I am studying a standard flyback converter. For the circuit we are talking about please refer to (I took it from www.power-eetimes.com) Flyback converter Circuit.

Now, I know the law of inductors, so when the switch turns on (it allows current to pass), there is a fixed voltage across the inductor. The current ramps up linearly and the slope is dependent on the value of the voltage source and the inductance.

Now, when the switch turns off and the current is decreasing in a very short amount of time, according to the law of the inductor, the voltages is inverted and therefore the voltage also changes polarity on the secondary side and current can flow on the secondary.

To my question: What determines the speed that the current is decreasing on the primary? I would suspect that it depends on the value of the impedance of the secondary side. The idea would be that a low impedance can accept (or let's say demand) a bigger amount of current and therefore allow a faster teardown of the field. I would suspect, that on the secondary side, current is the "driving" force and the exact voltage value is a result of impedance and current? So is it correct that the secondary impedance determines how fast the energy of the magnetic field is transferred to the secondary or is it "just" dependent on the output voltage instead?

  • 1
    \$\begingroup\$ In the future, please limit your questions to one specific question, that can be answered in a few sentences. You are very welcome to break up your problem into a few different question posts here. (I.E. Question 1 could be posted as one question, Question 2 as a second post, etc.) That way, answers to each specific question you're having are easier to find for others when searching this site. \$\endgroup\$ Commented Aug 28, 2016 at 15:00
  • \$\begingroup\$ Ok! Thanks for the info. I did not want to "spam" around ;-) \$\endgroup\$
    – Junius
    Commented Aug 28, 2016 at 15:40

1 Answer 1

  1. With a transformer (which a flyback converter's coils technically are a transformer), you don't look at each set of windings as a separate inductor, but instead as parts of the same inductor. When voltage is applied across the primary, current through the primary increases and a shared magnetic field forms across all windings (primary and secondary). The same with disconnecting power to the primary, it generates a shared magnetic field. As long as the secondary windings are connected to a path electricity can flow through (i.e. are not open-circuit), the current in the primary winding can instantly drop to 0 & the voltage across either set of windings will only spike high enough to force the transformed current through the load attached to the secondary side.

  2. The formula Vsw=Vin+Vout*(Np/Ns) is showing you a calculation for the voltage at the switching transistor when it opens the circuit. In any inductor, when a current-source is disconnected, the voltage will rise until a path is found to dissipate current through. Here, your switching transistor will see the "natural" open-circuit voltage of the power source it is switching, along with the additional voltage from the breaking-down field in the transformer. That additional voltage is dependent on the transformer's turns ratio, and the impedance of the load attached to the secondary coils. "How fast the current is decreasing" depends on the impedance of the path the current leaving the coils finds. Current through any inductor that is discharging will always be decreasing as a function of inductance versus back-emf (the voltage across the load), so measuring/calculating the output voltage gives you your breakdown speed.
    To calculate Vo at the moment the switching transistor "opens" (considering the output capacitor's reactance as part of Rload), Vo=Ip*(Np/Ns)*Rload. Meaning that the output voltage will rise until the load resistance can be forced to accept the current that just stopped flowing through the primary windings (current-transformed by the turns ratio of the transformer).

  3. The zener diode path (clamping circuit) limits the maximum spike voltage to protect your switching transistor from exceeding its maximum voltage in case your load impedance is too high, or the output is open-circuit. Once Vo*(Np/Ns) passes the avalanche voltage of the zener (plus the voltage-drop of the series diode), current will pass through the "clamping circuit," rather than the voltage being allowed to continue building until another part (usually the switching transistor) failed.
    Under normal operation, the zener diode path should appear as an open-circuit to the transformer, only allowing current to pass in "emergency situations" (when something goes wrong, and power has to be dissipated to avoid catastrophic failure).
  • \$\begingroup\$ @2.) Is it possible, that in Vo=Ip*(Ns/Np)*Rload, Ns and Np should be exchanged? \$\endgroup\$
    – Junius
    Commented Aug 28, 2016 at 15:36
  • \$\begingroup\$ @2.) "Current through any inductor that is discharging will always be decreasing as a function of inductance versus back-emf (the voltage across the load)". Can you provide a formula for this statement? For me this is kind of a chicken-egg problem. the rate of decrease in current depends on the back-emf? But does not the back-emf depend on the rate of decrese in current? \$\endgroup\$
    – Junius
    Commented Aug 28, 2016 at 15:39
  • \$\begingroup\$ on your first comment, yes...I think I did mess that up by using the turns-ratio calculation for voltage in a step where I was computing current. I'll change that as soon as I get done grabbing an exact formula to answer your second comment. \$\endgroup\$ Commented Aug 28, 2016 at 15:42
  • \$\begingroup\$ Hmm, thinking about it, I think you missed the crucial 3rd element in the current-decrease/back-emf/**load-impedance** triangle. The inductor wants to maintain its current passage, but the load impedance produces back-emf when current passes through it; then the back-emf causes the current through the inductor to decrease (then the cycle repeats). \$\endgroup\$ Commented Aug 28, 2016 at 15:52
  • \$\begingroup\$ Ok, I haven't come up with an exact formula yet (sorry, haven't had enough coffee for that calculus this Sunday), but here's a link to a page that should explain it better than me anywise: allaboutcircuits.com/textbook/direct-current/chpt-16/… The page explains time constants for charging circuits, but if you redraw the L/C circuit with a closed switch & no battery, the inverse formulas apply for discharching. \$\endgroup\$ Commented Aug 28, 2016 at 16:01

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