This effect of increasing reservoir size on diode and transformer current should be born in mind during any servicing operations replacing the reservoir capacitor with a larger value than in the original design "to reduce mains hum" may seem like a good idea, but could risk damaging the rectifier diode and/or the transformer. There is an advantage therefore in reducing the value of the reservoir capacitor, thereby allowing an increase in the ripple present, but this can be effectively removed by using a low pass filter and regulator stages between the reservoir capacitor and the load. This means using a higher current rating for the diodes and the transformer than would be necessary with a smaller reservoir capacitor. Both the input transformer and the rectifier diodes must be capable of supplying this current. If the capacitor is very large, its voltage will hardly fall at all between charging pulses this will produce a very small amount of ripple, but require very short pulses of much higher current to charge the reservoir capacitor. Therefore the shorter the charging time, the larger current the diode must supply to charge it. The charge (Q) on a capacitor depends on the amount of current (I) flowing for a time (t). The formula relating charge, time and current states that: This current partly discharges the capacitor, so all of the energy used by the load during most of the cycle must be made up in the very short remaining time during which the diode conducts in each cycle. The capacitor supplies the load current for most of the time (when the diode is not conducting). To obtain the least AC ripple and the highest DC level it would seem sensible to use the largest reservoir capacitor possible. Adding the capacitor increases the DC level of the output wave to nearly the peak value of the input wave, as can be seen from Fig. The DC output of the rectifier, without the reservoir capacitor, is either 0.637 Vpk for full wave rectifiers, or 0.317 Vpk for half wave. Typically the peak-to-peak amplitude of the remaining AC (called ripple as the AC waves are now much reduced) would be no more than 10% of the DC output voltage. The higher the load current, the more the discharge, but provided that the current drawn is not excessive, the amount of the AC present in the output is much reduced. The amount by which the reservoir capacitor discharges on each half cycle is determined by the current drawn by the load. At some point during the next cycle of the mains input, the rectifier input voltage rises above the voltage on the partly discharged capacitor and the reservoir is re-charged to the peak value Vpk again. Of course, even though the reservoir capacitor has large value, it discharges as it supplies the load, and its voltage falls, but not by very much. The load circuit is now supplied by the reservoir capacitor alone (hence the need for a large capacitor). Once the input wave passes Vpk the rectifier anode falls below the capacitor voltage, the rectifier becomes reverse biased and conduction stops. At some point close to Vpk the anode voltage exceeds the cathode voltage, the rectifier conducts and a pulse of current flows, charging the reservoir capacitor to the value of Vpk. During each cycle, the rectifier anode AC voltage increases towards Vpk. The action of the reservoir capacitor on a half wave rectified sine wave is shown in Fig. during the gaps between the positive half cycles when the rectifier is not conducting. This very large value of capacitance is required because the reservoir capacitor, when charged, must provide enough DC to maintain a steady PSU output in the absence of an input current i.e. The reservoir capacitor is a large electrolytic, usually of several hundred or even a thousand or more microfarads, especially in mains frequency PSUs. The rectifier diode supplies current to charge a reservoir capacitor on each cycle of the input wave. 1.2.1 shows an electrolytic capacitor used as a reservoir capacitor, so called because it acts as a temporary storage for the power supply output current.
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