Help!!!!! Leakage Current!!!!

Discussion in 'The Projects Forum' started by newbie, Nov 5, 2006.

  1. newbie

    Thread Starter New Member

    Nov 5, 2006
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    0
    HI all. Thx for taking a look at this thread. This is my first post. I am currently doing a project and I am stuck at this:

    Notice the red arrows in the picture attached. It is connected to an equipment directly.

    I'll get straight to the point. How do I prevent leakage current flowing in the direction of the red arrows, so no leakage current can escape through tt direction
    Limitations are not changing any values of the components in the picture.

    The Inputs are 0.25V analog signals which is suppose to trigger a separate circuit. But the the trigger circuit may introduce leakage current into the equipment.

    What I thought was adding a diode for each input to protect the equipment. Will it work?? If not what can I do??

    THX in advance!!!!
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    If I understand your concern, it's about preventing negative currents going out through the input resistors. If that's the case, just crank on the 10K pot until the input voltage is always positive, even with the .25 volts inputs (you didn't sate the polarities or nature of these inputs). Do recall that in real-world electronics thre are always leakage currents.

    Your circuit seems a bit elaborate. You have the one input pinned at 6 volts, and the other with the potentiometric adjustment to also be 6 volts. Since the 6 volt term will be common mode, why not just do away with the dividers and use 0 volts instead? The pot and fixed resistor to ground mess with input impedance. The signals are applied single-ended, so the instrumentation amplifier has no real function. You could do as well with an ordinary op amp.
     
  3. newbie

    Thread Starter New Member

    Nov 5, 2006
    3
    0
    Hi,



    First of all, thx for the reply.

    With response to beenthere, the .25V inputs are positive. In other words, the polarity is opposite of the red arrows.

    The .25V is a analog step input. The .25V goes high for maybe for example 0.5s, then goes low again. I'm trying to use this spike to trigger the LED on the LMC555 side using its Monostable Operation. And Monostable Operation is only activated when Trigger(Bar) input voltage is 1/3Vs.

    <http://www.national.com/ds/LM/LMC555.pdf>

    In other words, if the common mode is 0V (top part of the voltage divider would not be able to maintain a voltage lower than the bottom), I cant have a change in polarity at the Instru Amp (the .25V causes the Instru. Amp. to give a 0V output when the top part of the voltage divider exceeds the bottom 6V with for example 6.1V) side to cause the drop of in Trigger(bar) voltage. If you would look at the original circuit, you would notice that I swap the polarity pin 3 with pin 2 of Instru. Amp. (please look at original image)

    <http://focus.ti.com/lit/ds/symlink/ina122.pdf>

    Please correct me if I'm wrong.

    I have a question, beenthere. Do you mean tt I just increase the resistance at the Input to decrease the leakage current to the minimal??


    THX in advance!!
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    I think the instrumentation amp is still out of place. From your further description, a voltage comparator would do the trick. I've attached a circuit that will hold the LM311 output high with around 100 mv positive, and will swing the output to 0 when any of the .25 mv step signals come in.

    I might have got a value wrong on the 555.
     
  5. newbie

    Thread Starter New Member

    Nov 5, 2006
    3
    0
    Hi,


    I think I'm seeing the big picture.

    If I'm not wrong, u mean tt the 100mV positive will hold the output of the comparator at close to 12V, with the negative input 0V.

    When the negative input administors 250mV, the output will be driven close to 0V.

    Doing this solves the leakage current problem, and it can also simplify the circuit.

    Please correct me if I'm wrong.


    THX THX
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    That shoud be just the way it works.
     
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