Resistor

[Memphis]

Hello,

I am currently building a small curcuit, battery is a 12v lithium which runs to a 5v regulator, used digi meter and the output is 5v... perfect. ok so i want to connect an LED here, but i will need a resister?? i have several resistors at hand. a 330ohm 0.125w and some 330ohm 0.25w... how can i lower the voltage however to 1v for the LED? after hooking up either resisters the meter always tells me there is 5v through. thanks, hope someone can help

[rollingrobot]

Resistors lower current, not voltage. Basically your 330 ohm resistor any) would work. Remember, voltage isn't the same as current

[Memphis]

ok, thank you for fast response, sorry in advance for any more n00b questions, im very new to this whole type of thing

[madsci1016]

If i don't feel like writing out the math, I use this:

http://led.linear1.org/1led.wiz

To find what resistor is needed for any led or source voltage.

[Soeren]

Hi,

Resistors lower current, not voltage. [...] voltage isn't the same as current

But multiply the two and you might get a hint that it takes one to have the other... So yes... The resistor lowers the voltage as well.

[Memphis]

Hi soeren, which values would i multiply? and how come the digimeter still shows that there is 5v? thanks.

[definitionofis]

Why did you not use madsci1016's link ?

Is his diagram/link not like your circuit ?

Maybe you do not know the current and voltage requirement of your led ?

Maybe you put the meter where the led goes and measured 5v without the led installed ?
That is because meters have a different resistance than leds; 20,000 ohms or more.
You have to install the led and put the voltmeter across the led.

But if madsci1016's link is not your circuit, then ignore everything I said.

http://led.linear1.org/1led.wiz?VS=5;VF=1;ID=13

[Soeren]

Hi,

Hi soeren, which values would i multiply? and how come the digimeter still shows that there is 5v? thanks.

That was a comment regarding the current contra the voltage, which goes hand in hand.

With your 5V, you need to drop the voltage not taken by the LED.
Different LEDs have different voltage drops.
- Infrared 1.5V
- Red 1.8V
- Green and yellow  2.0V
- Blue, white, pink and ultraviolet around 3.6V
These are just approximate values but good enough for calculating a voltage dropping resistor.

Say you want to use a red LED. Then you need to drop 5V - 1.8V = 3.2V
An appropriate max. current for a regular LED is 20mA (0.02A)
The resistor then need to be 3.2V / 0.02A = 160 Ohm
Select the nearest standard value, either 150 Ohm or 180 Ohm and calculate what the current will be
With 150 Ohm:  3.2V / 150 Ohm = 21.3 mA
With 180 Ohm: 3.2V / 180 Ohm = 17.8 mA
Either will work!

The current in a LED can be as low as 1 mA and it will just glow dimly or up to eg. 30 mA with a somewhat shorter life.


The reason you are measuring 5V with just the resistor is, that your meter has got a high impedance ("resistance") at least 1 MOhm, probably more, so it won't pull the voltage down that much.
If you place a resistor of 91 Ohm where you wan't the LED to go, you will get something a bit closer to what the LED will drop, but it's a waste of time - just pop in the LED and if you want to know the precise voltage it drops (which doesn't really matter), just measure over the LED.

[Memphis]

hi sorry to confuse you with my question, i already followed what was said and put the 330ohm resistor between my LED, after that you had mensioned that it also lowers voltage (the resister) i am just interested in understanding exactly what is going on for furthur reference, many books i have read i do not understand, i however do understand from trial and error and people's response + testing.... i do not want to see the voltage drop after adding the LED..... my LED requires a max of 3v, i can run it at 1v, so i have added as i said the 330ohm resister between the 5v regulator, now i'd assume the resister would take the voltage down to a lower voltage for the LED requirement, however using the digi meter this is not the case it still shows me 5v, this is what i am trying to understand.... hence i asked about the values you refered to. hope you understand.

[rollingrobot]

Maybe you are putting your probes is the wrong places 

[waltr]

Do you have from +5 volt the resister then the LED to ground, series connections (as in the diagram in the resistor calculator) and you are still measuring 5 Volts across the LED leads then try reversing the LED. An LED only conducts current in one direction, in the other direction it will appear as an open circuit.

It is also possible that the LED is 'burned out' if you did connect it to 5 volts without a resistor. If reversing the LED didn't help try a new one (BTDT).

LEDs are current devices that require a certain voltage threshold to start conduction as Soeren stated in his post. So even with different value resistors in series the voltage across the LED will be nearly the same but the current through the LED will be different. That is why the resistor calculator in the link posted earlier has LED current and LED Forward voltage (the conduction threshold voltage) as parameters. The current adjusts the brightness of the LED.

To measure the current of the LED measure the voltage drop across the series resistor (the current is the same in every series connect part) and use Ohm's law to find the current (I = V/R).

Keep experimenting, asking questions and you'll get it.

[Memphis]

The LED is working fine, everything is working fine, just the output from the resister to the LED still says 5v that is what i am trying to understand.

[definitionofis]

Say how you are measuring it.

[Memphis]

hi ok i have done a quick layout and to show how i am testing via MS-Paint lol sorry i've not finished my PCB design software yet still developing it... anyways as u will see i have labeled the items and with digi meter sample lol

ok so the output of +5.00v on my digi meter, shouldnt the voltage be much lower? since the current has been limited? or am i misunderstanding the whole current and voltage thing.


@Soeren: just re-reading your post several times about the ohms... so if my LED will run at 1v, my input is 5v, then...

1V / 0.02A = 50 Ohm

 if i used a 50ohm resister, it should work? 1 / 50 = 0.02 which is 20mA ? if so then i think i understand that a little better.

[Weird Fishes]

You're diagram looks a little off, but I think your logic is getting close.

KVL says the voltage around a loop is 0. So this means that the voltage across your LED and across your resistor will be 5V. If you want your LED to run at 1V, which I doubt you actually do, read Soeren's post about what the voltage likely should be, then you'll want to use 5-1=4V for your current calculation. Then by Ohm's Law: V=iR R = V/i R = 4/.02= 200Ω.

Read up on KVL/KCL/Ohm's law and try applying it to some other circuits. Then draw out your circuit and apply it to that, solving for the resistance you need.

Hope this helps. Good luck.

[Soeren]

Hi,

hi sorry to confuse you with my question,

The only thing that confuse me right now is your reason for not reading what I wrote about it?

[...] after that you had mensioned that it also lowers voltage (the resister) i am just interested in understanding exactly what is going on for furthur reference, [...] now i'd assume the resister would take the voltage down to a lower voltage for the LED requirement, however using the digi meter this is not the case it still shows me 5v, this is what i am trying to understand.... hence i asked about the values you refered to. hope you understand.

You have a voltage source of 5V, a resistor and the impedance of your meter (which you can see as another resistor).
You create a voltage divider with these two resistors of say 180 Ohm and 1 MOhm
The voltage will then be 5*1000000/(180+1000000) = 4.999V which your meter will show as 5.00V if it's a 3½ digit meter.

If the impedance of your meter is higher, which is quite likely, say 2M to 10M, the measured voltage will just be even closer to 5V (you have way more tolerance in your voltage regulator btw.).

When you pop in the LED, you create a current divider - as I mentioned, a 91 Ohm resistor will behave much like a red LED.

Now if you still don't understand what's going on, I'd recommend reading a couple of beginners books on the subject, because then you must be lacking most of the underlying basic knowledge needed.

[waltr]

Memphis,
   Your circuit is OPEN and will draw no current therefore the voltage will be 5 volts.

Look again at the circuit in the resistor calculator that was posted earlier. The circuit must be completed, +5V to the resistor to the LED to ground for any current to flow. Refer to Ohm's Law, I = V/R.

For your learning pleasure try this: Get two resistor, one about 390Ohm and the other about 1k Ohm. Connect them in SERIES from the +5 volt and ground.
Now measure the voltage across each resistor. Each will have a different voltage and both will be lower than 5Volts. This is a basic voltage divider. Now use Ohm's law to calculate the current through each resistor (they should be very close to the same). Try some different value resistors and measure the voltage drops then calculate the current until you can predict the results.

just re-reading your post several times about the ohms... so if my LED will run at 1v, my input is 5v, then...

1V / 0.02A = 50 Ohm

 if i used a 50ohm resister, it should work? 1 / 50 = 0.02 which is 20mA ? if so then i think i understand that a little better.

This is not correct. If the LED is to have 1 volt across it then the resistor needs to have 4volts across it. See this is a voltage divider (but for some subtly different reasons). So to calculate the resistor value needed for 20mA, 4V/.02A = 200 Ohm.
If you used a 50 Ohm resistor to drop 4Volts then the current would be 4V/50Ohm = .08A.

This diagram is how the resistor's voltage drop needs to be measured.

[Memphis]

waltr: thanks for that explanation, i understand now, just the maths was the wrong way around.