Square wave (+ and -) coil driving?

Discussion in 'General Electronics Chat' started by electronice123, Feb 15, 2012.

  1. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Hi everyone.

    I am trying to drive a small xmfr with a square wave and get a square wave out. I need the square wave to change polarity like an AC wave does meaning that I need it to have equal positive and negative polarity. Basically I'm trying to make a push pull converter.

    I made a simulation in multisim using a 555, a counter, a rs flip flop, and a 4 transistor setup across the primary (not CT) coil similiar to a push pull driver. The simulation will not work though. Do I have to have a CT primary or is there something wrong with my circuit design?
     
    Last edited: Feb 15, 2012
  2. praondevou

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    please post the schematic as a picture. (and the multisim file as well if you want.)
     
  3. electronice123

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    Oct 10, 2008
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  4. praondevou

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    This is a Fullbridge. Commonly the term push-pull is used for a split-winding transformer and 2 switches.

    The pnp transistors don't have base resistors.

    Your power supply is 12V but your digital logic are 10V types. I don't know a workaround to solve this but the power supply voltage will need to be equal to the digital logic IC voltage (5V,10V or 15V types).

    What does the simulator say?

    Can you post the ms file?
     
  5. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    The simulator showed an errror at first, I allowed the multisim to fix the problem but I am getting a distorted square wave.

    I missed some pretty basic stuff....Thanks for pointing it out to me.

    I will fix those problems and go from there.

    What I'm really wanting to know is will this circuit work in the real world?
     
  6. praondevou

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    The 4049 can provide enough current for the bipolar transistors, the specified current for the 4043 is much lower though. You may either put a buffer in between (2x 4049 or a 4050) or use MOSFETs.

    Pin5 of the 55 is commonly connected to Gnd via 10nF. Unused inputs of logic ICs have to be tied to either VCC or Gnd.

    The output switching will be :

    1 time period ON
    1 time period OFF
    1 time period ON inversed
    1 time period OFF

    Is this what you wanted? It will not be a square wave because the time it stays at zero is the same time as it's positive or negative pulse width.
     
  7. praondevou

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    If you really want a square wave, take the 555 or any other oscillator, pass it through a FF to get 50% duty cycle, then pass the signal through an inverter.

    You now have two inverted drive signals. Add deadtime to both signals and drive the fullbridge with the signals as you do now.
     
  8. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Yeah that is what I am wanting as far as switching goes...

    What do you mean by FF?

    I thought the counter will make the output a 50% square wave?

    BTW, thanks for the help and suggestions.
     
  9. praondevou

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    FF = flipflop

    I meant something like this. Don't bother the component values, it's just an example.

    [​IMG]

    Of course it would not be clever to use one gate of the 40106 and one of the 4049. However for this oscillator you need a schmitt trigger gate. A 555 also does the job. The FF divides by two, duty cycle 50%.

    If using bipolar transistors you need something that provides the necessary base current. The 4049 can handle higher currents. If using MOSFETs normal gates can be used.

    This circuit has no deadtime yet. If needed I can draw it tomorrow.
     
  10. Wendy

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    Mar 24, 2008
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    All of the logic shown should handle 3V-15VDC for the power supply, except the 555, which is 4.5VDC minimum. CMOS 555's are also 3VDC - 18VDC for the power supply

    Base resistors are indeed needed.

    A CMOS 555 in that configuration will produce a analog 50% square wave. A FF will make it a digital, which is more precision, but possibly overkill.

    To create a inverted drive signal one inverter is needed. Two inverters would create the original drive signal. Why the extra components?
     
  11. electronice123

    Thread Starter Senior Member

    Oct 10, 2008
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    Praondevou,

    How can I add dead time to the circuit?

    Also, I want the frequency to be adjustabel from about 500Hz to 2kHz....How can I do this with the ckt you provided earlier....Just add a pot?
     
  12. praondevou

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