I have a Roll of LED Lights.
And it is Rated 12 Volts at 4. Watts Per Foot.

So I Cut 1. Foot Off of the Roll.

So this will work on 12 Volts at 4. Watts.

But if I want to Find Out the Current it needs I will use the PIE Formula.

P Divided By E = Amps.

So I will do 4. Divided By 12 = 0.333333333 Amps.

So can I just take this to the Third Number like this 0.333?

And would this be 0.333 mA?

Or do I not Understand this?

 

mA is thousandths, .333A is 333mA.

 

The distinction is that Ohm's law is for conductors. While a resistor could be used as a model for a semiconductor LED, it is not the best model.
Semiconductors PN junction have been used as a resistor but does not exactly have the same I/V charecteristic needed over an entire range for Ohm's law. Some ESR meters might be more accurate on an LED than a ohm meter. The direct proportionality of Ohm's law for a resistor under same test
will not contend with voltage drop so the invariance of a good resistor, the V/I charecteristics differ from an LED.

When we build an LED circuit we can use Ohms law to find the correct value for the series resistor for a recommended current range.
But there could be an application where critical values may include the PN junction however this would be done where the percentage error
is within reasonable specs sometime needing instruments which are calibrated using Ohms law.

To show the V/I curve there is a chart at 2:13 and some of the later LEDs also have better thermal properties in extremes than the older process.

 

I have a Roll of LED Lights.
And it is Rated 12 Volts at 4. Watts Per Foot.

So I Cut 1. Foot Off of the Roll.

So this will work on 12 Volts at 4. Watts.

But if I want to Find Out the Current it needs I will use the PIE Formula.

P Divided By E = Amps.

So I will do 4. Divided By 12 = 0.333333333 Amps.

So can I just take this to the Third Number like this 0.333?

And would this be 0.333 mA?

Or do I not Understand this?

Ohm's (E = IR), and Watt's (P = IE) Laws work in Amps. So if your result is decimal less than 1, it's fractional, not whole.

333mA = 0.333A

 

Hello,

The use of SI-prefixes is quite common in electronics and other scienses.
Here is a list of prefixes:

Bertus

 

I am sorry this is how I should have Asked.

I know if you See 25 mA you will Write it like this 0.025 A.
And if you See 325 mA you will Write it as 0.325 A.

So if my 1. Foot LED Strip is Rated 12 Volts at 4. Watts I do this to See the Current it will need.
4. Divided By 12 = 0.333333333 A.

So we can just take it to the Third Number so we have 0.333 A.
And would we not Say this is 333 mA?

 

Yes, you are correctly calculating the current from the wattage. It is one third of an amp, which is 333mA. Three digits of precision is more than enough, two is usually sufficient when calculating parameters in a circuit.

 

It is not good practice to run LEDs at their rated maximum current, if you want them to have a long life. The brightness difference would probably be imperceptible to the human eye if you ran them at 250mA intead of 330mA

 

Ok thanks for all the help.

Now that I know I am doing the Math Right to Find the Current for my 1. Foot LED Strip I am going to Ask if I have this Right about the Current?

If my 1. Foot LED Strip will take 333 mA witch is 0.333 Amps and if I have 4. of them in Parallel the Current will be
4. X 0.333 = 1332 Amps.

So is this a little over 1. Amp?

 

Ok thanks for all the help.

Now that I know I am doing the Math Right to Find the Current for my 1. Foot LED Strip I am going to Ask if I have this Right about the Current?

If my 1. Foot LED Strip will take 333 mA witch is 0.333 Amps and if I have 4. of them in Parallel the Current will be
4. X 0.333 = 1332 Amps.

So is this a little over 1. Amp?

You missed a decimal point…

 

So is this a little over 1. Amp?

1.33A is more than a little over 1A and may be enough of an overload to upset your power supply if that's rated at 1A.

 

Ok thanks for the help.

Now I did get this LED Strip
https://www.amazon.com/dp/B00X5MEHGS?psc=1&ref=ppx_yo2_dt_b_product_details

This is 16 Feet and is Rated 12 Volts at 3 Amps and 40 Watts.
So if I want to Find out how many Amps 1. Foot will take I do this Formula.
40 Divided By 12 Right?

 

That's only the start of the calculation. 40W/12V = 3.33A for the whole 16.4 ft strip. So for 1ft, current should be 3.33A/16 = 0.21A (absolute maximum).
Allowing that the rated current might be somewhat 'optimistic', ~0.15A would be my choice of operating current.

 

It is very common for these LED strips to be wired three LEDs in series with a series resistor to set the LED current. So I base my calculations on the current required for a group of three LEDs and multiply by the number of the three series LEDs in a segment. This number must represent the total number of LEDs per segment divided by three where the quotient is an integer and the remainder is zero.

 

Ok I Looked at my LED Strip.

It Says 12 Volts at 3.5 Amps.
And it is 16.4 Feet Long so lets Say 16.5 Feet.

So I did 3.5 Divided By 16.5 = 0.212 and to Check it I did this.
0.212 X 16 = 3.392 so this is getting Close.

So would I be Right 1. Foot will take 0.212 Amps?

 

In theory, yes. But I wouldn't run the LEDs at their maximum rating.

 

I think I am getting the Hang of the Math.

My LED Roll is 16.4 and it Runs on 12 Volts at 3.5 Amps.
So 3.5 Divided By 16.4 = 0.213414634

So I Read this as 1. Foot will need 0.213414634 Amps am I Right on this?

And I check this by doing this.
0.213414634 X 16.4 = 3.499999998 and this will come out to 3.5 so I know I am Right.

So did I do all the Math Right?
I just want to See if I have the Math Right for all of this?

 

The maths is right, but in the real world you don't need that many significant figures in the values. Two digits after the decimal point would be enough.

 

The only question I have is does 1 foot contain a multiple of 3 LEDs? If not, you cannot cut it at 1 foot. Count the number of LEDs in the strip you cut (which must be a multiple of 3) in ratio to the total number of LEDs in the 5m (16.4 ft) strip (which is 300)

So the current would be:

3.5A * numleds / 300

Bob

 

So if 1. Foot will Take 0.333mA then 4. Feet will Take 1,332mA am I Right?