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Homework Help: 555 timer IC - Calculating resistance needed

  1. Apr 29, 2015 #1
    Voltage = 9.0 V, Current = 20 mA, Capacitor (C) = 10 microF, Resistance before (too high) [R_A = R_B] = 10 k Ohms.

    In a a-stable oscillator, calculating the resistance needed for two resistors (R_A = R_B) to discharge and charge a capacitor (C) so that a LED does not blow. Currently my circuit has much too much resistance for testing and it works with the LED very dimly.

    Here is the worksheet: http://i.imgur.com/nx6bnkw.jpg

    a-stable oscillator: http://i.imgur.com/xWSC25R.jpg

    Relevant equations
    I know questions I have to do shortly require a few calculations with the resistance and capacitor, but I need other values first.
    t_charge = (R_A + R_B)C sec

    The attempt at a solution
    I assume V_cc is the input voltage, so that is 9V. I have measured that the amplitude from V_low to V_high is 9V. I thought V_low should be zero, but the diagram has it separate and I don't know whether to follow that or not.

    I know that V_D to V_U has amplitude 3V, but I have no idea what the actual values are.

    This is probably very wrong, but I could assume that the distance between zero and V_D and V_CC is 9-3 = 6 V and that each then must be 3 V.

    Time to charge is 150 ms, time to discharge is 65 ms.

    Summary of measurements: http://puu.sh/hvONo/851b30c03f.png [Broken]

    I'll keep editing this post as I work on it, but I've been stuck for days without making much progress.

    Other questions:
    Reading the questions ahead, I don't feel very confident with them either, so some tips or telling me what equations/variables to use would be very helpful.

    Scary questions: http://i.imgur.com/Mf6iBij.jpg
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 30, 2015 #2


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    Staff: Mentor

    From memory, doesn't the datasheet for the 555 indicate that the internal comparator voltages VD and VU are derived from a string of 3 equal-valued resistors between VCC and ground? So for VCC of 9v, these switching levels you see on the capacitor curve will be 3V and 6V, resp.

    Are you determining the charging/discharging durations using the design equations provided in this chip's datasheet, or are you attempting to derive these for yourself from the exponential relationship?

    Do you have the 555 datasheet (probably about 15 pages) as a pdf that you can constantly refer to?

    This implies that you are attempting this lab investigation without looking at the datasheet .... kPdWZ.png

    If that is the case, I don't think we can help you.

    P.S. your thread title was uninformative, I edited it
    Last edited: Apr 30, 2015
  4. Apr 30, 2015 #3


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    Science Advisor
    Homework Helper
    Gold Member

    There is a 470R in series with the LED. If the LED was ON all the time the current in the LED would be about... (9-2)/470 = 15mA (Assuming the voltage drop across the diode is about 2V). Many LED will work just fine at 20mA so the circuit might be fail safe (eg LED won't blow even if the oscillator stops in the ON state). What does the data sheet for your LED say?

    Your "summary of measurements" sheet suggests the ON:OFF ratio is 75:150 so the LED is ON about 75/225 or 33% of the time. That drops the average LED current to about 0.33 * 15mA = 5mA. May explain why it's dim?

    At this point I have to ask... What are you trying to achieve?

    You could reduce the 470R to make the LED brighter but this might stop it being fail safe.

    You could increase the ON/OFF ratio to make the LED brighter. The data sheet for the 555 explains how to calculate the ON/OFF ratio. See 7.4.2. pages 10 and 11.

  5. May 4, 2015 #4

    Thanks, sorry, I'm not used to this and keep forgetting to check data sheets and other information, thanks.
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