Principle rectifier (rectifier) The simplest is shown in Figure-1 below. Transformer (T1) is required to lower the AC voltage from the grid at the primary coil becomes smaller AC voltage in secondary coil.
In this circuit, diode (D1) serves only to change from AC to DC and continue the positive voltage to the load R1. This is called half-wave rectifier (half wave). To obtain a full wave rectifier (full wave) is required transformer with center tap (CT)
Positive voltage of the first phase forwarded by D1, while the next phase which is passed through D2 to the load R1 with transformer CT as a common ground .. Thus the load R1 gets a full wave voltage supply such as the picture above. For some applications such as for example for downloading a small dc motor supply or dc incandescent lamp, the shape of this voltage is sufficient. Although the voltage ripple seen here from the second series of the above is still very large.
series of half-wave rectifier with filter capacitor C is parallel to the load R. Apparently with this filter discharge voltage waveform can be flat. Figure-4 shows the DC voltage output from circuit half-wave rectifier with capacitor filter. Bc line is approximately straight line with a certain slope, which in this state to the load current supplied by voltage capacitor R1. Actually the line bc is not a straight line but exponential in accordance with the nature of discharging the capacitor.
The slope of the curve bc depends on the large current (I) which flows to the load R. If the current I = 0 (no load) then the curve will form a horizontal line bc. However, if the load currents greater, the slope of the curve bc will be more sharp. The voltage that comes out will sawtooth-shaped with a large ripple voltage is:
Vr =-VL VM
and dc voltage to the load is Vdc = VM + Vr / 2
The rectifier is a good circuit has a ripple voltage (Vr) the smallest. VL is the voltage discharge or discharging the capacitor C, so it can be written:
VL = VM e -T/RC
If equation (3) disubsitusi to the formula (1), then diperole
Vr = VM (1 - e -T/RC)
If T <<> -T/RC 1 - T / RC
so if this disubsitusi to the formula (4) to obtain a simpler equation:
Vr = VM (T / RC)
VM / R is nothing but the burden of 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 voltage value.
Vr = I T / C
This formula says, if I load the greater the flow, then the ripple voltage will be even greater. Conversely, if the greater the capacitance C, voltage ripple will be smaller. For simplification is usually considered T = Tp, ie a period of one sine wave from the grid of frequency 50Hz or 60Hz. If the grid frequency of 50Hz, then T = Tp = 1 / f = 1 / 50 = 0.02 sec. This applies to half-wave rectifier. For full wave rectifier, of course, the wave frequency doubled, so that T = 1 / 2 Tp = 0.01 sec.
Full wave rectifier with filter C can be created by adding a capacitor in the circuit of Figure 2. Can also use a transformer without CT, but by assembling the four diodes
For example, you design a full wave rectifier circuit of the power supply 220V/50Hz grid to supply the load of 0.5 A. What is the value capacitors that are necessary so that this circuit has a voltage ripple of no more than 0.75 Vpp. If the formula (7) are turned upside down then obtained.
C = I.T / Vr = (0.5) (0.01) / 0.75 = 6600 UF
For this size capacitor lots available elco type which has a maximum working voltage polarity and specific. Working voltage capacitors that are used must be greater than the output voltage power supply. You probably now understand why you created a series of audio buzzing, check back the power supply rectifier circuit that you created, if the voltage ripple is quite disturbing. If the market is not available such a large capacitor, certainly could with two or three pieces paralel capacitor
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