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 v_{i} 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 v_{i} be given by,

…..(1)

### Working of Half Wave Rectifier Circuit

Assume the diode to be ideal. During the positive half-cycle of the input sinusoid, the positive v_{i} will cause current to flow through the diode in its forward direction. It follows that that diode voltage V_{v }will be very small – ideally zero. Thus, the circuit will have output voltage V_{O} will be equal to the input voltage v_{i}. on the other hand, during the negative half-cycles of v_{i}, the diode will not conduct. Thus the circuit will have the equivalent and V_{O} will be zero. Thus, the output voltage will have the waveform shown in figure 2.

Let the amplitude V_{m}>> V_{v} where V_{v} is the cutin voltage for the pn diode. Hence we may assume that V_{v} = 0. Further let the diode be idealized with resistance R_{f} in the ON state and open circuit (R_{r} = ∞) in the OFF state. Then the current i through the diode and the load resistance R_{L} is given by,

for …..(2)

for …..(3)

Where

And the peak current I_{m} is given by

.…(4)

Figure 2. shows the wave forms of voltage v_{i} fed to the diode, current I through the diode and the output voltage v_{0}. Current I flows through the diode and the load resistor R_{L}. only during the positive half of the applied input voltage v_{i}. this load current I flowing through the load resistor R_{L} produces the rectifier output voltage v_{o}.

**Average or DC Current I _{dc}**

Average or DC output current of half wave rectifier is given by

…..(5)

Where i is given by equation (2) and (3). Current i = 0 i.e. no current flows during the period radians. Hence,

…..(6)

Or,

……(7)

**RMS Current**

The RMS current I is given by,

…..(8)

Substituting the value of i from equation (2) and (3) into equation (4) we get,

…..(9)

…..(10)

**Average or DC Output Voltage (V _{dc})**

Average or DC output voltage of half wave rectifier is given by

…..(11)

…..(12)

**RMS Output Voltage (V _{O})**

RMS value of voltage across the load resistor is given by

…..(13)

**DC Output Power (P _{dc})**

The output dc power across the load resistor R_{L} forms the useful output power and is given by,

…..(14)

**Total AC Input Power**

Out of the total a.c. power P_{i} from the a.c. voltage source, a part P_{d} is dissipated at the junction of diode and rest of the power P_{r} is dissipated in the load resistance R_{L}. Since the rectifier itself is assumed to be ideal, dissipated P_{d} takes place in the resistance R_{f} of the conducting diode. Then we get,

…..(15)

…..(16)

Total a.c. input power

…..(17)

**Rectifier Efficiency**

It is defined as the ratio of the dc output power P_{dc} to the a.c. input Power P_{i} and is, therefore, given by,

Rectifier Efficiency

Or ……(18)

% 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,

…..(20)

Where I_{1}, I_{2}, I_{4} etc are the rms values of fundamental, second, fourth etc harmonics and I_{sc}^{2} is the sum of the square of the rms values of the ac components.

Hence the ripple factor is,

…..(21)

Where F is the form factor.

For half wave rectifier,

Hence

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 V_{m}, the peak value of applied voltage. However on using a capacitor filter, the PIV ration of the diode increases to 2 V_{m}.

**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.

i.e. …..(22)

Hence, …..(23)

But,

Hence, …..(24)

If

**Regulation**

By regulation is meant variation of output voltage with change in load current. Thus,

Percentage Regulation …..(25)

Where V_{nL} is the no-load output voltage i.e. voltage with zero load current, and V_{L} is the output voltage with normal load current.

For half wave rectifier, combining equation (7) and (13) we get,

…..(26)

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 ) R_{0} = R_{f} as shown in figure 3. From this model, called Thevenin’s model, we note that dc output voltage V_{dc} decreases linearly with the increase of dc output current I_{dc}.

**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.

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