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ZW-L AMP DRAW

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ZW-L AMP DRAW
Posted by Koala on Friday, January 11, 2013 3:09 PM

Just got a few of the new ZW-L transformers.  The secondary is rated at 18volts and 620 VA.  I am trying to figure out how many of these things  I can plug into a 110 volt 15 amp circuit.  I came up with 5.2 amps on the primary side of the transformer.  I guess 2 would be correct or 3 if I use a 20 amp circuit.  Any ideas.

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Posted by ADCX Rob on Friday, January 11, 2013 7:04 PM

The ZW-L will put about a one horsepower load on the AC mains, but only with at the full 620 watt output, which would be a rare occurrence.  In practical use, 2 on a 15 amp circuit would be fine.

Rob

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Posted by Laurastom on Friday, January 11, 2013 9:28 PM

When calculating primary current almost all US electrical systems are 120V. The efficiency of the ZW-L also needs to be considered. I do not know what it is at high current output but I would start with 90%. If  the ZW-L's are operating at full load two is the limit. 620W➗120V➗.9=5.75A pimary current draw. Having more tham one putting out maximum power simultaneuosly is highly unlikely. Using a 50% duty factor would allow four to be connected to one 15A  circuit. 

Tom

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Posted by phillyreading on Thursday, January 17, 2013 9:25 AM
While I don't know all the information on the new ZW-L, I would say not to use more than two on a 15 amp 120 volt circuit. If there is any over-current draw by the ZW-L's it might pop-off the circuit breakers using three ZW-L transformers. Also most residential circuits have other things on the circuit such as lights or TV sets, so you can not go with a true figuire unless you add in all the current draw items for that circuit.
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Posted by lionelsoni on Thursday, January 17, 2013 9:43 AM

Does this ZW-L use phase control (like the CW-80)?  If so, the power factor may be rather low, meaning that the RMS current is greater than the power divided by the line voltage.  You need to divide the apparent power (usually expressed in volt-amperes) rather than the real power (usually expressed in watts) to get the RMS current, which is what heats the wires and trips the circuit breaker.

Bob Nelson

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Posted by Laurastom on Thursday, January 17, 2013 9:25 PM

Really interesting point, Bob. Lionel's video says this transformer is chopping the leading edge of the waveform. An oscilloscope would provide that answer. It ultimately depends on the transformer load. If it is fully loaded with incandescent lights then two is all a 15A circuit will supply. We're these mine I would just plug them all in the same circuit and see if the breaker tripped. If it did then it is time to install a new circuit from the panel. Probably not the best advice from a EE but sometimes the quick, simple approach works. 

Tom

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Posted by webenda on Friday, January 18, 2013 2:51 AM
CW-80-10 CW-80-50 CW-80-100 LIONEL 4150-90

 ..........Wayne..........

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Posted by Laurastom on Friday, January 18, 2013 10:20 AM

Thanks. First time I have seen the output waveforms. It demonstrates Lionel's statements that these power supplies will result in greater smoke output in Legacy engines than a PW transformer. Despite the output waveform shape at full power setting I would expect the primary side to not see any of this. The VA rating on the primary should be a good indicator of the current the house 15A breaker will see. 

Tom

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Posted by lionelsoni on Friday, January 18, 2013 1:22 PM

Except for a small component due to the magnetizing current (which in any case is likely not to be very sinusoidal in a small transformer), I would expect the primary-current waveform to match the waveform of the secondary current.  But these pictures are of output voltage, not current.  I agree that the volt-ampere specification, if any, is the number to use to calculate RMS primary current.

Bob Nelson

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