Calculating t1 and t2 on a 50% duty 555 astable

Discussion in 'General Electronics Chat' started by tracecom, Mar 30, 2013.

  1. tracecom

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    Apr 16, 2010
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    I found a circuit in the National datasheet for the 555 that is supposed to be a 50% duty cycle for a 555 astable. I built the circuit and it seems to work as expected. So I thought I would try to calculate the t1 (HIGH) and t2 (LOW) time for the output according to the formulae given. The HIGH calculation was pretty easy, and I think I got it right, but the LOW calculation is wrong. My work is attached; where did I go off track?

    Thanks.

    Never mind. I found my error.
     
    Last edited: Mar 30, 2013
  2. tracecom

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    Here are my (hopefully) correct calculations. As you can see, the duty cycle is not 50%; it's close, but not exact. Is that due to a calculation error that I have not found?

    Thanks.
     
  3. WBahn

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    Mar 31, 2012
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    I can't open the png file you attached to your last post and I don't see any attachment to your first post. So I don't even have any idea what the circuit is you are looking at.
     
  4. Jony130

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    Feb 17, 2009
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    The circuit diagram and the equation looks like this

    [​IMG]

    But there is a one think that I don't understand.
    The book give as the equation for discharge a capacitor.

    \Large V_c = V_s*(e^{\frac{-t}{RC}})

    And if we solve this for time we get this

    \Large t = RC *ln(\frac{V_s}{V_c})

    In this case Vs = 2/3Vcc and Vc = 1/3Vcc and R = RA||RB

    So we end up with

    t = RC•ln ( 2/3 / 1/3) = ln2•(R•C) = 0.693•RC

    So why this equation don't work in this case?
     
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  5. tracecom

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    The circuit in question is the one posted by Jony130, and my calcuation sheet is attached to this post as a PDF. My question is, "I found a circuit in the National datasheet for the 555 that is supposed to be a 50% duty cycle for a 555 astable. I built the circuit and it seems to work as expected. So I thought I would try to calculate the t1 (HIGH) and t2 (LOW) time for the output according to the formulae given. As you can see, the duty cycle is not 50%; it's close, but not exact. Is that due to a calculation error that I have not found?"

    Thanks.
     
  6. WBahn

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    Because your equation is for a capacitor that is discharging toward 0V. This one isn't. It is discharging to the voltage set up by the Ra,Rb voltage divider.

    The general form of a first order response is

    v(t) = Vf + (Vo-Vf)e^{\frac{-t}{\tau}}
     
  7. WBahn

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    My first thought is that it would be somewhat remarkable if two values from the 5% resistor sequence would give exactly a 50% duty cycle.

    My second thought is that I would not be surprised if you couldn't get closer to 50% that what these appear to give.

    I used to keep a spreadsheet that had all the ratios of the values in the E24 series sorted so that I could easily look up the closest pair to whatever ratio I needed. That was several computers ago.
     
  8. Jony130

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    Ok I think I understand now.

    So the solution is

    t = RC*ln {\frac{Vo - Vf}{Vc - Vf}

    When

    Vo - initial voltage across capacitor. In our example Vo = 2/3Vcc.

    Vf - finale voltage across capacitor, Vf = Vcc * RB/(RB +RA) = Vcc *22/73

    Vc - Voltage level we want to discharge our capacitor.

    Vc = 1/3Vcc in this case.


    So we have

    t = RC * ln * ( 2/3 - 22/73)/(1/3 - 22/73) = RC * ln (80/7) = RC * 2.436.
    So for RB = 22kΩ; RA = 51kΩ and C = 100μF we have
    t = RB||RA * 100μF * 2.436 = 15.369kΩ*100μF * 2.436 = 3.7442s
     
  9. tracecom

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    So, I guess that means that my calculation for t2 was correct since I arrived at the same answer?
     
  10. Jony130

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    Yes your calculation was correct.
    To get t1 = t2 for RA = 51kΩ and C1 = 100μF you need RB = 21.59015269142032kΩ
     
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