10.2.3.3 Rectification Ratio

2π 0 10.2.3.3 Rectification Ratio

The rectification ratio, which is a figure of merit for comparing or

the effectiveness of rectification, is defined as

Half-wave V L =

= 0.5V m

P dc V dc T dc

(10.17) P = L V L I L In the case of a full-wave rectifier, v L (t ) =V m | sin ωt| for both the positive and negative half-cycles. Hence Eq. (10.6)

In the case of a half-wave diode rectifier, the rectification can be re-written as

ratio can be determined by substituting Eqs. (10.3), (10.13), (10.8), and (10.14) into Eq. (10.17).

V L = 2 (V m sin ωt ) d(ωt )

( 0.318V m ) 0 2

( 0.5V m )

or

In the case of a full-wave rectifier, the rectification ratio Full-wave V L = √ = 0.707V m

(10.10) is obtained by substituting Eqs. (10.5), (10.15), (10.10), and

2 (10.16) into Eq. (10.17).

The result of Eq. (10.10) is as expected because the rms m value of a full-wave rectified voltage should be equal to that of 2 ( 0.636V )

Full-wave σ

= 81% (10.19) ( 0.707V m ) 2

the original ac voltage.

10.2.3.2 Current Relationships

10.2.3.4 Form Factor

The average value of load current i L is I dc and because load R The form factor (FF) is defined as the ratio of the root-mean- is purely resistive it can be found as

square value (heating component) of a voltage or current to its average value,

(10.20) The rms value of load current i L is I L and it can be found as

In the case of a half-wave rectifier, the FF can be found by

substituting Eqs. (10.8) and (10.3) into Eq. (10.20). In the case of a half-wave rectifier, from Eq. (10.3)

0.5V m

Half-wave FF =

0.318V m 0.318V m

Half-wave I dc =

In the case of a full-wave rectifier, the FF can be found by and from Eq. (10.8)

substituting Eqs. (10.16) and (10.15) into Eq. (10.20).

0.5V m

0.707V m

Half-wave I L =

Full-wave FF =

0.636V m

10 Diode Rectifiers 153

Therefore, the TUF of a half-wave rectifier can be obtained The ripple factor (RF), which is a measure of the ripple by substituting Eqs. (10.3), (10.13), (10.29), and (10.30) into content, is defined as

10.2.3.5 Ripple Factor

V dc Half-wave TUF =

where V ac is the effective (rms) value of the ac component of load voltage v L .

The poor TUF of a half-wave rectifier signifies that the trans- former employed must have a 3.496 (1/0.286) VA rating in order to deliver 1 W dc output power to the load. In addition,

V ac 2 = 2 V L −V dc (10.24)

the transformer secondary winding has to carry a dc current that may cause magnetic core saturation. As a result, half-wave

Substituting Eq. (10.24) into Eq. (10.23), the RF can be rectifiers are used only when the current requirement is small. expressed as

In the case of a full-wave rectifier with center-tapped trans- former, the circuit can be treated as two half-wave rectifiers

2 RF operating together. Therefore, the transformer secondary VA =

−1= FF −1

V dc rating, V s I s , is double that of a half-wave rectifier, but the out- put dc power is increased by a factor of four due to higher

In the case of a half-wave rectifier, the rectification ratio as indicated by Eqs. (10.5) and (10.15). Therefore, the TUF of a full-wave rectifier with center-tapped

Half-wave RF = 1.57 2 (10.26) − 1 = 1.21 transformer can be found from Eq. (10.32) In the case of a full-wave rectifier,

Full-wave TUF =

Full-wave RF = 1.11 2 − 1 = 0.482

In the case of a bridge rectifier, it has the highest TUF

in single-phase rectifier circuits because the currents flowing The transformer utilization factor (TUF), which is a measure in both the primary and secondary windings are continuous

10.2.3.6 Transformer Utilization Factor

of the merit of a rectifier circuit, is defined as the ratio of the sinewaves. By substituting Eqs. (10.5), (10.15), (10.29), and

dc output power to the transformer volt–ampere (VA) rating (10.31) into Eq. (10.28), the TUF of a bridge rectifier can be required by the secondary winding,

= 0.81 (10.34) where V s and I s are the rms voltage and rms current ratings of

TUF =

Bridge TUF

the secondary transformer. The transformer primary VA rating of a full-wave rectifier is equal to that of a bridge rectifier since the current flowing

in the primary winding is also a continuous sinewave.

V s = √ = 0.707V m

The rms value of the transformer secondary current I s is the

10.2.3.7 Harmonics

same as that of the load current I L . For a half-wave rectifier, I s Full-wave rectifier circuits with resistive load do not produce can be found from Eq. (10.14).

harmonic currents in their transformers. In half-wave recti- fiers, harmonic currents are generated. The amplitudes of the

0.5V m

harmonic currents of a half-wave rectifier with resistive load,

Half-wave I s =

relative to the fundamental, are given in Table 10.1. The extra loss caused by the harmonics in the resistive loaded rectifier

For a full-wave rectifier, I s is found from Eq. (10.16). circuits is often neglected because it is not high compared with other losses. However, with non-linear loads, harmonics can

0.707V m

Full-wave I s =

(10.31) cause appreciable loss and other problems such as poor power

factor and interference.

154 Y. S. Lee and M. H. L. Chow TABLE 10.1 Harmonic percentages of a half-wave rectifier with

In the case of a full-wave rectifier with center-tapped resistive load

transformer, from Eq. (10.5),

Harmonic 2nd 3rd

21.2 0 4.2 0 1.8 0 1.01 Full-wave V RRM = 2V m =

= 3.14V dc (10.38) 0.636

In the case of a bridge rectifier, also from Eq. (10.5),

10.2.4 Design Considerations

V dc

= 1.57V dc (10.39) 0.636 In a practical design, the goal is to achieve a given dc output

Bridge V RRM =V m =

voltage. Therefore, it is more convenient to put all the design It is important to evaluate the I FRM rating of the employed

parameters in terms of V dc . For example, the rating and turns diodes in rectifier circuits.

ratio of the transformer in a rectifier circuit can be easily deter- In the case of a half-wave rectifier, from Eq. (10.13), mined if the rms input voltage to the rectifier is in terms of

I dc Denote the rms value of the input voltage to the rectifier

the required output voltage V dc .

Half-wave I FRM =

= 3.41I dc (10.40)

0.318 as V s , which is equal to 0.707V m . Based on this relation and

Eq. (10.3), the rms input voltage to a half-wave rectifier is In the case of full-wave rectifiers, from Eq. (10.15), found as

I dc

Full-wave I FRM =

= 1.57I dc (10.41)

Half-wave V s = 2.22V dc (10.35)

The important design parameters of basic single-phase rec- Similarly, from Eqs. (10.5) and (10.29), the rms input tifier circuits with resistive loads are summarized in Table 10.2. voltage per secondary winding of a full-wave rectifier is

found as