Timeline for Reasonable maximum exciting current 60Hz step up transformer
Current License: CC BY-SA 3.0
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| Jul 19, 2016 at 17:44 | comment | added | FiddyOhm | If your think of the reactive part of the current as current that "bounces back" from the inductor, it will be easy to understand that it represents a certain amount of energy that the inductor at first absorbed and then returned to the source. Since your battery cannot re-absorb this returned energy it is indeed lost energy and must be debited against the efficiency of the converter. Depending on the type of battery, you may be harming the battery with these continuous "recharge" pulses. You should provide some way (e.g. diodes) to prevent them from reaching the battery's terminals. | |
| Jul 19, 2016 at 13:52 | comment | added | user80875 | A battery can not simply be recharged a little bit during each cycle. The inverter circuit would need to have some capacitance to accommodate the reactive component of an inductive load. Any harmonic content in the inverter output would reduce the transformer efficiency as described. Losses associated with the reactive current would reduce the inverter efficiency. If you want to ask questions about inverter design, post them separately. Generally, we have one issue per question here and try not to expand questions and answers into discussions. | |
| Jul 19, 2016 at 11:47 | comment | added | Ron Groove | Sorry for another comment. I was thinking here, maybe the current we see from the battery regards only to the real losses? If we had only the reactive power, due to the inductance, we would see current FROM the battery but also current TO the battery. I know the current is not only DC but has an alternate component. So, in this case, the inverter would be pumping current from and to the battery? This would account for the reactive power part? | |
| Jul 19, 2016 at 11:35 | comment | added | Ron Groove | Thanks for the comments! I'm using pure sine wave to measure the current since my inverter is based on pure sine wave. One thing that puzzles me is that I know the exciting current is shifted from the voltage, so not consuming real energy. But in the inverter condition, it will consume current from the battery so it will really waste real energy. The theoretical efficiency I meant would consider this exciting current. So if I have a transformer that consumes 3% of the current, when fully loaded this system would be 97% efficient if no other losses were considered. Is that makes sense? | |
| Jul 19, 2016 at 3:49 | history | answered | FiddyOhm | CC BY-SA 3.0 |