What is your experience RE: LED bulbs, relays and dimmers; and heat?
This is a little esoteric, but I don't know where else to start looking, and it is for darkroom consumption.
I'm hadn't been happy with the general work lights (white lights) in my darkroom, and am now on the 3rd iteration, which is two bare 150 watt light bulbs in porcelain ceiling fixtures. The light is now fine, but that's 300 watts and a surprising amount of heat! Since fluorescent is out (we've discussed this to death - suffice it to say I've seen the glow ...) I've thought about substituting LED bulbs, which would reduce the wattage consumption. However, LEDs with any comparison is actual light output to high wattage tungsten have heat sinks! Do they also put out heat? If they put out no more heat, or less, than the equivalent tungsten, then the point is moot. However, tungsten bulbs don't come with built in heat sinks! :blink:
I've also considered replacing the tungsten bulbs in the safelights with LEDs, but am concerned about the heat AND the warnings with many LED bulbs to not use them with relays or dimmers. OK, dimmers is not an issue, but what about the relay circuits in enlarger timers? What is going to be damaged: the LED bulb, or the timer?
OK, let's stay on topic. i.e., do LEDs produce heat like tungsten, and do they damage relay circuits? Does it make a difference if the relays are mechanical or electronic?
More than you probably wanted to know about biasing LEDs
It's been a long time since I did any work with LEDs (back in the '90s when I was an Electrical Engineering student), but typically you have a power supply of X volts (let's say 6V for sake of argument). A typical red LED has a forward voltage drop of 1.7V. Typically you choose a current value of 50% of the LED's rated current (Imax). We'll say for sake of argument that this is 20mA.
The voltage drop that will occur across the resistor is 6V - 1.7V, or 4.3V (power supply voltage - forward voltage drop of the LED). Ohm's law for resistors says R = V/I. R = 4.3 V / 20mA, which works out to 215 ohms. Pick the closest resistor you have and wire it in series with the LED.
Since P=IE (also Ohm's law), the power dissipated by the LED (including light emitted) is 1.7V * 20mA = 0.03W. The power dissipated by the resistor is 0.086W. Note that in the case of an AC powered LED, there are other electronics involved as well (such as a transformer and rectifier circuit) which will likely produce more heat than the LED and resistor itself.