Heating Element for Outdoor Camera System

[megahurts]

Hi,

I am building an outdoor allsky camera, am almost finished with the project, but struggling with the last part.

In order to keep dew and frost off of the camera dome we have to create a heater using 50 watt resistors and also attach a fan to blow the wind heat toward the top of the dome. A thermostat is used to regulate the temperature to 100 degrees F.

Here's some pictures of the setup:
Sentinel III Photos

Conceptually I'm having a hard time figuring out how to wire and power the heaters and the fan. I currently have a 12v DC power line coming into the enclosure for the purpose of powering the camera.

Is it possible to splice the camera's 12v DC power supply and run wires to the 2 resistors and the fan? Or will I need to have more than 1 power line coming into the enclosure? If I need an additional power, what kind?

Thanks for the advice.

 

[Dr_Doggy]

if it is all 12volts, you need to add up current required by devices VS power supplied,

DEPending on your thermostat output you may use a transistor-relay config or maybe be able to wire it direct

 

[ronv]

Pretty nice!
It is hard to tell from the pictures how everything is hooked up. The real question is what are the resistor values. This will tell us what voltage the design was made for. I can tell they are in parallel so if they are made for 12 volts i would guess them at about 5 ohms each or 2.5 ohms total. This would dissipate 28.8 watts each. If this is the case and your power supply can supply 5 amps or so at 12 volts you can use the same one. Ground to one side of the resistors, +12 to one wire of the thermostat, the other wire from the thermostat to the resistors. If the fan is to be switched it should be in parallel with the resistors. As Doggy stated the themostat must be able to switch 5 amps in this case.

 

[RadioRon]

We have to estimate how much heat you will need. If it is freezing outside, and you are using a thermostat, perhaps about 50 watts total is enough. One way to test that is to put a 50 watt light bulb inside the tube and power it with the tube all closed up and watch and see how hot it gets inside vs the temperature of the air outside. I'm guessing you want about a 25 degree C rise above ambient.

You need to understand ohm's law and the relationship to power. You can work backwards from the power you need, which I"m guessing is 50 watts. You multiply 50 watts x one half the value of one resistor (because they are hooked up in parallel) and then take the square root of the answer to get volts. You have to put this amount of volts (ACrms or DC) across the resistors to get that amount of heat. So tell us, what are the resistor values?