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EXP NO: 5 Full Wave Rectifier Information section Rectifier circuits are used to convert an ac voltage into a dc voltage. The current always flows in the same direction through the load resistor of a rectifier circuit. However, its magnitude changes: 1. Two-pulse midpoint circuit M2U The two-pulse midpoint circuit is made up from two single-pulse midpoint circuits. It uses both half-waves of the input voltage. A transformer with center tapping (fig. 1) is required for this purpose. The current is pulsating dc. Singlepulse midpoint circuit M1U From fig. 1 it is clear that the two component voltages VTr / 2 of a two-pulse midpoint circuit are opposed with respect to the center tapping. Either V1 or V2 conducts depending on the polarity of the input voltage (positive resp. negative halfwave). The current IL through the load resistor RL is made up of the two component currents through V1 and V2. These component currents are called branch currents. 1 2. Two-pulse bridge circuit B2U The two-pulse bridge circuit (fig. 2) is the simplest and most commonly used full-wave rectifier circuit (fullwave circuit - the load current flows through the primary winding in both directions) for small power classes (up to approx. 2 kW). It consists of a parallel connection of two series circuits (each containing two diodes). The external terminals of the diode pairs are connected together to form the dc terminals of the rectifier set. The center terminals of the diode pairs are connected to the ac voltage source. The diodes V1, V4, resp. V2, V3 are alternatively connected in forward and reverse directions in the bridge circuit (fig. 0.9). The effective value (r.m.s. value) of the signal which is: Vr (rms) 0.308Vm (full wave) The r.m.s. value of the ripple voltage can be determined approximately using the expression The average value of the output voltage VA in case of full- wave rectification, this is given by: Vo av 2 / 2 Vrms 0.636Vm Vdc 2 3. Smoothing and filtering Ripple voltage and hum frequency all rectifier circuits supply pulsating dc voltages, i.e. the dc voltage formed behind the mains rectifier is superimposed by an ac voltage. This ac voltage is designated the ripple voltage Vripple (Vripple = rms value). The frequency of the ripple voltage is called the hum frequency f hum. The maximum value of Vout is given by: F , Pmax , out The value of the ripple voltage is dependent on - the rectifier circuit used (circuit constant k), - the magnitude of the load current IL and - the capacitance of the load capacitor CL The load or smoothing capacitor in fig 3 is charged via the bridge rectifier and discharged via the load resistor as soon as the capacitor voltage V1 is greater than Vac - 2·VF. It serves as a voltage source, i.e. it smoothes the voltage at the load resistor and increases the mean dc voltage, during intervals when no current flows through the diodes. CL is calculated using the following expression: 4. 3 4. Filtering Should the load capacitor be insufficient to smooth the output voltage, a filter element (low-pass) can be connected between the load capacitor and load resistor to reduce the ripple voltage at the load resistor. An LC filter is used for large load currents and an RC filter for small currents (fig.4). The filtering factor S is given by the ratio of the ripple voltages at the input of the filter to that at it’s output: The resistor RS (coil LS) and the capacitor CS form a frequency-dependent voltage divider. The capacitor has a low reactance value for the ripple voltage. The ripple current flowing through the voltage divider causes a voltage drop across RS (LS). Thus the voltage across the load resistor is reduced by the voltage drop across RS (LS). RC Filtering The following is valid for XCS << RL 4 Full-Wave Rectification Objectives: To calculate and draw the DC output voltages of half-wave and full-wave rectifiers. Without smoothing capacitor and with smoothing capacitor Equipment: AC power supply or Function Generator 2 Voltmeters Function Generator Oscilloscope Components Diode: Silicon 4×(D1N4002) or Silicon Bridge rectifier Resistors: 10 kΩ, Capacitor :( 0.47 F) Capacitor :( 4.7 F) Electrolytic Capacitor :( 2×100 F) Fig. 5 1. Construct the circuit of Fig.5. Set the supply to 10 V p-p with the frequency of 50 Hz. Put the Channel 1 of the oscilloscope probes at function generator (trainer board) and sketch the input waveform obtained. 2. Put Channel 1 of the oscilloscope probes across the resistor and sketch the output waveform obtained. Measure and record the DC level of the output voltage using the (Dc bottom in the C.R.O.) oscilloscope. 3. Keeping the x-y button of the C.R.O. at outside position i.e. do not press it. 4. Adjust the power supply unit at 10 V, then a sinusoidal or another type from function generator input voltage Vi will be displayed on the screen of the C.R.O. 5. Measure the input signal voltage Vinput and the Voutput signal voltage Vo using both the voltmeter and C.R.O. V input = 10 volts With voltmeter With oscilloscope V (input) (Volt) 10 Table 1 5 Va.c Vp-p VA (output) (Volt) Vd.c VP-P 6. Draw the input waveform, Vi : and the output waveform, Vo : 7. Calculate the effective value of the input signal Vrms which is give by: Vrms = Vpp / (2√2) = Vm / √2 8. Transfer the graphs into a diagram , and determine the amplitude (peak value) Vin and the frequency f of the Vin(t). 9. Determine the amplitude and the frequency of the output voltage. 10. The second of Kirchhoff's Laws is used to calculate the output voltage: VA (T) = V MAX INPUT – 2× 0.7 V V(th): threshold voltage across the diode for Si diode=0.7 V Smoothing and filtering Objectives Representing the ripple voltage on the load voltage Determining the ripple voltage as a function of the charging capacitor and the load resistor Measuring and calculating the r.m.s. value of the ripple voltage Fig. 6 1. Assemble the circuit as shown in Fig. 6 and apply an a.c. voltage 10 V, f = 50 Hz. 2. Use channel 1 of the oscilloscope to measure the voltage Vo across the load resistor. 3. Record the settings of the oscilloscope: Y1 = --- volts/div (dc) t, X = --- ms/div. trigger switch to line. 4. Insert smoothing (filter) capacitor C = 0.47F, 4.7F,100 F; in the circuit to terminal parallel to the load one after other record the output voltage as shown in Fig.5, 5. determine the ripple voltage as a function of the charging capacitor and the load resistor 6. Connect an ammeter between terminals 1 and 3 in Fig.6 to measure the d.c. current. 6 7. Measure the d.c. component of the load current IL and the peak-to-peak value of the ripple voltage Vr pp for the combination of charging capacitor and load resistor as a function of the capacitance value of the smoothing capacitor C=0.47F, 4.7F, 100 F and at the same time measure the ripple voltage Vr pp using CRO and Record the values. 8. The r.m.s. value of the ripple voltage can be calculated using expression Vr ( rms ) I dc 2.4 I dc 2.4Vdc . C RL C 4 3 fC Where Idc in mA, C is in F, and RL is in k 9. Calculate the ripple frequencies for the values given in table 5 and enter your results into table 2. V (rms) 1 1 ( ) 10. Calculate the ripple factor r using the following equation: r r Vdc 2 3 F R LC 11. Draw the output signal voltage in each case of using C values. 12. Comment on the results you obtained. C (F) 0.47 4.7 V2 (voltmeter) (Volts) Vp-p (C.R.O. volts) I dc ( mA) T (msec) F = l/T (Hz) 100 Table 2 Questions: 1. Discuss the mechanism of operation of the Full-wave rectification using Si-bridge rectifier. 2. What do you notice from the values of the average output voltage in both; the half- and full wave rectification circuits? Comment. 3. Describe the ripple voltage dependence on the charging capacitor and the load resistor R L. 4. Is it possible to display both input and output voltages simultaneously with channel 1 resp. channel 2 give reasons for your answer in Full wave rectifier circuit (fig. 6) that you used? 5. Determine the peak reverse voltage across diode V1in full wave rectifier circuit (fig. 6) that you used. 7