Need Help Understanding this Inverter Circuit

Discussion in 'Homework Help' started by gronkle, Aug 27, 2012.

  1. gronkle

    Thread Starter New Member

    Jun 2, 2012
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    (I have been through every thread on the matter and still require amplification, in other words, if someone could break out the Big Purple Crayon to walk the dog here, it would help)

    This may look simple to everyone with an EE degree, however...here is a diagram for a simple split coil transformer whose oscillating current is driven (somehow) by two transistors. The inverter portion of the circuit is obviously the portion containing coil P1, coil P2, R (x2), and transistors Q (x2). Easy enough, right? How does this circuit control alternating switching of the transistors that alternates the direction of the current? (To the uneducated, it looks like the transistors would simply fire together, producing a collision)

    (Please see attachment - it will not display here for some reason:mad:)

    [​IMG]

    To best illuminate the question, I fully understand the concept of the mechanical switching depicted in the top image here that will produce a square AC-like wave when switched rapidly, but the bottom image's concept of operation eludes me:

    [​IMG]
     
    Last edited: Aug 27, 2012
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    Do u know tht the two windings supplying base drive are opposite to each other.

    One goes up and other down. So No collision the way I see it..

    And to make things more confusing, will the darn circuit even work ?
     
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  3. crutschow

    Expert

    Mar 14, 2008
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    OK, looking at the middle diagram:

    Assume that the top of the transformer windings all have the same polarity, i.e. if the top of one winding is positive, the top of all other windings are also positive (polarity is critical to proper circuit operation).

    Assume the circuit is in the process of oscillating with the top transistor being on and current is flowing through the collector winding from the power supply.

    The direction of current through the transformer and the Vcc polarity means that all the bottom winding connections of the transformer have a positive polarity and thus the base-emitter voltage of the top transistor is forward biased, keeping the transistor on.

    At the same time the transformer polarity is such as to reverse bias the base-emitter junction of the bottom transistor, keeping it off.

    This continues until the transformer core starts to saturate due to the increasing current flow through the top transistor collector. When this occurs, the base voltage from the transformer starts to drop and the top transistor starts to turn off.

    The turn-off of the top transistor causes an abrupt reversal of the transformer voltage due to the transformer inductance (The top transistor collector voltage becomes more positive then Vcc).

    This voltage reversal rapidly shuts off the top transistor and turns on the bottom transistor, reversing the current through the center tap winding.

    The bottom transistor now stays on until the core saturates in the reverse direction, completing the cycle.

    Make sense?
     
    Last edited: Aug 27, 2012
  4. gronkle

    Thread Starter New Member

    Jun 2, 2012
    18
    2
    Thank you very much Crutschow, finally some enlightenment, I think I get it, please let me describe it back to you to make sure I got it...

    From your description, it appears my primary failure was to:

    1) understand that the direction of current flowing from the first coil (e.g. P1) will act electromagnetically on the second coil (P2) (just like the S1 coil) to cancel the current via inducing a current directional bias until saturation occurs. (This effect acts like a temporary switch in way - cool in itself), and...

    2) (ureka) forgetting that the movement of electrons through the coil in a DC manner will spike this electromagnetic effect as the field builds...but eventually the field saturates and stabilizes, thus eliminating the production of current in the second coil because there is no rate of change in the magnetic field. (for the sake of knowledge creation to any other newbies reading this, look here for a supporting explanation on DC current in coils). The bias created in the second coil then is eliminated and the circuit oscillates back and forth due to this effect.

    Correct?
     
  5. crutschow

    Expert

    Mar 14, 2008
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    If I understand you correctly, you are correct. ;)
     
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  6. gronkle

    Thread Starter New Member

    Jun 2, 2012
    18
    2
    R!f@@

    I have not made this circuit, so I do not know if it functions as intended or not (hmmm... do you know of a good online virtual circuit tester?). The inverter portion of the circuit apparently takes 3V DC and turns it into a 400V square wave coming out of coil S1.

    The diode/capacitor ladder circuit on the other side of the transformer is a voltage multiplier that takes the 400V "AC-like" square wave and turns it into 1000V DC as the now-stored charge ratchets and grows on its way up the ladder. As I understand it (please correct me if I am off target) - the diodes cut the half of the positive portion of the square wave off and then prevents current flow in the opposite direction. The capacitors discharge in concert to amplify the charge up the ladder as the square wave continues, charging the capacitors in sequence.

    The circuit is intended to be used as a hand-held bug zapper. I am assuming that since the capacitors have their positive ends oriented towards one electrode and the other electrode is further down the voltage ladder, the electrodes will arc across the bug that completes the circuit.

    The weird thing to the newbie is that it looks like the DC charge of 1000V will then dump into the 400V AC-like square wave lower down the ladder.

    I get that there is a voltage potential between where the electrodes are placed, but how exactly does this work? (you would appear to be dumping a ~1000V DC spike to one of the lower AC-connected capacitors via the circuit completed by the fly)?

    Since the amplified voltage is again DC now, couldn't it just use a common ground to the battery?

    There is also a weird ground arrow on the inverter circuit that makes no sense to me at all. Since the device is hand held with not cables, what is heck that with respect to normal battery negative terminal?
     
  7. gronkle

    Thread Starter New Member

    Jun 2, 2012
    18
    2
    The strange ground artifact appears to be due to the requirements of the electronics simulator:

    "The concept of a ground in a circuit simulator is similar but not identical to the concept of an electrical ground in the physical world. In real life, ungrounded battery-operated circuits work just fine, because to the circuit, only relative voltages matter. However, inside a circuit simulator (or even when solving a circuit on paper!), we have to pick one node to be our reference in order to calculate voltages at other nodes"
     
  8. crutschow

    Expert

    Mar 14, 2008
    13,021
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    I prefer to use the term "common" for circuit ground since it often has nothing to do with earth ground. As gronkle noted, it's just the 0V reference node for the other circuit voltages.

    The high voltage output could be connected to circuit common, but there's no reason to do so.

    If you are interested in doing analog circuit simulations, the free LTspice program is fairly good and is used by many forum members. It does have a significant learning curve, but once you are past that, you will find that it's a great advantage to simulate your circuits to verify their operation before you build them. Saves a lot of breadboard modifications to correct errors.
     
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