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Powering a relay

Discussion in 'Electronic Basics' started by Captain Dondo, Sep 14, 2005.

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  1. OK, I have a real basic question....

    I have an embedded computer. It had digital I/O. Per the manufacturer,
    "When the DIO pins are configured as outputs, they can source 4 mA or
    sink 8 mA and have logic swings between 3.3V and ground."

    OK, I prefer to work as source. So... How do I get 4 ma @ 3.3V to
    control a 5V relay that needs 100 ma to operate?

    ISTR something about transistors, but it's been a few years since my
    electronics classes....

    Right now I need to be able to power LEDs, but eventually I need to
    siwtch 120 VAC and 24-48 VDC at a few amps load, and I thought I could
    use relays for that. But this has thrown a wrench into the works....

    Any suggestions?

    Thanks,

    --Yan
     
  2. Mark VB

    Mark VB Guest

    So, you'll have to use an NPN-transistor with Vce>5 V, Ic>100ma, Hfe>25.
    Don't forget to put a flyback-diode over the coil.


    HTH,
    Mark Van Borm
     
  3. Yan,

    A transistor solution shown below. The NPN are general purpose transistors
    like BC547. Don't forget the diode otherwise your transistor will be blown.
    An 1N4148 or similar will do. A much simpler solution is using a FET.
    A BSS295 or similar will do.


    +---+----+------+5V -+-----+-
    | | | | |
    | - _|_ | |
    .-. ^ |_/_|- - _|_
    | | | | ^ |_/_|-
    330| | +----+ | |
    '-' | | |
    | | +-----+
    | |/ |
    +------| NPN |
    | |> |
    | | |-+
    ___ |/ | ----->| BSS295
    -|___|--| NPN | |-+
    5k6 |> | |
    | | |
    | | |
    ----------+--------+-------- --------+-----
    GND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de


    petrus bitbyter
     
  4. ehsjr

    ehsjr Guest

    Nice. Another possibility is a darlington:



    +----+------+5V
    | |
    - _|_
    ^ |_/_|-
    | |
    +----+
    |
    ___ |/
    --|___|----| TIP120
    5k6 |>
    |
    |
    Gnd -----------------+--------

    Ed
     
  5. I think the OP was asking that when the I/O is _sourcing_ 4mA, that
    this is when the relay should be turned on. I may have misunderstood,
    though.

    Jon
     
  6. That's more like what I imagined for the OP, since that turns on the
    relay when the I/O is sourcing. However, assuming that the TIP120 is
    a Darlington, this also tosses away some extra of the very few volts
    available at +5V for the relay.

    So, what about:

    : +5 +5
    : | |
    : | |
    : \ |
    : / R1 |
    : \ 10k |
    : / |
    : | |<e Q2
    : +---------| 2N3906
    : | |\c PNP
    : | |
    :3.3V I/O R3 |/c Q1 |
    : CONTROL------/\/\---+----| 2N3904 |
    : SIGNAL 10k | |>e NPN +---------,
    : \ | | |
    : / R4 | )| 5V |
    : \ 100k | )| RELAY |
    : / \ )| --- D1
    : | / R2 )| / \ 1N914
    : | \ 1k | ---
    : | / | |
    : | | | |
    : gnd | | |
    : gnd gnd gnd


    Jon
     

  7. You're right. The transistor solution shown above inverts the output signal
    which might not be wanted. The FET solution does not invert. As you have 4mA
    source available, you may use a single transistor as shown below. You - or
    the OP - only need to keep an eye on the type of transistor choosen. Its
    current amplification should be high enough to drive the transistor fully
    into saturation at the 100mA collectorcurrent required. So not every general
    pupose transistor will do. In both transistor solutions the transistor that
    drives the relay should have not too high a VCEsat, so to keep enough
    voltage for the relay.

    +5V----+-----+-
    | _|_
    - |_/_|-
    ^ |
    | |
    +-----+
    |
    |
    ___ |/
    ---|___|--|
    680 |>
    |
    |
    |
    GND---------+--
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    petrus bitbyter
     

  8. May work but my main objection is the 2N3904 is used as a lineair amplifier.
    The base current though the 2N3906 may not be high enough to drive the
    transistor into saturation. Used in another - simpler - circuit, the 2N3904
    may do it all on its own. (See my other posting.) I'd remove R4 as it has no
    use and lower R3 to let's say 1k so to use the available control current.
    Then I'd remove (short) R2 but place a similar or even somewhat lower
    resistor between the base of Q2 and the R1/Q1 node to limit the base current
    for Q2.

    petrus bitbyter
     
  9. I agree. For a really low Vcesat at pretty high currents, the TIP42A
    is decent. But in this case at 100mA, I think the common 2n3904 won't
    be too bad.

    What bugs me and what drove me to offer two BJTs as a solution is that
    the OP probably has no real idea what the output voltage is at when
    sourcing 4mA. The OP sees the specification, but I wouldn't be
    surprised to see a drop of almost a volt at that current (200 ohms in
    the FET, or even more, isn't unusual.) So I'd generally rather add
    another BJT just to keep it as "no problem, at all" and just not worry
    about the question. Plus, it does put that nasty relay inductor just
    a little bit further away from the I/O. :) Well, that's how I was
    thinking, anyway.

    Jon
     
  10. Okay. I think that's a valid criticism to consider. I happen to like
    keeping Q1 out of saturation, myself. But let's look at the details
    and see where that goes.

    The arrangement of R3 and R4 yields about 90% of the 3.3V (current
    required is minimal and won't pull down the I/O pin) or say 3V at the
    base of Q1. This means that about .6V or .65V less than this will
    appear at the emitter of Q1. That alone suggests about (3 - .65)/1k
    or 2.35mA collector current for Q1. With the collector voltage no
    more than the harder-driven Q2 Vbe below 5V, or let's say about 4.15V,
    this means about 0.85V/10k or 85uA through R1. This leaves most of
    the 2.35mA or let's say at least 2.2mA for base drive on Q2.

    With 100mA drive for the coil, even if 200mA for a short time, the
    2.2mA will probably be okay as base drive for it. It's a relay though
    with inductance, so it probably will be a ramp up on the current until
    the DC ohms takes over.

    It seems okay.
    I think I agree. I just don't like pulling the mA out of the I/O, I
    guess.
    I added it for the specific scenario that the circuit might be tested
    without an I/O actually connected. I just wanted to make sure there
    was a little bit of pull-down to keep Q1 off in such a case.
    Well, not unless you also kill R2.
    Ah, there you killed it. But then you are sucking that current out of
    the I/O. I suppose that's okay. I just prefer overkill. For a
    one-off (which is all I ever try and make), anyway.
    A problem you created by deleting R2, which I had in the circuit in
    the first place and yielded a nice, poor-man's controlled current sink
    for the base drive.

    Jon
     
  11. ehsjr

    ehsjr Guest

    I got to thinking some more about this. Why are we restricting
    Vcc to 5 volts? It may be an artifical restriction. Later on he
    will have 120VAC and 28-48 DC that he wants to switch with a
    relay. Nothing says he can't use those (or any other) sources to
    power the relay. He can use a single bjt/fet/darlington as
    a driver stage. If there's no common ground, or if the 5V is
    all that's available, he could drive an opto with the driver stage.
    Opens up the choice of relays, too.

    Ed
     
  12. Kitchen Man

    Kitchen Man Guest

    Check out this paper:

    http://www1.jaycar.com.au/images_uploaded/relaydrv.pdf

    and realize there are integrated circuit relay drivers that will
    vastly simplify your design requirements, for instance:

    http://www.ee.washington.edu/stores/DataSheets/cd4000/74c908.pdf
     
  13. Jasen Betts

    Jasen Betts Guest


    NPN transistor (eg BC337) emitter to ground colector to relay coil
    other end of relay coil to +5V supply, diode (eg 1N4001) with the
    cathode (banded end) to +5V and the other end to the collector,
    base of the transistor goes to a 1K resistor and the other end of the
    resistor goes to the chip's output pin.
    for switching the 120AC it may be easier to use a triac and a MOC3010
    opto-isolator.

    for the DC 48V is kind of high for a commonly available relay to switch
    (most 240VAC relays are only good to 30VDC) a power MOSFET may be better
    suited.

    Bye.
    Jasen
     
  14. Jasen Betts

    Jasen Betts Guest

    at 4ma source and 8ma sink capacity he can drive pretty much any opto
    coupler directly.
     
  15. ehsjr

    ehsjr Guest

    Perhaps - but connecting an opto directly? What protects
    his embedded computer? Use a driver! The 4K7 (or whatever)
    base resistor protects the computer from shorted LED or
    bjt, and keeps the current minimal during normal operation.

    Ed
     
  16. Richard

    Richard Guest

    How many outputs would you have total? You can use a darlington ULN2804?
    has 8 of these in a single chip which will drive loads of 500 ma per output,
    with freewheeling diodes available internally, and very easy to interface
    to. You can drive the LED's off this, or the relays, or both.

    I use these all the time for TTL to Relay Drivers. My control board Outputs
    are rated at 6ma.

    Richard
     
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