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Discussion in 'General Electronics Discussion' started by shanan, Nov 24, 2014.

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  1. shanan

    shanan

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    Nov 21, 2014
    Hi

    my original idea seemed quite hard with my vague information

    can this circuit be changed, so input is 12v and the speed in which the circuit flashes can be increased and variable, it doesn't need to be exact per side a variable per side is fine, there won't be an LED it will be an induction coil

    I realise this is still quite vague info but I can work with what ever I just need a starting point

    thanks

    The-flip-flop-in-action-complete.gif
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    12,447
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    Nov 17, 2011
    No problem at all.
    The circuit will work at 12V as it is (note that the electrolytics need to be rated for at least 12V, e.g. 16V or 25V types). However, even at 9V and more so at 12V the frequency will not match what you calculate using the ubiquitous equations for this type of multivibrator. This is due to the reverse breakdown of teh base-emitter diode of the transistors when off (the base-emitter breakdown voltage is ~6V for these). Place an additional diode in series with the base connection to each transistor to avoid this type of breakdown.

    The frequencya can be adjusted using potentiometers instead of the 10k resistors.

    Note that when you use an inductor as load (instead of the LED) you should place a flyback diode across the inductor to protect the transistor from high voltage spikes.
     
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  3. shanan

    shanan

    16
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    Nov 21, 2014
    cool thanks Harold sorry I'm a real amateur

    when you say place a flyback diode "across" the inductor from your example its literally flying back in parallel how do I work out what diode to use ?

    thanks
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You can use almost any diode. 1N914 or 1N4148 are the workhorses of the trade.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    The astable multivibrator is not a particularly appropriate circuit to use to drive an inductor. If you describe your whole project what you want to achieve, we can give much more specific advice.
     
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  6. shanan

    shanan

    16
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    Nov 21, 2014
    I want to have 2 coils wound around the same torus that are reversed in opposite direction, and to variably change the speed.

    alternatively or probably better option I can have 1 coil with 12dc to 12ac inverter however it needs to be a square wave switching between 12+ to 12-. and needs to be variable frequency.

    alls that im trying to achieve is to quickly reverse the induction pole

    this is a complete experiment working with eddy currents so nothing needs to be exact or perfect just what ever.
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    OK, you have a toroid and you want to generate a magnetic field using one or two coils of wire so you can measure eddy currents. You want the injected magnetic field to alternate polarity at an adjustable frequency. You want the drive signal to be a bipolar square wave.

    Your best option would be a circuit called an H-bridge. This can be powered from a single supply voltage and can generate a symmetrical, bipolar drive signal to the coil. You can drive the H-bridge from a square wave oscillator with variable frequency, and you have lots of options there.

    I can recommend some components if you answer these questions.

    What is the frequency range?
    How much current do you want flowing in your coil?
    What is the DC resistance of the coil?
     
  8. shanan

    shanan

    16
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    Nov 21, 2014
    I haven't built the coil as I was wanting to know the limitations of the circuit so don't know the resistance I can build that based on what you can come up with.
    also same with the frequency and current, pretty much as fast as possible (but variable) and as much current as possible.
    but it will be powered from a 12v battery.
    sorry thats vague I just wanted to start somewhere.
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Well, I'll give you some vague suggestions then. Check out the following H-bridge ICs:
    http://www.digikey.com/product-detail/en/TA7291P(O)/TA7291PO-ND/1730026 (1A rated)
    http://www.digikey.com/product-detail/en/BA6219B/BA6219B-ND/658358
    http://www.digikey.com/product-detail/en/TLE5206-2S/TLE5206-2S-ND/1283081 (5A rated)

    These are only three of a number of possibilities. I specifically chose THT parts because I don't know whether you can handle SMT. Here's a filter for the whole set of alternatives including SMT parts: http://www.digikey.com/product-sear...803c8,3fc0059&stock=1&quantity=1&pageSize=250

    These ICs will drive the coil. You will need to limit the current according to the specification of the device you choose, by either adding a series resistor, or making sure that your coil's DC resistance is high enough to limit the current.

    You need to provide control signals into the H-bridge IC. There are lots of options, including a 555 and a CD4046. If you need a duty cycle of exactly 50% (i.e. exactly equal positive and negative durations) you can use a flip-flop (e.g. half of a CD4027) to divide the oscillator frequency by two and provide true and complement outputs.
     
  10. shanan

    shanan

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    Nov 21, 2014
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Actually the BA6219 isn't very easy to use; I shouldn't have suggested it. Here's a circuit that uses the Infineon TLE5206-2, which is rated for 5A output current.

    271480.001.GIF

    U1 is a standard astable (oscillator) based on a 555. The frequency is determined by RT, which I've specified as a potentiometer so you can adjust it, and CT, and is roughly equal to 0.72 / (RT × CT).

    The oscillator's output feeds the clock input of the first half of U2, a 74HC73 dual JK flip-flop, which divides the frequency by two to ensure exactly equal high and low times, and provides two complementary outputs which drive U4.

    U3 is a 7805 linear regulator that provides a clean and steady 5V supply rail to U1 and U2. The supply rail to the right of U3 is noisy because of the heavy currents flowing in the coil; U3 prevents this noise from interfering with U1 and U2.

    U4 is the H-bridge IC. It is driven with two complementary signals from U2 and it outputs drive coil L1 through current limiting resistor RSER. RSER plus the DC resistance of L1 must be high enough to prevent overcurrent in the output circuits of U4. Example values for RSER:

    4.7Ω/50W (http://www.digikey.com/product-detail/en/6-1625984-4/A102171-ND/2056014) for a current of about 2.5A

    6Ω/50W (http://www.digikey.com/product-detail/en/KAL50FB6R00/KAL50FB6R00-ND/1646199) for a current of about 2A

    The LED is provided to indicate various types of errors that can be detected by U4. See the data sheet for more information.

    All capacitors named CDn are decoupling capacitors and should be connected as close as possible to their relevant IC as directly as possible between the power and 0V pins.
     
  12. shanan

    shanan

    16
    0
    Nov 21, 2014
    hey Kris this looks bloody awesome thank heaps

    CT what value cap do you suggest I use there? or is that dependant on inductor.

    I think I owe you a box or maybe a crate it is crate day this week end.
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
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    Nov 28, 2011
    No worries.
    It depends on the frequency you want. Choose a frequency and calculate the cap using
    f = 0.72 / (RT CT) or
    CT = 0.72 / (f RT)
    LOL :) Shall I PM you my address? LOL
     
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