Calculating the current through an LED based on voltage across LED and resistor?

[realflow100]

Calculating the current through an LED based on voltage across LED and resistor?

How?
LED is getting 2.64v while the resistor has 0.230v across it
Dont know what the power supply voltage is.

I cant figure it out.. google was no help either trust me i tried every search term of "led current resistor calculator" wording combination with LED included and without LED included and couldn't find exactly this situation

Even an LED resistor calculator site couldn't help me and forced me to enter a voltage power supply when i dont KNOW the power supply's voltage.
but its got to be under 100mA because i set my multimeter to 200mA range and put my leads across the LED and only got 93 to 94mA or so.

 

[druidaudio]

What is the value of the resistor?
Once you know that, you can calculate current through the LED and resistor (assuming they are in series) by Ohm's Law.

I = V/R

 

[realflow100]

the resistor is 15 ohms or 14.9 as my multimeter reads

 

[kaputt]

15.3mA

 

[realflow100]

Thanks!!!
Can you explain why this calculation works and how I can do it myself in the future?

I cant quite figure it out. and google isnt helping for some reason.
Do you think I could figure out the supply voltage this way too?
its an LED in a USB phone charging powerbank and I wanted to know how much current its drawing.
so that if I wanted I could choose my own resistor to bring up the current and replace the LED with a much more higher power brighter one. since I have some same-size LED's that are rated for about 20mA current to 100mA maximum each that are extremely bright
I'm thinking of going for 50mA but what resistor should I use for that?

 

[JonSnell Electronic]

the resistor is 15 ohms or 14.9 as my multimeter reads

Measure the voltage across the resistor and using Ohms Law that will give you the current; I = V/R or 0.23 / 15 = 0.015333 or 15.333 mA

 

[kaputt]

Google "Ohm's law". The voltage drop across a resistor is depending on current. If you know two of the three you can calculate the third.
Also have a look at a data sheet of an LED. Check the curves. They have a fixed voltage drop that is (relatively...) independent of the current running through it.

 

[realflow100]

ok thanks!!! Will the same amount of current be going through the LED as well as the resistor? Does that mean if they're both getting 20mA for example 20mA would be dissipated in the resistor?
Does that mean 40mA is being used and 20mA is being wasted and 20mA is going to the LED? Or is only 20mA being used, twice? How does that work?

 

[kaputt]

Current can't be "dissipated". Keyword "current loop". At every point in a current loop the amount of current is the same. Voltage can be quite different also depending on the chosen ground potential but current can't.

 

[kaputt]

Or easier (hopefully): it's 20mA through resistor and LED. The same 20mA of current.
Imagine you hook up your resistor and LED to the positive and Negative terminal of a battery. 20mA come out of the positive terminal and need to return to the negative terminal.

 

[lcsaszar]

Do you think I could figure out the supply voltage this way too?

The voltages are adding together, the supply voltage is the sum of the individual voltages measured in a series circuit. You can also measure the supply voltage directly.

 

[realflow100]

ok thanks!!!!!!! I was unable to measure the supply voltage directly as i cannot find the other circuit connection plus when the LED is on or off it changes.

 

[scottjoplin]

I've often found imagining electricity flowing through a circuit is like water flowing in a pipe can help. The voltage is like the pressure, the resistance is the size of the pipe and the current is the amount of water that flows. Unless there is a leak the flow has to be the same

 

[realflow100]

I got either 4.97v or 2.87v
when I added the voltages together in 2 different ways.
Which would be more correct?

4.97 is the USB voltage on the output of the USB phone charging power bank

but the battery is 3.68v with no load.

so somewhere i'm getting something wrong because there must be more resistors in series or something. cause neither of those voltages are very accurate.
with no load the USB powerbank outputs 5.15v and under about 1 amp its at 4.98
and surely the LED doesnt draw 1 amp. so something strange is going on here.

Is a leak in the pipe similar to a resistor

Or is a resistor more like a partially closed valve instead?

I feel like it could be either. but a partially closed valve doesnt waste much energy at all compared to a leak.

 

[rif]

There's a nice android app called electrodroid that helps you calculate such things. You still need to know what you're doing. It also has some other nice things such as common standards for pinouts, schematic symbol translation. Check it out, it's free.

Also, in your first post, I'm pretty sure that's not the correct way to measure current in the led. I'll let others chime in.

 

[scottjoplin]

A resistor is like a partially closed valve, a tap is a variable resistor

 

[DF96]

LED is getting 2.64v while the resistor has 0.230v across it

Then the voltage supplying them is 2.87V (=2.64+0.23). Google Kirchoff's voltage law.

the resistor is 15 ohms

Then the current is 15.3 mA (= 0.23/15). Google Ohm's law.

Will the same amount of current be going through the LED as well as the resistor? Does that mean if they're both getting 20mA for example 20mA would be dissipated in the resistor?
Does that mean 40mA is being used and 20mA is being wasted and 20mA is going to the LED? Or is only 20mA being used, twice? How does that work?

Google Kirchoff's current law. Power is dissipated; current is never dissipated (it just goes round and round).

 

[jan.didden]

I got either 4.97v or 2.87v
when I added the voltages together in 2 different ways.
Which would be more correct?

4.97 is the USB voltage on the output of the USB phone charging power bank

but the battery is 3.68v with no load.

so somewhere i'm getting something wrong because there must be more resistors in series or something. cause neither of those voltages are very accurate.
with no load the USB powerbank outputs 5.15v and under about 1 amp its at 4.98
and surely the LED doesnt draw 1 amp. so something strange is going on here.

Is a leak in the pipe similar to a resistor

Or is a resistor more like a partially closed valve instead?

I feel like it could be either. but a partially closed valve doesnt waste much energy at all compared to a leak.

The USB thing is an active circuit that 'transforms' the battery voltage up to close to 5V. There is not a simple relationship between battery current and USB current.

But in a simple series chain of a resistor and LED current in = current out. So what comes out of the top of USB charger flows through the R, through the LED, and then back to the bottom of the USB charger.

Current in a loop cannot 'disappear' unless you split the loop.

Voltages in a loop add up. So if you measure say 5V between the top and bottom of the USB charger which feeds a series chain of R and LED, and you then measure the voltages across the R and the LED separately, they must add up to what you measured on the USB charger.

Jan

 

[realflow100]

I havent got a smartphone. my phone ive got doesnt even have a camera and its like one step up from a walkie talkie. texting takes like 5 minutes for one short sentence because you have to press the same button 4 times for a certain letter for exmaple.

is there a windows/pc program or online thing I can use instead?
and I dont want a windows 10 "app" thats not what im looking for.

I'm good with electronics somewhat and I can solder decently. (would be easier to solder with something to hold tightly onto what im working on!)

 

[scottjoplin]

Google "online ohm's law calculator" This one's quite good Ohm's Law calculator

I use big blobs of blu tack to hold stuff while I'm soldering