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Discussion in 'Electronic Basics' started by Flygel, May 12, 2005.

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  1. Flygel

    Flygel Guest

    HI all

    I have made circut B at this scheme.
    http://ourworld.compuserve.com/homepages/Bill_Bowden/printer2.gif

    it somewhat works, and when there is current on base of transistor 1 (from
    the left)
    the output is 9 volts.. witch is great.

    BUT, when there is no base on trans 1 , there is a output of 4 volts..
    what causes this ?

    it makes the relay hit when there is base current, but not release when the
    base current ends :(

    Thanks for any help

    Chris
     
  2. Brian

    Brian Guest

    I would put something like a 10K resistor between the base and emitter of
    the 2N2905 transistor. If that doesn't do the job, then you must have a bad
    transistor. What kind of circuit do you have turning on the 2N3904
    transistor? How much current does the relay require?

    Brian
     
  3. jgreimer

    jgreimer Guest

    The circuit isn't a good design for the very problem you had. Apparently
    the 2N3904 has a little leakage current which isn't too unusual.
    Furthermore leakage increases with temperature so if you put the circuit
    inside a warm printer, it will get worse.

    The first thing to do is bring the input to the base of 2N3904 to the normal
    low voltage during operation, for instance zero volts. Then measure the
    voltage across the 300 ohm resistor. This will tell you the leakage current
    (V(Re) / 300). Let's say you measure 0.15 V giving you a leakage current of
    0.15 / 300 or 500 uA. Then bring the input to the normal high level voltage
    and measure the emitter current. For instance if the normal high voltage is
    5 V then the voltage across the 300 ohm resistor will be about (5 - 0.7) or
    4.3 V and emitter current is 4.3 / 300 or about 14.3 mA.

    There's a really obvious answer at this point. If the leakage current
    causes 0.15V across a 300 ohm resistor at the low input level and the high
    input level causes 4.3 V across the 300 ohm resistor, then if you also use a
    300 ohm resistor from the collector of the 2N3904 to +12 V, you should get
    the same voltages across that resistor, if the 2N2907 weren't connected. The
    0.15 V due to leakage current won't be enough to turn on the 2N2907 and the
    14.3mA - (.7 / 300) =12 mA, is more than enough to drive the 2N2907.
     
  4. Chris

    Chris Guest

    Hi, Chris. I don't believe the originator of your circuit B knew what
    he was doing -- you got bad advice. As others have mentioned, the
    leakage current of the first transistor is partially turning on the
    second one. Most relay manufacturers say the voltage across the relay
    has to go well below 20% of nominal to ensure turnoff.

    Here's something close that will only require one more resistor and
    will work well (view in fixed font or Notepad):

    ` VCC
    ` +
    ` |
    ` .-. VCC
    ` 300| | +
    ` | | |
    ` '-' |
    ` | ___ |<
    ` o-|___|--|
    ` | 300 |\
    ` ___ |/ |
    ` o-|___|--| .--o
    ` 2K |> | |
    ` | | C|
    ` === - C|
    ` GND ^ C|
    ` | |
    ` ======
    ` GNDGND
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Both transistors are being used as pure switches. The pullup resistor
    to Vcc will swamp out any leakage current from the first transistor.
    As a practical matter, this should get your project going.

    Good luck
    Chris
     
  5. Bill Bowden

    Bill Bowden Guest

    Hi, Chris. I don't believe the originator
    Well, I would use 1K resistors in place of the 300 ohm resistors to
    reduce the load on the 2N3904 transistor.

    If you use 300 ohm resistors as shown, they will essentially be in
    parallel and the total load will be 150 ohms at 12 volts which equates
    to 80mA of collector current for the 2N3904.

    I'm not sure a 2N3904 tansistor is capable of that current with
    a 1mA input. You need a gain of 80 at 80 milliamps which
    is a little optimistic for a little 2N3904.

    Besides that, you also have a power dissipation problem, since the 3904
    may turn on to it's limit at say 40mA and drop 6 volts, which is around
    1/4 watt of heat. This may cause more problems.

    If you use 1K resistors, the total load will be 500 ohms, or about
    24 mA and the input to the PNP transistor will be half that, or 12mA.
    You only need a gain of 10 for the PNP to supply 120 mA which should be
    enough for most 12 volt relays.

    So, it's not as simple as it looks, there are many factors to consider.
    But I will change the circuit to your suggested configuration using 1K
    resistors.

    Thanks,

    -Bill Bowden
     
  6. Bob Monsen

    Bob Monsen Guest

    I'd say you probably aren't connecting the circuit and parallel port
    grounds together. The circuit itself looks fine.
     
  7. Chris

    Chris Guest

    Hi, Bill. Yep, at least the 300 ohm pullup. That could be changed to
    1K without any problem.

    With stuff like this, you sometimes wonder what the person who did the
    original circuit meant. I was guessing he felt he needed the 300 ohm
    resistor for base drive on the second transistor (we don't know coil
    resistance), and I was going with that.

    Your idea of changing both resistors sounds great, and is an
    improvement which would save power.

    Good luck
    Chris
     
  8. Rich Grise

    Rich Grise Guest

    For the 2N3904's collector resistor, you could use a much higher-value
    resistor - all it's doing is wasting power anyway. I'd be comfortable
    with as much as 47K. You don't really have to worry all that much about
    thermal noise or phase response, after all. :)

    Cheers!
    Rich
     
  9. Bill Bowden

    Bill Bowden Guest

    With stuff like this, you sometimes wonder what the person
    I am the person who did the original circuit, and I was using
    a relay with a coil resistance of 120 ohms which worked
    well. I suspect the original poster is using a relay of much
    higher resistance which may not turn off if there is any
    small leakage current from the first stage.

    But, if that is the case, and he is using a relay with a
    coil resistance of say 500 ohms, he only needs 24mA
    of current, which is easily obtainable from a single stage.
    No need for the second PNP transistor at all.

    Thanks,

    -Bill
     
  10. Flygel

    Flygel Guest

    Hey.

    Cool to see youre actually here .

    First i want to say thanks for a very informative and nice webpage!

    I use a 100 ohm coil on the relè .

    chrisitian
     
  11. Brian

    Brian Guest

    Cool to see you are here, too. It would have been nice to hear from you
    sooner. :>)

    Brian
     
  12. Bill Bowden

    Bill Bowden Guest

    chrisitian,

    That circuit should work as shown, If you get a voltage
    across the relay with the input open, or at ground, you
    may have a bad transistor.

    But, you can also use Chris's version with the extra
    resistor, but I would change the 300 to 1K resistors.

    -Bill
     
  13. Bill Bowden

    Bill Bowden Guest

    It would have been nice to hear from you sooner. :>)

    Brian,

    Yes, well I don't read all the newsgroup postings every day,
    so I didn't discover the thread until it was 4 days old.

    Sorry about that,

    -Bill
     
  14. Brian

    Brian Guest

    No Bill, I wasn't talking about you, I was talking about Chris (who posted
    the original question). I had asked him what kind of circuit was driving
    your circuit and how much current his relay required (there was no reply).
    I'm sure that some of the people who tried to help him, would've like that
    information too.

    Brian
     
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