need help on full bridge converter

[y6n1212]

Hi all, i am revising on the basic full bridge dc-dc converter.

It's actually pg 618, equation 14.40 of Muhammad H Rashid's book- "Power Electronics Circuits, Devices and Applications 3rd edition".

I've think for such a long time to derive an equation. How can i get Pi=Vs Ip(avg) k? (where Vs= primary voltage and Ip(avg)= average primary current, k=duty cycle)


Please help.
Thanks!!

 

[ahsan_i_h]

May be Ip(avg) is the average of the peak two levels of the current wave. For example for a square wave(0A then 2A then 0A, then 2A like this), we get 2A just after rising edge and also 2A just before falling edge. Now what if, after rising edge it is 2.3A and before falling edge it is 1.9A. Then we have to make an average of 2.3A and 1.9A like (2.3A+1.9A)/2=21A to express the peak current level of that squarewave. Then we will treat it as it is a squarewave like 0A then 21A then 0A then 21A then 0A etc.

 

[y6n1212]

May be Ip(avg) is the average of the peak two levels of the current wave. For example for a square wave(0A then 2A then 0A, then 2A like this), we get 2A just after rising edge and also 2A just before falling edge. Now what if, after rising edge it is 2.3A and before falling edge it is 1.9A. Then we have to make an average of 2.3A and 1.9A like (2.3A+1.9A)/2=21A to express the peak current level of that squarewave. Then we will treat it as it is a squarewave like 0A then 21A then 0A then 21A then 0A etc.

Thanks.
But how can i get the equation of input power? Pi=k * Vs * Ip(avg). Pi= average V * average Ip right? Where is k * Vs come from?

 

[ahsan_i_h]

For a square wave current 0A then 2.1A then 0A then 2.1A then 0A then...... like this, The average current is 2.1A*k.
Now we know that, power=voltage*average current.
So Pi=Vs*2.1A*k
This 2.1A is the Ip(avg) because we made the 2.1A by averaging 2.3A and 1.9A.
The name of the Ip(avg) should be average of the peak of primary current.

If you need more explanation, ask again.

 

[y6n1212]

For a square wave current 0A then 2.1A then 0A then 2.1A then 0A then...... like this, The average current is 2.1A*k.
Now we know that, power=voltage*average current.
So Pi=Vs*2.1A*k
This 2.1A is the Ip(avg) because we made the 2.1A by averaging 2.3A and 1.9A.
The name of the Ip(avg) should be average of the peak of primary current.

If you need more explanation, ask again.

Thanks so much!! understand totally, great explanation from you!!:-D