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[Reply #15 on:]

I see simple example circuits with LEDs, one resistor, an LED, and a power source. Why do these not have reverse voltage issues? What IS reverse voltage. Does it apply to a DC circuit *note, I have not gotten to AC circuits yet, so I know nothing of them. ARGH!!! *Tears hair out*

As long as you have the led wired in the correct direction (anode terminal towards voltage negative terminal), you will never have a reverse voltage problem with a DC voltage series circuit. However if you reverse the polarity of the voltage source (swap + and - terminals), or turn the led terminals around in the circuit, then the led is subjected to a reverse biased condition (turned off) and it's reverse voltage breakdown specification comes into play.

That make sense?

Lefty

[Reply #16 on:]

How about this then - forget I brought it up, and just make sure your anodes are connected to the high voltage side of your circiuit and the cathodes to the low voltage side as in the example circuits I posted.

If you have a single LED, the +5v can be an arduino output pin, the transistor is not needed, and the 'bottom' of the series LED/resistor can be ground. High output turns on LED.

Or the +5v can be +5v, and arduino connection can be made at the bottom of the LED/resistor. Low output turns on LED.

I would limit current to 20mA thru the arduino tho.

I'm off to bed....

[Reply #17 on:]

Thank you both for your help etc.

CrossRoads: Will be taking your advice on the Arduino outputs when it arrives. I still have much to learn.

Retro: Thanks for the clear explanation at the end.

I NOW understand. As I learn more I'm sure it will all make better sense.

I'm off to bed. Rest well.

[Reply #18 on:]

Simple LED resistor calculation:

(Vsupply - Vled) / Iled = R

So for a normal red/yellow/green LED, the forward voltage drop of 1.8V results in:

(6V - 1.8V) / .02 = 210 ohms. Nearest "stock" 5% resistor would be 220 ohms.

For a White or Blue LED with a higher drop, limiting the current more, and using a higher voltage, you would get:

(9V - 3V) / .015 = 400 ohms. You could go with a 390 ohm 5% resistor.

At 5V, 220 ohms will let 20mA through a monochrome LED; 470 limits it to roughly 10mA, and 1K gets it down to approximately 5 mA. Limiting the current saves battery power, but makes the LED proportionally dimmer. If you can remember those 3 resistor/current values, then adjusting upwards for higher voltages is pretty easy.

[Reply #19 on:]

Remember? Engineers don't need to remember - we just derive it as needed! And then approximate a solution using the parts on hand.