PHYS 536 DC Power Supply

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1 PHYS 536 DC Power Supply Introduction The process of changing AC to DC is investigated in this experiment. An integrated circuit regulator makes it easy to construct a high-performance voltage source using only four parts: a transformer, full-wave bridge rectifier, capacitor filter, and regulator. A zener diode regulator is also included as a simple illustration of regulator properties. Rectifier Figure 1: The bridge rectifier The proper arrangement of the 1N914 diodes for full wave rectification is shown in Fig. 1. The purpose of the rectifier is to convert the AC voltage from the transformer in to DC voltage. This is accomplished using the diodes. An ideal diode will pass current when forward bias, but no current when reverse biased. A real diode will, however, pass current when the breakdown voltage is exceeded. 1

2 However, this destroys the diode. A real diode will result in an 0.8 V drop across the diode, while an ideal diode would not drop any voltage. Because the diodes will only allow current flow in one direction, when the top AC input is positive with respect to the bottom input, current will flow through the top right diode, through the load attached to the right side of the rectifier and exit the load on the left side of the rectifier, and finally pass through the bottom left diode and leave through the bottom AC input. When the bottom AC input is positive with respect to the top input, current will pass through the bottom right diode, through the load, and out the top left diode. During both sequences, current always flows through the load in the same direction, and so the current provided is DC current. Rectification The peak voltage, V p, from the rectifier is slightly less than the peak transformer voltage, V T because voltage is dropped across it the diode, so V p = V T 2V d (1) V d is the voltage drop across the diode, typically 0.8 V, and in the bridge rectifier the current flows through two diodes. Since the voltage of the transformer secondary is always specified as an effective or RMS voltage, V e, and is related through the peak voltage by the following relation V T = 2V e Ripple Voltage If the rectified wave is applied to an RC circuit, it will take time to discharge the capacitor, and so smoothing of the rectified wave will occur, assuming that the time constant of the RC circuit is chosen such that RC 1 f, so that the time constant is large compared to the period of the rectified wave. For a capacitor, Hence, i = C dv dt dv = i C dt 2

3 The ripple voltage can be estimated, then, by V r = i c Δt (2) C For half wave rectification, the capacitor will continue to charge throughout the period (T = 1 f ), of the original wave, so the ripple voltage can be estimated as V r = i c fc For full wave rectification, the capacitor only discharges until the voltage begins increases again, so the capacitor only discharges for half the period of the original wave (T = 1 2f ), so V r = i c 2fC This relationship is, however, only approximate, since once the rectified voltage becomes larger than the capacitor voltage, the capacitor will stop discharging and instead increase its voltage as it gains more charge, so the capacitor is not discharging over the entire half period, so a better approximation to the ripple voltage is V r = 0.7i c 2fC (3) By choosing a large capacitance the ripple can be reduced to any desired level, however this brute-force approach has several drawbacks. First of all, the capacitors may be prohibitively large and/or expensive. Another drawback is that even though the ripple may be negligible line voltage variations are transferred to the output voltage. A better approach is to regulate the output voltage. The current drawn from the capacitor depends on the type of the regulator used. We will consider two types of regulators: IC and zeners. V o is the output voltage of the regulator and R L is the external load resistor connected to the regulator, i.e. it is the resistance of the device or load to which power is being transferred. For an IC regulator and for a zener I c = V o R L (4) I C = V c V z R 2 V A V Z R 2 (5) 3

4 where V c is the average capacitor voltage. Assuming V r V A,thiscanbe assumed to be V A. When the load resistor is attached directly to the capacitor without a regulator the current is given by the relation Zener Regulator I C = V c R L V A R L (6) Figure 2: Electrical characteristics of a zener diode A zener diode is similar to a regular diode with one exception, when a reverse bias is applied, the diode breaks down in a controlled fashion, so that the current through the resistor, i Z, is greater than zero and the voltage across the resistor is V Z V ZK as shown in Fig. 2. While ideally the voltage would remain unchanged, there is some small internal resistance in the zener, so the zener under reverse bias is best described as an ideal diode in series with a voltage source and small resistance r z as shown in Fig. 3. This proves useful because, so long as the zener diode is reverse biased, it provides a nearly constant voltage. The output voltage of the zener is given by r z R 2 V o = V z + V C I 0 (r z R 2 ) (7) R 2 + r z R 2 + r z Note that for R 2 r z and R L R z, V o V z 4

5 Figure 3: Piecewise Linear model of a zener diode Where r z is typically of the order of a few ohms. Note that under eq. 7, a ripple in the capacitor voltage will result in a ripple in the output voltage, and differentiating this equation, r z dv o = dv c (8) R 2 + r z The third term in eq. 7 also indicates that the output current affects the output voltage, as given by dv o = di o (r z R 2 ) If the load resistance draws too much current, the zener diode will drop out of conduction and act as an open circuit, at which point the zener will cease to act as a regulator. The maximum current that may be delivered is 5

6 I o,max = V c,min V o R 2 I z,min (9) Where V c,min is the minimum capacitor voltage (the peak voltage less the ripple voltage) and I z,min is the minimum current required by the zener diode to maintain conduction. IC Regulator When using a voltage regulator, a minimum voltage, V i,min must be maintained across the input pin and ground pin. For the 7805 regulator, the required voltage is 7.2 V. To insure that the voltage remains above the minimum required voltage, the capacitor voltage must follow the condition V c,peak ΔV V i,min (10) IC Regulator Ripple Rejection The ripple rejection for an IC regulator is specified in db of attenuation. ( ) ΔVo RR(dB) = 20 log ΔV i (11) Solving this for the output voltage ripple, ΔV o =ΔV i 10 RR 20 (12) IC Regulator Temperature Drift The power dissipated by the voltage regulator is given by P D = I o (V i V o )+V i I Q (13) where I Q is the quiescent current of the regulator, which is 6 ma for the 7805 regulator. The IC regulator s junction temperature is a function of this power dissipation. The junction temperature is T J = T A + θ JA P D (14) 6

7 For the 7805 regulator, θ JA =27.3 C W. Changes in the junction temperature will result in a shift in the regulator s output voltage. The shift in output voltage is given by The value of ΔVo ΔT ΔV o = ΔV o ΔT (15) ΔT mv for a 7805 regulator is -0.8 C. 1 Basic Power Supply Figure 4: Initial circuit Figure 5: Initial circuit with the capacitor 1. Begin by testing your 4 1N914 diodes. Do this using the multimeter. On the multimeter, there should be a symbol which looks like a diode symbol. Turn your multimeter to this setting. Attach the leads of the multimeter to each diode, alternating the position of the leads. If the diode is functioning properly, the multimeter should read 0.7 V in one 7

8 direction, and an overload in the other. If any diodes do not pass this test, discard them and replace them with another diode. 2. The desired peak voltage across resistor R 1 is 19.6 V. Calculate the necessary transformer voltage using Eq Do not connect the provided variac and 5:1 step down transformer to the circuit yet. Using the multimeter, set the voltage to the desired transformer voltage calculated in the previous step. Keep in mind that the multimeter measures RMS voltage, not peak to peak voltage. 4. Before building the circuit, draw the shape of the expected voltage across the resistor. 5. Build the circuit in Fig. 4. Use R 1 = 1 kω using the variac and 5:1 transformer as the source AC voltage. 6. Measure the voltage across the resistor. Is it the expected 19.6 V? Is the waveform as expected? If the voltage across the resistor is not 19.6 V, adjust the voltage until it is. 7. Calculate the expected ripple voltage for the circuit in Fig. 5 if R 1 =1kΩ and C 1 = 100 μf. 8. Build the circuit in Fig. 5 with R 1 =1kΩandC 1 = 100 μf. Be careful when inserting C 1, which is an electrolytic capacitor. For these capacitors, orientation matters, so be sure that the side connected to ground corresponds to the side of the capacitor marked with a strip and negative symbol. 9. While observing the voltage across the resistor, remove and replace the capacitor in the circuit. What effect does the capacitor have on the output voltage? 10. Measure the ripple voltage. How does it compare to the calculated value? 2 Zener Voltage Regulator A voltage regulator can also be built using a zener diode which is reverse biased. Under reverse bias, the zener provides a nearly constant output voltage. The zener diode does have a small resistance, however, r z. For the computations in the section, assume r z =10Ω. 1. Initially, no load resistance is used. With no load resistance, V C =17V, R 2 = 1 kω, and the assumed r z = 10 Ω, calculate the capacitor ripple voltage (ΔV ), and the output ripple voltage (ΔV o ). Because the ripple voltage will be small, it can be assumed that V C V C,peak. 8

9 Figure 6: Zener Voltage Regulator 2. Build the circuit in Fig. 6 using V C =17V, R 2 =1kΩ. Donoconnect any load resistor initially. 3. Measure the ripple voltage ΔV and ΔV o. From these measured values, calculate r z. 4. Calculate the change in V o that would result from an output current of I o =5.2 ma. 5. Use the multimeter to measure the output voltage V o without a load resistor.insert R L =1.2 kω. Measure the output voltage again with the load resistor in place. How did the output voltage change? 6. Calculate r z using the measured I o and ΔV o from the previous step. 7. Calculate the maximum I o that will leave a minimum of 2 ma of current flowing through the zener diode. If the zener diode current falls below this value, the diode will drop out of conduction. 8. If R L is replaced with a 470 Ω resistor, the zener diode will drop out of conduction. Calculate the expected V o with this load resistor. 9. Using R L = 470 Ω, measure V o. 3 Power Supply with an IC Voltage Regulator Figure 7: DC power supply using an IC voltage regulator 9

10 1. The 7805 IC voltage regulator requires an input voltage of 7.2 V to function correctly. If the voltage supplied by the capacitor drops below this voltage, the regulator will not work correctly. Calculate the minimum peak capacitor voltage that will keep the input voltage above 7.2 V. Assume C 1 = 100 μf andr L = 100 Ω. The output voltage for the 7805 regulator is 5 V. 2. Build the circuit in Fig. 7 using C 1 = 100 μf andr L = 100 Ω. Adjust the variac until V C,peak is 2 V larger than the value calculated in the previous step. 3. Measure the ripple voltage across the capacitor, ΔV. 4. With both V C and V o shown on the oscilloscope, decrease the variac voltage until you see a ripple voltage appear in the output voltage. When the ripple appears in the output voltage, the regulator has stopped functioning as a regulator. Measure the V C at which this occurs. 5. Estimate the change in the V o caused by changing the input voltage from 7 V to 25 V with a load resistance R L =1kΩ. 6. Replace the current load resistor with R L = 1 kω. Measure V o with an input voltage of 7 V, and with an input voltage of 25 V. Take both measurements as quickly as possible to minimize temperature drift caused by ohmic heating of the regulator. How does the change measured compare to that calculated in the previous Required Components Diodes: (4) 1N914, (1) 1N4375 Capacitors: (1) 100 μf, (1) 0.1 μf Resistors: (1) 100 Ω, (1) 470 Ω, (1) 1 kω, (1) 1.2 kω ICs: (1) 7805 voltage regulator 10

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