1. Single Phase Half Wave Controlled Converter

Ex. No. 1 14.July.2015

Date:

Single Phase Half-Wave Controlled Converter Aim :

 Desi Design gn a sing single le phas phasee half half-w -wav avee cont contro roll lled ed rect rectif ifie ierr supp supply lyin ing g powe powerr to a load load of 

[ 100 + j 31.4 ]

 ohm. A 230 V, 50! single phase voltage source is employed to deliver this

 power to the rectifier circuit. Derive an e"pression for

i ( ωt ) , the average current, the power 

a#sor#ed #y the load and the power factor for the gating angle of

45

o

 $omment on delay angle if a source of %00V dc is connected in series with the &-' load and derive the

(

i ωt 

)

.

Components Required: S.No

% 2 3 / 5 

Desription

(hyristor )*$&+, ( Diode, ener diode &' 'oad &esistors $apacitors :solation (ransformer

Ciruit Diagram:

Speifiation

!t".

2/, 212/ %1/004,%2V /.6,%.5m %006, 3.376, 2206, 50078 pot 0.%u9 %;%

% 2 % 3, %, %, % % %

#ig.$ schematic diagram of open loop control of single phase half-wave controlled converter 

#ig.%schematic diagram of closed loopcontrol of single phase half-wave controlled converter 

&heoretial 'aveforms:

Fig.3 Output waveforms of synchronized ramp signal based pulse Generation circuit with dierent stage

Anal"sis: Case: $ R-( (oad (he current flowing to the controlled half- wave rectifier is sum of the forced and natural response therefore the current is given #y

i ( ωt ) =i f  ( ωt )+ in ( ωt )=

V m

−ωt /ωτ 

sin ( ωt −θ ) + A e

(1)

(he constant  A is determined from the initial condition

i ( α ) =0

i ( α )=0 =

 A =

[

V m Z 

V m Z 

sin

Z 

− ωt / ωτ 

sin ( α −θ ) + A e

]

( α −θ ) =e−α /ωτ ( 3)

(2 )

*u#stituting for A and simplifying

¿ V m ( α  ωt ) ωτ  sin ( ωt −θ )−sin ( α −θ ) e − / for α ≤ωt≤ β Z  (4 ) 0 otherwise

¿

i ( ωt )=¿ (he e"tinction angle

 β  is defined as the angle at which the current returns to !ero, as in case of the

uncontrolled rectifier.
ωt = β ,

i ( β )=0 =

V m Z 

sin ( β −θ ) sin ( α −θ ) e

( α − β ) /ωτ 


Fig.4 Typical wave forms of half-wave controlled converter

9rom 9ig/, (he average )dc+ output voltage is  β

V m 1 V dc =V o= V m sin ( ωt ) d ( ωt )=  ( cos α −cos β ) ( 6 ) 2 π  α  2 π 

∫

(he average current is computed from

(5 )

γ  is

 β

1  I o= i ( ωt ) d ( ωt ) ( 7 ) 2 π  α 

∫

(he rms current is computed from  β

1 2  I rms = i ( ωt ) d ( ωt ) ( 8) 2 π  α 

∫

Case: %R-( Soure (oad A controlled rectifier with a series resistance, inductance, and dc source is shown in 9ig.5. (he analysis of this circuit is very similar to that of the uncontrolled half-wave rectifier. (he ma=or difference is that for the uncontrolled rectifier, conduction #egins as soon as the source voltage reaches the level of  the dc voltage.

Fig. 5 Typical wave forms of half-wave controlled converter

 9or the controlled rectifier, conduction #egins when a gate signal is applied to the *$&, provided the *$& is forward-#iased. (hus, the gate signal may #e applied at any time that the ac source is larger than the dc source; −1

α min= sin

( )( ) V dc V m

9

(he current is e"pressed with

α   specified within the allowa#le range

¿ V m V  ωt /ωτ  sin ( ωt −θ )− dc ∓ A e for α≤ ωt ≤ β Z   R (10 ) 0 otherwise

¿ i ( ωt )=¿


 A =

[

V m Z 

sin ( α −θ ) +

 ]

V dc  R

Simulation Ciruit:

Simulation Waveform:

−α /ωτ 

e

( 11)

Hard'are Ciruit:

Hard'are Waveform)R (oad*

Result: *ingle phase half wave converter designed and implemented.

Conlusion: