LABORATORY 10 TIME AVERAGES, RMS VALUES AND THE BRIDGE RECTIFIER. Bridge Rectifier

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1 LABORATORY 10 TIME AVERAGES, RMS VALUES AND THE BRIDGE RECTIFIER Full-wave Rectification: Bridge Rectifier For many electronic circuits, DC supply voltages are required but only AC voltages are available. Then, the required DC voltages are obtained from the AC voltages by rectification and filtering. In the rectification process, an AC current is converted to a time-varying current that flows in a single direction, and so it is a time-varying DC current. Subsequent filtering smoothes out the variations to produce an almost constant dc current and voltage. The diode bridge rectifier circuit is shown in Fig The operation can be understood by considering the polarity of the applied voltage V i at the top node a, with respect to the bottom node b. When this voltage is positive, diodes D 1 and D 2 are forward biased and therefore conduct, thereby causing the output voltage V o to be equal to the input voltage V i. (For simplicity, the small voltage drops across the conducting diodes are neglected.) When, however, V i is negative, node b is positive with respect to node a and thus diodes D 3 and D 4 conduct, making V o = -V i. Since, however, V i is negative, V o is positive. Thus, V o is always positive and is equal to the magnitude of V i. That is, V o = V i. Fig Diode Bridge Full-Wave Rectifier Suppose that the input voltage is V i = 10 sin ωt V. Then, if V T = 0, the input and output voltage waveforms will be as shown in Fig by the dotted curve. Actually, though, for a real bridge rectifier, the output waveform will be shifted down by 2V T because of the diode voltage drops. This is shown by the solid curve in the bottom figure Also, because of this shift, the output waveform will be zero for a short time between the rectified half cycles. This type of

2 rectification is called full-wave rectification because both half cycles of a cycle of the input contribute to the output. Fig Now, suppose that in addition to a sinusoid, the input voltage contains a DC component, as can be obtained from the sources shown in Fig What then is the output voltage V o? Fig Again, the key to understanding the circuit operation is the polarity of the voltage drop from node a to b, which is V ab = 10 sin ωt + 3 V. When V ab is positive, diodes D 1 and D 2 conduct. When this voltage is negative, only diodes D 3 and D 4 conduct. Thus, the input and output voltage waveforms are as shown in Fig

3 Fig Finally, suppose in the circuit of Fig. 10.1, that a capacitor is placed across the load resistor. What would the V o waveform be then? The answer is that when V i increases for the first time to its positive peak, the capacitor would charge to this peak value minus 2V T. Then, when V i started to decrease, the capacitor could not discharge through the diodes because that would require reverse diode currents. The capacitor would, however, start discharging through the load resistor R L. Then, if the R L C time constant was much greater than the duration of a half cycle of the sinusoidal input voltage, the capacitor voltage would not decrease as fast as the sinusoidal voltage, and thus would reverse bias the diodes. The diodes would remain reverse biased until the input sinusoidal voltage (minus 2V T ) exceeded the capacitor voltage. In the meantime, though, the load voltage would be the slow exponentially decaying capacitor voltage as is shown in Fig Clearly, the resulting load voltage is smoother than a full-wave rectified sinusoid. Fig Procedure: 1. Construct the circuit shown in Fig Be careful to connect the diodes so that they conduct in the directions indicated. Diode casings usually have a stripe at one end to designate the end corresponding to the bar in the diode circuit symbol (the cathode end). The 2.2-kΩ resistor is included to limit the current to a safe value. Apply a 5-V peak, 60-Hz sine wave and, using an oscilloscope, observe the voltage across the a-b terminals. Examine the waveform, including amplitude and time values.

4 Fig Place a 0.1-μF capacitor across (in parallel with) R L, and reexamine the voltage waveform across the a,b nodes. 3. Add a 1.5 V offset voltage to the function generator. This has the same effect as the configuration in Fig Using the oscilloscope, again observe the voltage waveform across the a,b nodes. 4. Repeat step 2. Fig. 10.7

5 Time Averages and RMS Values In electrical circuits, one frequently encounters periodically varying current and voltage waveforms. If the waveform x(t) repeats itself every T seconds, then x(t) = x(t + NT) where N is an integer and T is called the period. Figures 10.8 (a)-(d) show several periodic waveforms. Figure 10.8 (a) sinusoid, (b) triangular, (c) square, (d) half-wave rectified waveforms The average value of a periodic waveform is defined mathematically as X ave = < x(t) > = { 0 T x(t) dt}/t. The integration represents the area under the curve during one period. Thus, it can be seen that the average of a sinusoid is zero. Prelab Question: 1. Determine the time averages of waveforms in Figures 10.8(b), (c), (d). The Root-Mean Square of a Periodic Waveform: We saw earlier that the time average of one of the most important periodic waveforms, the sinusoid, is zero. However, the average itself is of little importance. Since power absorbed or provided by any element in a circuit depends on the square of the current or voltage, average power can be computed by knowing the time average (mean) of the square of the periodic waveform.

6 < x 2 (t) > = 0 T x 2 (t) dt/t. The root-mean square (RMS) value is simply the square root of the mean of the square. X rms = < x 2 (t) > Note: AC voltmeters and ammeters measure the RMS value, not the peak value. The RMS value is sometimes also referred to as the effective value. Prelab Questions: 2. Show that the RMS value of the sinusoid x(t) = A cos(ωt + θ) is A/ 2. Hint: Use trigonometric identity: 2(cosα) 2 = 1 + cos2α in evaluating the integral. 3. Determine the RMS value of the waveform shown in Figure 10.8(d). 4. What would be the RMS value of a full-wave rectified waveform? Give your answer in terms of the amplitude of the input sinusoid. Ripple Filtering: The full-wave rectified waveform shown in the bottom curve of Figure 10.2 presents two problems: 1. The time-average value (the DC value) of the output is much lower than the amplitude of the input sinusoid. 2. The output oscillates around the DC value, an undesirable feature for most applications (Imagine it running the DC motor in a portable shaver). The full-wave rectified waveform can be considered as made up of two parts: the DC part and the oscillating part. The oscillating part is called the ripple. where V DC = < v o (t) > v 0 (t) = V DC + v ripple (t). The ripple can be reduced (filtered) by using an appropriate capacitor connected across the load resistor R L (refer to Figure 10.6 for the load resistor). This is explained in the text preceding Figure The filtered waveform is shown in Figure 10.5, along with the unfiltered waveform. The filtered waveform is much smoother.

7 DC Value of the Filtered Waveform and Ripple Factor: At t = T/4, the capacitor is fully charged (to the peak voltage of the input sinusoid, V p ). Then, it discharges via the load resistor with a time constant τ = R L C. Since the capacitor is chosen to ensure that τ >> T, the exponential decay of the capacitor voltage continues almost until t = 3T/4. Therefore, in the interval T/4 t 3T/4, the waveform is given by The minimum occurs at t = 3T/4 and is given by v 0 (t) = V p exp{ (t T/4)/τ}. V min V p exp( T/2τ) V p (1 T/2τ) The ripple factor (rf) and the DC voltage are given by the following equations: rf = (V p V min )/V DC V DC = (V p + V min )/2 = V p (1 T/4τ) = V p (1 1/4fR L C) Prelab Question: 5. Determine the ripple factor for f = 60 Hz, R L = 1 MΩ, and C = 0.1 μf. Procedure: For the circuit in Fig 10.6 with a capacitor added in parallel to the load resistor: 1. Measure the peak value and the minimum value in the rectified output. 2. Measure the average voltage (V DC ) of the rectified output using the Oscilloscope. 3. Determine the ripple factor. 4. Compare the DC voltage and ripple factor with the respective theoretical values.

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