# Flyback voltage in Flyback Converters

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) .

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?

• 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. Commented Aug 28, 2016 at 15:00
• Ok! Thanks for the info. I did not want to "spam" around ;-) Commented Aug 28, 2016 at 15:40