Operation of single-phase half-wave rectifier with active load
Conditions: La = 0, Ld = 0, a=0
The a peak rectified current of the load and a peak thyristor current are
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Direct component of a rectified voltage is
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Direct component of rectified current and current of the thyristor are
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RMS current of the thyristor is equaled to RMS current of a secondary winding of the transformer
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A peak inverse voltage
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Let's define 1-th harmonic Ud and id
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It is an even function.
Then
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A RMS rectified voltage
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A ripple factor of rectified voltage Ud for 1-th harmonic is
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A ripple factor of rectified voltage Ud is
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RMS value of EMF of secondary winding is
.
Instantaneous value of i1 we should define from a magnetic balance condition of the transformer by a variable component. Magnetic voltage drop is not taken into account. The direct component is not transferred into a primary winding.
Then
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Let’s define rms value of i1
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Let’s define the total powers of the transformer’s windings.
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