Please help with resistor LED calculation

[justinpugh]

Hi All

I have a NEWTRENT IMP500 battery. The battery has output of 5V,600mA. I use it to run a an LED conected via a bucktoot 7027. The buck toot ensures a steady output of 350mah. The problem I have is the battery has a sensor on it that switches of if the led is switched of. The LED has a switch on it and when switched of there is no draw from the battery. If left of for 30secs the battery itself switches off and when the led is next required you have to switch the LED on and the battery. It sounds simple but is a pain as the battery switch needs to be depressed for 2 secs to come.

To overcome this problem I have fitted a resistor and LED to the main led unit switch. This LED ensures that there is a constant draw from the battery if if the main LED is off.

Can anyone help me calculate the resistor and LED I need to ensure a) the resistor doesnt heat up and b) the LED doesnt burn out overtime.

I believe that the minimum draw from the battery is .03mA before the battery switches of..

Please can you show the workings so that I know how to do it. Thanks very much fro your help...
Justin

 

[audioguru]

Most 5mm ordinary LEDs burn out if their current is over 30mA (0.03A).
You want to discharge the battery at a little more than 30mA so it does not turn off.
So calculate a resistor that gives the LED 20mA and a second resistor by itself that draws 15mA is 5V/15mA= 333 ohms so use 330 ohms which is the nearest standard resistor value. Its heating is 5 squared/330 ohms= 0.076W so any little 330 ohm resistor will be cool.

The resistor for the LED is calculated with the difference between the forward voltage of the LED (which you did not say) and 5V.
So if your LED is 2V then the resistor gets 5V - 2V= 3V. The resistor value is 3V/20mA= 150 ohms for a 2V LED. Its heating is 3 squared/150 ohms= 0.06W so any little 150 ohm resistor will be cool.

 

[justinpugh]

Thanks for your help, Im still confused. Ive spent all day on this problem. I have just read your reply and that great. However I now know that the battery needs a minimum draw od .05A to stay on. However I have bought a pack of various LEDs. I have one red one that I think needs 2v and draws .1A. (Ive tried it on my variable power supply??)

I`m not an electronics man, so Im just confused by what you mean when you say "a second resistor by itself that draws 15mA", does this mean I fit a second resistor?? and if yes where does it go?? I have already put a 33ohm resistor in series with the LED. All works and the battery stays on, but the LED seams to be very bright!!! If I increase the resistor size the battery goes off!!

I appreciate your help,,, this is something that has been a nightmare for months for me!!!
Thanks

 

[audioguru]

The max allowed current in an ordinary 5mm diameter LED is 30mA but 20mA is frequently used for long life. You said that the battery needs more than 30mA of current so I had the LED use 20mA and a second resistor use 15.2mA for a total current of 35.2mA.

Now you say the battery needs a current of 50mA of current so it does not turn off and you found an LED that survives a current of 100mA.
You "think" the LED is 2V then the 33 ohm resistor has 3V and the current is 3V/33 ohms= 91mA.
I think you will soon burn out the LED. Look at its datasheet.

 

[justinpugh]

Yes, I understand, Thank you for your help. I have been having problems with the LEDS burning out after 6-8 hrs use. I wasn`t sure how I should place the resisistor. You have helped me understand the maths.

I placed one resistor in series with the LED and I tested the draw it was 20mA, this wasn't enough to keep the battery on and it went off as predicted in 30 secs. I placed the 2nd resistor across the + and - of the power supply. This meant that I now had a current draw from the battery of 60mA, the resistors I used were both 150ohm brown,green,brown. I left the unit on for several minutes and there was no noticeable heat and the LED was not to bright.

Im sorry I sound a bit vacant!! but I`m struggling to get to grips with all this.

My last question is, Have I made a mistake,,,,is it ok to put the second resistor across the battery terminals like this or have I misunderstood you and I should have also put this in series with the other resistor.

I have attached a photo to show what I have done, the led is lit but I put my thumb over it as the red glow made the resistor colours look different.

 

[audioguru]

Don't buy cheap Chinese LEDs that don't have a detailed datasheet.
LEDs burn out because you have exceeded their maximum allowed current that is shown by the manufacturer in the datasheet.

You have a 2.0V red LED in series with a 150 ohm current-limiting resistor and powered by 5VDC. Then the current is 20.0mA.
You have another 150 ohm resistor across the 5.0V so its current is 33.3mA. A 1/4W resistor will get warm.
The total current is 53.3mA.
If you need more current then add more LEDs with resistors or more 150 ohm resistors across the 5V.

 

[justinpugh]

"Don't buy cheap Chinese LEDs that don't have a detailed datasheet".???? I bought the Led from Maplin, it was part of a mixed pack of LEDS. It probably does have a data sheet, but I wouldn`t understand it...

I am interested in knowing how the maths work and thanks to you and your clear explanation, I now do..

Thanks for all your help, reading my post`s and your kind replies...

Cheers

Justin