555 monostable question

Discussion in 'Homework Help' started by TsAmE, Nov 5, 2010.

  1. TsAmE

    Thread Starter Member

    Apr 19, 2010
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    0
    Draw a circuit diagram of a 555 edge-triggered monostable, using a 1μF timing capacitor.

    The output pulse length should be approximately 3.3 seconds.

    Include a potentiometer across the supply to allow the output pulse-width to be varied by changing the threshold.

    Indicate component values and 555 pin numbers (other than the supply pins).

    If the potentiometer is set to deliver a voltage equal to one third of the supply voltage, estimate the output pulse-length T.

    I am not sure how to calculate the output pulse-length T. I know that since the potentiometer delivers a voltage = 1/3 of the supply voltage then the new upper threshold is Vsupply/3 and the lower trigger level = (Vsupply/3) / 2 = Vsupply/6. But then what would you do?
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    This give you a complete explanation, along with a theory of operation.

    555 Monostable

    Break it down. After you read this what are you having problems with?

    Pin 5 allows you to change one of the limits of the 555. A 555 uses 1/3 and 2/3, pin 5 is the 2/3 voltage setting point.

    Some major math required, since this pertains directly to the RC time constant and RC charge curve.

    RC Precision Drawing

    Basically you have to calculate how much percentage of an RC time constant the two new set values will take.
     
  3. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    I know that the old period of the monostable use to be 1.1RC = 3.3 seconds. Now since a control voltage (Vc) is applied to pin 5 this changes the timing period. Pin 5 changes the upper threshold from 2/3 of the supply voltage (V+) to Vc and the lower trigger level to Vc/2.

    Now the info in the problem given is that Vc = 1/3V+, so this is the new upper threshold and the lower trigger level is Vc/2 = (1/3V+)/2 = V+/6. So all this should modify the timing period from 1.1RC to kRC, but I am not sure how to get this k.
     
  4. TsAmE

    Thread Starter Member

    Apr 19, 2010
    72
    0
    I figured it out a part of it:

    First I worked out R from the old pulse given:

    3.3 = 1.1RC
    3.3 = 1.1R(1μF)
    R = 3.3 / (1.1 x 1μF)
    = 3 x 10^6 Ω

    The pulse changes from 1.1RC to kRC so in order to find k:

    t/RC = k
    =-ln(1 - Vc/V+)
    =-ln(1 -V+/3V+)
    =-ln(1 - 1/3)
    = 0.4

    t = kRC
    = 0.4 x (3 x 10^6)(1μF)
    = 1.2s

    but the correct answer in my notes said:

    t = 0.4 x 0.33
    = 0.13s

    This means that my resistor value is probably wrong, but I cant see what I did wrong.
     
  5. Wendy

    Moderator

    Mar 24, 2008
    20,765
    2,535
    You have to remember where the 1.1 in the formula is coming from. It is the amount of time it takes for the cap to charge to the 66.6% of the RC curve (2/3 of the power supply voltage), shown below.

    [​IMG]

    Move the percentage (the 2/3 value, which is 66.7%) and the new number read on top of the graph becomes the equation.

    What is the new percentage (voltage on pin 5)? It can not go below 33%, and it can not be 100%.
     
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