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.
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.
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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?
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?
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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
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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.
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.
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?
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?
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.
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.
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.
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.
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.
Attachments
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
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!)
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!)
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
I use big blobs of blu tack to hold stuff while I'm soldering
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