A diode bridge rectifier is supplied from a 230 V, 50 Hz single phase ac source. If the load draws a constant current of 16 A, the conduction angle of diode, $D_{1}$ in degree will be

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Solution

Due to source inductance,

Average reduction in output voltage,

$Δ V_{d 0} = 4 f L_{s} I_{0}$

$= 4 \times 50 \times (12 \times 10^{- 3}) \times 16 = 38.4 V$

$Δ V_{d 0} = \frac{V_{m}}{π} [c o s α - c o s (α + μ)]$

As $α = 0^{o}$ for diodes

$38.4 = \frac{V_{m}}{π} [1 - c o s μ]$

$38.4 = \frac{230 \sqrt{2}}{π} [1 - c o s μ]$

$c o s μ = 1 - 0.37$

$μ = c o s^{- 1} 0.63$

$μ = {50.95}^{o}$

$∴$ Conduction angle of diode

$= 180^{o} + {50.95}^{o}$

$= {230.95}^{o}$

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A diode bridge rectifier is supplied from a 230 V, 50 Hz single phase ac source. If the load draws a constant current of 16 A, the conduction angle of diode, D1 in degree will be

Due to source inductance, Average reduction in output voltage, ΔVd0=4fLsI0 =4×50×(12×10−3)×16=38.4V ΔVd0=Vmπ[cosα−cos(α+μ)] As α=0o for diodes 38.4=Vmπ[1−cosμ] 38.4=230√2π[1−cosμ] cosμ=1−0.37 μ=cos−10.63 μ=50.95o ∴ Conduction angle of diode =180o+50.95o =230.95o

A diode bridge rectifier is supplied from a 230 V, 50 Hz single phase ac source. If the load draws a constant current of 16 A, the conduction angle of diode, $D_{1}$ in degree will be