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Operation of single-phase half-wave rectifier with active load

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Conditions: La = 0, Ld = 0, a=0

 

The a peak rectified current of the load and a peak thyristor current are

 

Direct component of a rectified voltage is

 

Direct component of rectified current and current of the thyristor are

RMS current of the thyristor is equaled to RMS current of a secondary winding of the transformer

 

A peak inverse voltage

Let's define 1-th harmonic Ud and id

 
 

 

 


It is an even function.

Then

A RMS rectified voltage

A ripple factor of rectified voltage Ud for 1-th harmonic is

A ripple factor of rectified voltage Ud is

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

Let’s define rms value of i1

Let’s define the total powers of the transformer’s windings.

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