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Not understanding pull-down resistors and why they work

  1. Apr 6, 2013 #1
    basics4.gif

    I'm understanding that their purpose is to keep the pin connected to a stable potential (the ground), thus giving it a stable LOW voltage reading when the switch is not closed, but wouldn't the electric noise still affect the voltage reading, since the point at the switch closest to the pin is still susceptible to electric noise?

    Could someone explain to me where I'm going wrong?
     
    Last edited: Apr 6, 2013
  2. jcsd
  3. Apr 6, 2013 #2
    Without the pulldown you can think of it as a pencil balancing on its tip; only a small amount of force (noise) will tip it one way or the other. With the pulldown in place, you take an active choice and say that the pencil shall lie in a particular direction (GND). Whether or not the circuit is still susceptible to noise after the introduction of the pulldown resistor is a different subject.
     
  4. Apr 6, 2013 #3

    vk6kro

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    Science Advisor

    No, you are not wrong.

    A resistor to ground is not as perfect as a real ground, but it is vastly better than nothing.

    Closing the switch certainly pulls the gate input high, but opening it does not pull the input low without the resistor.

    It is a design complication that the noise on the input should be low enough to not affect the circuit operation. It is a digital circuit, so relatively large noise levels would have no effect.

    In practice, noise levels would be nowhere near large enough to affect circuit operation.
     
  5. Apr 6, 2013 #4
    Thank you gentlemen, that makes sense. I appreciate it!
     
  6. Apr 6, 2013 #5
    Also - in the circuit you have shown - what is the alternative? With no resistor - and no connection to ground, you have no voltage reference when the switch is open and this would be very noisy & you obviously can not connect the terminal directly to ground.

    In general practice, all circuit elements should have a reference to something (voltage, ground etc) - an not be left "floating"
     
  7. May 24, 2013 #6
    Hi,
    I want to ask some questions here to check my understanding.
    When pull-down resistor is added, we still affect the noise. If so how is it improved?
    With pull-down resistor R1, portential will be dissipated P = I^2*R1 , right?
     
  8. May 24, 2013 #7
    Im a bit unsure of what you mean, but let me try.

    With the pull-down resistor, of course we are dissipating energy given the formula

    [itex]P = \frac{V^{2}}{R}[/itex]
    or
    [itex]P = I^{2}\cdot R[/itex]

    So of course, for a system containing lots of these pull downs, we would like to keep the power consumption as low as powerful (we want to be green right? ;) ). This means we would need to have a high resistance.

    However, the problem with a high resistor is, as you say, noise. Now, noise is a completely different idea and is it only enters the scene when we do not deal with ideal circuits anymore.

    When the switched is opened (the pin is having a ground potential) radiated EMI (Electromagnetic Inteference aka. noise) can make the potential of the pin go up or down. If it goes up, in order to get down to the ground potential again, current will have to flow through the resistor in order to lower the potential again. Now, if the resistance is very high, this will take a long time (and here by long time I mean microseconds perhaps). With a small resistance, a larger current will flow.

    So high resistances will make the stuff more susceptible to noise, yes, but there is not much to do about it right there. The EMI will have to be removed by other means, like shielding.

    Normally, resistor values in the kilo-ohm range are used.

    I hope this helps.
     
  9. May 24, 2013 #8
    Hi Runei,
    I don't quite understand this part. Could you explain a bit more?
    When pull-down resistor is connected, the input voltage at the pin always zero as switch open and free noise?
    When switch open and enviroment has noise, it will induce a voltage at the pin, right?
    And the voltage doesn't depend on the value of resistor?
     
  10. May 24, 2013 #9
    basics4.gif

    In the picture shown the switch is open. In an ideal environment the voltage at "1" will be 0. Thus, it will have the same potential as the ground.

    If we are in a noisy environment, we have to imagine that there is a small capacitance in parallel with the resistor to ground. This is called a parasitic capacitance, since it is an unwanted but always present nuisance.

    attachment.php?attachmentid=58994&stc=1&d=1369418281.png

    When this capacitor is there, and we have incoming noise, the voltage may rise. Lets say the voltage suddenly rises to 1 volt because of incoming electromagnetic interference. Now there is a voltage across the resistor to ground, and current will flow. As the current flows, the capacitor will be discharged and return to 0 volts. The rate at which this happens depends on the capacitor value (very very small) and the resistor value. If the resistor value is very very high, then there will only be a small current flowing, and the potential will decrease more slowly, than if the resistor was small.

    That is the first part of it. Now, the voltage rise (from the noise) is actually dependent on the resistor value.

    attachment.php?attachmentid=58994&d=1369417994.png

    If we imagine a new wire connecting to point "1". This wire is a "virtual" one, and is only there to signify the incoming noise. The noise is modeled as an incoming current that lasts for lets say 2 microseconds.

    Now, if the resistor is small, more current will flow through the resistor, and the voltage at "1" after the incoming current is over is less than if the resistor had been big.

    The conclusion of this is that a small resistor will help you against noise, but it will also increase power consumption.
     

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  11. May 24, 2013 #10
    Thank you for the prompt and thorough answer! :smile:
     
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