Energy recovery (reset) winding on isolated forward converter - Page 1
Forum section: electronics / beginners

author: utn56
author_rank: Newbie
timestamp: December 21, 2012, 08:29:47 pm
content: Hi all,I´m studying for my Power Electronics Uni course, have a small doubt with Forward converter. The topology is the same as in:http://en.wikipedia.org/wiki/Forward_converter.Of course primary winding has a magnetizing inductor which is not drawn there.The question is: During the time the switch is off, the energy stored at the core (magnetizing inductor) is "recovered" or sent back to the source by means of the auxiliary reset winding. Is this correct?But if the circuit is powered by a non-rechargeable battery, I don´t think that´s a good idea. How is it implemented in that case? I think a diode and large capacitor between the battery and the circuit will help (because one can recharge a capacitor, indeed), but really don´t know. Am I right?Thanks in advance.

author: PA0PBZ
author_rank: Super Contributor
timestamp: December 21, 2012, 09:04:01 pm
content: Quote from: utn56 on December 21, 2012, 08:29:47 pmThe extra winding + diode is to demagnetize the transformer, not to recover energy. But I see your point, it has to dump the energy somewhere and since the switch is off when this happens it has to be dumped into the battery. It will not be charged however, because the energy you get out of it will always be lower than what you put in, it's not another infinite energy resource.

author: utn56
author_rank: Newbie
timestamp: December 22, 2012, 02:42:29 am
content: Quote from: PA0PBZ on December 21, 2012, 09:04:01 pmYes, I understand that. During demagnetization the energy stored in the transformer´s core is sent back to the voltage source.Quote from: PA0PBZ on December 21, 2012, 09:04:01 pmNo, please don´t think I´m one of those weirdos trying to get the perpetual motion. I understand that only the energy stored in the transformer core is recovered. The rest is sent to the load through that buck-like circuit and disipated there. So the total balance is that the source is giving energy, not receiving.But my question is pointing to the fact that during demagnetization (when the switch is off indeed) there is a small portion of current flowing backwards through the source. I don´t know 100% how a non-rechargeable battery works (chemics), but I suppose one cannot recharge that kind of battery even with that small pulse of current. At best chance that energy will be disipated in the battery, but not recovered.So I assume a diode in series and a large capacitor in parallel with the circuit, would help, because the capacitor can be recharged. But I´m not sure if this is the usual way.Thanks.

author: SeanB
author_rank: Super Contributor
timestamp: December 22, 2012, 04:29:12 am
content: Batteries can be charged bit it is not going to be very efficient. A slight charge will not harm them though. The energy returned to the supply will be stored by the input capacitor you need there, and will just raise the voltage slightly if you are using a diode to feed the input ( mains transformer to provide the input) by a small amount. The next cycle will use this energy again. This then shows up as the input ripple voltage.I have charged zinc carbon cells using a low current pulsed DC, as the radio they were in was used mostly on mains, but often on battery, so the prolonging the life of the cells by putting charge back in them was useful. It more than trebled the life of the cells, and over the years they did not leak at all. Alkaline cells though do not like the charging.