# Flyback discontinuous mode, double voltage over the switch

Considering the following simplified flyback schematic.

Which has the following voltage current graphs

Why is there 2*Vin over the switch when the switch is turned off? And why is that change to Vin and 0 so intense?

• What is the output voltage? What is the transformer voltage? What is the input voltage plus transformer ratio times output voltage? – winny Dec 29 '18 at 16:02
• It is just a conseptual design. So the transformer has just a 1/1 ratio the Vin is just an undefined voltage and there are no losses due to latency or parasitic effects. – J. Joly Dec 29 '18 at 16:05
• Actually V(s1) does not look like this at all and is wrong. Since there is a low ESR load, there is no 2Vin but there is V=LdI/dt and Is will get the load current reflected back (transformed and added) into I s1. T1 also has L[H] so there are 2nd order ringing too. – Tony Stewart Sunnyskyguy EE75 Dec 29 '18 at 16:30
• But if you put a high impedance low pass filter on switch it will attenuate the real results. to obtain 2Vin – Tony Stewart Sunnyskyguy EE75 Dec 29 '18 at 16:40
• Like a snubber? – J. Joly Dec 29 '18 at 16:42

When the power switch closes, the primary side of the transformer "sees" $$\V_{in}\$$ if you neglect all the drops. During this on-time, the secondary-side diode sees its anode biased at $$\\frac{V_{in}}{N}\$$ while its cathode is biased at $$\V_{out}\$$. The peak inverse voltage or PIV of the diode is thus $$\\frac{V_{in}}{N}+V_{out}\$$. You select the diode breakdown voltage based on this reverse-bias condition with some margin applied.
When the power switch opens, the primary-side current is scaled by the transformer turns ratio and circulates in $$\D_1\$$. As this diode conducts, the secondary side of the transformer sees $$\V_{out}\$$ as long as $$\D_1\$$ conducts. This voltage "flies" back to the primary side of the transformer hence the name flyback converter. The switch now sees the series combination of $$\V_{in}\$$ plus the reflected voltage: $$\V_{sw}=V_{in}+NV_{out}\$$. You have $$\2V_{in}\$$ at the switch terminal if $$\NV_{out}=V_{in}\$$ and this is a very weird example in my opinion.
When the primary inductance is fully demagnetized within the switching cycle (discontinuous conduction mode or DCM), $$\D_1\$$ stops conducting and the switch voltage returns to $$\V_{in}\$$. These are idealized waveforms as parasitic inductors and capacitors ring during transitions. You have more explanations on the flyback converter in this seminar.