Question regarding flyback converters, transformers, and a circuitfrom Art of Electronics

[Michael]

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

 

[Mark]

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).

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

 

[[email protected]]

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?

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.