is this lighting circuit overloaded?

[marcosgue]

Hello friends, I'm come with this data in each luminaries: 46W 1.02A 0-10V dimming
Vin/Vent:120-277Vac
Freq/Frec:50/60Hz
Iin/Ient:0.46-0.20A
PF/PF:>0.90
Iout/Isal:1.02A
Vout/Vsal:30-45Vdc

This's 277V lighting circuit and the circuit has 50 luminaries this type plus emergency fixtures buckeye types and some recessed can luminaries.
The panel schedule call for 20A breaker and #12 awg branch circuit.
Is this lighting circuit overloaded? someone can help me with this question. Thanks in advance

 

[Little Bill]

If you go by the 1.02A listed in your description, then yes it's overloaded.
If you go by the watts divided by the voltage (46/277=0.17A) then x 50 lights=8.50A. then no. Even with the 125% (8.50x1.25 =10.63) its good.
Also would depend on the current used by the can lights.
So, I don't know if the input or output current would be used for calculation.

 

[hbiss]

Looks like Iin= 0.2A at 277V. That's what you go by. That's a total of 10 amps for 50 luminaries.

-Hal

 

[marcosgue]

Thanks Bill for your response, this's my point there's some where in the nec that I don't remember where's recommend use the amperage instead the watts for the calculation, but like you I don't know which current would be use.
about the can lights they 're only 10 and very little current, I'd like to focus in the 50 luminaries because is the major portion of the circuit load and I believe if 20A ocpd and #12 awg wire support this load, the rating would be ok for the entire circuit.
question, regarding to this continuos load circuit, instead of 20A the amperage should not be more than 16A in this circuit?

 

[marcosgue]

thanks hbiss, all your comments are very helpful like little bill and many others knowledgeable colleagues in this forum, but I'd like to go a little deep in this type of driver to understand why we use the Iin instead the Iout for the branch circuit calculation, their relationship and how the driver works, of course if this possible. Thanks

 

[synchro]

This is a constant current driver that provides DC power to the LED load. The Vout of 30-45VDC is the range of LED load voltages that it can accommodate while providing a constant current of 1.02A. The DC voltage depends on the voltage drop developed across a chain of series connected LEDs when driven by 1.02A. The more diodes there are in series the higher the DC voltage will be, up to the driver's capabilty of 45V maximum. And so the maximum DC power output that the driver can provide is 45V x 1.02A = 46W. Dimming will be accomplished by either reducing the amount of current from the 1.02A, either by supplying a reduced level of continuous DC current or by pulse-width modulating the 1.02A.

Note that the above comments are only about the DC output that the driver provides to the LEDs, and not about the AC input that you need to consider. As Hal mentions, the maximum AC input current feeding the driver will be 0.2A at 277V which is 55.4W. And so the driver has a 46W/55.4W = 83% efficiency under these conditions. When the driver is supplied with 120V, the maximum input would be 0.46A or 55.2W, which is very close to the same input power as at 277V. And so the driver maintains its efficiency over a wide range of AC input voltages.

 

[infinity]

I agree with Hal at 277 volts each driver is rated at .20 amps.

 

[texie]

Always use amps input as that accounts for PF. The branch circuit sees amps not just watts. Amps out is not relevant for branch circuit size.

 

[kwired]

Thanks Bill for your response, this's my point there's some where in the nec that I don't remember where's recommend use the amperage instead the watts for the calculation, but like you I don't know which current would be use.
about the can lights they 're only 10 and very little current, I'd like to focus in the 50 luminaries because is the major portion of the circuit load and I believe if 20A ocpd and #12 awg wire support this load, the rating would be ok for the entire circuit.
question, regarding to this continuos load circuit, instead of 20A the amperage should not be more than 16A in this circuit?

All the art 220 calculations are normally using VA, which would include any power factor on top of watts. Divide by voltage and result is your current.

 

[marcosgue]

if I understand this topic in the lights LEDs technology the drivers can handle a wide range of input currents which are relevant to size the branch circuit, and also there's a constant current which limit the power of the light that the driver can feed. that's correct?

 

[kwired]

if I understand this topic in the lights LEDs technology the drivers can handle a wide range of input currents which are relevant to size the branch circuit, and also there's a constant current which limit the power of the light that the driver can feed. that's correct?

The constant current is the output current to the LED itself, that needs to be same to maintain same light output.

Input current will vary depending on input voltage on those drivers that accept a range of input volts. (usually 120-277)

 

[marcosgue]

thanks for all very helpful responses and feedbacks

 

[infinity]

Match the voltage to the listed nameplate driver current and that's all that you need. The other numbers in involving the output are irrelevant.