Single phase half controlled converter in the discontinuous conduction mode
Single phase half controlled converter in the discontinuous conduction mode (a) π – θ ≤ β ≤ π (b) π ≤ β ≤ π α :
When T1 is fired at ωt = α the output voltage (instantaneous value) is larger than the back emf. Therefore, the load current increases till vo becomes equal to E again at ωt = π – θ. There, onwards the load current starts decreasing. Now if io becomes zero before T3 is fired at ωt = π α the conduction becomes discontinuous. So clearly the condition for continuous conduction will be i0|ωt= α≥0
- When this equation doesn’t hold good, the conduction becomes discontinuous.
When the conduction becomes discontinuous two cases arises.
- π – θ ≤ β ≤ π
- π ≤ β ≤ π α
- In the case of (π – θ ≤ β ≤ π),
V0= Vi for α ≤ ωt ≤ π
V0 = 0 for π ≤ ωt ≤ β
Vo = E for β ≤ ωt ≤ π α
Questions of this topic
A single phase half controlled converter charges a 48v 50Ah battery from a 50v, 50Hz single phase supply through a 50mH line inductor. The battery has on interval resistance of 0.1Ω. The firing angle of the converter is adjusted such that the battery is charged at C/5 rate when it is fully discharged at 42 volts. Find out whether the conduction will be continuous or discontinuous at this condition. Up to what battery voltage will the conduction remain continuous? If the charging current of the battery is to become zero when it is fully charged at 52 volts what should be the value of the firing angle.