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need help with a speed variator

Discussion in 'General Electronics Discussion' started by ziphlox, Jun 19, 2013.

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  1. ziphlox

    ziphlox

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    May 20, 2013
    hi everyone,
    i'm working on a little project of a speed variator for low power motors.
    i've done a full study and analysis to this circuit, how it works and the role of each component, etc.
    i'm just finding a hard time figuring out why should we use the capacitor C2, so i would be very grateful if you can help me with that :)
    i have another question which is more important. the motor that is used in the output of the circuit is a 12V DC motor. the only information i know is that i should use a 12V DC motor, but i wanna know which power range of motors can be used or controlled by this circuit ? is there any method to know ? how can i determine the maximum motor power that this circuit can handle without any problems ?
    thank you in advance.
     

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  2. duke37

    duke37

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    Jan 9, 2011
    C2 is to smooth the power supply to the 555 so that false triggering is reduced.

    The TIP33 is rated at 10A.

    R5 will determine the base current of the TIP33 which will be the limitation. You could take output current to be at least ten times the base current.
     
  3. ziphlox

    ziphlox

    15
    0
    May 20, 2013
    thank you so much.
    i have another question please. how can i know the maximum power of the motor i can use ?
    & how can i measure the number of tours of the motor ?
     
  4. duke37

    duke37

    5,364
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    Jan 9, 2011
    If you chose 5A as the current with 24V then the power will be 120W.

    I assume you wish to measure the speed of the motor.
    There are many ways of doing this, you can get revolution counters which can be pushed on the end of the motor shaft. Hold it there for a minute say.
    You can put a light/dark patch on the shaft and use a light detector which is fed to a digital counter.
    For rough estimates, you can use a led fed from AC to illuminate the patch. If fed from the mains the motor speed can be adjusted so that the patch is stationary.
    A rotating magnet can be used to switch a reed relay, I have used this up to 1000rpm. You will need a counter.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    With the 220 ohm base drive resistor, the TIP33 should be able to switch 0.5 amps, maybe 1A. That corresponds to 12W to 24W for a 24V motor.

    Modern designs would use an N-channel MOSFET instead of the TIP33. These have lower losses, and are easier to drive, than bipolar transistors. You can use exactly the same circuit; just replace the transistor with the MOSFET: base becomes gate, collector becomes drain, and emitter becomes source. There are literally thousands of suitable MOSFETs. The IRF540 (http://www.digikey.com/product-detail/en/IRF540NPBF/IRF540NPBF-ND/811869) is fairly common and not too expensive, and it's rated for 33 amps! Overkill, sure, but it's a common, cheap part.
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    I agree with Kris.

    It's interesting that R6 has been specified to prevent the voltage rising close to a typical Vgs(max). It's almost as if that circuit were designed for a mosfet...
     
  7. ziphlox

    ziphlox

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    May 20, 2013
    thank you guys so much for the information, you have been so helpful.
    i just would like to ask about the diodes used in that circuit, they are 1N4004 diodes, i read somewhere that they are the most suitable for this circuit because they have what's called a "high surge capacity" and they are "catch diodes". the problem is, i'm not familiar with those concepts and i would be very grateful to you if you can explain to me what that means ?
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    https://en.wikipedia.org/wiki/Flyback_diode

    Yes, the 1N400x diodes do have a high surge capacity. Although they're only rated for 1A continuous forward current, they can conduct much higher currents for a short time, i.e. a surge, without damage, which may be important for D4 in your diagram.

    Edit: I think it will depend on how inductive the motor is. If its inductance is low, its back EMF pulse will be short, and can be considered to be a "surge". If its inductance is high, the back EMF pulse will be longer. Also, there's a capacitor across the motor, which will slow things down slightly. I think!

    (The D1 and D2 positions are not critical at all, and I would normally use a much smaller diode such as the 1N914/1N4148. I assume the designer specified 1N4004s for those positions to simplify the parts list. That's fine.)

    Speed is not important for D4 since the PWM frequency is quite low. This is just as well, because the 1N400x diodes are slooow! They're designed for 50/60/400 Hz mains supply rectification.

    That reminds me. What is the PWM frequency? The capacitor in the oscillator (C1) is 1 uF, which will give a pretty low PWM frequency I think. If the motor makes a low-pitched buzzing noise under load, or when running at low speed, you could reduce C1 to, say, 0.22 uF to increase the PWM frequency.
     
    Last edited: Jun 21, 2013
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The max current through the catch diode will be the motor current.

    The reason for these diodes is that after your circuit switches power off to the motor, current still wants to flow in the inductor (the electromagnet inside the motor is an inductor).

    If you don't provide a path for that current, the voltage will rise until *something* fails and allows the current to flow. In your case it would typically be the transistor.

    The avalanche breakdown of the transistor is a bad thing which will cause it to get quite hot and eventually fail (maybe in milliseconds, seconds, minutes, hours, days... it all depends on the transistor and the degree of overload)

    The diode is there to provide a nice safe path for that current to flow until resistive losses in the motor and the small voltage drop across the now forward biased diode dissipate the energy.

    In this application the diode catches the current, redirecting it safely and prevents a high voltage from being thrown at the poor (almost) defenceless transistor.

    Other options can include a zener diode from collector to base to allow the transistor to turn on slightly (rather than suffer avalanche failure of the CB junction). This will still make the transistor heat up, but at least it is prevented from suffering the indignity of an avalanche failure :) I have seen this technique employed in transistors designed for automotive applications.

    Surge capacity is just the ability of the diodes to take a pulse of current that is far higher than they can withstand continuously. The actual figures will be in the datasheet and they vary depending on the duration of the pulse, and how often it is repeated. You may find very surprising figures for a single very short pulse. However in this application the pulses will occur at the switching frequency of the 555, and the duration is hard to measure. The best bet is to rate the diode for the full motor current where this is practicable.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Excellent advice Steve! I nominate that post for "valuable tutorial" status!
     
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