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-   -   Magnetizing Current in Push-Pull Converter (http://www.diyaudio.com/forums/power-supplies/50414-magnetizing-current-push-pull-converter.html)

scottieman 27th January 2005 10:32 AM

Magnetizing Current in Push-Pull Converter
 
Dear All,

I need to build a step-up DC-DC converter using push-pull topology. The following link just shows you which topology exactly I am using. (http://home.netvigator.com/~blessyou/pushpull.html)

I have read alot of documentation about push-pull converter. However, it seems no one cares about the magnetizing current of the transformer (or the current due to the inductance of transformer).

The following is my problem (please refer to the above link):

When T1 switched on, the secondary side of transformer action forward biased diode D2 and the inductor current iL ramps up. At the same time, the magnetizing inductor is charged through T1 too. When both switches (T1 and T2) are off, the inductor (L) in secondary side is freewheeling with the help of diode D1 and D2.

However, where DOES the current of magnetizing inductor go?? CAN anyone helps me to understand this issue.

Thanks alot
Scottie

bjorge 27th January 2005 05:51 PM

The current in an inductor isn't interesting per se, it's the currentchanges. Whenever current changes, there is a voltage generated whose polarity would cause a current of opposite direction. This causes the inductor to "delay" a sudden current raise, and - opposite - generate a voltage that keeps up current when the external voltage that cause the current is gone.

In this circuit, the inductor charge will either be delivered to the load or the capacitor. If the inductance is large relative to capacitance, you will have a voltage spike. When "beautifully harmonized", the inductor and capacitor will both help smoothening the output voltage.

Claude Abraham 27th January 2005 09:25 PM

Re: Magnetizing Current in Push-Pull Converter
 
Quote:

Originally posted by scottieman
Dear All,

I need to build a step-up DC-DC converter using push-pull topology. The following link just shows you which topology exactly I am using. (http://home.netvigator.com/~blessyou/pushpull.html)

I have read alot of documentation about push-pull converter. However, it seems no one cares about the magnetizing current of the transformer (or the current due to the inductance of transformer).

The following is my problem (please refer to the above link):

When T1 switched on, the secondary side of transformer action forward biased diode D2 and the inductor current iL ramps up. At the same time, the magnetizing inductor is charged through T1 too. When both switches (T1 and T2) are off, the inductor (L) in secondary side is freewheeling with the help of diode D1 and D2.

However, where DOES the current of magnetizing inductor go?? CAN anyone helps me to understand this issue.

Thanks alot
Scottie


The magnetizing current takes the path through D4, then the lower half of the transformer primary, then through V1, recharging V1. The cathode of D4 goes one forward drop *below ground*, and the lower half of the transformer primary is subjected to a voltage of V1 + Vfwd. In other words, the magnetizing current that was present in T1 and the upper half of the transformer primary when T1 was on, continues through the other half of the primary winding, and the diode across the other MOSFET after T1 turns off. I hope I've helped. Best regards.

scottieman 28th January 2005 06:47 AM

Quote:

The magnetizing current takes the path through D4, then the lower half of the transformer primary, then through V1, recharging V1. The cathode of D4 goes one forward drop *below ground*, and the lower half of the transformer primary is subjected to a voltage of V1 + Vfwd. In other words, the magnetizing current that was present in T1 and the upper half of the transformer primary when T1 was on, continues through the other half of the primary winding, and the diode across the other MOSFET after T1 turns off. I hope I've helped. Best regards.
Thanks Claude. However, I have the following questions:

If the magentizing current really goes through the body diode of switch T1 or T2 (depends on the previous state) when both switches, T1 and T2, are off, it means there are voltage drop across the transformer primary side. The magnitude of the voltage acorss transformer equals to Vin (without accounting diode drop), and the time for this voltage to exist is D*T (D is duty ratio, T is the period of switching frequency).

Since any voltage across one winding of a transformer, this voltage will reflect to the other windings acrroding to the turn ratio. Hence, the transformer secondary sees a voltage pulse with 2*D on-time rather than 1*D.

If what I guess is correct, it makes me confusing @_@

Scottie


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