been milled flat similar to a commutator.

A carbon brush makes contact with the windings. When the brush is moved across the windings, the turns-ratio changes, which changes the output voltage. This type of autotransformer provides a very efficient means of controlling AC voltage. Autotransformers are often used by power companies to provide a small increase or decrease to the line voltage. They help provide voltage regulation to large power lines.

The autotransformer does have one disadvantage. Since the load is connected to one side of the power line, there is no line isolation between the incoming power and the load.

This can cause problems with certain types of equipment and must be a consideration when designing a power system.

LABORATORY EXERCISE

Name ______ Date _

Materials Required:

480-240/1volt, 0.5-kVA control transformer AC voltmeter 2 AC ammeter, in-line or clamp-on. (If the clamp-on type is used, a 10:1 scale divider is recommended.) 4 100-watt lamps In this experiment the control transformer will be connected for operation as an autotransformer.

The low-voltage winding won't be used in this experiment. The two high-voltage windings will be connected in series to form one continuous winding. The transformer will be connected as both a step-down and a step-up transformer.

1. Series connect the two high-voltage windings by connecting terminals H2 and H3 together. The H1 and H4 terminals will be connected to a source of 208 VAC. Connect an ammeter in series with one of the power supply lines, as shown in Ill. 6.

2. Turn on the power supply and measure the excitation current. The current will be small, and it may be difficult to determine this current value.

I_(EXC) _________ amp(s)

3. Measure the primary voltage across terminals H1 and H4.

E_(PRIMARY) _______ volts

4. Measure the secondary voltage across terminals H1 and H2. (Note: It is also possible to use terminals H3 and H4 as the secondary winding.) E_(SECONDARY) ___ volts

5. Determine the turns-ratio of this transformer connection.

Ratio ____

6. Turn off the power supply.

7. Connect an AC ammeter in series with the H2 terminal and a 100 watt lamp as shown in Ill. 7. The secondary winding of the transformer will be between terminals H2 and H1.

8. Turn on the power supply and measure the amount of current flow in the secondary winding.

I_(SECONDARY) _____ amp(s)

9. Measure the voltage drop across the secondary winding with an AC voltmeter.

E_(SECONDARY) ____________ volts

10. Turn off the power supply.

Ill. 6 Connecting the high-voltage windings as an autotransformer.

Ill. 7 Connecting a load to the autotransformer.

11. Calculate the primary current using the turns-ratio. Since the primary voltage is greater than the secondary voltage, the primary current will be less than the secondary current. To determine the primary current, divide the secondary current by the turns-ratio and add the excitation current.

12. If necessary, reconnect the AC ammeter in series with one of the power supply leads.

13. Turn on the power and measure the primary current. Compare this value with the computed value.

I_(PRIMARY) ____________ amp(s)

14. Turn off the power supply.

15. Connect another 100 watt lamp in parallel with the existing lamp ( Ill. 8). Ill. 8 Adding load to the autotransformer.

16. If necessary, reconnect the AC ammeter in series with the secondary winding of the transformer.

17. Turn on the power supply and measure the secondary current.

I_(SECONDARY) _______ amp(s)

18. Calculate the primary current using the turns-ratio.

I_(PRIMARY) ______ amp(s)

19. Turn off the power supply.

20. If necessary, reconnect the AC ammeter in series with one of the power supply leads.

21. Turn on the power supply and measure the primary current. Compare this value with the computed value.

I_(PRIMARY) ____________ amp(s) 22. Turn off the power supply.

23. Reconnect the circuit as shown in Ill. 9. Terminals H1 and H2 will be connected to a source of 120 VAC. Connect an AC ammeter in series with terminal H2.The entire winding between terminals H1 and H4 will be used as the secondary.

24. Turn on the power and measure the excitation current of this transformer connection. I_(EXC) ______ amp(s)

25. Measure the voltage across terminals H4 and H1.

_____ volts

26. Compute the turns-ratio of this transformer connection. Ratio ______

27. Turn off the power supply.

28. Connect an AC ammeter and four 100 watt lamps in series with terminals H4 and H1, as shown in Ill. 10.

29. Turn on the power and measure the secondary current. I_(SECONDARY) ____________ amp(s)

30. Turn off the power supply.

31. If necessary, connect the AC ammeter in series with one of the primary leads.

Ill. 9 The autotransformer connected for high voltage.

Ill. 10 Adding load to the secondary winding.

32. Compute the value of primary current using the turns-ratio and the measured value of secondary current. I_(PRIMARY) _______ amp(s)

33. Turn on the power and measure the primary current. Compare this value with the computed value. I_(PRIMARY) ____________ amp(s)

34. Measure the voltage across terminals H1 and H4. E_(SECONDARY) _____volts

35. Turn off the power supply.

36. Disconnect the circuit and return the components to their proper place.

QUIZ:

1. An AC power source is connected across 325 turns of an autotransformer and the load is connected across 260 turns. What is the turns-ratio of this transformer?

2. Is the transformer in question 1 a step-up or step-down transformer?

3. An autotransformer has a turns-ratio of 3.2:1. A voltage of 208 volts is connected across the primary. What is the voltage of the secondary?

4. A load impedance of 52 ohm is connected to the secondary winding of the transformer in question 3. How much current will flow in the secondary?

5. How much current will flow in the primary of the transformer in question 4?

6. The autotransformer shown in Ill. 3 has the following number of turns between windings: A-B (120 turns), B-C (180 turns), C-D (250 turns), and D-E (300 turns).

A voltage of 240 volts is connected across B and E. Find the voltages between each of the following points:

A-B _____ A-C _____ A-D _____ A-E _____ B-C _____ B-D _____ B-E _____ C-D _____ C-E _____ D-E _____

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