Help with LEDs ohms laws

Hello everyone,

This is my first post, so its a welcome and help thread.

I drew a simple circuit that I need help with



Am I understanding this correct, the LEDs are in parallel so volts is the same through each LED, but current will be divided among the 10 LEDs? LEDs are essentially a "short circuit" so they will allow as much current as possible, and pop? So I will need to lower the volts and current. the next image is my understanding of how to do this.



I want R1 to use .8A, so using Ohm's Law (V=IR) I would go 5V/.8A=6.25Ω. That will leave me with 5V .2A going to the other part, now I need to lower the volts so I would go (Vs-Vf) 5-3=2 2V/.2A=10Ω.
Let me know if I did any of this wrong. Before you ask there are other strands of LEDs that will require more current than this example, that's why I am using 1A.

Well, I appreciate the fact that you actually tried something, as opposed to a lot of the knuckleheads that post here just demanding answers without doing ANYTHING.

However, you're circuit is not quite right. Here's how it works: If you want 20mA per LED (which sounds like a lot to me) you need to put a resistor in SERIES with each led. The LEDs are NOT short circuits, they have a fairly constant 'forward voltage' of about 2 volts when they are conducting (this varies for different devices, but just assume that for now). Thus, if you want to put 20mA through the LED, you would calculate the resistor by:

(5V-2V)/R=20mA==> R=150 ohms.

This method will require you to use 10 resistors. You COULD use just a single resistor in series with all the LEDS, but this presents a problem-not all the LEDS are exactly the same, so some will carry more current than the others. This may not matter, but you should be aware of it. If you choose to do it that way, then you would use a 15 ohm resistor (5-3/)(.02*10))

You do not need to waste 0.8A in a very hot resistor R1. The 5V/1A power supply will be cooler and very happy with a 0.2A load.

Ok, explain to me how and when a resistor lowers volts and when it lowers current?

LEDs:
Manufacturer: Cree, Inc.
RoHS: RoHS Compliant Details
Illumination Color: Cool White
Luminous Intensity: 14000 mcd
Wavelength/Color Temperature: 5000 K
Viewing Angle: 25 deg
Forward Current: 20 mA
Forward Voltage: 3.2 V
LED Size: 5 mm
Packaging: Ammo
Lens Dimensions: 5 mm
Lens Shape: Dome
Maximum Operating Temperature: + 95 C
Minimum Operating Temperature: - 40 C
Mounting Style: Through Hole
Type: 5 mm Round LEDs

that's right from the supplier.

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Audioguru said:

You do not need to waste 0.8A in a very hot resistor R1. The 5V/1A power supply will be cooler and very happy with a 0.2A load.

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I guess you didn't read my whole post, try again and pay extra attention to the very last line.

After only 2 posts, you are already getting snarky.

I guess you didn't read my whole post, try again and pay extra attention to the very last line.

Click to expand...

Anyway, there is no need for R1 to absorb the remaining current from your power supply, more LEDs or not, to make your Ohm's law equation happy. Your power source will supply 5V up to a maximum current of 1A. It will work fine supplying 0.2A. Think of it this way, a car battery can supply 13.8V at 700A (or so). That does not mean you need to have 700A of load on the battery to make it happy.

As for when a resistor lowers volts and when it lowers current, in the case of your circuit, it only lowers current because the LED has a fixed voltage drop (3.2V) when forward biased. If you replace the LED with another resistor and work your Ohm's Law calculations, you can change the voltage drop across the replacement resistor by adjusting the current through the circuit which is done by changing the value of the original series resistor.

Snarky Josh,
Your 5V/1A power supply can light ONE 20mA LED with a series current-limiting resistor or light TEN 20mA LEDs each with a resistor or light FIFTY 20mA LEDs each with a resistor. You do not need to WASTE current in R1 because R1 is not needed and it is not wanted. Get rid of R1.

If your LEDs all have EXACTLY the same forward voltage (but they won't because they are all a little different) then you can connect some LEDs in parallel and use a single resistor to limit their total current.

Audioguru said:

Snarky Josh,

Click to expand...

spudboy488 said:

After only 2 posts, you are already getting snarky.

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First I want to apologize for being snappy, yesterday was a long day.

I know current is "taken" not "pushed", I read somewhere that LEDs will draw as much current as they can which is why I thought of R1. how do I figure out home many ohms are needed for R2? Vs-Vf 5-3=2, 2V/.02A=100Ω?

Joshp689 said:

First I want to apologize for being snappy, yesterday was a long day.

I know current is "taken" not "pushed", I read somewhere that LEDs will draw as much current as they can which is why I thought of R1. how do I figure out home many ohms are needed for R2? Vs-Vf 5-3=2, 2V/.02A=100Ω?

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See post #2.

Cree do not make a white LED with the spec's you listed. Maybe it is an old obsolete one. What is its part number?

LEDs have a range of forward voltage but you listed only 3.2V. Other Cree white 5mm LEDs have a range of minimum (?), typical (3.2V) and maximum (4.0V).
Because the voltages vary a lot you cannot connect them in parallel unless you measure the voltage of each LED then use only the few that have identical voltages.

Audioguru said:

Cree do not make a white LED with the spec's you listed. Maybe it is an old obsolete one. What is its part number?

LEDs have a range of forward voltage but you listed only 3.2V. Other Cree white 5mm LEDs have a range of minimum (?), typical (3.2V) and maximum (4.0V).
Because the voltages vary a lot you cannot connect them in parallel unless you measure the voltage of each LED then use only the few that have identical voltages.

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Here is the link to the LEDs I am using, I could use other LEDs but I havent looked for new ones and this is what I bought. The data sheet link is broken so I cant look at that.

How do you test LEDs for their voltage?

To test LED forward voltage: Connect +5V to 220 Ohm resistor (which will give ~10mA of forward current). Connect other end of the 220 Ohm resistor to the Anode of the LED. Connect the cathode of the LED to return leg of your power source. Use a DVM and measure the voltage across the anode (+) to cathode (-). This is the forward voltage of the LED.

Like most other LEDs, the datasheet of your white LED from Cree shows its forward voltage at 20mA: minimum = (?), typical= (3.2V) and maximum= (4.0V).

I think I might be using ohms law wrong, when figuring ohms do I use Vs, Vd, or Vf?

The way you were calculating R2 looks correct.

Assuming the power supply voltage is 5V, and the LED forward voltage is 3.2V, then the voltage across the resistor is 5 - 3.2 = 1.8V.
If the resistor is 100Ohms, then the current will be 1.8V / 100Ohms = 18mA.

The LED voltage may be a bit higher or lower than 3.2V, so the current may be a bit lower or higher than 18mA.

Here's the datasheet: www.cree.com/~/media/Files/Cree/LED%20Components%20and%20Modules/HB/Data%20Sheets/C512A%20WNS%20WNN%20942

Note that 20mA is the maximum current for the LED, so aiming for a slightly lower current is probably a good idea. e.g. 150 Ohms (as someone suggested earlier) may be a better choice than 100 Ohms, unless you need maximum brightness.

godfreyl said:

Note that 20mA is the maximum current for the LED

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No.
The maximum allowed continuous current is 25mA. All the spec's are at 20mA like most other 5mm LEDs.

As stated previously, putting LEDs in // without series resistors is begging for failure. Unless 'perfectly' matched one LED will draw more current, and then more and finally burn out, and then the next LED will do the same.

The only hope in the second circuit is to have R2 limit the total current to under the max current for ONE LED.