Faculty: Subject

FAKULTI KEJURUTERAAN ELEKTRIK : MAKMAL Review KEJURUTERAAN Release Date ELEKTRIK Last Amendment Subject Code : SEE 3732/3742 Procedure Number

:1 : 2003 : 2009 : PK-UTM-FKE-(0)-10

SEE 3732 / 3742 FACULTY OF ELECTRICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA SKUDAI CAMPUS JOHOR

INDUSTRIAL ELECTRONIC LABORATORY VOLTAGE REGULATOR

PART A VOLTAGE REGULATOR

OBJECTIVES: 1. To Measure the effects of line and load changes 2. To observe current limiting

BASIC INFORMATION A zener diode is low current device used for voltage regulation. For voltage regulators capable of handling large currents, we need to combine the zener diode with negative-feedback amplifiers. In this experiment you will test a transistorized voltage regulator. Zener-Diode Regulator Figure 1-1 shows a bridge rectifier driving a zener-diode regulator. The dc voltage and ripple across the filter capacitor depend on the source resistance, the filter capacitance and the load resistance. But as long as Vin is greater than Vz, the zener diode holds the output voltage constant. The limitation on a zener – diode regulator is this. Changes in load current produce equal and opposite changes in zener current. The changes in zener current flowing through the zener impedance produce changes in the final output voltage. The larger the changes in zener current, the larger the changes in output voltage. If the changes in zener current are only a few milliamperes, the changes in load voltage may be acceptable. But when the changes are tens of milliamperes or more, the changes in load voltage become too large for most applications.

Vin 240V

R

Vout

RL

Fig. 1-1: Bridge supply drives zener-diode regulator 1

Zener Diode and Emitter Follower The simplest way to increase the current-handling ability of a zener-diode regulator is to add an emitter follower as show in Fig. 1-2. The load voltage still equals the zener voltage (less the VBE drop of the transistor) but the changes in zener current are reduced by a factor of β. Because of this, the regulator can handle larger load currents and still maintain an almost constant load voltage.

Vin R

Vo Vz

RL

Fig. 1-2 Emitter increases current capacity This circuit is an example of a series voltage regulator. The collector-emitter terminals are in series with the load. As a result, the load current must pass through the transistor, and this is the reason the transistor is often called a pass transistor. The voltage across the pass transistor equals. VCE = Vin - Vout

(1-1)

And its power dissipation is PD = (Vin – Vout) Iout

(1-2)

Negative feedback Figure 1-3 is an example of a negative-feedback regulator. Transistor Q2 acts like emitter follower as before. Transistor Q1 provides voltage gain in a negative-feedback loop. Here is how the circuit operates. Suppose the load voltage tries to increase. The feedback voltage VF will increase. Since the emitter voltage of Q1 is held constant by the 2

zener diode, more collector current flows through Q1 and through R3. This reduces the base voltage of Q2. In response, the emitter voltage of Q2 decreases, offsetting almost all the original increase in load voltage. Similarly, if the load voltage tries to decrease, the feedback voltage V F decreases. This reduces the current through Q1 and R3. The higher voltage at the base of Q2 increases the emitter voltage of Q2 and this almost completely offsets the original decrease in the load voltage. Vin

R

3

Q2

Vout R

R

2

RA

Q1

VF

R1

VZ

Fig. 1-3: Negative-Feedback voltage regulator Therefore, any attempted change in load voltage is compensated by negative feedback. The overall effect is to produce an almost rock-solid load voltage, despite changes in load resistance. A mathematical analysis leads to this expression for the output voltage: Vout = ACL (VZ + VBE)

(1-3)

Where VZ is the zener voltage and VBE is the base-emitter drop of Q1. Also, the closedloop gain is ACL = (R2/R1) + 1

(1-4)

This means we can use a zener voltage around 6 V where the temperature stability of the zener diode is optimum. By adjusting the ratio of R2 to R1, we can produce a regulated output voltage with essentially the same stability as the zener voltage. R A in Fig. 1-3 allows us to adjust the output voltage to the exact value required in a particular application. In this way, we can adjust for tolerance in zener voltage, VBE drops, and feedback resistors. 3

Current Limiting The voltage regulator of Fig. 1-3 is a series regulator. As it now stands, it has no short-circuit protection. If we accidentally place a short circuit across the load terminals, we get an enormous current through Q2. Either Q2 will be destroyed or a diode in the power will burn out, or both. To avoid these possibilities, regulated supplies usually include current limiting. Figure 1-4 shows one way to limit the load current to safe values even though the output terminals are accidentally short-circuit. For normal current, the voltage drop across R4 is small and Q3 is OFF; under this condition, the regulator works as previously described. If excessive load current flows, however, the voltage across R 4 becomes large enough to turn ON Q3.The collector current of Q3 flows through R3; this decreases the base voltage of Q2 and reduces the output voltage to prevent damage. Since R4 is 1 Ω, current limiting begins when load current is in the vicinity of 600 to 700 mA. By selecting other value of R4, we can change the level of current limiting. V in

R3 Q2

Q3 R4

R

Vout R2

Q1

R1

VZ

Fig. 1-4: Voltage regulator with current limiting

4

SUMMARY 1) A zener diode can be used to regulate voltage. 2) The limitation on a zener-diode regulator is the amount of current it can handle, typically in the milliamperes. 3) One way to increase the current capacity of a regulator is to add an emitter follower. This increase the current capability by a factor of β. 4) Temperature effects in zener diodes are minimum in the vicinity of 6 V. 5) By using a negative-feedback amplifier and a zener diode, we can build a voltage regulator with excellent voltage stability. 6) Current limiting is needed to avoid the damage that results from accidentally short-circuiting the regulator output terminals.

5

SELF TEST (Prelab) Check your understanding by answering these questions. 1. The limitation on a zener diode regulator like in Fig. 1-1 is its _______________ capacity. 2. When the changes in load current are tens of milliamperes or more in Fig. 1-1, the changes in load __________________ became too large for most applications. 3. To increase the current capacity of a regulator, we can add __________________ emitter follower. This increases the current capacity by a factor of ____________. 4. If R2 = 30 kΩ and R1 = 10 kΩ, then the closed-loop gain of Fig. 1-3 equals _____. 5. If ACL = 3, VZ = 6V, and VBE = 0.7V, then the regulated output voltage of Fig. 1-3 is _______________V. 6. The current-limiting resistor R4 of Fig. 1-4 is changed to 4.7Ω. Current limiting will occur for a load current between _______________ and ________________.

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PROCEDURE MATERIALS REQUIRED Power Supply Equipment Resistors Potentionmeter Rheostat One zener Diode Transistor

: : : :

Adjustable from 0V to 30V Digital multimeter, Milli ammeter 1Ω (2W), 3.3Ω, 3.3 kΩ, 4.3 kΩ, 10 kΩ, 47 kΩ 5 kΩ

: :

IN753 (or any with VZ near 6.2 V) Two 2N3904 (or almost any small-signal NPN silicon Transistor), One TIP 121 Power Transistors.

Calculations Referring to Fig. 1.5(a), derive the expressions for Vout and Io(max) and state any assumptions made during the course of their derivation. Substitute all the values of R 1 and R2 for ACL, VBE2(ON) and VZ for Vout and VBE3(ON) and R4 for Io(max). Fill in Table 1-1. *assume VZ = 6.2V and VBE2(ON)= 0.7V

4.3k

Q2 Q3 V in

3.3

47k I1

3 .3k

Q1 VZ

RA

5k

10k

Fig. 1.5(a): Experimental circuit

7

Vout

Biasing Condition Refer to Fig. 1.5(a), set Vin =20V and adjust RA to get Vout (open circuit) = 10V. Measure the base, the collector and the emitter voltages of Q1, Q2 and Q3 with respect to ground. Calculate the value of VBE, VBC and VCE for each transistor. Fill in Table 1-2.

Adjustment Range Refer to figure 1-5(a), set the supply voltage, Vin= 20V. Measure the output voltage when the potentiometer RA is adjusted to way down minimum and also when it is adjusted to way up to maximum. Fill in Table 1-3.

Effect of Line Change Refer to Figure 1-5(a), set Vin=20V and Vout (open-circuit) to 10V. Plot a graph of Vout (open-circuit) versus Vin. For that purpose you need to measure Vout (open-circuit) for some values of Vin. Adjust Vin to 0V and measure Vout (open-circuit). Repeat the measurement for every increment in Vin until reaches 30V. Fill in Table 1-4.

4.3k

Q2 Q3 3.3

47k I1

3 .3k

Q1 VZ

RA

5k

Rheostat

V in

Vout

10k

Fig. 1.5(b): Experimental circuit

8

1

Effect of Load Change Refer to Fig. 1-5(b). Set Vin = 20V and Vout (open circuit) = 10V, connect the load (i.e. rheostat connected in series with 1Ω resistor) to the output of the regulator. One end of Rheostat terminal must be connected to the regulator output post and one end of the one ohm terminal must be connected to ground post of the regulator. Plot Vout (V) versus Iout (mA) and PC2 (mW) versus Iout (mA). For the purpose you need to measure Vout (V) and calculate the power in transistor Q2 for some measured value of Iout (mA). Given that: PC2 = (Iout+I1) [Vin-Vout-(Iout+I1)R4] (mW) Begin with the highest value of the load resistance. Try with some suitable value of Iout (mA) but not necessarily similar to the suggested values as in Table 1-5. Use Table 1-5 as a guide only. You must do the measurement of Iout (mA) and Vout (volt) from 10V down to almost 0V. To prevent from burns and damage to regulator circuit one should turn off the supply immediately when Vout (volt) reaches 0V. Note: Iout (mA) is measured by measuring the voltage across the 1Ω resistor.

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PART B

THREE-TERMINAL IC REGULATORS

OBJECTIVES 1. To experiment with a fixed Ic regulator 2. To connect an adjustable IC regulator 3. To test a current regulator BASIC INFORMATION Early IC Regulators Figure 2-1(a) shows the pin diagram of the LM300 (metal-can package). By connecting the device into the circuit of Fig. 2-1b, it can be used as a variable-voltage regulator. The external resistor connected to pin 1 is used to set the level where the current limiting occurs. Pin 2, the booster output, is where an external pass transistor is connected to increase the load-current capability. The unregulated input goes to pin 3 while pin 4 is grounded. Pin 5 is for an external capacitor to by pass the zener diode; this reduces the noise it generates. Feedback voltage VF goes to pin 6. We need to connect a compensating capacitor to pin 7; this prevents oscillations. (Oscillation is undesirable ac signals that are internally generated). The regulated output voltage comes out of pin 8.

LIMIT 1 BOOSTER 2

8 OUPUT

INPUT 3

7 COMPENSATION

6 FEEDBACK

GROUND 4 5 BYPASS

Fig. 2-1(a) LM300 pin out

10

1M 47 F 1 8 7 2

V out

6 3

4

5

0.1 F

Vin

Fig.2-1(b) Simple voltage regulator The disadvantage of these IC regulators is the need for external components, plus eight terminals or pins that have to be connected in various ways to get what we want. Ideally, we should not have to connect many external components. Furthermore, an IC regulator thought to be a simple three-terminal device; one pin for the unregulated input, another for the regulated output, and the third for common (ground).

Three-Terminal IC Regulators The second-generation IC regulators are three terminal devices that can supply load currents from 100 mA to more than 3 A. These new IC regulators, available in plastic or metal packages, are easy to use and are virtually blowout-proof. The LM340 series is typical of second-generation IC regulators. Figure 2-2 shows a block diagram. A built-in reference voltage drives the no inverting input of an amplifier. The feedback voltage comes from an internal voltage divider, preset to give output voltage from 5 to 24 V. Specifically, the following voltage are available in the LM340 series: 5, 6, 8, 10, 12, 15, 18 and 24V. The chip includes a pass transistor that can handle more then 1.5 A of load current. Also included are current limiting and thermal shutdown. Thermal shutdown occurs when the internal temperature reaches 175˚C. At this point the regulator turns off and prevents any further increase in chip temperature. Thermal shutdown is a precaution against excessive power dissipation. Because of the thermal shutdown and the current limiting the LM340 series is almost indestructible.

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V in

THERMAL SHUTDOWN ANDSHUTDOWN THERMAL CURRENT LIMITING AND CURRENT LIMITING

+

Vout

-

Fig. 2-2: Equivalent circuit for the LM340 series

A Simple Regulator Figure 2-3 shows the LM340 in its simplest configuration, a fixed-voltage regulator. In this case we have a bridge rectifier driving an LM340-5, a device with a 5-V output (tolerance is typically ± 2 percent). Pin 1 is the input, pin 2 is the output and pin 3 is ground. A circuit like this not only regulates output voltage, it attenuates ripple. The LM340-5 has a typical ripple rejection of 80 dB, equivalent to 10,000. In other word, any ripple at the input is 10,000 times smaller at output.

Vin 240V

1

LM340-5

2

Vout

3

RL

Fig. 2-3: Bridge circuit drives IC regulator

When the IC is more than a few inches from the supply filter capacitance, the inductance of the connecting lead may produce unwanted oscillations. For this reason, a bypass capacitor C1 is often used on pin 1. Typical values for this bypass capacitor are 12

from 0.1 to 1μF. To improve the transient response of the LM340, an output bypass capacitor C2 maybe connected; it is typically around 1μF. An LM340-5 will regulate over an input range of 7 to 20 V. On the other hand, an LM340-24 holds its output at 24 V for an input range of 27 to 38 V. In general any device in the LM340 series needs an input voltage at least 2 to 3 V greater than the regulated output; otherwise, it stops regulating.

Adjustable IC Regulator Figure 2-4 shows external components added to an LM340 to get an adjustable output voltage. The common terminal of the LM340 is not grounded but rather is connected to the top of R2. This means the regulated output VREG is across R1. A quiescent current IQ flows into pin 3. Therefore, the total current through R2 is I2 = IQ + (VREG/R1)

(2-1)

The output voltage is Vout = VREG + [IQ + (VREG/R1)] R2

Vin

(2-2)

LM340

Vout

IQ

R1 I2

R2

Fig.2-4. Adjustable voltage regulator For the LM340 series, IQ has a maximum value of 8 mA and varies only 1 mA over all line and load changes. Often, IQ is negligible compare to VREG/R1, and Eq (2-2) reduce to Vout = (R2/R1 + 1) VREG

(2-3)

Note how the relationship (R2/R1) +1 is similar to the voltage gain of a noninverting amplifier. 13

Current Regulator Figure 2-5 is another application for the LM340. Here load resistor takes the place of R2. As before, the quiescent current IQ flows through RL. Also the current through R1 flows through RL. Therefore the load current is Iout = IQ + (VREG/R1)

(2-4)

Where IQ is negligible, the foregoing reduces to Iout = VREG/R1

(2-5)

As an example, suppose VREG = 5V and R = 10Ω. Then EQ. (2-4) gives an Iout of 508 mA, while Eq. (2-5) gives 500mA. This output current is essentially constant and independent of RL. This means we can change RL and still have a fixed output current.

Vin

LM340 VREG

IQ

I out

R1 RL

Fig. 2-5: Current Regulator SUMMARY 1. The second –generation IC regulators are three terminal devices, which can hold the output voltage constant. 2. The LM340 series is typical of second-generation IC regulators. The regulated voltage LM340 series are from 5 to 24 V. 3. LM340 devices include current limiting and thermal shutdown. 4. When an IC regulator is more than a few inches from the supply, it may be necessary to connect a bypass capacitor across the regulator input. 5. The input voltage to an LM340 device should be at least 2 or 3V greater than the regulated output. 14

SELF TEST (Prelab) Check your understanding by answering these questions. 1. The second-generation of IC regulators has only ___________ pins or terminals. 2. The _____________ series is typical of the second-generation IC regulators. 3. An LM340 device not only regulates the dc output voltage, it also attenuates any __________ at the input. 4. If an IC regulator is more than a few inches from the unregulated supply, a __________capacitor may be needed across the regulator input to prevent oscillations. 5. The ___________voltage to an IC regulation ought to be at least 2 to 3 V greater than the regulated output. 6. If R2 = 100Ω and R1 = 50Ω in Fig. 2-4 then an LM340-6 would produce a Vout of ___________ V. 7. If R1 = 50Ω in Fig. 2-5, then an LM340-8 would produce an Iout of ___________mA.

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PROCEDURE MATERIALS REQUIRED Power supply Resistors Capacitor Equipment IC regulator

: Adjustable from 0-V to 25-V : 1-kΩ ½-W, Rheostat : 1μF/50-V : Digital multimeter, Analog mA meter : LM340-15 (or equivalent – MA7815)

The Effect of Line Change on the Output Voltage

7815 V in

V out

V REG

IQ

R1(1k )

Fig. 2-6(a) Refer to Fig 2-6(a). Output should be open circuit. You are going to plot Vout versus Vin. For that purpose you must measure some input and output voltages and fill on Table 2-1. Table 2-1 serves as a guide only. You may choose more suitable values as the need arises.

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The Effect of Load Change on the Output Voltage

V in IQ

V REG

RHEOSTAT

V out

7815 R1(1k )

1

Fig. 2-6 (b) Refer to Fig. 2.6(b). You are going to plot Vout (V) versus Iout (mA). For the purpose set Vin = 20V and for Iout = 0 mA measure Vout then connect the load across the output of the regulator (i.e. Rheostat in series with 1Ω resistor). Measure the output voltage (Vout) and the output current (Iout). Calculate the power dissipated in IC7815, PC (mW). The output current (Iout) is measured by measuring the voltage drop across the 1Ω resistor. Given that: PC = (Iout)(Vin-Vout)(mW) Begin with the highest load resistance take the measurement and then slide down the value a little bit and take another measurement. Do this bit by bit until the output voltage goes down to almost 0V. You may use Table 2-2 as a guide only. Choose your own suitable output current as it deems appropriate. To prevent from burns and damage to regulator circuit make sure you immediately switch off the power supply when the output voltage reaches 0V.

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