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
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'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.
The added benefit is that they look awesome at dusk!
Just keep the case temperature below 90-95degC ( die <120degC ), otherwise your LEDs will die.
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.
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
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.
When you say 50 degrees, do you mean Fahrenheit or Celsius?
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