Dual power supply HF coil driver

Discussion in 'General Electronics Chat' started by DC_Kid, May 6, 2016.

  1. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    hi all,

    a basic push-pull circuit.
    any reason why this would not work? the inductor is around 1.0uH, plan to run oscillator in the 10k-200kHz area.

    the power supply T1 can handle ~200A w/o blinking an eye, but to drive fet gates I need about 10v to get the fet to saturate.

    i know some components are missing, but the center-tap has me a tad confused. i suspect T2 does not need center-tap. so how does the output on 555 look like +12v and -12v to the fet gates (what's the reference point)? i suspect my 555 pwer supply is just a +24v supply as there is nothing referencing that center-tap.


    [​IMG]
     
    Last edited: May 6, 2016
  2. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I think this is what you want instead of the bridge rectifiers
    upload_2016-5-6_19-31-57.png
    That would give you a 1 volt and a 12 volt supply. You will need to regulate the 12 volt supply because the 555 is only good for 15 volts.
     
  3. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    doesnt that give me +V only, one way current? i need split rail so that the load hanging on the two fet's can be AC
     
  4. AnalogKid

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    Aug 1, 2013
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    A few things...
    A 555 cannot run on 24 V.

    A 100 A power MOSFET has a very large gate capacitance. Driving the FET fast enough that it doesn't spend much time "unsaturated" can take 5 or 10 *amps* of dynamic gate current. Linear Tech et. al. make special gate drive chips for this.

    At Digi-Key, a 200 A MOSFET has a gate capacitance of 12 nF. If you start with a 200 kHz frequency, that is a period of 5 us and a half-cycle period of 2.5 us. If the transistor ramps up or down for 10% of that, and stays saturated or off for the other 90%, the risetime is 250 ns. Using the linear ramp approximation:
    EC=It
    I = (E x C) / t = (9 V x 12 nF) / 250 ns = 0.44 A

    Note that a 10% risetime might be way too slow to prevent transistor overheating.

    Do you intend to filter either of the rectified AC outputs?

    ak
     
  5. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    i add some clarity.
    my power supply can run 200A no problem.
    in this circuit the impedance of coil will limit the current through fet to <=15amps (6.28fL)

    lets disregard the amps part for now. if the T1 on top side is +-1v how do i get a 555 to have proper rail-to-rail output so i can drive fet gates?
     
  6. ronv

    AAC Fanatic!

    Nov 12, 2008
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    So what you want is a + and - 1 volt DC supply that you PWM with the 555 at some higher frequency?
     
  7. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    running the 555 at 50% duty cycle in a push-pull split rail like i have shown yields equal time AC on the load. i would be using 555 to adjust frequency only.
     
  8. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    so how about this single rail schematic using +12v
    Pin1 - square wave PWM 50% duty cycle, +12v(hi) 0v(lo)
    Q2/Q3/Q6 nFet, Q1/Q4/Q5 pFET
    hi side PWM - Q2/Q4/Q6 conduct, Q1/Q3/Q5 block
    lo side PWM - Q2/Q4/Q6 block, Q1/Q3/Q5 conduct
    L1 is small henry but hi freq yields hi impedance
    R2/R3 are small ~50ohm
    D1/D2 zeners to clamp flyback
    assumption is that the Q6/Q4 & Q3/Q5 pairs (Q6 drives Q4, Q5 drives Q3) have same rise/fall times as Q1 & Q2

    does this work?


    [​IMG]
     
    Last edited: May 24, 2016
  9. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    It's very difficult to say. Please redraw the schematic with all +12V's across the top pointing up, and all GND's across the bottom, pointing down. I'm not being overly-picky here. Making the schematic show the overall energy flow will go a long way to "seeing" how the circuit functions. You are trying to show what is called and H-bridge. Hint: place the inductor horizontally in the center; it is the crossbar of the H.

    ak
     
    Sinus23 likes this.
  10. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    you just want me to rotate the coil 90 degree and then drag connectors so that the +12v is on one side with all the GND's on the other? i can do that, but the positions of the fets will remain the same.

    as is, with a "1" the current is 12v-->Q4-->L1-->Q2-->GND
    with a "0" the current is 12V-->Q1-->L1-->Q3-->GND
    Q5 and Q6 are just fets to flip the logic around to make it work (pFet and nFet have their challenges, etc)
     
  11. AnalogKid

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    OR, move the FETs so they are correctly positioned with respect to the other components and symbols, to present a clear image of the energy flow through the circuit.
    http://electronics.stackexchange.com/questions/128127/mosfet-h-bridge-design
    http://forum.arduino.cc/index.php/topic,16654.0.html
    https://upload.wikimedia.org/wikipedia/commons/5/5e/Bipolar_Stepper_Motor_H-bridge.png

    ak
     
  12. DC_Kid

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    Feb 25, 2008
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    is this one easier to view
    [​IMG]
     
  13. AnalogKid

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    I'm curious, and this is not a facetious question - does it really make sense to you to draw the schematic that way?

    ak
     
  14. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    schematic makes perfect sense to me, and so did the 1st one. is this 2nd one hard for anyone to understand the flows?
     
  15. BR-549

    Well-Known Member

    Sep 22, 2013
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    A print needs structure like a sentence. The power or signal flows from left to right.

    High voltage potentials are at the top and low potentials are at the bottom.

    This gives everyone the same reference.
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    ... for the last 108 years.
     
  17. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    well, its too simple to keep flipping it around, but thanks for the suggestions. i'll run some tests on my bboard, then move it to some hefty fets.

    in this specific case, it is AC across the inductor, so saying that the power flows left to right would not be accurate, because it would in fact also flow right to left.
     
  18. AnalogKid

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    It's AC, so electrons flow in both directions.

    ak
     
  19. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    i agree. but not sure why i am replying. can you elaborate on your comment?
     
  20. AnalogKid

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    Semantically, I'm not comfortable with power flowing through a resistor. Electrons (not current) flow; power is dissipated. But the main reason I responded is that I misread your post and missed the word "also". oops.

    ak
     
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