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Easy LED Sonar Heater

Posted by Xerravon on February 21, 2013 at 1:30pm

They say the sonar modules will not work under 50 degrees Fahrenheit (10 degrees Celsius). I have tried using the 4, 10 ohm resistors but this seems to work much better, and is easier to install for me This is how I keep my sonar module warm in the winter, all you need is a small piece of the strip of red LED's, this is where I got mine

http://www.quadroufo.com/product_info.php?products_id=78&osCsid=h227449ro31sm8hvhoo3ge3sv4

make sure to get red, they are much hotter than other colors. (red LED's usually use more power (watts) than other colors)

you only need a piece with 3 leds o it, it wraps around the sonar perfectly (currently cost .46 cents)

Just remove the film for the adhesive and wrap around your sonar, solder the wires to 12v and use a tie wrap to hold it securely. I hooked mine into my lighting so I can switch them on and off.

here is some pic's.

led, heater, sonar

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Comments

Craig Fraser February 22, 2013 at 2:54pm

Here's what we commonly use in "hobby astronomy" (as an alternative): 330 ohm 1/2W resistors. The last bag of 100 I got at digi-key was ~$3. This is for 12V/3s use only, adjust resistor value for other voltages. Parallel as many resistors as you want/need, each produces ~1/2W of heat (100% efficient! lol).

Typically we lay out two bare copper conductors, fasten the ends down, then solder the resistors across the conductors every 1/2 to 3/4 inch. Usually put a sheath over the contraption, some just use duct tape. Wrap heater around whatever you want to keep warm, put down pants, etc.
Richard Warrender February 22, 2013 at 4:17am

The added benefit is that they look awesome at dusk!
Nick Arsov February 22, 2013 at 3:11am

Hi guys,

Just keep the case temperature below 90-95degC ( die <120degC ), otherwise your LEDs will die.

Cheers
Chris Card February 21, 2013 at 10:50pm

Mark you are correct about the efficiency. I was just using easy numbers to illustrate that visible light leds will provide heat in a thermal form not a radiated form like other types of lamps.

cheers
Xerravon February 21, 2013 at 8:56pm

Hi,

I should have said the red LEDs use more power (watts) not current. this is because of the lower forward voltage. for a 1 watt red led the typical forward voltage is 2.2v for a blue 1 watt it is 3.5v. to calculate the value of the resistor for current limiting you subtract the forward voltage from the supply voltage then divide the left over voltage by the current you need for the diode (led) to calculate power P = I x E

so for a 1 watt red running at 350ma, 12v - 2.2v = 9.8v... for power 9.8v x 350ma = 3.43 watts

for a 1watt blue led at 350ma, 12v - 3.5v = 8.5v... power 8.5v x 350ma = 2.975 watts

more power, more heat
Craig Burden February 21, 2013 at 8:11pm

The reason the Red LED's are so much hotter is that they run off the lowest voltage of all the colours. This is because it takes less energy to create a red light wave than any other visible colour as it has the lowest frequency.
Randy Brazeau February 21, 2013 at 5:25pm

50 degrees C would be really hot.
Randy
LanMark February 21, 2013 at 5:11pm

Chris it is more like 90/10 not 20/80.. I think you are thinking incandescent bulbs not LEDs.. which are very efficient.
Ben Carson February 21, 2013 at 5:07pm

When you say 50 degrees, do you mean Fahrenheit or Celsius?
Chris Card February 21, 2013 at 4:35pm

Nice Idea! And its FUN.

The thing about LED's is that all of the "wasted" energy is thermal. So if an LED is 20% efficient at producing visible photos, then 80% of the consumed energy is lost as thermal energy and in this case it is going to some use.

Wrap some thin translucent foam around them for better heating