Electrical – Flyback output voltage

I did debate about this answer being a comment so don't bust a gut voting down (or up) but I had too much to say to fit a comment.

Can you get 3A from those batteries? For now I'll assume you can...

Assuming that you are producing no-more than 80V on the primary (because the FET will break if you are doing a 100v) and given your turns ratio of 1:160, and given the "imperfect coupling factor" (that I pull out of the experienced side of my brain) I'd say you should produce about 5kV.

The FET is getting warm and this isn't a worry but it does rather hint at your 12V supply rail falling to somewhat less when it is powered on. Maybe you can measure it. My suspicion is that it might be falling to about 6V or 8V and this will not efficiently turn the fet on and this could be causing the FET to warm. If this is true then maybe you might only be producing 1kV.

Of course, there is a strong possibility that you damaged your FET with over voltage the very forst time you used it. I note on your earlier threads/questions you were advised to use a 200V device with lower on-resistance.

I'll also point out that a lot of these circuits are from unprofessional sites that might have (with luck) got something to work and easily misconstrued their own circuit diagram thus misleading everyone who tries to build one. Also notable about the website that contains the circuit is the lack of information about the transformer and how to wind it.

The original design came from: -

http://www.geocities.com/CapeCanaveral/Lab/5322/fbt2.htm

But don't bother trying the link because geocities is no more.

Looking at your pictures and knowing how difficult it is to get a step-up transformer working with ratios over 100 I have to point out that there is no hope for the way you have it connected. 555 needs to drive the transistor which provides voltage to the FET and the FET source, emitter of the transistor and 0V of the 555 have to be "rigidly connected". The currents you might be inducing in 0V connections could easily mean you get less than 200V out of this design.

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).