# Red LED from 12v supply

Discussion in 'General Electronics Chat' started by Xufyan, May 8, 2011.

1. ### Xufyan Thread Starter Member

Aug 3, 2010
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How much resistance should i use in series with REd LED if the supply is 12volts , how can i measure the current of 12volts supply ?

2. ### tom66 Senior Member

May 9, 2009
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What is the forward current and forward voltage of the LED?

For a small, 5mm LED forward current (known as If) is about 20mA.

For red and low-current yellow, forward voltage is about 1.8V.
For low-current green, forward voltage is about 2.2V.
For newer green LEDs, forward voltage is 2.8V - 3V.
For blue and white LEDs, forward voltage is from 3.2V - 4.0V.

Check the datasheet to be sure!

A very quick "hack" with 12V is to use a 560 ohm resistor, this limits the current for even low voltage drop LEDs. However, the LED will not be as bright as it can be.

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Apr 5, 2008
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4. ### Xufyan Thread Starter Member

Aug 3, 2010
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the forward voltage is 1.83v and forward current is 20mA , the supply is 12volts

i am doing this,

12volts/20mA = 600ohms

is it the correct way to calculate ??

Apr 5, 2008
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Last edited: May 8, 2011
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6. ### Xufyan Thread Starter Member

Aug 3, 2010
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Hey thanks alot !!!!!
a really great addition to my knowledge !

but, if instead of LED i have a CMOS ic (4078BP 5volts) and the supply remain 12volts then how much resistance should i use to limit the voltage to 5 volts to operate the IC ?

Last edited: May 8, 2011
7. ### tom66 Senior Member

May 9, 2009
2,613
218
Generally, you don't use resistors to regulate voltage for things like logic gates - this is because their current draw varies with what state they are in, and this causes changes in the voltage. In this case, you would use a voltage regulator, like a 7805. Make sure you add the capacitors as recommended by the datasheet. A voltage regulator reduces the voltage by dissipating it as waste heat. Drawing too much current will cause it to heat up, and you may need a heatsink in some cases. It basically works as a resistor which continuously adjusts for the changing load. For low power logic like 4078's, there should be no need for a heatsink.

Although, why do you need to run it on 5V? The CD4000 series is rated up to about 18V.

8. ### Wendy Moderator

Mar 24, 2008
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I suspect the OP is mixing logic types, which may work, but usually doesn't.

A quick word, don't be afraid to ask specific questions, we don't judge here, and most of us are here to help. We like to help.

Back to the subject. LEDs are not linear, the amount of light they produce isn't 1:1 to the amount of current they get. This is a good thing, it means that you don't have to go for maximum brightness if all you need is a indicator.

A circuit I came up with a long time ago will allow a circuit with 9V or more to drive an LED with very little loading on the logic gate.

One of the reasons I don't use TTL much anymore is the 5V limit. It makes it harder to drive a lot of electronics, and it isn't what I would call a friendly voltage (batteries don't you know). Having said that, it is a good and cheap logic, and a 6V battery power supply with a diode in series works.

9. ### ke5nnt Active Member

Mar 1, 2009
384
16
Something not mentioned yet is power dissipation across the resistor. Using Ohm's Law to calculate a 10.17 volt drop across the resistor at 20mA, we find that the resistor will dissipate ~200mW. Using a 1/4 Watt resistor only gives you 50mW of head room, you'll probably be safer using a 1/2 Watt resistor.

10. ### Wendy Moderator

Mar 24, 2008
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Good point, if for whatever reason you can't get a ½W resistor you can always use 2 ¼W resistors and split the load. That or you can use less current for the LED as I suggested earlier. There is no law that says an LED must use the maximum current they are rated for. They will last much longer if you don't.

11. ### tom66 Senior Member

May 9, 2009
2,613
218
In this case though, make sure each resistor is double the size. For example 560 ohms -> about 1.2k each.

12. ### ke5nnt Active Member

Mar 1, 2009
384
16
I assume this is for resistors in parallel. You could run 2 resistors in series that are each ~1/2 the 560 Ohm total. About 280 Ohms each in series would give you the total 560, and each would dissipate 1/2 the power. 280 is probably not a standard value, so just use the next highest.

13. ### Wendy Moderator

Mar 24, 2008
21,414
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I'd use one 270 and one 300 myself, or 2 270Ω. Give it a while, you will memorize the values.

May 28, 2009
506
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Adding to what Ke5nnt said, If you do parallel resistors to split the load, you will also split the resistance. Calculate resistors in parallel by

R1xR2/R1+R2 for two resistors in parallel

1/(1/R1)+(1/R2)+(1/R3)+ etc... for three or more resistors in parallel

Last edited: May 11, 2011
15. ### nbw Member

May 8, 2011
36
10
The beautiful thing about resistors - series and parallel them in every way to get the resistance and wattage rating you need. And, many applications aren't overly fussy if they get a 9K1 or 10K for example. Accuracy-wise, 1% metal film packs are so cheap now I tend to just use them instead of 5% carbons.