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Operation of the half-wave rectifier with active - inductive load and limited inductance

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Conditions: a=0, ra=0, La=0, 0 < Ld < ∞


Figure 1.5

 

Ld is connected in series with a load for smoothing a rectified current.

Equivalent resistance and inductance of a circuit are

According to second Kirchhoff’s law

where

Characteristic equation is

From here

We should search out solution as i=iss+ifr; where ifr – a free component, iss - a steady state component.

Where

Constant A we could be found from initial conditions:

At

Then

from here

Let's designate

Then

 

 

Figure 1.6

 

Let's find the energy reserved in inductance L by period

Energy which inductance accumulates

Energy which inductance takes out

Thus, in the time of period

A direct component of the rectified voltage is

where l - a pulse load current duration.

A direct component of a rectified current is

It is possible to determine λ from the condition: at u = λ Þ id =0

 

If Ld would increase, λ also increases and Idm goes down, hence, Kr decreases. Thus, it is possible to use Ld as the filter of a load current.

 


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