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4-8

Three Phase Full Converters (H.Rashid):

3-Φ Full converters are used in industrial applications upto 120 kw level, where two quadrant operation is required.

Consider the 3-Φ full converter (figure 4-9a), with a highly inductive load.

The thyristors are fired at an interval of

The frequency of output voltage is 6 f s & the filtering requirement is less than that of three phase semi converter and half wave converters.

At ω t = (

) ≤ωt≤( ) , thyristor T 1 & T 6 is conducts and the line-toDuring the period ω t = ( line voltage V ab (= V an – V bn) appears across the load.

At ω t = (

T 6 is turned off due to natural commutation.

During the interval (

/ 3.

) thyristor T 6 is already conducting and thyristor T 1 is turned on.

) thyristor T 2 is fired and thyristor T 6 is reversed biased immediately.

) ≤ωt≤ (

) , thyristors T 1 & T 2 conduct and the line-to-line

voltage V a c appears across the load.

If the thyristors are numbered as (fig: 4-9 a), the firing sequence is 12, 23, 34, 45, 56 and 61.

Figure 4-9 b shows the wave forms for input voltage, output voltage, input current and the currents through the thyristors.

1

If the three line-neutral voltages are defined as; =V

m

sin ω t

=V

m

sin (ω t – (

))

=V

m

sin (ω t + (

))

The corresponding line-to-line voltages are;

Where V

m

=

–

=

–

=

–

=

√ V

m

sin (ω t + (

=

√ V

m

sin (ω t – ( ))

=

√ V

m

sin (ω t + (

))

))

is the peak phase voltage of Wye-connected source.

2

The average output voltage is found from; V

d.c

=

=> V

d.c

(

∫

√

(

(

(

√

∫ =

) )) (

)

) - - - - - - - - - (4-57)

For maximum average output voltage; That occurs at a delay angle of V

dm

√

=

=>

V

d.c

( =

= 0 ° is V

dm

;

) √

(

√

=>

)

( )

=>

V dm =

√

In this way average normalized average output voltage is obtained as; V n = V d c / Vd m =>

Vn=(

√

)) / (

(

√

)

=>

Vn=(

) - - - - - - - - - (4-58)

So that the r.m.s output voltage is found from; V

r.m.s

=[

V r.m.s = √

∫ xVm[

2

. Sin 2 (ω t + ). ( (

√

))]

½

) cos 2 )] ½ - - - - - - - - (4-59)

3

Example 4 – 8 (H.Rashid): A 3-phase Full converter (figure 4-9 a) is operated from a 3-Phase Y-connected 208 v, 60 Hz supply and the load resistance is of R = 10 Ω. It is required to obtain an average output voltage of 50 % of the maximum possible voltage. Calculate; (i). The delay angle = ? (ii). R.m.s & average output currents ? (iii). R.m.s & average Thyristor currents ? (iv). Rectification efficiency ? (v). T.U.F ? (vi). Input power factor ? Solution: From the given data, the Phase voltage is obtained as; V S = 208 /√

=>

V S = 120.1 v

So that V m can be obtained as; V

m

=√ xV

=>

S

V

m

= √ x 120.1

=>

V

m

= 169.8 v

Hence the maximum output voltage can be obtained as; V

dm

=( √

xV

V

dm

=( √

x 169.8 ) / 3.142

m

)/

(when

= 0, so cos 0 = 1) => V

dm

=

The average output voltage can be obtained as; V n = Vdc / Vd m So that from the given condition of problem; V n = 0.5 V n = Vdc / Vd m (a).

0.5 = Vd c / 280.81

R = 10 Ω => Vd c =

From eq 4-58, Vn=( =>

(b).

=>

&

) - - - - - - - - - (4-58) 0.5 = (

)

= cos – 1 (0.5)

=>

=>

= 60 °

The average output current is I

dc

=V

dc

/R

=>

140.45 / 10

=> I d c =14.045 Amps

The r.m.s output voltage can be obtained as; V r.m.s = √

xVm[

(

√

) cos 2 )] ½ - - - - - - - - (4-59)

4

=>

√ x 169.83 [

=>

(

√

) cos 2 x 60 °)]

x 169.83 [

=>

[

½

cos 120 °)] ]

½

=>

[

½

(

)]

½

=>

V

=> I

r.m.s

=> I

DT

= 4.68 Amps.

=> I

RT

= 9.2 Amps.

r.m.s

=

And the r.m.s load current is obtained as; I

r.m.s

=V

(c).

r.m.s

/R

=>

159.34 /10

=

The average current of Thyristor is given by; I

=I

DT

dc

/3

=>

14.045 / 3

Similarly the r.m.s current of Thyristor is obtained as; I

RT

=I

r.m.s

x√

=>

15.93 x √

=>

15.93 x 0.577

(d).

The rectification efficiency is obtained as; η = Pd c / Pa c - - - - - - - - - - (2 – 44)

Pd c = V

dc

xI

Pa c = V

r.m.s x

dc

I

r.m.s

η = Pd c / P a c (e).

=>

(140.45) (14.045)

=> Pd c = 1973.3 watts.

=>

(159.29) (15.93)

=> Pa c = 2537.5 watts.

=>

1973.3 / 2537.5

=> η = 0.778 or 77.8 %.

15.93 x

=> I S = 13

The r.m.s input line current is; I S= I

r.m.s

x√

=>

Similarly, the input volt-ampere rating is obtained as; VI=3VSx I Now

3 x 120.1 x 13

=> V I = 4683.9

T.U.F = Pd c / V I

=> (V (f).

=>

S

dc

xI

d c)

/ (3 V S x I S)

=> (140.45 x 14.045) / (4683.9)

=>

T.U.F = 0.421

The output power is given by; P O = (I

r.m.s

)2xR

=>

P O = (15.93) 2 x 10

=>

253.765 x 10

=> P O = 2537.65

The input power factor is given by; P.F

input

= P O/ V I

=>

2537.65 / 4683.9

=> P.F

input

=

Note: The Power factor is less than that of three phase semi converter, but higher than that of 3-Φ half wave converter.

5

View more...
Three Phase Full Converters (H.Rashid):

3-Φ Full converters are used in industrial applications upto 120 kw level, where two quadrant operation is required.

Consider the 3-Φ full converter (figure 4-9a), with a highly inductive load.

The thyristors are fired at an interval of

The frequency of output voltage is 6 f s & the filtering requirement is less than that of three phase semi converter and half wave converters.

At ω t = (

) ≤ωt≤( ) , thyristor T 1 & T 6 is conducts and the line-toDuring the period ω t = ( line voltage V ab (= V an – V bn) appears across the load.

At ω t = (

T 6 is turned off due to natural commutation.

During the interval (

/ 3.

) thyristor T 6 is already conducting and thyristor T 1 is turned on.

) thyristor T 2 is fired and thyristor T 6 is reversed biased immediately.

) ≤ωt≤ (

) , thyristors T 1 & T 2 conduct and the line-to-line

voltage V a c appears across the load.

If the thyristors are numbered as (fig: 4-9 a), the firing sequence is 12, 23, 34, 45, 56 and 61.

Figure 4-9 b shows the wave forms for input voltage, output voltage, input current and the currents through the thyristors.

1

If the three line-neutral voltages are defined as; =V

m

sin ω t

=V

m

sin (ω t – (

))

=V

m

sin (ω t + (

))

The corresponding line-to-line voltages are;

Where V

m

=

–

=

–

=

–

=

√ V

m

sin (ω t + (

=

√ V

m

sin (ω t – ( ))

=

√ V

m

sin (ω t + (

))

))

is the peak phase voltage of Wye-connected source.

2

The average output voltage is found from; V

d.c

=

=> V

d.c

(

∫

√

(

(

(

√

∫ =

) )) (

)

) - - - - - - - - - (4-57)

For maximum average output voltage; That occurs at a delay angle of V

dm

√

=

=>

V

d.c

( =

= 0 ° is V

dm

;

) √

(

√

=>

)

( )

=>

V dm =

√

In this way average normalized average output voltage is obtained as; V n = V d c / Vd m =>

Vn=(

√

)) / (

(

√

)

=>

Vn=(

) - - - - - - - - - (4-58)

So that the r.m.s output voltage is found from; V

r.m.s

=[

V r.m.s = √

∫ xVm[

2

. Sin 2 (ω t + ). ( (

√

))]

½

) cos 2 )] ½ - - - - - - - - (4-59)

3

Example 4 – 8 (H.Rashid): A 3-phase Full converter (figure 4-9 a) is operated from a 3-Phase Y-connected 208 v, 60 Hz supply and the load resistance is of R = 10 Ω. It is required to obtain an average output voltage of 50 % of the maximum possible voltage. Calculate; (i). The delay angle = ? (ii). R.m.s & average output currents ? (iii). R.m.s & average Thyristor currents ? (iv). Rectification efficiency ? (v). T.U.F ? (vi). Input power factor ? Solution: From the given data, the Phase voltage is obtained as; V S = 208 /√

=>

V S = 120.1 v

So that V m can be obtained as; V

m

=√ xV

=>

S

V

m

= √ x 120.1

=>

V

m

= 169.8 v

Hence the maximum output voltage can be obtained as; V

dm

=( √

xV

V

dm

=( √

x 169.8 ) / 3.142

m

)/

(when

= 0, so cos 0 = 1) => V

dm

=

The average output voltage can be obtained as; V n = Vdc / Vd m So that from the given condition of problem; V n = 0.5 V n = Vdc / Vd m (a).

0.5 = Vd c / 280.81

R = 10 Ω => Vd c =

From eq 4-58, Vn=( =>

(b).

=>

&

) - - - - - - - - - (4-58) 0.5 = (

)

= cos – 1 (0.5)

=>

=>

= 60 °

The average output current is I

dc

=V

dc

/R

=>

140.45 / 10

=> I d c =14.045 Amps

The r.m.s output voltage can be obtained as; V r.m.s = √

xVm[

(

√

) cos 2 )] ½ - - - - - - - - (4-59)

4

=>

√ x 169.83 [

=>

(

√

) cos 2 x 60 °)]

x 169.83 [

=>

[

½

cos 120 °)] ]

½

=>

[

½

(

)]

½

=>

V

=> I

r.m.s

=> I

DT

= 4.68 Amps.

=> I

RT

= 9.2 Amps.

r.m.s

=

And the r.m.s load current is obtained as; I

r.m.s

=V

(c).

r.m.s

/R

=>

159.34 /10

=

The average current of Thyristor is given by; I

=I

DT

dc

/3

=>

14.045 / 3

Similarly the r.m.s current of Thyristor is obtained as; I

RT

=I

r.m.s

x√

=>

15.93 x √

=>

15.93 x 0.577

(d).

The rectification efficiency is obtained as; η = Pd c / Pa c - - - - - - - - - - (2 – 44)

Pd c = V

dc

xI

Pa c = V

r.m.s x

dc

I

r.m.s

η = Pd c / P a c (e).

=>

(140.45) (14.045)

=> Pd c = 1973.3 watts.

=>

(159.29) (15.93)

=> Pa c = 2537.5 watts.

=>

1973.3 / 2537.5

=> η = 0.778 or 77.8 %.

15.93 x

=> I S = 13

The r.m.s input line current is; I S= I

r.m.s

x√

=>

Similarly, the input volt-ampere rating is obtained as; VI=3VSx I Now

3 x 120.1 x 13

=> V I = 4683.9

T.U.F = Pd c / V I

=> (V (f).

=>

S

dc

xI

d c)

/ (3 V S x I S)

=> (140.45 x 14.045) / (4683.9)

=>

T.U.F = 0.421

The output power is given by; P O = (I

r.m.s

)2xR

=>

P O = (15.93) 2 x 10

=>

253.765 x 10

=> P O = 2537.65

The input power factor is given by; P.F

input

= P O/ V I

=>

2537.65 / 4683.9

=> P.F

input

=

Note: The Power factor is less than that of three phase semi converter, but higher than that of 3-Φ half wave converter.

5