I dont know how where to enter what can you help me.
Where would I put in the LED's voltage into the calculator?? I cant just enter it as the voltage because that would make no sense. and make the rest invalid to put in.
I havent got blue tack.. and I'm usually working at really odd angles and its hard to hold what im working on down. I usually need it lifted up in the air at some weird angle so that I dont melt the object behind what im soldering on or touch the carpet cause i have carpet flooring in my room.
I'm also desoldering components a lot as well.
Where would I put in the LED's voltage into the calculator?? I cant just enter it as the voltage because that would make no sense. and make the rest invalid to put in.
I havent got blue tack.. and I'm usually working at really odd angles and its hard to hold what im working on down. I usually need it lifted up in the air at some weird angle so that I dont melt the object behind what im soldering on or touch the carpet cause i have carpet flooring in my room.
I'm also desoldering components a lot as well.
I've put in 0.23V and 15ohms and pressed calculate and it tells me the amps Ohm's Law calculator
This looks handy http://www.ebay.co.uk/itm/PCB-Holde...564332?hash=item5b31fb1c6c:g:7ccAAOSwTglYkFJC
This looks handy http://www.ebay.co.uk/itm/PCB-Holde...564332?hash=item5b31fb1c6c:g:7ccAAOSwTglYkFJC
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Unless someone slept through all their junior school maths classes they should not need to use an online calculator for Ohm's law. Pocket calculator or pencil and paper should suffice. Even the online calculators require some understanding; if you have enough understanding to use one then you don't need one.
An app to divide voltage by resistance? You want an algebra app??
The important thing is that you understand the relationship V = I * R. As mentioned before, if you know 2 out of three you can calculate the other. You need to understand that, no app is going to do that for you.
And once you do, you only need a calculator.
Example: V across R = 10V, R = 100 ohms. If V = I * R then obviously I = V / R. So I = 10 / 100 = 0.1A = 100mA.
Jan
ok i finally figured it out
i just didnt know how to enter the things
i didnt know if i needed to know the voltage of the LED as well as the voltage across the resistor and resistance of the resistor in order to do the calculation
or if i only needed the resistance of the resistor and the voltage across the resistor
and it seems when its fully charged up the USB power pack the LED gets about 19-25mA of current which seems perfectly reasonable. so I might not need to change the resistor. just a more brighter more efficient LED with a lower voltage rating then it'll be enormously brighter
its just a small project. swapping out the LED with a brighter one.
i just didnt know how to enter the things
i didnt know if i needed to know the voltage of the LED as well as the voltage across the resistor and resistance of the resistor in order to do the calculation
or if i only needed the resistance of the resistor and the voltage across the resistor
and it seems when its fully charged up the USB power pack the LED gets about 19-25mA of current which seems perfectly reasonable. so I might not need to change the resistor. just a more brighter more efficient LED with a lower voltage rating then it'll be enormously brighter
its just a small project. swapping out the LED with a brighter one.
Modern LEDs are already pertty bright with just a few mA so replacing it, you might get more brightness than you want. In that case, just increase the resistor value until you get down to the brightness you like.
Is this on a front panel?
Jan
How about this:
If you have an LED and a power supply, you need a series resistor to limit current. If we have a 5v supply, we look at the LED. Whatever LED you use will have a typical forward voltage. Let us say 1.2v. SUbtract that from 5 and we get 3.8v has to drop across the mystery resistor. We also have a target LED current. I will arbitrarily chose 10ma. The more current, the brighter the LED. SInce current is constant through a series circuit, I have to calculate what resistance I need for 10ma to drop 3.8v. 10ma is 0.01A.
So R=V/I 3.8v/0.01A = 380 ohms. 390 is the closest standard value, so I chose that.
You might find that 10ma is more than enough, so try 5ma. 3.8/.005 = 760 ohms
If you have an LED and a power supply, you need a series resistor to limit current. If we have a 5v supply, we look at the LED. Whatever LED you use will have a typical forward voltage. Let us say 1.2v. SUbtract that from 5 and we get 3.8v has to drop across the mystery resistor. We also have a target LED current. I will arbitrarily chose 10ma. The more current, the brighter the LED. SInce current is constant through a series circuit, I have to calculate what resistance I need for 10ma to drop 3.8v. 10ma is 0.01A.
So R=V/I 3.8v/0.01A = 380 ohms. 390 is the closest standard value, so I chose that.
You might find that 10ma is more than enough, so try 5ma. 3.8/.005 = 760 ohms
I want it stupidly blindingly bright so I have to choose an LED with a low forward voltage even when its under max current since its REALLY HARD to replace the resistor on the PCB since its surface mount type. resistor and i dont know where it goes to (the pcb trace disappears under another surface mount component thats bridged over the end of it)
so maybe some high power focused LED chip would work.
then I could just pop it in place. solder it and be good
it just needs to fit.
I'm looking for around 20mA range or maybe slightly more even.
I'll test some bright LED's that I have laying around and see if I can find some that will fit that spec and be bright enough.
one thats 3.6v would barely light for example.
and one thats 2.2v would probably work great and be super bright
usually LED's can tolerate a decently sized current range around their maximum as long as its not continuous. and the current slowly decreases over time depending on the battery voltage. to around below 20mA so thats completely fine for me
so maybe some high power focused LED chip would work.
then I could just pop it in place. solder it and be good
it just needs to fit.
I'm looking for around 20mA range or maybe slightly more even.
I'll test some bright LED's that I have laying around and see if I can find some that will fit that spec and be bright enough.
one thats 3.6v would barely light for example.
and one thats 2.2v would probably work great and be super bright
usually LED's can tolerate a decently sized current range around their maximum as long as its not continuous. and the current slowly decreases over time depending on the battery voltage. to around below 20mA so thats completely fine for me
This may be a bit harder, but imagine that you have a LED with no voltage drop. That means all the voltage available from the supply is across the R. That thus determines the max current you can get with that resistor: max voltage / R.
Jan
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the usb port died so now its permanently shorted and wont work anymore.
one of the metal prongs that hold the usb male connector in got shoved down to one of the contacts and now its shorted and wont work.
time to buy another one from walmart next month i guess..
I desoldered and saved the batteries at least so now I have two more 2,200mAh LI ION cells. along with 3 more. so I could make a 10,000mAh powerbank if I wanted. I just need the USB charging circuit board with the USB plug.
one of the metal prongs that hold the usb male connector in got shoved down to one of the contacts and now its shorted and wont work.
time to buy another one from walmart next month i guess..
I desoldered and saved the batteries at least so now I have two more 2,200mAh LI ION cells. along with 3 more. so I could make a 10,000mAh powerbank if I wanted. I just need the USB charging circuit board with the USB plug.
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