The filter

Chapter The filter

Teach Yourself Electricity and Electronics Third Edition Book
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Teach Yourself Electricity and Electronics Third Edition Book

  • This circuit subjects the diodes to a PIV of 2.8 times the rms ac input voltage.Therefore, they should be rated for PIV of at least 4.2 times the rms ac input voltage.In this circuit, each capacitor charges to the peak ac input voltage when there is noload (the output current is zero). As the load draws current, the capacitors will havetrouble staying charged to the peak ac input voltage. This isn’t much of a problem aslong as the load is light, that is, if the current is low. But,for heavy loads, the output volt-age will drop, and it will not be smooth dc.The major difference between the voltage doubler and the supplies discussed pre-viously, besides the increased output voltage, is the fact that the dc output is filtered.The capacitors serve two purposes: to boost the voltage and to filter the output. Addi-tional filtering might be wanted to smooth out the dc still more, but the circuit of Fig.21-7 is a complete, if crude, power supply all by itself.The filterElectronic equipment doesn’t like the pulsating dc that comes straight from a rectifier.The ripple in the waveform must be smoothed out, so that pure, battery-like dc is sup-plied. The filter does this.Capacitors aloneThe simplest filter is one or more large-value capacitors, connected in parallel with therectifier output (Fig. 21-8). Electrolytic capacitors are almost always used. They are po-larized; they must be hooked up in the right direction. Typical values range in the hun-dreds or thousands of microfarads.390 Power supplies21-8A simple filter. Thecapacitor, C, should havea large capacitance.The more current drawn, the more capacitance is needed for good filtering. This isbecause the load resistance decreases as the current increases. The lower the load re-sistance, the faster the filter capacitors will discharge. Larger capacitances hold chargefor a longer time with a given load.Filter capacitors work by “trying” to keep the dc voltage at its peak level (Fig. 21-9).This is easier to do with the output of a full-wave rectifier (shown at A) as comparedwith a half-wave circuit (at B). The remaining waveform bumps are the ripple. With ahalf-wave rectifier, this ripple has the same frequency as the ac, or 60 Hz. With afull-wave supply, the ripple is 120 Hz. The capacitor gets recharged twice as often witha full-wave rectifier, as compared with a half-wave rectifier. This is why the ripple is lesssevere, for a given capacitance, with full-wave circuits.