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Question regarding flyback converters, transformers, and a circuit from Art of Electronics

Hi there - I was looking at the high voltage video flyback converter (page 374, figure 6.55A) in art of electronics. I feel like I understand most of it, but I want to clarify a couple things just to be sure:

First of all, I'm only really comfortable with transformers being used to take AC in and spit AC out. That's really all I've ever been taught to do with them. Their full functionality was never really explained, nor their makeup, or anything else. Essentially I was taught Vs/Ns = Vp/Np. Thinking about it though, am I right in thinking that a transformer is essentially an inductor that has multiple paths for current to flow? Or in other words, when one inductor in a transformer has a certain amount of energy stored, that energy is shared between both the primary and secondary and can be used to create a current through either the primary or the secondary.

Like in figure 6.55A - this is what I'm thinking: Q1 turns on and the primary coil reaches some high current and then Q1 turns off. A significant amount of energy has been stored in the transformer, so it does what it can to continue a current flow. It can't make any current flow in the primary, but the secondary does have a path for current, so current flows through there. Would the voltage across the secondary then be Ls*(dip/dt)? (where Ls = inductance of secondary, and dip/dt is the derivative of the current through the primary). Since Ls is much larger than Lp, this is going to be a very large voltage.

Additionally, if C1 were already at a very high voltage, D1 is there to protect Q1 so that the collector of Q1 won't see above about 300VDC (as the voltage across the primary will increase trying to get current to flow through it). D2 is there so that the transformer can only charge C1, not drain it as well. D3 is there to protect the collector from going too far negative when Q1 is first turned on.

How far off am I?

Thanks!

-Michael

Reply to
Michael
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think of a flyback "transformer" as two coupled inductors. when the pri current shuts off, the current flows in the secondary and works like a current source. The value of the current source is determined by what the pri current was and the turns ratio and the inductance. The voltage will be determined by the load. In most applications there is a voltage feedback loop that continusously adjusts this "current source" to obtain the desired output voltage. But the flyback "transformer" inherently is a current source.

Mark

Reply to
Mark

There's a neater description.

If your transformer has two coupled windings, you can describe its behavior in terms of the the inductance of the first winding L1, the inductance of the second winding L2 and the mutal inducatance of the two windings M

The voltage V1 across the first winding is then given by

V1 = L1. di1/dt + M. di2/dt

while V2 across the second winding is

V2 = M. di1/dt + L2. di2/dt

where i1 and i2 are the currents through the first and second windings.

In a perfect transformer M^2 = L1.L2

In real transformers M^2 can get pretty close to the product of L1 and L2. If you've got a really low inductance core it may drop as low as 95% of the product of L1 and L2. If you haven't got a core at all it can be quite a lot less.

-- Bill Sloman, Nijmegen

Reply to
bill.sloman

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