Re: Flyback transformer design confusion!!

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)

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.

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)



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.

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


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.



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.

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? 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.

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).

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.



snipped >

lemonjuice



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