Re: Flyback transformer design confusion!!

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

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!

you wrote:

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.

which is of course integration.




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.

I can. I have doubts about your literacy though.

and yeah, sure, changing L changes the depth (or lack thereof) of CCM.


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!!!!!!!!!!!

true, but your design equation is WRONG.

try (without snipping them) to answer these three questions (or rebut my answers):

1) what is the energy stored in the core, just after the switch turns on

Estart = 0.5*L*Imin^2

2) what is the energy stored in the core, just before the switch turns off

Efinish = 0.5*L*Ipeak^2

3) what is the change in energy in the core during the on time

delta_E = Efinish - Estart = 0.5*L*(Ipeak^2 - Imin^2)


which is of course rather different from *your* so-called design equation, delta_E = 0.5*L*(Ipeak - Imin)^2



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.

its quite possible, I've done so many times. but not using your incorrect "design" equation, which is my point.

the examples I have used were chosen specifically to highlight the error in your equation (although the above calculation shows it quite clearly).





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.

Dude, I'm one of them, and I DONT use that equation, I use the correct one.

OK, assume operating at fixed F, CCM.

know Vin(min - max), Vout, Pout, F

choose max duty cycle Dmax

calculate turns ratio from Vout + Vdiode = Vinmin*(Ns/Np)*Dmax/(1-Dmax)

choose the load current Idcm below which the psu operates in DCM

during the interval Toff = (1-Dmax)/F, all of that energy is transferred to the load, in a single sawtooth pulse. The charge used by the load over one cycle (no charge is transferred to the load during the on time) is:

Qload = Idcm*Tsmps = Idcm/F.

This charge must be transferred during the off time, which is the integral of the secondary current:

Qsec = 0.5*Toff*Is_peak = 0.5*Is_peak*(1-Dmax)/F

equating the two:

Qload = Qsec: Idcm/F = 0.5*Is_peak*(1-Dmax)/F

cancel the F's and re-arrange:

Is_peak = 2*Idcm/(1-Dmax)


We know the off time and the secondary voltage, which allows us to solve for the secondary inductance:

(Vout + Vdiode) = Lsec*(Is_peak - Is_min)/Toff

now because this is the boundary between CCM and DCM, Ismin = 0, so the change in secondary current is Is_peak. solving for Lsec:

Lsec = (Vout + Vdiode)*Toff/Is_peak

substituting in the ecpressions for Toff and Is_peak gives:

Lsec = (Vout + Vdiode)*[(1-Dmax)/F]/[2*Idcm/(1-Dmax)]

Lsec = (Vout + Vdiode)*(1-Dmax)^2/(2*F*Idcm)

and then reflect that to the primary to get the primary inductance:

Lprim = (Np/Ns)^2*Lsec

Lprim = (Np/Ns)^2*(Vout + Vdiode)*(1-Dmax)^2/(2*F*Idcm)

if we want, we can use the relationship between Vin and Vout to simplify this further:

Vout + Vdiode = Vinmin*(Ns/Np)*Dmax/(1-Dmax), so

Lprim = (Np/Ns)^2*(Vinmin*(Ns/Np)*Dmax/(1-Dmax))*(1-Dmax)^2/(2*F*Idcm)

simplifying:

Lprim = (Np/Ns)*Vinmin*Dmax*(1-Dmax)/(2*F*Idcm)


voila. as one moves the degree of DCM (choosing a bigger or smaller Idcm) the required primary inductance moves. so what does the current waveform look like at full load?

Pin = Pout/n (n = efficiency, IOW include losses)

the input current up-slope is now fixed (we have chosen L,F,Vin):

Vinmin = Lprim*delta_Ip/Ton

delta_Ip = Vinmin*Ton/Lprim

we know what Ton is:

delta_Ip = Vinmin*(Dmax/F)/Lprim

so:

delta_Ip = (Np/Ns)*Vinmin*(Dmax/F)/[Vinmin*Dmax*(1-Dmax)/(2*F*Idcm)]

delta_Ip = (Np/Ns)*(2*Idcm)/(1-Dmax)

or, in other words, delta_Ip = (Np/Ns)*delta_Is, which it should.

That the equations prove each other is a strong indication they are right - this sort of cross-checking is useful as a bull*** detector.


OK, now we get to the fun bit, where your supposed design equation arises - what is the trapezoid offset at full power:


energy transferred per cycle:

delta_E = Pin/F


energy at start of on-time:

E_start = 0.5*Lprim*Ip_min^2


energy at end of on-time:

E_finish = 0.5*Lprim*Ip_max^2

Ip_max = Ip_min + delta_Ip

E_finish = 0.5*Lprim*(Ip_min + delta_Ip)^2

expanding:

E_finish = 0.5*Lprim*(Ip_min^2 + 2*Ip_min*delta_Ip + delta_Ip^2)


by definition, delta_E = E_Finish - E_start

(this is where you go wrong)

delta_E = 0.5*Lprim*[(Ip_min^2 + 2*Ip_min*delta_Ip + delta_Ip^2) - Ip_min^2]


simplifying:

delta_E = 0.5*Lprim*[2*Ip_min*delta_Ip + delta_Ip^2]


solving for Ip_min:

2*delta_E/Lprim = 2*Ip_min*delta_Ip + delta_Ip^2

simplifying:

2*delta_E/Lprim - delta_Ip^2 = 2*Ip_min*delta_Ip

so

delta_E/(Lprim*delta_Ip) - 0.5*delta_Ip = Ip_min

but delta_E = Pin/F thus:

Ip_min = Pin/(F*Lprim*delta_Ip) - 0.5*delta_Ip


voila, everything is now calculated. but we can simplify this equation further, as we know:



delta_Ip = (Np/Ns)*(2*Idcm)/(1-Dmax)


thus:

Ip_min = Pin*(1-Dmax)/(2*F*Lprim*(Np/Ns)*Idcm) - (Np/Ns)*Idcm/(1-Dmax)

although thats hardly an improvement.




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.


during the trapezoid (Ton for primary, Toff for secondary) the average value is (Imin + Imax)/2.

rather than the method I used above, I can directly calculate the secondary trapezoid at full power, by equating charges again:

charge delivered to load over one cycle:

Qload = Imax/F where Imax = max output load current


charge removed from secondary during Toff:

Qsec = [(Is_min + Is_max)/2]*(1-Dmax)/F


equating the two and simplifying:

Imax/(1-Dmax) = 0.5*(Is_min + Is_max)


but Is_max = Is_min + delta_Is, so:

Imax/(1-Dmax) = Is_min + 0.5*delta_Is

so

Is_min = Imax/(1-Dmax) - 0.5*delta_Is


but Imax = Pout/Vout, so:

Is_min = Pout/[Vout*(1-Dmax)] - 0.5*delta_Is


- actually, use Pin, and "pretend" the power is delivered to the output, even though not all of it gets there (theoretical analyses just assume 100% efficiency):

Is_min = Pin/[Vout*(1-Dmax)] - 0.5*delta_Is



now I already derived the equation for Ip_min:

Ip_min = Pin/(F*Lprim*delta_Ip) - 0.5*delta_Ip



so if I reflect the equation for Is_min thru the transformer, these should equate:


Ip_min = (Ns/Np)*Is_min

substituting in the expression for Is_min:

Ip_min = (Ns/Np)*Pin/[Vout*(1-Dmax)] - 0.5*(Ns/Np)*delta_Is



and we know, by definition, that delta_Ip = (Ns/Np)*delta_Is, so:

Ip_min = (Ns/Np)*Pin/[Vout*(1-Dmax)] - 0.5*delta_Ip


which ought to equal the original equation:

Ip_min = Pin/(F*Lprim*delta_Ip) - 0.5*delta_Ip


the 0.5*delta_Ip is in both equations, so equating the two gives:

Pin/(F*Lprim*delta_Ip) = (Ns/Np)*Pin/[Vout*(1-Dmax)]


re-arranging and simplifying:

F*Lprim*delta_Ip = Vout*(1-Dmax)*(Np/Ns)


and we know what delta_Ip is:

delta_Ip = (Np/Ns)*(2*Idcm)/(1-Dmax)


so:

F*Lprim*(Np/Ns)*(2*Idcm)/(1-Dmax) = Vout*(1-Dmax)*(Np/Ns)

thus:

F*Lprim*(2*Idcm) = Vout*(1-Dmax)^2


and we also know that:

Lprim = (Np/Ns)*Vinmin*Dmax*(1-Dmax)/(2*F*Idcm)


therefore:

F*[(Np/Ns)*Vinmin*Dmax*(1-Dmax)/(2*F*Idcm)]*(2*Idcm) = Vout*(1-Dmax)^2

simplifying:

(Np/Ns)*Vinmin*Dmax/(1-Dmax) = Vout


which, of course, it does! Therefore the two expressions for Ip_min are in fact equal.

[OK, ignoring diode drop. If I assume all of the loss is in the diode, the equations line up exactly, but it gets a little bit messier]





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

what, no duty cycle?

appropriate variation of pulse duty cycle works to maintain Vout
constant during operation.

during a single switching cycle however, Vin is pretty much constant.




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.

it is also true to say that, assuming the smps can in fact drive the load, the *LOAD* sets the load current.


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.

yep, and its right


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


yep, and its wrong.

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


really?

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


LOL


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? .


although this dissertation contains pretty much everything needed to design flybacks, for the purpose of our argument (your incorrect equation), its irrelevant. all that matters is:


1) what is the energy stored in the core, just after the switch turns on

Estart = 0.5*L*Imin^2

2) what is the energy stored in the core, just before the switch turns off

Efinish = 0.5*L*Ipeak^2

3) what is the change in energy in the core during the on time

delta_E = Efinish - Estart = 0.5*L*(Ipeak^2 - Imin^2)


and that is a thorough dis-proof of your equation.


Till later

lemonjuice


Cheers
Terry
.