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 v_o becomes equal to E again at ωt = π – θ. There, onwards the load current starts decreasing. Now if i_o 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

Therefore,

or,

• 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= V_i                   for α ≤ ωt ≤ π

 V₀ = 0                   for π ≤ ωt ≤ β

 V_o = E                     for β  ≤ ωt ≤ π α

 

 

 

 

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