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Voltage spikes through voltage regulator?

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gregmcc

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I've built a simple circuit with (3 LED's in series with a 100R resistor) x 3. These are ultra bright LED's which are mounted to a quad bike - run from the 12V battery.
I'm running the circuit through a 7812 as the battery is more like 13.5V than 12V. My question is, is it possible when starting the bike the battery voltage could spike, if so could this get through the volt. reg and damage the LED's? Is there anything else I should be using to protect against spikes?
 
Stick a zener diode across them - but the 7812 won't be doing much anyway, as you don't have enough extra voltage for it to work - use a Low Drop Out (LDO) regulator instead (or, probaby better, use a lower voltage to feed the LED's).
 
If you're connected from the battery then it will act as a capacitor and absorb any spikes. The battery itself will not spike.

A current regulator rather than as voltage regular should be connected in series with LEDs (- a 78xx can be used in a current regulating circuit)
 
I need to drive 9 LED's so I thought using a 3 x 3 setup with 100R on each of the 3. This is fed from 12V. Should I drop the regulator and rather use a zener (12V?) , or is there a better config I should look at. They're standard ultra bright LED's - 20mA, 3.3V

The first time I build the circuit I setup the LED's for 12V and didnt use a reg. I plug it in to the battery and forgot it wasn't exacly 12V so it didnt work. The LED's just went extra bright and turned off, luckily they didnt blow, hence I thought a 12V reg would be a good idea.

Mmm - a current regulator. Didnt think about that at all. Thanks - I'll look into it.
 
If you look at the datasheet for the LM7812 you'll find that the input voltage needs to be between 2V - 3V higher than the output for it to regulate properly, so your battery will need to be around 14V -15V.

Drop the regulator and the zenner idea, you don't need a stable voltage for LEDs all that's required is a resistor to limit the current to a safe level.

Use ohm's law to calculated the required resistor value. You need a separate series resistor for each LED; it isn't aceptable to use one resistor for the three.

What's the minimum battery battery voltage?

You'll need to factor this into your calculation.

Calculate the resistor for the LEDs' maximum current at the battery's maximum voltage, then work out the current at the end of battery life, also look at the discharge graph on the datasheet. This will give you an idea of how long your battery will last and how dim the LEDs will be when the battery is flat.
 
its a motor bike battery - so i guess it ranges from 12-14V? It'll never go flat as its charged via the alternator.

Thanks for the idea - I guess its back to the basics.
 
(3 LED's in series with a 100R resistor) x 3.
Is perfectly OK. Just calculate the resistor for the highest voltage from the battery (probably 14.4 or so). It won't be as bright (about 3/5 the current) when the engine is stopped, I hope that's OK.

Say the LEDs are 3.0V each, three in series is 9V. (14.4 - 9) = 5.4. For 20 mA, you use 270 ohms on each leg. They should be quarter or half watt resistors.

If you put a 20 mA current regulator on each leg, the brightness can be more constant.

In either case, I would put a reverse-biased diode across each string to protect them from negative voltages (sometimes referred to as "load dump").
 
Led

I always find it good practise to put a small signal diode e.g. 1 N914 or 1N4448 in antiparrallel with the LED to avoid getting a negative spike across the LED (s).

So in your case you can put a diode like above mentioned antiparrallel across each LED or put one across the 3 LED's inseries but after the series resisitor.

Keep the LED running current below 20 mA's and as already said forget the 7812. It won't work correctly.

Calculate the resisitor for the highest voltage possible 14.4 volts minus the volts drop across the LED's. and current at ± 18 mA.
 
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