Electrical – How to alternate 2 LED strings using 555 IC

Do you have in place the 10 nF input coupling capacitor? Perhaps what is going on here is an input impedance effect.

If we consider the comparator inputs on the 555 to be basically like open-loop op-amps, that means that their input impedance depends on the device used (CMOS versus BJT). Negative feedback isn't there to help with impedance.

The CMOS inputs will have a high input impedance and so the input of your circuit is basically dominated by the two 100K biasing resistors, which create an impedance of 50K. In combination with the 10 nF coupler, the input passes through everything above around 300 Hz. (The -3dB corner frequency is 318 Hz). This is well below the input signal frequency.

But what if the input impedance is somehow seriously degraded with the NE555?

Just a thought.

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.