The half wave rectifier utilizes alternate half cycles of the input sinusoid. Figure 1 shows the circuit of a half-wave rectifier circuit. The a.c. voltage to be rectified is applied to the input of the transformer and the voltage vi across the secondary is applied to the rectifier. The circuit consists of the series connection diode D and a resistor R. Assuming sinusoidal waveform, let the input voltage vi be given by,
Working of Half Wave Rectifier Circuit
Assume the diode to be ideal. During the positive half-cycle of the input sinusoid, the positive vi will cause current to flow through the diode in its forward direction. It follows that that diode voltage Vv will be very small – ideally zero. Thus, the circuit will have output voltage VO will be equal to the input voltage vi. on the other hand, during the negative half-cycles of vi, the diode will not conduct. Thus the circuit will have the equivalent and VO will be zero. Thus, the output voltage will have the waveform shown in figure 2.
Let the amplitude Vm>> Vv where Vv is the cutin voltage for the pn diode. Hence we may assume that Vv = 0. Further let the diode be idealized with resistance Rf in the ON state and open circuit (Rr = ∞) in the OFF state. Then the current i through the diode and the load resistance RL is given by,
And the peak current Im is given by
Figure 2. shows the wave forms of voltage vi fed to the diode, current I through the diode and the output voltage v0. Current I flows through the diode and the load resistor RL. only during the positive half of the applied input voltage vi. this load current I flowing through the load resistor RL produces the rectifier output voltage vo.
Average or DC Current Idc
Average or DC output current of half wave rectifier is given by
Where i is given by equation (2) and (3). Current i = 0 i.e. no current flows during the period radians. Hence,
The RMS current I is given by,
Substituting the value of i from equation (2) and (3) into equation (4) we get,
Average or DC Output Voltage (Vdc)
Average or DC output voltage of half wave rectifier is given by
RMS Output Voltage (VO)
RMS value of voltage across the load resistor is given by
DC Output Power (Pdc)
The output dc power across the load resistor RL forms the useful output power and is given by,
Total AC Input Power
Out of the total a.c. power Pi from the a.c. voltage source, a part Pd is dissipated at the junction of diode and rest of the power Pr is dissipated in the load resistance RL. Since the rectifier itself is assumed to be ideal, dissipated Pd takes place in the resistance Rf of the conducting diode. Then we get,
Total a.c. input power
It is defined as the ratio of the dc output power Pdc to the a.c. input Power Pi and is, therefore, given by,
% rectifier efficiency …..(19)
From equation (18) we conclude that the rectifier efficiency increases as the ratio reduces. Further from equation (19) we find that the theoretical maximum value of rectifier efficiency of a half wave rectifier is only 40.6% and this is obtained when .
Frequency Components in the Rectifier Output Analysis shows that current waveform of halfwave rectifier contains
- dc components
- ripple components
- Component of fundamental frequency f of peak value . This is the lowest frequency components.
- Even harmonic components of frequencies 2f, 4f etc.
Harmonic components are found to have progressively diminishing amplitudes.
Ripple Factor (
This is defined as ratio of the effective value of te a.c. components of voltage (or current) to the direct or average value.
The RMS value of the load current is given by,
Where I1, I2, I4 etc are the rms values of fundamental, second, fourth etc harmonics and Isc2 is the sum of the square of the rms values of the ac components.
Hence the ripple factor is,
Where F is the form factor.
For half wave rectifier,
Hence ripple factor
Peak Inverse Voltage (PIV)
It is the maximum reverse voltage which the rectifier is required to withstand during nonconducting period. In half wave rectifier, PIV equals Vm, the peak value of applied voltage. However on using a capacitor filter, the PIV ration of the diode increases to 2 Vm.
Transformer Utilization Factor (TUF)
It is defined as the ratio of dc power delivered to the load and ac rating of the transformer secondary winding.
By regulation is meant variation of output voltage with change in load current. Thus,
Percentage Regulation …..(25)
Where VnL is the no-load output voltage i.e. voltage with zero load current, and VL is the output voltage with normal load current.
For half wave rectifier, combining equation (7) and (13) we get,
Equation (26) indicate that the half wave rectifier functions as if it were a constant voltage source in series with an internal resistance (output resistance ) R0 = Rf as shown in figure 3. From this model, called Thevenin’s model, we note that dc output voltage Vdc decreases linearly with the increase of dc output current Idc.
Advantage of Halfwave rectifier
- Simple Circuit
- Low Cost
Disadvantage of Halfwave Rectifier
- Low rectifier efficiency
- High Ripple Factor
- Low TUF
- DC saturation of transformer core results in magnetising current and hysteresis losses and production of harmonics.