Picture 1
Principle rectifier (rectifiers) The simplest is shown in shown below. Transformer is needed to lower the AC voltage from the grid at the primary coil becomes smaller AC voltage on the secondary coil.
In this circuit, diodes just continue to contribute to a positive voltage to the load RL. Which is called a half-wave rectifier (half wave). To obtain full-wave rectifier (full wave) is required transformer with center tap (CT) as shown below.
Picture 2
Positive voltage of the first phase was succeeded by D1, while the next phase is passed through D2 to the load R1 with transformer CT as a common ground. Thus R1 burden gets a full wave voltage supply such as the picture above. For some applications such as for supply to small dc motors or dc incandescent lamp, the shape of this voltage is sufficient. Although the voltage ripple seen here from both series are still very large.
Picture 3
Picture 3 is a series of half-wave rectifier with filter capacitor C is parallel to the load R. Apparently with this filter discharge voltage waveform can be blended. Fig, 4 shows a DC voltage output from circuit half-wave rectifier with capacitor filter. Bc line is approximately straight line with certain slope, which in this state for a load current supplied by voltage capacitor R1. Actually the line bc is not a straight line but in accordance with the nature of exponential capacitor discharge.
Picture 4
Bc curve slope depends on the large current I flowing to the load R. If the current I = 0 (no load) then the curve will form a horizontal line bc. But if the load currents greater, the slope curve will become sharper bc. The voltage that comes out will be shaped with a sawtooth ripple voltage of the magnitude is:
Vr = VM-VL ... ....... (1)
and dc voltage to the load is VDC = VM + Vr / 2 ..... (2)
Good rectifier circuit is a circuit which has the smallest ripple voltage. VL is a discharge or a discharge voltage capacitor C, so it can be written:
VL = VM e -T/RC .......... (3)
If equation (3) disubsitusi to formula (1), it is obtained:
Vr = VM (1 - e -T/RC) ...... (4)
If T <
so if this disubsitusi to the formula (4) can be obtained from a simpler equation:
Vr = VM (T / RC) .... (6)
VM / R load is none other than I, so that with this visible relationship between the load current I and the value of capacitor C to the ripple voltage Vr. This calculation is effective to obtain the desired ripple tengangan value.
Vr = I T / C ... (7)
This formula is to say, if I load current was higher, the greater the voltage ripple. Conversely, if the capacitance C was higher, the voltage ripple will be smaller. For simplification is usually considered T = Tp, ie a period of one sine wave of the grid frequency of 50Hz or 60Hz. If the grid frequency 50Hz, then T = Tp = 1 / f = 1 / 50 = 0.02 sec. This applies to half-wave rectifier. For full wave rectifier, of course, the tuner frequency doubled, so that T = 1 / 2 Tp = 0.01 sec.