Re: Flyback transformer design confusion!!

On Fri, 17 Nov 2006 03:20:32 +1300, Terry Given <my_name@xxxxxxxx>
wrote:

lemonjuice wrote:
On Thu, 16 Nov 2006 14:05:17 +1300, Terry Given <my_name@xxxxxxxx>
wrote:


lemonjuice wrote:

On Wed, 15 Nov 2006 07:52:22 +1300, Terry Given <my_name@xxxxxxxx>
wrote:


sniped>



Why decrease Rl to increase output power?

Que?!

assuming we are talking about a regulated output voltage (we are),
reducing the load resistance is the ONLY way to increase output power



It is not the only way man ...There is a whole lot of things you can
do.

can you read?

OK, show me then! When applying a constant DC voltage across a resistive
load, just *how* can I increase the power into that load, other than by
changing the resistor? I cant.

(assuming of course that the power supply is regulating the voltage)



1) The power given by the expression above is that of the inductance
at the primary coil. To find the power dissipated in the load you
would have to calculate the average current flowing through the load.


indeed. and unless something is horribly wrong with your smps, the two
should be roughly equal - losses mean the "inductor power" dE*F is
somewhat larger than the load power.


and of course for a DC supply with a fixed load, the integration is
pretty darned easy. Pout = Vout*Idc_out = Vout^2/Rl

eg
1) Integration???Didn't you know its Ohms law that says Idc_out =
Vout/Rl ?

duh. you went on about integration (which although technically correct
is trivial with DC)
I went about it ...????
Check any of my posts ...I never mentioned it!


2)I hope you haven't forgotten that also from Ohm's law
Pout=Iout^2*Rout. So changing the output current(through the load)
Iout will change the power output.

Right. I can do that by changing the voltage and/or the resistance. but
its a power supply, so the voltage, by definition, is not supposed to
change (at least not by very much). Therefore the only way to change
Iout is to change the resistor.

damn You seem to have lost the whole sense of this thread then. The
op was
asking why different inductance values could work for a given power
output.
He was not talking about varying Resistive load differences ... man
learn how to read.

try a real-world example: a 300W 5V supply with a 10 Ohm load resistor.
Pload = 2.5W. Now try changing the current thru that resistor. you cant,
its connected across a fixed voltage source.....

if I then change it to a 1 Ohm resistor, its still 5V, but Pout is now
25W (and Iout = 5A)

Totally irrelevant. Op and I were talking about design equations for
a Power supply to be designed. Not an already built power supply
with fixed unchangeable components!!!!!!!!!!!
You don't think about increasing (possible though) or maintaining
constant the power capabilities of an already built power supply which
is what i've been talking about all the time.
Hecks I hope you understand at least that.




2) You don't have to vary Rl when you change the load power demands.

this must be a semantic issue; Rl = load resistance. changing Rl is the
only way to change load power if the output voltage remains constant.

It wasn't clear whether you meant R*I or RLoad however
as I pointed out above Pout=Iout^2*Rout would still apply.

The flyback transformer regulates itself over a large value of input
voltages and currents. If the load demands are below its Power
capabilities the flyback transformer operates in discontinous mode by
shutting itself inbetween cycles.


this covers a wide range of phenomena - load reducing, moving from CCM
into DCM (at which point duty cycle begins reducing) into pulse
skipping, where D has reduced to the minimum value, and the associated
minimum quanta of energy is greater than the actual load, so the output
voltage overshoots, and the error amp prevents pulses until the output
voltage has (slowly, there is very little load) decayed below the
reference (by an amount governed by the error amplifier compensation
components).

I wasn't talking about the typical SMPS voltage regulating effect but
a different effect limited to flybacks.


which one? sounds like a blocking oscillator, or some form of
self-oscillating flyback converter.



I specifically used the example of CCM to *prove* your equation
0.5*L*(Ipeak - Imin)^2 was wrong.


Now now .. you didn't follow my hint below on understanding why
you are wrong.
My formulae is the universal one all Flyback transformer designers in
the world use!
I enjoy the fact that you can't see why it is right. This is your last
chance ... to derive it.
I hate telling people the answers as I prefer them to get to the
right conclusions by reasoning on it themselves ... IMO it is more
dignifying. Think about this ...
Do you know exactly what is the current flowing through the
primary inductor?

yes of course I do. its a trapezoid
You've been yelling trapezoid before ... so that is a rather obvious
answer.
So when you calculate the power you insert the values of a trapezoid
in the formula? Tell me what values ?
What is the average current?
I already asked you this before.


Hint: Ipeak ? I min? or a function of the two? If you can tell the
value then you can easily see why my formula is right and yours
isn't.My formula was written for the limiting condition of DCM but it
works also in CCM with a factor of 1/2 applied. Let me know if you
still don't get the point.

give a worked example. your formula is just plain wrong.

So you think your formula P = 0.5 L (Ip^2 - Imin^2) is right?



In 1 cycle the average current flowing through

I never wrote this. therefore you must have.

the input voltage sets the current
up-slope


I did write this, and its true.



average current flowing through input voltage ???? I assume you mean
through the input inductor?

you wrote the rubbish about average current


The triangular or trapezoidal (as you fancied calling them) upslope
(as you call it) on the current - time waveform of the inductor is
set by the ratio of the input voltage and the primary inductance.
BECAUSE
Ipeak - Imin = Vin * ton /L (1)
=> slope is Ipeak-Imin/ton = Vin/L
Vin input voltage , ton time transistor is on , L inductance of
primary


yeah, thats right.



Contrary to what you say the "average current" is a completely
different entity and SETS no slope.

I never said that. I dont know where you got that line of text from, but
*you* wrote it, not me.

damn strange ... I've never corrected my own self ... hey it could be
the first time!

The current-time waveforms of the inductors are made up only of
instantaneous currents! Write the mathematical expression for the
average current and you will see that.
Even from eqn.1 it is quite evident that by reducing L you can
increase Ipeak - Imin which increases the permissible power. They are
magnetic and physical limits to this but that is another story.

off on a tangent



and the output voltage sets the down slope, so Ipeak - Imin is
constant, regardless of load (as long as the load is sufficient to
ensure DCM operation).

Wrong man ... totally inaccurate. So according to you I can't inject
whatever current I want into the transformer ... even after changing
the primary inductance?

since when does one change the primary inductance of a flyback
converter?
Yes you have to decide that when designing it.
the input and output voltages, power, frequency and mode of
operation all combine to select the primary inductance value. design it,
build it, it stays constant.

Though what you are considering is not the case I and the Op were
working on I remind you that in an SMPS
Input voltages are not necessarily constant ...
Vout = ton/toff * Vin
appropriate variation of pulse duty cycle works to maintain Vout
constant during operation.

Neither is the mode of operation it all depends on the
limiting current flowing through the load which is a function also of
the turns ratio. Now I would have to write so much on the limiting
current ... but I'll stop here.
How does that sound to you?
Hint: what are the current steady state interrelations? How are these
related to the output/input voltage in the flyback transformer?

Double the output power (by halving Rl) and the
trapezoid gets taller - Ipeak and Imin *both* increase - but the slopes
dont change (nor does the duty cycle).

Now whoever suggested halving RLoad? you ? Its unnecessary. Just
increase the input current and your trapezoids get taller.


wow, you can increase the input current of your power supply (which
presumably regulates the output voltage) but dont change Rload! where
does the input current go to?

oh, thats right, according to your maths, the power stays the same
regardless of the height of the trapezoid, it only depends on (Ipeak -
Imin).

weren't you rambling about increasing power? Now you claim I'm
saying power stays constant. I'm starting to give up understanding
you.

Your formula for Power has exactly the same variables as mine does
Yr formula is
P = 0.5 L (Ip^2 - Imin^2) Now don't say I invented that ... you
wrote it.

Mine is 0.5*L*(Ipeak - Imin)^2 CCM

Common sense would warn you that something is wrong with yours ...

I'm trying to help you see your errors. I could give you the answers
immediately but stimulating your thinking is best IMO.

you cant even grasp a simple energy balance, which shows your maths to
be wrong. I'm picking you dont design flyback converters for a living.

Energy balance ... wheeeeeee . Where is it? .



snipped >

lemonjuice



ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM


Till later

lemonjuice



ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM
.

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