Dimmable led min 60% - use power resistor?

[grasfulls]

I will be installing LED undercabinet lighting where the driver requires a minimum 60% load. Not a problem EXCEPT one standalone cabinet on its own switch and using only 4.18 watts, min driver size = 10W. Manufacturer wants me to use a Lutron synthetic load (lottsa $ just to dim), I want to use a power resistor.
Can I do this?
What value (watts and ohms)?
Does it get wired in parallel to the load?

When I looked for power resistors I see several with 5 watt ratings but different resistance values, I do not understand that. Ignorance is not bliss.

LED tape = Diode LED Arhc24, 2.09Wperft
https://www.diodeled.com/avenue-24-premium-24v-led-tape-light.html

Driver = Omnidrive with 24vdc output
https://www.diodeled.com/custom/download/productFile/filename/DI-TD-Cut Sheet.pdf/

 

[iwire]

It might be cheaper just to hide some extra LEDs away.

 

[grasfulls]

Hide LEDs

Hide LEDs

It might be cheaper just to hide some extra LEDs away.

sheesh....totally enclosed top of cabinet, easy place to do just that, except light may bleed through crown molding....I can box them in ..
Thanks, that is a good idea.

 

[drcampbell]

A 2400-Ohm resistor will dissipate 6 watts at 120 volts.
http://www.resistorguide.com/ohms-law/

A resistor's power rating is based on a certain temperature rise and opportunity for heat to escape - it might get mighty hot in a confined space.

 

[Smart $]

A 2400-Ohm resistor will dissipate 6 watts at 120 volts.
http://www.resistorguide.com/ohms-law/

A resistor's power rating is based on a certain temperature rise and opportunity for heat to escape - it might get mighty hot in a confined space.

The load is required to be equal or greater than 60% of the driver output rating. The output rating of the driver is 10W @ 24V. LED tape cut to length will be a 4.18W load. OP'er needs to increase the load at least 1.82 watts to achieve 60% of 10W.

24V² ÷ 1.82W = 316 ohm resistor maximum, 2W minimum

24V² ÷ (10W – 4.18W) = 99 ohm minimum, 6W resistor minimum

24V² ÷ resistor ohm rating* = minimum resistor wattage dissipation rating

*Resistor rating has a tolerance. Use lower range value... and always round up.

 

[iwire]

I put forth adding a resistor to the circuit runs afoul of 110.3(B).

 

[Smart $]

I put forth adding a resistor to the circuit runs afoul of 110.3(B).

I agree the instructions do not say one can install a resistor, but they do not say you can't. In fact, in such a predicament they "almost" say you can or rather have to in a footnote:

Ensure to load the driver at least 60% the labeled load for proper
dimming performance (required for dimmable installations only).

 

[cadpoint]

Well, it does happen and it wasn't in this original wiring diagram.

This is 120V actuator for cooling lines, the resistor limited the slam that the power would do
to the actuator ,extending the life of the actuator, per the HVAC tech.

We were hired by the HVAC Contractor on original install of all wiring for exact temperature
controlled refrigerators, .1 of a degree. HVAC contractor supplied the resistor.

I don't know why you don't want to use what you were told to use, besides it's California.
www.lutron.com/sensors

 

[drcampbell]

That's an RC snubber, not a resistor. A .01-microfarad capacitor and 100-ohm resistor in series.

 

[winnie]

The OP was told to use a Lutron synthetic load...which doesn't really make sense; the synthetic load is device intended to operate on the 120V side to provide a minimum load for the 120V dimmer; it is not a device intended to operate on the DC side of the LED driver...unless there is another sort of Lutron synthetic load.

If the minimum load on the LED driver is only to permit proper function of the upstream 120V dimmer, then simply having multiple LED sections with the separate drivers all connected to the same dimmer is probably sufficient.

If this stand alone section of LED is being used with its own dimmer, then I'd simply put a couple more feet of strip on that driver, and let the dimmer be adjusted as needed.

I'm in with the idea that the simplest way to add proper load is to use a bit more LED strip. A simple resistor won't match the function of the LEDs and might make that section dim differently than the rest; you would need a properly sized resistor connected in series with rectifying diodes to better mimic the characteristics of the LED strip, at which point you might as well simply use the LEDs.

IMHO your best approach is to figure out where you can actually _use_ another few watts of LED strip. Since this is a stand-alone section, it might very well tolerate being a bit brighter than the rest of the installation, or perhaps could use a special 'lit shelf' in the cabinet. I am bugged by the idea of an LED strip hidden away, but that might be best .

-Jon

 

[cadpoint]

That's an RC snubber, not a resistor. A .01-microfarad capacitor and 100-ohm resistor in series.

That's fine, if you zoom in on the first picture one can see the diagrammed line work stamped on the object.
Thanks for the clarification, I was told it was what it was, and was required to install...

 

[paulengr]

LED's are current devices. So normally when we build a power supply the goal is to regulate voltage while allowing the current to fluctuate as needed. In contrast for an LED power supply the goal is to regulate current while allowing voltage to fluctuate as needed. For larger power LED's I believe they even have nasty problems like lower resistance as temperature increases. The voltage drop from red/yellow LED's is about 3 V and that from the green/blue LED's is around 6 V. "White" LED's are blue ones with a phosphor coating that emits broad spectrum light (with an 80% efficiency).

Thus if your LED strip isn't "long enough" you have to make it up with more "LED's" although as stated this could be a resistor. The easiest way is with more LED's.

 

[winnie]

The 'led strips' used for lighting are usually designed with a combination of LEDs and resistors, and are intended for operation on a constant voltage supply. The resistor lowers efficiency, but acts to allow the system to be used with a constant voltage supply.

This is important because the LED strips consist of parallel sets of 2 or 3 LEDs in series with a resistor. If you didn't have the resistor and were depending on current drive, you would also have to depend on much better matching of the individual LEDs.

-Jon