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What should be a resonable exciting current for a 720VA 60Hz transformer? Primary 12V/60A and secondary 110V. I'm measuring about 2A (around 3% of maximum rated current) in the primary while the transformer is idle (secondary open circuit). This is the step up transformer of my inverter project.

Another question: when the transfomer is fully loaded @ 720VA, suppose for simplification a pure resistive load, does this exciting current still represent a loss? I.e, are these 2A still wasted? I understand part of this current is just reactive and part is effectivelly loss. I'm trying to figure out the maximum theoretical efficiency of a system like this.

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Your measured 2 A open-circuit current seems reasonable. You are measuring 24 VA, but the watts loss may be about 1/3 of that or 8 watts. You can assume that will be the iron loss and that it will not change with load. Once you load the transformer, you will see copper loss added to that. If you want to determine the efficiency of the transformer, you will need a wattmeter. If you search the internet, you should be able to find the procedure for performing open-circuit and short-circuit transformer tests and calculating copper losses.

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In a typical 60 Hz power transformer intended for sinusoidal waves an excitation of between 5 and 10% is acceptable. So, the 3% you measure (assuming you are measuring it correctly and under correct conditions) is reasonable. However, if you are driving the primary with square waves as in a typical inverter topology, the classic sinusoidal excitation rules of thumb are not quite applicable because of the high frequency components of the square wave excitation.

The exciting current is not considered a loss because it is in quadrature (90 degree phase shift) from the excitation voltage. Under open circuit secondary conditions the primary and the core essentially form an inductor. There are both copper losses due to the excitation current flowing thru the primary winding, as well as iron losses due to magnetic flux flowing in the core, even when the secondary is unloaded. If you are driving the primary with square waves, a normal 60 cycle iron core will loose more energy to heat than it would with sinusoidal drive due to the higher harmonic content of the square wave. So, you should choose a "better" material for the core to improve efficiency and avoid overheating the core.

"Maximum theoretical efficiency" would be 100% if the theory assumes lossless iron and Ohmless copper! Not likely in the harsh, cruel world of electronics reality in which we exist.

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  • \$\begingroup\$ 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? \$\endgroup\$ – Ron Groove Jul 19 '16 at 11:35
  • \$\begingroup\$ 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? \$\endgroup\$ – Ron Groove Jul 19 '16 at 11:47
  • \$\begingroup\$ 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. \$\endgroup\$ – Charles Cowie Jul 19 '16 at 13:52
  • \$\begingroup\$ 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. \$\endgroup\$ – FiddyOhm Jul 19 '16 at 17:44

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