# CAPACITOR FILTERS Electronics Help

The frequency of the fundamentaL component of a half-wave-rectified waveform is the same as the frequency of its original (unrectified) ac waveform. Since the ac power source used in most de supplies has frequency 60 Hz, a half-wave-rectified  waveform contains a 60-Hz fundamental, plus harmonic componentmultiples of 60 Hz, plus’ de, or average value, component. A ft. l-wave=rectified waveform has one-half the period of a half-wave-rectified waveform, so it has twice the frequency. It therefore contains a 120-Hz fundamental, plus harmonics of 120 Hz, plus a de component that is twice that of the half-wave-rectified waveform, as we have seen. A low-pass filter is connected across the output of a rectifier to suppress the ac components and to pass the dc component.

A rudimentary low-pass filter used in power supplies consists simply of a capacitor connected ac oss the rectifier output, that is, in parallel with the load, as illustrated in Figure 17-10. Here, we show a simple half-wave rectifier with capacitors that are integer C connected in parallel with Rl.’ The forward resistance of the diode is small in comparison to RL, so during positive half-cycles, the capacitor charges to the peak value of chc ac input.
h i:- lot convenient to 31)”iyze the circuit it’ Figure 17-10 in terms of filter theory because the nonlinear operation of the diode effectively changes the circuit resistance as the diode is alternately forward and reverse biased. Instead, we analyze It from u c stand, _'”t of the transient voltage across the capacitor. Figure 17-11 (a) ‘;1(IWS how the capacitor charges and discharges during a full cycle of the at. input. Notice that the capacitor charges and its voltage rises with the input voltage when the input is large enough to forward bias the  iode. The  capacitor discharges through RI. when the input falls to a level below which the diode is rf”’erf>(: biased. The smaller the R1.C time constant, the further the capacitor voltage -!.:.’ } before another positive pulse arrives and recharges the capacitor.

Compares the output waveforms that result when large and small time constants are used. The voltage fluctuation in the filtered waveform is called the ripple voltage, which in most applications should be kept as small as po sible. Figure 17-ll(b) shows that a heavy load (small Rd will result in an undesirably -large ripple voltage.

A full-wave bridge rectifier with a capacitor filter. Also shown is the” form. In this case, positive [.’1/1 t’ are present during every half-cycle of input, so the capacitor voltage does not decay as far a” it does in the half-wave circuit before another pulse is available to recharge it. As a consequence, the peax-to-oeak ripple voltage is smauer, ior a given RLC time constant, than it is in the half-wave circuit. Note carefully that the r-umerator is the rms value of just the ripple (1\C components) in the rectified waveform, not the rm,; value of the total was· form . Let us first compute the percent ripple in unfiltered half- ” «d full-wave-rectified waveforms. Using calculus, it can be ‘shown that the fins values of the (7C components of half- and full-wave-rectified voltages are

where VI’H i–the peak value of the rectified waveform in each case. (Note carefully that equations 17-7 and 17-8 give the rms values of the ac components only, not the rms values of the waveforms themselves.) ThenThese ripple values are quite large and demonstrate the need for a Iilter in most power-supply applications. If a filter is not used, or if it is improperly designed, the ac components of the ripple are superimposed on the signal lines in the device that receives power from the supply. Power-supply ripple is therefore a soarce of noise in an electronic system. It is one of the principal causes of 60-Hz ( r 120-Hz) hum in an audio amplifier.We

will now derive an expression for the ripple in the output of a rectifier
having a capacitor filter and load resistance Ri. (Figures 17-11 and 17-12). Since ;\ knowledge of ripple magnitude is important in only those applications where rl; pic affects the performance of a system using the power supply, w e assume that the filter is well designed and that, as a consequence, the ripple voltage is small compared to the de component. In other words, we assume that the capacitor voltage does not decay significantly from its peak value between the occurrences of the rectified pulses that recharge the capacitor. This circumstance is called light loading because the charge supplied to RL by the capacitor is small compared to the total charge stored on the capacitor. To simplify the computations, we can assume that the ripple voltage in a lightly loaded filter is a sawtooth wave, as illustrated.

The waveform in a lightly loaded capacitor tilter across the output of a full-wave rectifier. The derivation that follows is applicable li both half-wave and full-wave rectifiers, provided the assumption of light loadin, IS valid for both cases. Note that the period Tshown in the figure is the fundamental period of the rectified waveform, typically 1/60 s for half-wave rectifiers and 11120 s for full-wave rectifiers. The light-loading assumption in the half-wave case is more restrictive than in the full-wave case because there is a longer time between pulses, during which the capacitor voltage can decay. The ripple waveform approximated by a sawtooth voltage with peak-to-peak value V/,/” This approximation is equivalent to assuming that the capacitor charges instantuncolOly and that its voltage decays IinearJy jn.sleao of exponenlj;)JJy. A.f.Iumingtha.<:(ne vo(fage decays linearly is equivalent to assuming that the discharge current is constant and equal to V”..IRI., where Vdc is the de value of the filtered waveform. A~ shown ill (c), the total change in capacitor voltage is VI•I• volts, and this change OCL:lllS over the period of time T. There fore. since ~Q = 11f,

Since T = 1/f,. where f, is the frequency of the fundamental component of the ripple (typically 60 Hz or 120 Hz), equation 17-11 can be writtenIt is clear that the de value cannot exceed VI’R volts, and that it equals V”R when R, = 00 (i.c., when the load is an open circuit).The rills value of a sawtooth waveform having peak-to-peak value VI’/’ is known Equation 17-18 confirms our previous analysis of the capacitor filler :A large R,.C time constant results in a small ripple voltage, and vice versa. The light-load assumption on which our derivation is based is generally valid for percent ripple less than about 6.5%. From a design standpoint, t he values of I, and R,. are usually fixed, and the designer’s task is to select a value of C that keeps the  ripple below a prescribed value

A full-wave rectifier is operated from a 60-Hz line and has a filter capacitor connected across its output. What minimum value of capacitance is required if the load is 200 n and the ripple must. be no greater than 4%?

Solution Using the decimal form of r (r = 0.(4), we find, from equation 17-IH, The rectifier shown in Figure 17-10 is operated from a 60-Hz, 120-line, It has a 100-J.LF filter capacitor and a I-kfl load.
1. What is the percent ripple?
2. What is the average current in RL?

Solution

1. We must first assume that the filter is lightly loaded, then perform the computation based on that assumption, and then verify from the result that the assumption is valid. From equation 17-18, The full-wave rectifier in Figure 17-6 has RI. = 120 1. A 100-J.LFfilter capacitor is connected in parallel with RI.’ The input is a 60-Hz sine wave .having peak value 30 V. The filter does not satisfy the lightly loaded criterion •-so the equations for approximating the ripple voltage are not applicable. Use SPICE to deterrnind ; the peak-to-peak ripple voltage.  Solution. Figure 17-14(a) shows the SPICE circuit and input data file. Note that VIN is not connected to ground (node 0), since it must be applied across nodes 1 and 2. One-half the period of the 60-Hz input is 8.33 ms, so the .TRAN and .PLOT statements will produce a plot of the output extending over the first two half-cycles of input (0 to 17 ms). The results are shown in Figure 17-14(b). We see that the minimum and maximum values of the second pulse are 17.98 V and 28.40 V, respectively, so the peak-to-peak ripple voltage is 28.40 – 17.98 V = 10.42 V.

Posted on November 19, 2015 in Power Supplies and Voltage Regulators