Connect with us

Electric Calculations

Discussion in 'General Electronics Discussion' started by Mr. Oddly Fox, Jul 13, 2015.

Scroll to continue with content
  1. Mr. Oddly Fox

    Mr. Oddly Fox

    19
    0
    Jul 10, 2015
    I have the formulas handy, but I can not quite get the numbers I require.
    (it's been a while since I've actively used all of the formulas)
    I have a 5V/2.5A AC adapter. I need to step it down to the area of 3.7V.
    I expect that each of the 0.5W LEDs on the circuit to take around 120mA of current.
    How many ohms should the resistor be? How much current will I be reduced to if I use that resistor?

    Thank you for your time.
     
  2. Martaine2005

    Martaine2005

    3,331
    910
    May 12, 2015
    Hi mr oddly,
    It could crudely be done by putting two diodes in series. .6v Vf on each.
     
    Gryd3 likes this.
  3. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Please confirm your adaptor is AC, and not DC...
    It's most certainly AC input, but you need to ensure you have the proper output type as well.
    Why do you need the 3.7V ? Is this for the LEDs or for a microcontroller?

    You can use a voltage regulator to give you a nice constant 3.7V, otherwise if it's for the LEDs you just need to do a little math.

    Additionally... the color of the LED makes a difference :p (Or better yet, the rated 'forward voltage' on the spec sheet for the LED)
    This value will be anywhere from 1.8 to 3.3v usually.

    So, time for the math!
    Voltage = Source voltage - (LED Forward Voltage * Number of LEDs in series)
    Resistance = Voltage / Current
     
  4. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    Hello
    Please supply the details of the PSU.
    Thanks
    Adam
     
  5. Mr. Oddly Fox

    Mr. Oddly Fox

    19
    0
    Jul 10, 2015
    The forward voltage of the LED type is 3.6V. I was aiming for that area of 3.2-3.7V.

    The AC adapter specs were listed.
    5V/2.5A DC. I'm not sure if that's too much for a 3.6V device.
     
  6. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    Ah the reason I asked is because you said AC which makes a difference. Now we know it's DC then that's easier.
    Adam
     
  7. davenn

    davenn Moderator

    13,802
    1,941
    Sep 5, 2009
    hi there
    welcome to EP :)

    have a read of our how to drive LEDs resource, it should answer all your questions ......
    https://www.electronicspoint.com/resources/got-a-question-about-driving-leds.5/

    Dave
     
  8. BobK

    BobK

    7,682
    1,688
    Jan 5, 2010
    We still need the current of the LED at 3.6V in order to calculate the resistor.

    Bob
     
  9. Martaine2005

    Martaine2005

    3,331
    910
    May 12, 2015
    Agreed, and the amount of leds too.

    Martin
     
  10. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    You have a couple of options.

    1) Use a single resistor for each LED. You need to check what the forward voltage is at 120 mA if it's on the datasheet. So we will say it's 3.6 Volts for the minute. So (5 Volts - 3.6 Volts) / 120 mA = 11.6R (Use 12R) wattage is I^2*R = (0.12^2)*12 = 172.8 mW so go for 1 Watt resistors.

    2) Design a simple constant current circuit for each LED. This simplifies the calculations because you don't need to worry about the forward drop of the LED because you have set the current to 120 mA.

    Thanks
    Adam
     
    Martaine2005 likes this.
  11. AnalogKid

    AnalogKid

    2,482
    715
    Jun 10, 2015
    LED's are current-mode devices. Putting two diodes in series with a 5 V output will not provide any current limiting, and could damage the LEDs. The best way to size a series resistor is to determine what current you want through the LED, look up the corresponding forward voltage on the datasheet, subtract that from 5 V, and kick in Ohm's Law. LED forward voltage is not a rigidly controlled parameter, so the actual forward voltage could be off by 10%.

    For most high-brightness LEDs, the brightness is specified as a function of current, not voltage. For the most consistently repeatable brightness, drive the LED with a constant current source designed for the forward current on the data sheet (or less) rather than a resistor.

    ak
     
  12. Martaine2005

    Martaine2005

    3,331
    910
    May 12, 2015
    totally agree with you. I wasn't commenting on leds, just a crude way to drop .6v 0r .7v . From 5v to 3.7v. No leds were harmed in this conversation.

    Martin
     
    Arouse1973 and davenn like this.
  13. Mr. Oddly Fox

    Mr. Oddly Fox

    19
    0
    Jul 10, 2015
    Thank you all.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-