Dimming LEDs. Help please with resistor values etc.

[dasatcopthorne]

Hi Guys.

 

We are using some LED signals by Absolute Aspects but they are far too bright.

 

I guess they already have resistors built in for 12v operation which we use as the supply.

 

My question is as follows;

 

I have tried 4.7k resistor and it seems to be about right but my idea is to use a variable resistor on veroboard for each aspect so that I can adjust them individually.

 

I thought I'd get some 10k miniature versions seen on ebay but the power dissipation is often quoted as 0.1w.

 

Is this likely to be enough please?

 

Or do I need more? Say 8th or quarter watt?

 

Never could do the maths on this equation.

 

Cheers

 

Dave

[Ray H]

With my tin hat on . . . . . 

 

You can calculate the power rating (W = Watts) required by multiplying the volts (i.e. 12v in this case) by the current the LED takes (I believe the usual wisdom is that the current used by a standard LED is 0.03A) giving a power rating of 0.36w. You might be able to reduce this figure if you can measure the current the LED is taking.

 

Hopefully someone else will confirm my values.

[Gordon H]

With my tin hat on . . . . . 

 

You can calculate the power rating (W = Watts) required by multiplying the volts (i.e. 12v in this case) by the current the LED takes (I believe the usual wisdom is that the current used by a standard LED is 0.03A) giving a power rating of 0.36w. You might be able to reduce this figure if you can measure the current the LED is taking.

 

Hopefully someone else will confirm my values.

 

Yes and no...

The idea is to make the LEDs less bright - so 30mA is not the right sort of figure to be aiming at. More like 3mA.

Assuming an LED Vf of 1.5V, that leaves about 10.5V to drop across the resistor, so at 3mA that would require a series resistance of around 3.5k.

So, your choice of a 4.7k variable is reasonable. However, I would suggest that you include a fixed resistor in series with the variable, to prevent the possibility of turning its value down to zero ohms and blowing the pot and/or LED in doing so.

Add a 1k fixed resistor and you will get a minimum current of 1.8mA, and a maximum of 10.5mA. That should give you sufficient variability under almost all circumstances.

Power wise, the maximum in the fixed resistor would be 0.11W (so use a 1/4W). For the variable, is will depend on the setting, but set to 1k like the series resistor, the LED current will be 5.25mA and both the fixed and variable dissipations will be 27.5mW each. Higher settings of the variable will progressively lower the dissipation. Your 0.1W variable should therefore be OK.

[47137]

You may be able to get the effect you want with parallel combinations of resistors. Start with one resistor which is a bit larger than you need, say 4.7k. So the LED is a bit dim. Then connect two 4.7k resistors in series, and connect the combination in parallel with the first one.

 

If the LED is too bright with one 4.7k resistor, add another resistor in series. Supposing the LED is now too dim, home in on the brightness you want by adding a third resistor in parallel with the second one.

 

This approach lets you buy just resistors in just a few values, and make the actual resistance you need yourself. The calculator here may help. Do make sure there is always some resistance - say 1k, if not you will blow the LED.

 

- Richard.

[dasatcopthorne]

Thanks Guys for your replies.

 

However may I stress that the LEDs already have some resistors fitted as they are made to and are currently working on 12v.

 

I can't see the resistor value so I'm guessing at 1k.

 

So I shall be adding to this.

 

Cheers

 

Dave

[Gordon H]

 

I can't see the resistor value so I'm guessing at 1k.

 

 

If you measure the current it draws from 12V you will get a good idea.

[DCB]

First you need to realise it does not matter how much energy is dissipated or the maths but what matters is the LED dims.  There is no standard current for a LED, the red ones use less current and fail with less current than almost every other colour, nor do most 12 volt power units deliver 12 volts, 15 is more likely. but the smart solution is a handfull of resistors 4.7, 47, 470, etc and play with them adding them in series and parallel until until the colour is right, without knowing the actual supply voltage and the value of the internal resistor there really  is no other option 

[Gordon H]

First you need to realise it does not matter how much energy is dissipated or the maths but what matters is the LED dims.  There is no standard current for a LED, the red ones use less current and fail with less current than almost every other colour, nor do most 12 volt power units deliver 12 volts, 15 is more likely. but the smart solution is a handfull of resistors 4.7, 47, 470, etc and play with them adding them in series and parallel until until the colour is right, without knowing the actual supply voltage and the value of the internal resistor there really  is no other option 

 

That is what the maths is for, so you can work things out in advance like real engineers do. The aim here is to set the brightness, not the colour.

And while we're at it, energy is not dissipated - power is dissipated. Energy is merely converted from one form into another.

[DCB]

Gordon if you don't know the supply voltage, and many "12 volt" power units can deliver up to 19 volts, probably more, and and questioner doesn't know the internal resistance of the "12 Volt" LED there is not enough information available to come up with a mathematical answer that is any use.  The LED is apparently working within its design parameters but is a bit too bright, stick a resistor in series and see what happens, when you have tried a few and found one which makes the LED is too dim or simply fails to light then you can check its rating and do the maths.

[SHMD]

Gordan, I think I know where DavidC is coming from here.
There is nothing in Ohms law, data sheets and “best practice” to give a definitive answer to what looks right for me in this particular location under these conditions.

Best solution is to put a POT in series, (of suitable power or limit it with another series resister), and adjust the POT until it “looks right”.
Then either remove the POT, measure it and replace it with the nearest available resister or just leave it in situ.

(But I fear that the concept of series limiting resisters may be beyond some of RMwebs members…)


Kev.

[dasatcopthorne]

Thanks guys for all your answers. Most enlightening.

 

I think I have it sorted now.

 

Cheers

 

Dave

[Gordon H]

Gordon if you don't know the supply voltage, and many "12 volt" power units can deliver up to 19 volts, probably more, and and questioner doesn't know the internal resistance of the "12 Volt" LED there is not enough information available to come up with a mathematical answer that is any use.  The LED is apparently working within its design parameters but is a bit too bright, stick a resistor in series and see what happens, when you have tried a few and found one which makes the LED is too dim or simply fails to light then you can check its rating and do the maths.

 

Surely anyone wanting to get involved with this kind of thing for themselves ought to make a simple DMM one of their first investments?

Once you have the ability to measure the supply voltage and the current being drawn it is a simple matter to work out what is inside your 'black box' signal.

How does this 'not knowing the supply voltage' situation sit with your own philosophy of never needing current limiting resistors if you get the forward voltage just right?

 

Regardless of all this, the considerations involved for using a pot to establish the required brightness (in the absense of direct calculation) have already been given above.