Electronic – Power dissipation when connecting LED to +-15v

For a device you will often see a figured called \$\theta_{JA}\$. This is called thermal resistance.

This tells you that in a typical ambient environment for every watt dissipated, the device will heat up x°C above ambient. You must include ambient temperature into your calculation. In an open lab environment, it might be 25°C but in reality inside the casing of some electronics it can be much hotter.

If you add a heatsink you need to know \$\theta_{JC}\$ (junction-case resistance), \$\theta_{CI}\$ (case-insulator resistance, if any), \$\theta_{IH}\$ (insulation-heatsink resistance, if any), and finally \$\theta_{HA}\$ (heatsink-ambient resistance.) Like normal electrical resistance you can add these together to get a final figure for how much your device will heat up when it dissipates x watts.

1. Depends on what the LEDs are rated for. The left branch LEDs have a forward drop of 11.8V, and the right branch drops 10.8V. The limiting resistors at the bottom of each branch should restrict the current to 1A, so make sure your LEDs are able to handle that current.

2. A better way to drive this would be with a constant current LED driver. A switching regulator will be far more efficient and you won't be burning as much power as heat (if that matters for your application). The method you have here will work fine as long as your supply stays fixed. If it sags at all your LEDs will dim proportionally.

3. Depends on the LEDs. You need to read the specifications for whatever in particular you're using. See how much power the LEDs will dissipate thermally, and look at the environment you're running them in. If you're in a well ventilated space you could get away with just affixing them to a metal heat sink if the average temperature of your space isn't too high. If it's in a tight or enclosed space you might want to have the fan just to force some circulation over the LED heat sinks.

4. Depends on the specs of your components, the environment you're operating in, and how effective your passive heat sinks are.

5. Read the specifications of your supply. Generally you buy the supply to fit the loads. In this case you have 2A drawn by the LEDs (1 A in each of the parallel LED branches), plus whatever your fan draws which isn't indicated on the schematic.

6. Might be a good idea, there's the potential the supply could catch fire if overheated significantly, however it's more likely it will just burn out. Are you overdrawing the intended output current from it? If that's case you can keep it for a lighter load. If it's just getting hot while unloaded or at normal load recycle it.

7. No. There's a 100x difference between the two.

8. Switch is not an issue, just make sure it's rated for the current. Keep in mind if you just put the potentiometer in series with the LED supply whatever power doesn't get used by the LEDs will get dissipated as heat in the potentiometer...which might hurt. It's not an efficient way to do it. Better option would be to have an adjustable constant current regulator. Or you could use a MOSFET and a 555 timer to switch the lights at a fixed frequency and adjust the duty cycle to control the brightness.

9. No. You don't have the voltage to drive any more LEDs down either branch. I just saw the edit though. You could use the Q and /Q outputs of a 555 to drive some FETs to alternately run two separate circuits.