help with led's in series

Discussion in 'The Projects Forum' started by markosillypig, Aug 10, 2008.

  1. markosillypig

    Thread Starter Active Member

    Jul 21, 2008
    184
    0
    hi all i have made a strobe effect led circuit.
    but i have a problem when i connect the led's in series 2 different colours and different volltages they do not work.1 colour works perfect and doubled up they just don't is it becouse of the different voltages???
    the white led is 3.8 volt and the red is 2.1 volts
    i have enclosed a pick so you know what i'm talking about

    thanks all
    marko
     
  2. markosillypig

    Thread Starter Active Member

    Jul 21, 2008
    184
    0
    hi all
    i'm having problems with led's
    when i connect 2 led's in series they work nice.
    but when i connect a second lot different colour and voltage they just dont work together.
    is this becouse they are of a different voltage and how can i get around this problem

    thanks all
    marko
     
    Last edited: Aug 10, 2008
  3. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,346
  4. markosillypig

    Thread Starter Active Member

    Jul 21, 2008
    184
    0
    i want to use 2 white led's at 3.8 volt and 2 red led's at 2.1 volts
    and i need them to blink in same colour at same time

    thanks
    marko
     
  5. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    LEDs need a current-limiting resistor in series so that they don't blow up.
    If you want LEDs to blink then buy LEDs with a blinking circuit inside or make your own blinking circuit.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Do you mean that you want the two red LEDs to blink simultaneously, and the two white LEDs to blink simultaneously, but independent of the red LEDs?

    You say the white LEDs have a Vf of 3.8. What is their current rating?
    You also say the red LEDs have a Vf of 2.1. What is their current rating?

    What do you have available for a power supply?
     
  7. markosillypig

    Thread Starter Active Member

    Jul 21, 2008
    184
    0
    hi supply voltage is a pp3 9 volt through a led stroboscope circuit bought from maplin http://www.maplin.co.uk/Search.aspx?criteria=stroboscope&source=15&SD=Y
    i have removed the led's and attached wire's to the led's as pick shows the led's do not work simultaneously they work ok one colour at a time
    but not together
    yes i want the red and white led's to work together on the same output from the divice
    the red is 20ma
    and the white is 25ma

    pick attached to help i hope

    marko
     
    Last edited: Aug 11, 2008
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, that was very helpful.
    So, here is the schematic and manual for your device:
    http://www.velleman.be/downloads/0/manual_mk147.pdf
    If you tried to run the red LEDs in that circuit, you may have burned them out already.

    However, reinstall the white LEDs as they were originally.

    The white LEDs are 3.8v @ 25mA, so the 22 Ohm resistor R7 should have around 25mA going through it.
    E=IR (Voltage = Current x Resistance) so V(R7)= 0.025A x 22 Ohms = 0.55V
    0.55V + (3.8 x 2) = 8.15v. Hmm, they seem to be off a bit. 9v batteries tend to start off somewhere around 8.6v. No matter, perhaps they were counting on the (heavy) internal resistance of the battery. [eta - they were counting on the voltage drop across T3, which I'd forgotten about.] So let's work the other way 'round, and calculate a resistor for the red LEDs.

    You've said the red LEDs are 20mA @ 2.1V, so let's figure out the size resistor we'll need.
    2.1v x 2 = 4.2v, which is the total Vf across them.
    From the other result, we know that they're counting on 8.15v being available. So, let's subtract our LED's Vf from the total voltage.
    8.15 - 4.2 = 3.95.
    Now we need to know what resistance we need to get 20mA across 3.95v. Ohm's Law says:
    R = E/I (Resistance = Voltage/Current)
    R = 3.95/0.020
    R = 197.5
    The closest standard resistor is 200 Ohms. Let's see what our actual current will be:
    I = E/R
    I = 3.95/200 = 19.75mA - very close. Let's check the wattage necessary
    P = EI (Power in Watts = Voltage x Current)
    P = 3.95 x 0.01975 = 0.0780125, we double that for reliability: 0.156025 or 156mW - you can use a 1/4 Watt 200 Ohm resistor.

    So, connect one of the red LED's cathode (short lead) to where the white LED's cathode connects to T3's collector on the board.
    Connect the anode (long lead) of that LED to the cathode of the other red LED.
    Connect one lead of a 200 Ohm resistor to the anode of the 2nd red LED
    Now here's the tricky part - connect the other lead of the 200 Ohm resistor to the side of R7 that is next to the power switch.

    See the attached; it's just a small excerpt of the schematic in the manual showing where to connect the new LEDs and resistor.
     
    Last edited: Aug 12, 2008
  9. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    The picture in the OP seems to show half of the LEDs in backwards. There would be no current.

    Also, what is the voltage applied across the LEDs?
     
    Last edited: Aug 12, 2008
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You're right, I never really addressed that. However, with the OP's last response to the thread, I was able to track down Velleman's schematic for the kit, and demonstrated how to calculate the resistor for the red LEDs and how to connect them to his kit in my prior reply.

    I calculated it to be about 8.15v across the white LEDs and the 22 Ohm resistor; so between the battery's internal resistance and T3's ON drop must make up the 0.45 difference between the 8.15v and a fully charged 9v battery (which is actually 8.6v.) That's my best guess.
     
Loading...