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Problem switching a 5V mini relay from a low power LED (1,2v 2 mA) input

Discussion in 'Electronic Basics' started by [email protected], May 27, 2007.

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  1. Guest

    I am trying to make a 5V mini relay switch when the HDMI indicator LED
    in my DVD player lights up.

    I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via an
    optocoupler (4N33 - infrared/darlington) "piggybagged" on the
    indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
    slightly. Supplying an input of 1,5V from before the indicator LED
    resistor (220ohm) generates the same result. I even tried using the
    (normally unused) basis of the optocoupler as input, hoping it was
    more "sensitive" - but still only "humming"

    The setup works when "piggybacking" a standard 3V LED (>10mA), where
    the relay swithes fine. What have I overlooked ?

    Any suggestions for (simple) curcuitry I could use to trigger the 5V
    relay from the 1,5V 2 mA input (I only have 5V supply power
    available) ?
     
  2. Jamie

    Jamie Guest

    the opto isn't coming in because 1.2 volts is most likely not enough to
    fully.
    the LED in the unit is clamping down at a lower voltage.
    Disconnect the LED in side and use the same leds to drive
    the LED in the opto with no resistor.
    if you still need the LED as an indicator, connect the original
    LED to the COIL voltage of the relay via a resistor to protect it.

    If you don't want to do it that way.
    follow the lead wires of the internal LED back on the board to
    where there should be a Resistor that limits the current to that
    LED when on, connect your option with a Resistor on that side of the
    internal resistor..
    P.S.
    It's possible the internal driver for the LED may not be able to
    drive 2 LED's with out damage.
    so my first suggestion maybe better, use it's own driver circuit to
    drive the opto LED and rig up an indicator from the coil voltage..
    also, you need a reverse diode on the coil so that you don't damage the
    opto unless it already has one internally.
    Also make sure you have sufficient voltage to operate the relay.
     
  3. Guest

    Thank you for your reply.

    I think I already tried your second suggestion without success -
    "Supplying an input of 1,5V from before the indicator LED
    resistor (220ohm) generates the same result."

    I followed the lead wires of the internal LED back on the board to
    where there was a Resistor (220 ohm) that limits the current to that
    LED when on, and connected the optocoupler there. Unfortunately the
    result was still no switchin, which unfortunately also indicates to
    me, that your first suggestion might also fail.

    I think I need another alternative.
     
  4. It could be that the LED is being pulsed and is not a steady on
    condition. You would need a scopr of DVM with frequency input to
    verify it.
    Dave
     
  5. Chris

    Chris Guest

    You're almost there. The problem can be seen in the datasheet by
    looking at the "Current Transfer Ratio":

    http://www.fairchildsemi.com/ds/4N/4N33.pdf

    In the middle of page 3, it says I(CTR), or the DC current transfer
    ratio, is a minimum of 500%. That means you can't expect more than 5
    times the output current from the photodarlington than you're putting
    in at the optoLED input, and your 4N33 photodarlington won't be
    saturated at that current (which you'll need for driving a relay).

    Use a second generic NPN transistor like the 2N3904 to bump up the
    gain and give you a clean switch. View this in fixed font or
    cut&paste to M$ Notepad:
    |
    | VCC VCC VCC
    | + + +
    | 4N33 | | |
    | 1 | 5 | |
    | -----. .----o | | 1N4001
    | | | | C| -
    | | ~ |/ | RY1 C| ^
    | V ~ .-| | C| |
    | - | |> | | |
    | | | | |/ | |
    | 2 | | '--| o------'
    | -----' | |> |
    | 6| | |
    | | |4 ___ |/
    | | o---|___|---| 2N3904
    | .-. | 2.2K |>
    | 220K| | .-. |
    | | | 10K| | |
    | '-' | | |
    | | '-' |
    | | | |
    | === === ===
    | GND GND GND
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    Since your 4N33 only has to switch 2mA for the transistor base, you
    shouldn't have any problems.

    Good luck
    Chris
     
  6. Jamie

    Jamie Guest

    Did you disconnect the internal LED and connect the Opto to that
    same line?
    You must disconnect the internal LED, because it clamps down the
    voltage. that is where you're getting the 1.5 volts from./
    if you disconnect that LED then test the voltage when it's on, you'll
    find it much higher than 1.5 V
    most likely at least near what ever supply voltage you have in the unit.

    You can also place an milliamp meter in series with the OPTO LED to
    test it's current. You should be getting around 15 ma+ at least.
    If this is all passing..
    Then I would suggest that the output in the opto does not bias low
    enough to drive the relay, you would most likely need a help transistor..
     
  7. If you don't need isolation, you can use the simple circuit below, changed
    from what was suggested. It should work if the cathode of the LED is at
    ground. You might also need about a 10k resistor from the 2N3904 base to
    GND to make sure it turns off. Change the 2.2k to 1k if it does not turn on
    well enough.
    Paul
     
  8. Guest

    Any easy way to determine if the LED is being "pulsed", before trying
    something else ?

    Ole
     
  9. Rich Grise

    Rich Grise Guest

    If you didn't use a series resistor with your opto, you probably blew
    its LED.

    Either run another 220R resistor from the same spot you mentioned
    (switched Vcc), to the anode (probably) of the opto's LED, so there are
    two LED/R series sets running in parallel, or to save $0.003 on a
    resistor, unsolder the 220 from the LED and interpose the opto's LED in
    series, watching the polarity.

    If the opto has a good enough transfer ratio, that should do it.

    Another way which doesn't need an opto (i.e., if the grounds are in
    common), then take a 1K or so from the LED to the base of any NPN
    general-purpose transistor - if you only need 50 mA for your relay,
    almost any type should work. Or, better yet, connect the 1K to the
    high side of the 220, to be sure to saturate your switch.

    Have Fun!
    Rich
     
  10. Rich Grise

    Rich Grise Guest

    Switched + to LED/R o----||-----+---->|---o DC meter
    |
    _
    ^
    |
    GND

    That's any ol' cap, and 2x signal-type diodes - I suppose 1n400x would
    work. If you use a polar cap, positive goes to the right.

    Cheers!
    Rich
     
  11. Guest

    Indicator LED was indeed pulsed (which also explains the low power
    mesurements)
    100 uF capacitor in parallel with relay solved problem

    Thanks all,
    Ole
     
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