Half wave rectifier
- The Half wave rectifier uses a single diode and is the simplest form of diode.
- A step down transformer, diode and load is connected with an A.C supply.
- The diode connection is as shown in figure with diode and transformer and load.
- The connection allows the diode to conduct in only one cycle.
- There are two types of rectifiers, positive and negative rectifiers depending upon the cycle in which they conduct.
- The rectifier conducts only in the negative half cycle and produces negative sinusoidal pulses in the output.
- The rectifier conducts only in the positive half cycle and blocks the negative half cycle produces positive sinusoidal pulses in the output.
Half wave rectifier specifications
The ripple factor is given as
The DC current is given by,
The maximum DC load current is Imax
Output DC voltage (VDC)
The output DC voltage is
The Maximum secondary voltage is VSmax
The ratio of output DC power to the input AC power is rectifier efficiency.
The half wave rectifier rectifier offers 40.6% efficiency.
Root mean square (RMS)
The root mean square (RMS) value of half wave rectifier is
The root mean square (RMS) value of a half wave rectifier is
The ratio of RMS value to the DC value is form factor
F.F = RMS value / DC value
The form factor of a half wave rectifier is F.F = 1.57
- It is cheaper than other rectifiers as it uses only one diode.
- It is easy in terms of construction as it requires very few components.
- There is loss of power
- Pulsating direct current
- The output voltage produced is low.
Full wave Rectifier
Classification on the basis of construction
1. Center-tapped transformer
The circuit has two diodes connected to a single load (RL)and each diode conducts alternately. During the positive half cycle the diode D1 conducts
During the negative half cycle diode D2 conducts and current flows through resistor R in the same direction in both the half cycles.
The output waveform is as shown below
As both diodes conduct in their respective half cycles, the final output waveform obtained shows the continuous wave for every half cycle.
2. Bridge Rectifier
- The circuit has four diodes connected as a bridge as we can see from the above figure.
- The use of four diodes in the form of bridge does not require any center tap transformer. This also reduces the circuit cost. Hence transformer secondary is connected to one side of the diode bridge network and other side to the load
- There are four diodes but only two diodes conduct in each half cycle.
- The diodes D2 and D3 both conduct in the positive half cycle also D1 and D4 are reversed biased.
- The diodes D1 and D4 both conduct in the negative half cycle also D2 and D3 are reversed biased.
Full wave rectifier specifications
The ripple factor is expressed as
Therefore full wave rectifier offers γ = 0.48
The ratio of DC output power to the AC input power is rectifier efficiency.
η = output power PDC / input power PAC
Therefore full wave rectifier offers 81.2%. efficiency.
Peak inverse voltage (PIV)
The peak inverse voltage is the reverse bias condition in which the maximum voltage which a diode can offer.
Then peak inverse voltage (PIV) = 2Vsmax
DC output current
The current produced by D2 is Imax / π and then current produced by D3 is Imax / π.
So, the output current IDC = 2Imax / π
where, Imax = maximum DC load current
DC output voltage
The DC output voltage at the load resistor
VDC = 2Vmax
Root mean square (RMS) value
The root mean square (RMS) value of is
Then root mean square (RMS) value of full wave rectifier is
The form factor then becomes
F.F = RMS value of current / DC output current
The full wave rectifier offers F.F = 1.11