Using jumpers in a circuit

Discussion in 'General Electronics Chat' started by samoz, Jun 11, 2009.

  1. samoz

    Thread Starter New Member

    Jun 10, 2009
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    I'd like to design a circuit that uses jumpers to get options from a user. When the jumper is off, I want the data signal to be a logic 0. When the jumper is on, I want the data signal to be a logic 1. However, I'm a little confused about how to do this.

    I'm talking about a digital circuit, so a simple 1 or 0 is all I need.

    If I have one side of the jumper at Vcc and the other connected to my signal and I place a jumper across, this would bring the signal high. However, if the jumper was left off, what would happen?

    Would the data signal float to some middle, nondescript zone?

    Should I tie the data signal to ground and then put a current limiting resistor on either the data signal or Vcc? Would this do what I want?
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
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    You can always pull a level up with a resistor. Use the jumper straight to ground to pull the level low. But you can't pull up a low.

    If you need to invert the level, use an inverter IC.
     
  3. hobbyist

    Distinguished Member

    Aug 10, 2008
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    Heres one way it could be done.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    This is about as easy as it gets:
    [​IMG]
     
  5. samoz

    Thread Starter New Member

    Jun 10, 2009
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    Thanks very much both of you for the schematics! Those will both work for me!
     
  6. samoz

    Thread Starter New Member

    Jun 10, 2009
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    Ok, quick question. For the pull down resistor, if the output will be triggering transistors, should I use 470 Ohms or 10K ohms?

    I know you said TTL is 470 Ohms, but I just want to double check that this is what I should be using for only signalling transistors.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    That changes things.

    If you're operating from around 5v, then you could use this:
    [​IMG]

    If you're running from other than 5v, you'll need to adjust the values of R1/R2.

    If the circuit shown above is used as input to the base of a transistor, base current will be around 9mA. This is sufficient to put a 2N2904 transistor into saturation.

    If you are going to use other than 5v for Vcc (shown as Vdd), then divide the values shown by 5, and multiply those results by the actual Vcc you're going to be using.

    Note that in this circuit, the output voltage won't actually be Vdd if jumpered; it'll be about 4.77v with no load. That's plenty for either TTL or CMOS to recognize it as a logic 1.
     
  8. radiohead

    Active Member

    May 28, 2009
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    why not use DIP switches instead of jumpers?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    DIP switches would work the same as jumpers. They'd still need resistor/resistors to limit current through a transistor's base.

    The purpose of the 10k resistor is to ensure that the transistor is fully turned off. If PNP transistors were being used instead, R1/R2 would need to be swapped.
     
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