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Voltage loss over diodes/transistors

Discussion in 'Electronic Basics' started by Jan Nielsen, Jun 20, 2007.

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  1. Jan Nielsen

    Jan Nielsen Guest

    Hi,

    I have a circuit that runs off a 5VDC regulator.
    I need to drive 2 relays, and the ones I have are 6Vdc, but ofcause they
    switch fine at lower voltage.

    The problem is, to switch the relay I have a transistor, and then a
    diode to prevent back current, after the transistor and diode I am left
    with 3.8V which is very close to the limit of what the relay will
    accept, so its not that reliable.

    Is there a way to limit that loss, or do I have to run at a higher
    voltage to start with ?


    /Jan
     
  2. Rich Grise

    Rich Grise Guest

    Don't put the diode in series - put it in parallel with the relay coil,
    in reverse polarity, of course.

    Good Luck!
    Rich
     
  3. ian field

    ian field Guest

    Use an additional 8V reg just for the relay.

    Also do what Rich said, the transistor VCEsat should be around 0.2 - 0.4V so
    you should have way more than 3.8V left for the relay.
     
  4. Chuck

    Chuck Guest

    If you can use an adjustable regulator,
    you can sense the 5 volts at the relay,
    thus avoiding the voltage drops.

    Chuck
     
  5. Eeyore

    Eeyore Guest

    As has been mentioned, the diode should not be in series with the coil but
    reverse parallel.

    To keep the transistor voltage loss to a minimum, drive the base with plenty of
    current to as to saturate the transistor ( say use a ratio of Ic/Ib of 10:1 )
    and select a transistor with a low saturation voltage Vce(sat).

    What's the coil resistance btw ?

    Graham
     
  6. Jan Nielsen

    Jan Nielsen Guest

    Thanks all, I connected the diode across the relay coil legs and it works.

    Sounds right with the 0.2-0.4 loss from the transistor, plus the 0.7V
    from the diode = 3.8.

    I cant run the circuit at much higher voltage, since theres a mcu
    accepting at most 5.5V.

    But it works with the parallel diode, thanks.

    /Jan
     
  7. What voltage is available up stream of the regulator? Using
    that voltage (with an additional series resistor, if needed,
    to drop the extra voltage) will keep the regulator output
    cleaner and give you a certain activation.
     
  8. Jan Nielsen

    Jan Nielsen Guest

    Eeyore skrev:
    100ohm between the coil legs acording to my DMM.
    The base of the transistor is being driven from a mcu, so it can get
    atmost 20mA.

    /Jan
     
  9. Jan Nielsen

    Jan Nielsen Guest

    John Popelish skrev:
    Theres 9-14V or so before the regulator, from a 12V 20Ah battery drawing
    7.5A at peak, my circuit is only using about 60mA at peak, and <10mA at
    idle, which is most of its time.


    /Jan
     
  10. Jasen

    Jasen Guest

    you have the diode in the wrong place, (should be parallel with the
    relay in the non-conducting direction)
    if you can use the power from before the regulator to run the relay
    that would help.

    Bye.
    Jasen
     
  11. ian field

    ian field Guest

    The suggestion seems to be that if you run the relay off a separate supply
    path, you reduce the risk of switching transients on the mcu rail - a
    dropper resistor from the unreg supply would be fine and if you included a
    small electrolytic this would charge to the full supply when the relay is
    off giving a brief burst of higher voltage making the relay snap to more
    positively, you could also reduce the overall relay current to the minimum
    safe holding value without compromising pull-in. With the voltage margins
    you stated, my suggestion of a separate 8V regulator would require a LDO
    type.
     
  12. It would also eliminate about a half watt of heat into the 5
    volt regulator (move it to the series resistor), possibly
    eliminating the need to heat sink it.
     
  13. Eeyore

    Eeyore Guest

    Now that's useful.

    Such measures can help reduce the build-up of hot spots.

    Graham
     
  14. Eeyore

    Eeyore Guest

    Excellent advice.
    Yes, it can cause latch up.

    Graham
     
  15. Jamie

    Jamie Guest

    Config your circuit using a common emitter for the transistor.
    the collector goes to one side of the coil.
    the other side of the coil to your 5 VDC.
    the base with a resistor in series goes to your switching signal.

    Place a protection diode across the relay coil. , the cathode (line
    side) will go to the 5VDC+ of the coil, the anode to the collector side
    of the coil..
    Coils will release high voltage energy if allowed to discharge at a
    rapid rate. Coils discharge in reverse polarity and thus the diode makes
    for a component to do the job.

    Also, it's good practice to put a bypass diode from the input to the
    output of the regulator. many regulators do not like voltages much
    higher on the output side than on the input side. if you place the
    diode with it's cathode to the input and anode on the output. current
    will only flow when the output is higher than the input and thus force
    caps on the output to discharge. etc..
     
  16. Jan Nielsen

    Jan Nielsen Guest

    John Popelish skrev:
    My regulator dont need heatsink, its a small plastic version, rated at
    max 100mA, the relayas draws 50mA each, but wont be on at same time, and
    the rest draws a few mA, so it will do :)

    I have the bulky 7805 laying around too, that needs to be bolted to the
    copper.

    /Jan
     
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