help with AUTOMATIC WATER-LEVEL CONTROLLER project

Discussion in 'General Electronics Chat' started by spidermanIIII, Jul 19, 2014.

  1. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    i searched for AUTOMATIC WATER-LEVEL CONTROLLER project and i found this project but there are things i can't understand it first what is the usage of the capacitors 11 and 12 and how the ic 556 get triggered when the when the float interrupt IR sensor and how the output get high always
     
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  2. wayneh

    Expert

    Sep 9, 2010
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    I believe C11 and 12 act low pass filters to help quiet the output of the sensors. Only a valid event will cause triggering.

    The rest of your question seems to be about how the 555 works. There are more sources of info on this than anyone could ever read.
     
  3. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    the output of the sensor is DC not AC so how could it low passfilter
     
  4. wayneh

    Expert

    Sep 9, 2010
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    Noise. There's always noise. And if the sensor is some distance away, the noise increases.
     
  5. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    i don't think so
     
  6. shakilabanu

    Member

    Jul 8, 2014
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    Well you should...because that is what they are doing there.... as wayneh says..they are in place to ensure that some noise doesnot trigger your 555.
     
  7. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    lets say that they are low pass filter can you explain how ic 556 get triggered however the trigger pin connected to 7806 ic and the output of the sensor which is TSOP 1738 which output is pulses at 38k Hz freq if it the output of sensor will be one pulse due to capacitor
     
  8. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    As i see it the 38kHz is keeping the 556 re-triggered so both outputs stay high, when the float blocks the beam the monostable is allowed to time out,then goes low, flipping the cmos bi-stable, when the tank gets full the other output flips the bi-stable and the pump stops.


    If your going to build one i would use a simpler circuit using float switches.
     
  9. wayneh

    Expert

    Sep 9, 2010
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    The connection to Vcc is thru a 1MΩ resistor, which will be overwhelmed by the signal from the low impedance sensor. I think it's there to ensure start-up conditions, but in continuous operation I believe you could ignore that connection.

    The output of the receiver is not modulated at 38kHz. Read the data sheet again.

    [I just read it myself. The sensor is active low, so the 1M resistor is needed to pull the voltage high when the sensor doesn't see anything.]
     
  10. wayneh

    Expert

    Sep 9, 2010
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    I don't think so, since this is one of those sensors that goes back to "off" if it sees a continuous 38kHz signal. It won't stay on unless it sees data riding along on the carrier.
     
  11. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    can you explain more
     
  12. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    Yep, I was wrong just read datasheet sensor is high unless it receives a modulated signal,it has an internal 10k pull up.
     
  13. MrCarlos

    Active Member

    Jan 2, 2010
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    All started with:

    I can't understand it first what is the usage of the capacitors 11 and 12.
    And how the IC 556 get triggered when the when the float interrupt IR sensor.
    And how the output get high always.

    If you look at Fig. 2, contained in the document you enclose, You will see that the C terminal via the NE555, a signal is generated. Approx. 32KHz.
    This frequency can be adjusted in a certain range, via potentiometer VR1.
    This signal makes on and off, at this rate, the type IR LED1 and LED2 in the Capillary Tube. (Fig. 3).

    When the float does not cover the light from the LED1, that ray of light hit the RX1 IR sensor (TSOP1738).
    The IR sensor has 3 terminals:
    1 - GND.
    2 - Vcc.
    3 - Output.
    On its PIN 3 comes out a signal that reaches the NE556 PIN 6.
    This signal is the same frequency as that generated by the NE555. (Fig. 2)
    Because it comes from there, meaning through the IR LED1.
    Well, it goes out of the terminal F and arrives at the same terminal F to NE556 PIN 6. (Fig. 1)

    We have arrived where R8 and C11 are located.
    Remember that, at the junction of R8 and C11, a signal, Approx 32KHz. arrives.
    What does the circuit formed by R8 and C11 ??
    Well it is nothing more and nothing less than a filter for the frequencies. (32 KHz.).
    Then the frequency of 32 KHz. It is “Eaten” by the filter.

    Remember The RC constant ?.
    T = R*C.
    The RC constant with values​​:
    R8 = 1000000 Ohms.
    C11 = 0.000010 Farads.
    It is 10 Seconds.
    It means that in 10 seconds, the capacitor is charged to 63.3% of the applied voltage.
    But:
    The period of the signal of 32 KHz is 1/F.
    1/32000 = 0,00003125000 seconds, = 31,250 microseconds.
    So we're downloading and uploading to C11 at a rate of 31,250 microseconds.
    Therefore, the filter will “eat” that 32KHz signal.
    As if it came no sign, low level in the NE556 PIN 6.

    But what if the IR light is interrupted between LED1 and RX1 ?. (Fig. 3)
    Then RX1 maintain a High level at its PIN 3 (F). And the capacitor C11 will begin charging until it reaches the threshold level to trigger the NE556 which will change its output PIN 5,
    And,
    The RS type flip-flop (Formed by N1 and N2) will change its output state,
    Polarizing the transistor T1 to stop driving current in its joint collector-emitter so the relay RL1 will open its contacts stopping the pump.

    When IR light is interrupted between LED1 and RX1, (Fig. 3) It means that the tank roof is full, so the pump must be turned off.

    R9 and C12 do the same job, but for the other timer that the NE556 has. (Fig:1)
    But this other timer of the NE556 is used to turn on the pump.
    Because the float interrupts the IR light between LED2 and RX2. (Fig. 3)

    As someone said, R8, C11 and R9, C12, also ensure startup circuit conditions: At power-up circuit, pump off.

    R1, D4, C7 and C10, do the same thing as: R2, D3, C8 and C9. (Fig. 1).
    Are filters for power supplies, which polarize the IR sensors RX1 and RX2 (Fig. 3).

    Note that there are some things to improve on the circuit contained in Fig. 1.

    ! Ugh ! Still need to explain the system to the tank underground.


     
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  14. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    I should thank you for this fantastic explanation . i am new in studding electronics and i have some questions about the the filter which formed by C11 and R8 , does the voltage of 6 volt will affect the the pulses from IR sensor F .we learned in school that the low pass filter has the diagram of this attached image and it cames in different diagram in these pdf . i want to know how you calculate that capacitor will charge 63.3% of applied voltage please i want to know
     
  15. wayneh

    Expert

    Sep 9, 2010
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    No. Again, read the data sheet. The output does NOT contain the carrier frequency. It gets demodulated.
     
  16. wayneh

    Expert

    Sep 9, 2010
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    So I guess the two 1M resistors are there to protect against loss of connection to the sensors.
     
  17. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hi Everyone.

    SpidermanIIII Comments.
    1)- I am new in studding electronics and I have some questions about the the filter which formed by C11 and R8, does the voltage of 6 volt will affect the the pulses from IR sensor F.

    2)- We learned in school that the low pass filter has the diagram of this attached image and it cames in different diagram in these PDF.

    3)- I want to know how you calculate that capacitor will charge 63.3% of applied voltage please I want to know.


    My answers:
    1)- It could be if the voltage regulator output will vary. Noisy also affect.
    This voltage is regulated to 6 volts by the 7806 IC. (Fig.1).

    2)- If what I have written I mention something like: low pass filter, high pass filter, band pass filter.
    I apologize, I should not have.

    Yes, The circuit shown in the image that you enclose could be a low pass filter.
    But also it is a filter which I showed in my post #13. (Voltage Supply For TSOP1738.PNG).
    Only this is to eliminate the possible ripple on 6 Volts power supply.

    3)- Well, this is not something I invented or discovered.
    It would be much faster if you're looking for on Google.com by RC time constant.
    I found there this video which I think is good to start:
    http://www.youtube.com/watch?v=Ppqijqfuk7M

    Look: T = RC, T is also called tau.
    If the value of R in the formula you write in Ohms.
    If the value of C in the formula you write in Farads.
    The result will be T in seconds.
    The theory states that:
    The capacitor is charged at one time TAU to 63.2% of the applied voltage.
    There are no secrets in this.

    Wayneh Comments.
    No. Again, read the data sheet. The output does NOT contain the carrier frequency. It gets demodulated.

    My answer:
    Let me wait a while to see if someone else says something about yours.
     
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  18. spidermanIIII

    Thread Starter Member

    Nov 22, 2013
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    my question here what are you mean with "Eaten" is it high or zero. and shall we called capacitor 11 is low pass filter
     
  19. Digvesh solanki

    New Member

    Mar 23, 2010
    11
    0
    hi,
    Capacitor 11 & 12 are used for making the IC 556 to operate in monostable mode. actually when across these capacitors when negative going pulse come with amplitude less then 1/3*VCC come then this pulse give the trigger to IC's internal comparator so that output can go high.
    when water level goes high for example upto level D then water interrupts the transmitter & receiver sensors (IR LED1 & IR RX1) & pin no 3 i.e pin F goes lows below 1/3*VCC then it gives negative going pulse which triggers IC it will actuate the relay mechanism.
     
  20. MrCarlos

    Active Member

    Jan 2, 2010
    400
    134
    Hello spidermanIIII

    "Eaten" in that sentence.
    It is an idiom used in some countries.
    figuratively is like saying that disappears. because it eats the thing disappears. anything.

    We're going to do an experiment.

    We know from their 556 data sheets:
    That setting it as a mono-stable, the IC changes the state of its outputs on the negative transition of a pulse applied to any of its TR(2 or 8) inputs. -TRigger.
    Of course if we apply a pulse to the TR(6); Its output PIN 5 will change state.
    And if we apply a pulse to the TR(8); Its output PIN 9 will change state.

    Any of its outputs: PIN 5 or 9:
    The time remaining in that state, is determined by: RC time constant.
    This time constant is formed by a capacitor and a resistor connected to its input TH(2) -Threshold.
    The resistor goes to the Vcc, the capacitor goes to ground.

    In our particular case we have:
    R3 = 33000 Ohms.
    C3 = 0.0000001 Farads.
    So the T (tau) is 33 milliseconds .. Remember I changed the value of C3.

    Let's see if this is true.
    Familiarize yourself with the picture Testing NE556.PNG attached.
    Or the image that appears in Document Testing NE556.PDF attached.

    Now look at the picture Testing NE556 (2).PNG attached.
    As we can see, the oscilloscope is set as follows:
    Horizontally, 5 milliseconds per division.
    Vertically, all channels 1 volt per division.

    Look at the upper white arrows in the image:
    Each negative transition (yellow line)
    The Q on the 556 changes state. (Red line).
    A transition is jump because it is already in its second state, High.
    When tau time (blue line) Ends, the Q 556 returns to its stable state, Low.

    Taking the blue line, could you please measure the time and amplitude as the green arrows are pointing ?
    With the information you get. calculates the percentage of voltage to the blue line reaches the time the Q 556 returns to its stable state.
    Remember that the applied voltage is 6 Volts.
    Considering the first white arrow near the blue line.
    How much time does take the blue line to reach the voltage you have measured ?

    Surely the results you obtained are similar to those we have been discussing.
    The inaccuracies are the result of ISIS Proteus Simulator.
    Errors in out readings.
    Or the error rate of the value of the components used in the circuit.

    I hope I have made ​​mistakes in the above.
     
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