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my explaination

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by bananasiong, May 21, 2006.

  1. bananasiong

    bananasiong New Member

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    555

    Hi,
    If i don't connect the pin 5 (Voltage Control) with a capacitor to GND, what will happen? I've seen a circuit which uses that pin with a variable resistor to control the voltage, then what it the purpose of connecting it with a cap to GND? The 555 timer will still work well without it right?
     
  2. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Look at the datasheet for the 555. Every circuit has a capacitor from pin 5 to ground if pin 5 isn't used for sweeping the frequency.
    Without a capacitor, pin 5 can pickup interference, even from the power supply, that can vary the frequency.
     
  3. bananasiong

    bananasiong New Member

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    From the datasheet, the maximum power dissipation is 600mW, so the output shouldn't be more than that, P=VI, right? How if i short pin5 to GND without a cap? Then the 555 won't work right?
     
  4. dave

    Dave New Member

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  5. audioguru

    audioguru Well-Known Member Most Helpful Member

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    No, that is the rating for a tiny surface-mount package. A regular-size LM555 has a max dissipation of 1180mW.

    No. The power dissipation in a load can be much more than the power dissipation in the 555.
    15V supply, 200mA load. The output of the 555 when driving the load will be about 2.5V or 12.5V so it dissipates 500mW. The load dissipates 2.5W.

    Of course it won't work.
     
  6. bananasiong

    bananasiong New Member

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    U mean, let's say i supply 5v to the 555, the output is 3.3v (from the datasheet). 600mW is dissipated by the 555, and the rest of it by the load. Let's say the load is 330Ohm, 0.5A*5V=2.5W, 600mW dissipated by 555, 1.9W by the load.
    But the resistor used normally is only 1/4 watt, so we should increase the resistance??
     
  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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  8. bananasiong

    bananasiong New Member

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    ??
    I know the current is the division of the output and the resistor, 3.5/330. But how do u get the 555 dissipates 24mW and the resistor dissipates 56mW?
    P=VI=VV/R=IIR
    but i still can't get the 24mW and 56mW..

    the other question (look my attachment if u've forgotten the circuit):
    I'm doing a robot mower, i've tested the navigation system, it works well. When it goes to the multiple loops of wire (cat5), from the output of the 555, it will detect the electromagnetic field and turns to other direction.
    Without running the blades (powered by 2 DC motors), it works well. I didn't connect them to the supply because i thought that it is just blades, nothing more, and it is quite noisy when running, so i test the operation without turn it on.
    But when the time i try to turn on them, maybe because of the electromagnetic field produced by the DC motor, the LM393 circuit keeps response to it. Maybe is interference. The wires placed quite near.
    Can i use the alluminium foil to cover the wires from the tank circuit to avoid the interference (red color from the diagram)? Then one end of the foil connected to GND. Is that called shielding interference?
     

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  9. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The 555 and the 330 ohm load resistor are in series.
    I goofed on calculating the current.
    The 555 dissipates about 15.9mW and the 330 ohm resistor dissipates 37.1mW.
     
  10. audioguru

    audioguru Well-Known Member Most Helpful Member

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    You need to keep the motors away from the detector coil.
    If the motors are permanent magnet ones with brushes, then the brushes make sparks which transmits interference. A 0.1uF ceramic disc capacitor with very short wires across the motor will help reduce the interference.

    The supply also needs a 0.1uF ceramic disc capacitor across it.
     
  11. bananasiong

    bananasiong New Member

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    Yes!! I got it!!

    But their place are fixed..
    the capacitor with short wire connect with the motor in parallel? What about the supply?
     
  12. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Since the detector coil picks up the magnetic field from the motors then your project doesn't work. Can you mount the detector coil out on a stick in front of the robot?

    Yes.

    Connect a 0.1uF ceramic disc capacitor at the supply pins of the LM393.
     
  13. bananasiong

    bananasiong New Member

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    The tank circuit is already sticked in front of it, but still affected, i'm worried that, the wire connected from the tank to the input of the 393 will pick up the field.

    one end to the +ve and the other end to the -ve? means that in parallel?

    *Can i shield the wires or shield the motors with the alluminium foil to avoid the interference as i mentioned before?
     
  14. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Connect the metal covers on the motors to 0V with a wire.
    Use shielded cable like is used to connect stereo components together (cut off the RCA connectors) to connect to the tank.
     
  15. bananasiong

    bananasiong New Member

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    Yes.. Thanks.. The metal body of the motor won't cause any short circuit if connected to GND, will it?
    Then do i still need to include the bypass capacitor?

    What is shielded cable? How to get it? Can i make it myself? What is the RCA connector?

    *I need to use a diode to make to current to flow only one way, it is connected series with 2 DC motors. I was using 1N34A, but after it was turned on and off for once, the diode burned (i think) and cause open circuit. So i replace it with a 1N4001, now it works well.
    My question is, if i replace the 4001 woth a LED, does it use more current?
     
  16. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Probably not.

    Of course it is needed.

    In my country, the stereo tuner, CD player and amplifier are connected together with shielded cables with RCA plugs. I cut off the plugs and use the shielded wire.

    I don't know why current in your circuit doesn't flow one way without a diode. Is it AC?
    Look at how much current the motors use when they work hard. Much more current than the max current ratings for the 1N34A or an LED.
    Don't use an LED in series, connect a current-limiting resistor in series with the LED and connect the combination to the power source of the motors to indicate when they run.
     
  17. bananasiong

    bananasiong New Member

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    I've shielded the wires and the motor. I've tested using batteries which are fully charge and not fully charged (not full). When i use the lower power, my robot mower operates well, no interference. When i use the fully charged batteries, of course, the motor run faster and interference occurs.
    The motor should run as fast as possible for good mowing system, but the electromagnetic field produced by the motor is bigger.
    Can i shield the wires and the motor thicker (turn more round) to solve this? I use alluminium foil (for cooking) to shield them.
     
  18. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Shielded cable or aluminum foil will stop the sparking of the motors from radiating electrical interference. But will do nothing for the magnetic interference.
     
  19. bananasiong

    bananasiong New Member

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    !!?!???!?!?!??
    What should i do besides moving them far away to each other?
    Is there anything to remove or direct the unwanted electromagnetic field to somewhere else?
     
  20. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I guess using a coil for a sensor is not the correct thing to do in a machine that is powered by electric motors.
    Hee, hee. Diesel engines? Hee, hee.
     
  21. bananasiong

    bananasiong New Member

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    Oh no... you're kidding right? I mean the diesel engines.. haha.. i use only AA's.. I'll try to solve it..
    Anyway, you've helped me a lot, thank you very much.
     

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