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Flyback converter reflected voltage

  1. Sep 16, 2011 #1
    I am trying to understand when exactly the voltage is reflected from secondary back to the primary.

    1. The primary Switch turns ON, the primary inductor current ramps up
    2. The primary switch turns OFF, current is induced in the secondary inductor-->diode conducts -->capacitor charges

    From flyback waveforms, when the switch goes off, the voltage at the primary is
    Vin+Reflected voltage (from secondary to primary). Ignore leakage inductance effect.

    The moment the primary switch goes off, secondary diode starts conducting and simultaneously secondary voltage is reflected on to the primary coil?

    Since the secondary side is not conducting when the primary switch is ON, there's no mag field in the coil. So no energy stored. But when the switch goes OFF, where does the secondary get it's energy from?
  2. jcsd
  3. Sep 16, 2011 #2
    For an ideal transformer with total coupling between the primary and secondary, the voltage on one winding is always exactly some proportion of the voltage on the other winding, no matter what produced that voltage or where it came from.

    The coupling in a flyback is fairly good because of that iron core, so for a basic understanding you can assume primary and secondary waveforms will always look the same (but with different magnitude).

    So first think about what happens in the primary winding (i.e. its just an inductor). The plumbing analogy of an inductor is simply a long pipe, where the inertia of the water running through that pipe is notable. If you put a pressure difference across the ends of the pipe, that means there is a force on the water, and the water will accelerate (its flow will gradually increase over time because there is a constant force on it). If you then suddenly close a valve to stop the flow, that heavy moving water bangs into the closed valve with a sudden high burst of pressure. Its the inertia of the water that caused that.

    Likewise if you put a constant voltage across an inductor, then the current (flow of electrons) steadily increases. If you then, after a time, switch off the constant voltage across that inductor, then there is a sudden burst of voltage (electron pressure) across the inductor.

    When the switch turns off and the voltage suddenly increases, it forward-biases the diode. If the load after the diode is a high resistance then the secondary voltage will be allowed to increase to some high value even when the diode is forward biased, and this also puts a high-voltage pulse on that resistive load. However, if the load has notable capacitance (which is typical for many high-voltage applications driven by a flyback), then that capacitor will gain some charge from the pulse and its voltage will increase some, but if its a large capacitance then its voltage may not increase much at all with that single pulse. So it effectively limits (clips) the high voltage pulse while the diode is forward-biased. It flat-tops the pulse waveform.

    Since we're assuming perfect coupling between the primary and secondary, the waveform on the primary will also be clipped at the same point in the waveform.

    On a side note, a small capacitance is also placed across the inductor to lower its resonant frequency so that its not so fast that notable power is wasted in the core, and so it doesn't radiate as much EM. So this typically makes the pulse more of a ring than a pulse, and you'll probably see some of that in the waveforms.

    Also, in reality the coupling isn't perfect so there may be some difference in the primary and secondary waveforms (besides the magnitude).
    Last edited: Sep 16, 2011
  4. Sep 17, 2011 #3
    Even though the secondary is open (diode not conducting), this is true?

    Thanks for the explanation. But you did not say anything about reflected voltage.
    I am trying to understand when exactly reflected voltage occurs from sec to primary.

    So, the Secondary has some voltage built up during primary current ramp up.

    when switch is turned off,
    1. Secondary voltage polarity is reversed (diode conducts)
    2. At the same time, the secondary voltage is reflected onto the primary side?
  5. Sep 17, 2011 #4
    The statement is always true. That's what "ideal" and "total coupling" mean here.

    Googling 'transformer +"reflected voltage"' yields only 14K hits (and "reflected current" only 4K hits), and this conversation is on the first page of hits. And from those hits the definition of "reflected voltage" appears to be simply one-half of what I state here: If you've been told the voltage on a secondary, then the reflected voltage (on the primary) is the voltage on the secondary scaled by the turns ratio. There may be other useful definitions, but I'm not familiar with them.

    For an ideal transformer things are quite simple, and trying to separate voltages into "reflected" and "not reflected" going each way across the transformer seem to add complexity, IMO.
    Last edited: Sep 17, 2011