Single-Phase, Full-Wave Controlled Circuits









sin Φ sin Φ









+

( )

^{− ( − ) +}

α π

π ε π π α

i_L ^cotΦωt π2

−^cotΦ(^ωt−^α)

(3.59) The average and rms values of iL(␻t) can be obtained from the respective defining integrals in Eqs. (3.52) and (3.53).

3.2.2 Single-Phase, Full-Wave Controlled Circuits

Each of the circuit configurations ofFig. 3.4can be used for full-wave rectification of a series R-L load, as inFig. 3.17.The cost of a diode switch is much less than that of a controlled switch of the same rating, and it also does not require an

FIG.17 Single-phase, full-wave controlled rectifier circuits with series R-L load.

associated firing circuit. From the cost viewpoint therefore it is desirable to use a circuit with few controlled switches. The single-switch circuit of Fig. 3.17b is likely to be the most economic. For comprehensive control, however, including the facility to act as an inverter, it is necessary to use the fully controlled bridge circuit of Fig. 3.17d.

In the arrangement of Fig. 3.17 the two diodes freewheel the load current when one or both of the switches is in extinction. Continuity of the load current is enhanced, with R-L loads, by the use of a freewheel diode across the load impedance, as shown in Fig. 3.17b–d. In many applications the load-side induct-ance is made deliberately large to ensure a continuous flow of load current. It is also usually desirable that the load current be largely dc (i.e., the load current

should have the largest realizable average value) with low ripple content. The ideal load current ripple factor, Eq. (2.11) is zero, which occurs when the load current is pure dc.

The four-switch circuit ofFig. 3.17dhas the load voltage and current wave-forms shown in Fig. 3.18 for ␣ ⳱ 30, with large L. Commutation of the two conducting switches occurs naturally at the end of a supply voltage half cycle.

The load voltage is seen to be the full-wave equivalent of the corresponding half-wave voltage half-waveform ofFig. 3.16.From Fig. 3.18 it is seen that the average load voltage is given by

E E t d t E

av m

= ¹_π

∫

^{απ sinω ω} = _πm

(

¹+^cosα

)

^(3.60)

FIG.18 Voltage and current waveforms for the single-phase, full-wave circuit of Fig.

3.17d, large L,␣ ⳱ 30.

Also,

Iav⳱ Id (3.61)

The output power is dissipated in resistor R. With ideal switches and a lossless inductor L the output power also equals the input power

Pout⳱ Pin⳱ I²dR (3.62)

The input current is defined by the expression

i_s ωt I_d π I_d α

π

( )

^{= −} π α²_{+ (3.63)}

The rms value of the input current is given by

I i t d t When␣ ⳱ 0, Is⳱ Id. The fundamental component of the supply current is found to have the rms magnitude

I_s I_d

By inspection of the supply current waveform in Fig. 3.18, it is seen that the displacement angle is␺1is␣/2. The input power may therefore be alternatively defined as Note that real or average power P is associated only with the combinations of voltage and current components of the same frequency. Since the supply voltage is sinusoidal and therefore of single frequency, it combines only with the funda-mental (supply frequency) component of the input current. The combination of the fundamental voltage component with higher harmonic components of the current produces time-variable voltamperes but zero average power.

The power factor of the bridge circuit is obtained using Eqs. (3.64) and distortion factor components of the power factor can be obtained via the Fourier components a1b1 of is, (␻t), Sec. 3.13.

When a freewheel diode is not used the symmetrical triggering of opposite pairs of switches in Fig. 3.17d(or Fig. 3.4d) results in the waveforms of Fig.

3.19,for a highly inductive load. Compared with Fig. 3.16 the load current is unchanged and so therefore is the power dissipation. But note that the rms value of the supply current is increased so that the power factor is reduced. It is seen from Fig. 3.19 that the average load voltage is given by

E E t d t E

av m

= ¹_π

∫

^{απ α+ sinω ω} =²_πm ^{cosα (3.68)}

Also, the average voltage of the load inductor is zero so that

I I E

av d R

= = av _(3.69)

The rms value ELof the load voltage in Fig. 3.19 is given by E_L² 1 E_m^{2 2} t d t

The harmonic nature of the load current in Fig. 3.17 could be obtained by calculat-ing the Fourier series for the periodic function i_L(␻t), for an arbitrary R-L load,

FIG.19 Voltage and current waveforms for the single-phase, full-wave circuit ofFig.

3.17dwithout freewheel diode FWD, large L,␣ ⳱ 30.

with or without the diode. Alternatively, the various harmonic terms of the Fourier series for the periodic voltage eL(␻t) (Fig. 3.18or 3.19) can be applied since they are valid for any load impedance. The Fourier series of the load voltage may be written

e t E E n t

E E t E t

L av n

n

n

av

ω ω ψ

ω ψ ω

( )

^{= +}

(

⁻

)

= +

(

−

)

^{+ −}

∑

^{ˆ cos}

ˆ cos ˆ cos

1

2 2 2 4 4 ψψ

ω ψ

4

6 6 6

( )

+^{Eˆ cos}

(

^t−

)

^{+… (3.71)}

In the case of waveform e_Lin Fig. 3.19a,

ˆ cos

When the harmonic voltage terms of Eq. (3.71) are applied to the series R-L load ofFig. 3.17,the following series is obtained for the load current

i t I I n t

… A close approximation to the rms current ILfor any R-L load can be obtained by using the square law relationship of Eq. (2.12). If the load impedance is highly in-ductive, the harmonic terms in Eqs. (3.71)–(3.76) are negligibly small. The load current is then constant at the value ILor Id, which also becomes its rms value IL.

A further application of this bridge circuit in the form of a dual converter is described in Sec. 12.2.1.

3.2.3 Worked Examples

Example 3.5 A series semiconductor switching circuit,Fig. 3.12, has a load in which the resistance is negligibly small. Deduce and sketch the waveforms of the load voltage and current for a firing angle␣ smaller than the load phase angle⌽. What are the average values of the load current and voltage?

When⌽ ⳱ 90, cot⌽ ⳱ 0 and Eq. (3.46) reduces to seen that the current waveform is symmetrical about␲ so that extinction angle x is given by

x⳱ 2␲ ⳮ ␣ From Eq. (3.48), therefore,

␪c⳱ 2(␲ ⳮ ␣) (3.78)

Equation (3.52) is indeterminate for⌽ ⳱ 90, but a solution for the time average current can be obtained by integrating Eq. (3.77)

FIG.20 Voltage and current waveforms for the single-phase, full-wave controlled recti-fier circuit with highly inductive load;␣ ⳱ 30.

i E

Z t d t

E Z

E

Z t

av

m

m m

= c −

= −

+

−

∫

∫

1 2

2

2

π α ω ω

π α ω

α α θ

α π α

(cos cos )

(cos cos ) dd t E

Z

m

ω

π π α α α

=

[

⁽ − ^{) cos} +^sin

]

When␣ ⳱ 0, Iav ⳱ Em/␻L. The output average current is then constant at its maximum realizable value, which represents ideal rectifier operation. The voltage across the load inductor may be obtained by differentiating the current expression Eq. (3.77), noting that|Z| ⳱ ␻L:

The average value Eavof eL(␻t) may be obtained by the usual integration method and is found to be zero, as can be seen by inspection inFig. 3.20.

Example 3.6 In the series R-L circuit ofFig. 3.12R⳱ 25 ⍀, L ⳱ 150 mH. The supply voltage is given by e⳱ Emsin ␻t, where Em⳱ 400, V at a frequency of 50 Hz. Calculate the average load current for the SCR firing angles (1) 30 and (2) 120. be made from Eq. (3.51)

θc = + − +π Φ α ∆ 180= ^o+62^o−30^o+7^o (say) =219^o

The value␪c⳱ 219 is used as a first guess in Eq. (3.49). By iteration it is found that ␪c ⳱ 216.5. The accuracy with which ␪c can be read from Fig. 3.18 is sufficient for most purposes.

cos␪c⳱ ⳮ0.804

sin␪c⳱ ⳮ0.595

The characteristics of Fig. 3.14suggest that this figure is high and should be about 60. Iteration from Eq. (3.49) gives a value ␪c⳱ 64. In Eq. (3.52),

Example 3.7 A half-wave, controlled rectifier circuit has a series R-L load in which ⌽ ⳱ tan^ⳮ1 (␻L/R) 艐 80. The ideal single-phase supply voltage is given by e⳱ Emsin␻t. Explain the action for a typical steady-state cycle when

␣ ⳱ 60 and the circuit includes a freewheel diode (Fig. 3.15).What effect does the diode have on circuit power factor?

The load voltage and current are shown inFig. 3.16.The supply current is

(␻t) is given by the portions of the current curve between ␣ → ␲, 2␲ Ⳮ ␣ → 3␲, etc., noting that is (␣) ⳱ is(2␲ Ⳮ ␣) ⳱ 0. The nonzero value of the load current at␻t ⳱ ␣ is due to residual current decaying through the diode during the extinction of the thyristor switch. In the presence of the diode, the energy stored in the magnetic field of the inductor is dissipated in resistor R rather than being returned to the supply. Current and power flow from the supply to the load only occur during the conduction intervals of the thyristor. Because of the significantly increased rms load current (compare the load currents in Figs. 3.14 and 3.16) however, the load power dissipation is significantly increased. All of this power must come from the supply, although not at the instants of time in which it is dissipated. The supply voltage remains sinusoidal at all times.

The power factor, seen from the supply point, is PF P

=EI

The rms supply voltage E⳱ Em/兹2 is constant. The rms supply current I probably increases by (say) 20%. But the power dissipation can be assessed in terms of the rms value of the load current. Comparing iL (␻t) in Figs. 3.13 and 3.16 suggests that rms value IL is at least doubled and the power increases at least four times. The presence of the freewheel diode therefore causes the power factor to increase.

Example 3.8 A single-phase, full-wave bridge circuit, Fig. 3.21, has four ideal thyristor switches and a highly inductive load. The electrical supply is ideal and is represented by e ⳱ Em sin ␻t. Sketch waveforms of the load current, supply current and load voltage for␣ ⳱ 60 Calculate the rms value of the supply current and the power factor of operation, in terms of␣.

The waveforms of operation with a highly inductive load are given inFig.

3.19.The supply current is represented by the relation

i_s( t) I_d I_d , ω ,

α π α

π α

= ⁺ − α π + 0

2

This has an rms value defined by I_s = ₂¹_π

∫

^02πi_s^2(ωt d t^{) ω} Therefore,

I_s² I d t_d² I_d ² d t

0

1

=2  + −

 



+

+

∫

π

∫

_α^{π α} ω ^α_{π α}( ) ω

,

FIG.21 Single-phase, full-wave controlled rectifier circuit with highly inductive load.

The rms value of the negative parts of the wave is equal to the rms value of the positive parts so that

I_s² 1 I d t_d² 1I_d²

=_π

∫

^{απ α+ ω} =_π ^{(π α α}+ − ⁾ Therefore,

Is⳱ Id

As␣ varies, the waveform of is(␻t) is unchanged and so is its rms value, but the switchover from positive to negative value occurs at the firing points.

Power dissipation is determined by the rms value of the load current P=I R_L² =I R_d²

The power factor is found to be PF P But from Eq. (3.68),

I E The power factor is therefore

PF=2 2 π cosα

When␣ ⳱ 0, the PF ⳱ 2兹2 /␲, which agrees with the value for a full-wave diode bridge.

Example 3.9 In the single-phase, full-wave rectifier ofFig. 3.21the load consists of R⳱ 10 ⍀ and L ⳱ 50 mH. The ideal sinusoidal supply voltage is defined as es⳱ 240兹2 sin ␻t at 50 Hz. Calculate values for the average and rms load currents, the power dissipation and the power factor at the supply terminals if the thyristor firing angle␣ ⳱ 45.

The circuit of Fig. 3.21 is seen to be a topological rearrangement ofFig.

3.17d,without the freewheel diode. The load voltage has the segmented sinusoidal form ofFig. 3.19a for any R-L load if the load current is continuous. The load current will be of some waveform intermediate between that of Fig. 3.19b (which is only valid for highly inductive loads) and the corresponding segmented sinusoi-dal waveform (not given) obtained with purely resistive loads.

At supply frequency the phase-angle of the load impedance is given by

Since␣ ⬍ ⌽, the mode of operation is that of continuous load current and the equations of the present Sec. 3.2.2 are valid.

The average load current is given by Eqs. (3.68) and (3.69):

I E

In order to calculate the rms load current, it is necessary to calculate its Fourier harmonics. From Eq. (3.72), cos 2␣ ⳱ cos 90 ⳱ 0, and the peak values of the lower order load voltage harmonics are

Eˆ₂⳱ 56.92 V Eˆ₄⳱20.99 V Eˆ₆⳱ 13.27 V

Also, in Eq. (3.62) the following impedances are offered by the load at the speci-fied harmonic frequencies.

Z R L

The peak values of the current harmonics are therefore

ˆ .

From Eq. (2.12) the rms value of the load current is given by I_L² I_av² 1 I₂² I₄² I₆²

= +2

(

^ˆ + + +^{ˆ ˆ …}

)

and

I_L=

( )

^{+ }_

( )

⁺

( )

⁺

( )

^_

It is seen that the effects of the fourth and sixth harmonic current components on the total rms value are negligible. The load power is seen, from Eq. (3.64), to be

P_L =

(

^{11 68.}

)

^{2×10=1364 22.} W

and this is also the power entering the circuit terminals, neglecting rectifier ele-ment losses. The rms value of the supply current is equal to the rms value of the load current. The power factor at the supply point is, therefore,

PF P

This compares with the value cos⌽ ⳱ cos 57.5 ⳱ 0.537 for the load impedance alone.

PROBLEMS

In document william shepherd (Page 89-103)