555 PWM: Load Shedding PLC input. Frequency difficulties.

Discussion in 'The Projects Forum' started by 1rspn, Feb 12, 2013.

  1. 1rspn

    Thread Starter New Member

    Feb 12, 2013
    3
    0
    Hello there,
    I am building a 555 oscillator circuit for a college project and come across a small difficulty. It is a input circuit for a PLC which will implement Load Shedding.
    The 555 circuit is to simulate the pulse output from a Digital Supply Meter, where each pulse were to equal 1kWh say (this value is not important at this stage). So I have an astable 555 circuit arrangement, with a potentiometer to vary the frequency output. This output is connected to a counter within a PLC. The PLC will give an output, once the counter reaches a set value, that will cause the frequency of the 555 circuit to reduce by a set amount. (eg. 5Hz)
    I have tried configurations of increasing the values of the R1, R2, or C1 to create a decrease in the frequency by so much. But this reduction of frequency ends up proportional to the Potentiometer.
     
  2. 1rspn

    Thread Starter New Member

    Feb 12, 2013
    3
    0
    For example:
    if R1 was the potentiometer of 10KΩ,
    R2 = 10KΩ
    C1 = 100μF
    T = Time period

    T = ln(2)x0.0001x(10000+2x10000) = 2.08s

    Over 30mins: # of pulses would be 865

    When R1 is increased to 100kΩ,

    T = ln(2)x0.0001x(100000+2x10000) = 8.32s

    Over 30mins: # of pulses would be 216

    Now if I was to switch in an added resistor in series with R2, I would increase both the Time High and the Time Low, hence decreasing the frequency. (this is the effect the output of the PLC would have)

    New Outputs:
    R1 = 10kΩ
    R2 = 11kΩ
    C1 = 100μF

    T = ln(2)x0.0001x(10000+2x11000) = 2.22s

    Over 30mins: # of pulses would be 811

    When R1 is increased to 100kΩ,

    T = ln(2)x0.0001x(100000+2x11000) = 8.46s

    Over 30mins: # of pulses would be 213

    This means that when R1 = 10kΩ, the change of frequency, due to the increase of R2, is 865-811 = 54 Pulses. or 0.03 Hz

    When R1 = 100kΩ, the change of frequency, due to the increase of R2, is 216-213 = 3 pulses. or 0.0017Hz

    Could anyone suggest ideas or methods to change the circuit to ensure the same # of pulses drop, no matter how I simmulate the frequency of the Digital Supply Meter?

    Btw I find that Voltage Control Oscillator method, or trying to use the control pin, sill doesn't rectify my problem.

    I hope this made sense.
     
  3. 1rspn

    Thread Starter New Member

    Feb 12, 2013
    3
    0
  4. tracecom

    AAC Fanatic!

    Apr 16, 2010
    3,869
    1,393
    Have a look at this post.

    http://forum.allaboutcircuits.com/showpost.php?p=427432&postcount=2

    SgtWookie is one of our most experienced and trusted members.

    But to your question, I frankly don't understand what you are after. We have a good information source about 555 circuits here. http://www.allaboutcircuits.com/vol_6/chpt_8/1.html Maybe there is something there that will help.
     
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