The capacitor charges during the positive half-cycle, then discharges through the load during the negative half-cycle when we have no output. In figure 2 we inserted a capacitor across the load. Connect the same AC voltmeter across a clean DC source such as a car battery, one will read zero volts AC. This is normal as one is reading the "ripple" riding the unfiltered raw D.C. Switch the meter to AC, one will still read a voltage of some value. Put a DC voltmeter across the load above in figure 1, one will read about 5.66 volts. Pure DC, such as from a 12 volt auto battery, has none of the "ripple" and will be a real 12 volts. But the average (or measured) voltage DC is peak times.
To get peak we multiply 12.6 by 1.414, which equals about 17.8 volts. In the case above, we get very poor power transfer with the diode off during the negative half-cycle and the positive half-cycle changing constantly between zero volts and peak. See your text.) We must have voltage and current together to get power, so an open switch, broken wire, or a shut-off diode delivers no power. (I'm not going into all of Ohm's Law here. (Electric charges) Power is voltage times current. What is power? Voltage (in volts) is the "push" and the current (in Amperes) is what is being pushed. In the case of 60 Hertz it's 1/60 = 16.7 mSec.īuild a low voltage DC power supply Part 3 The period of a sine wave from 0 degrees to 360 degrees equals 1/F. Thus current will flow only of the positive half-cycle (0 to 180 degrees) and the diode will shut-off during the negative half-cycle from 180 degrees to 360 degrees. When the anode (A) is positive and the cathode (K) is negative current flow from positive to negative will flow through the diode, through the load, and back to the power supply. A diode is a solid-state device that conducts in one direction only. Many devices, in particular solid state electronics, must use DC or direct current.
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