Flyback transformer design confusion!!

Liar. I see them, but yours just went away.

-- Service to my country? Been there, Done that, and I've got my DD214 to prove it. Member of DAV #85.

Michael A. Terrell Central Florida

Que?!

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

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

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.

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 specifically used the example of CCM to *prove* your equation

it even works open-loop. Vin and Vout (and turns ratio) set the duty cycle. if Rl decreases, initially Vout will drop. but the secondary current downslope also drops (it is proportional to Vout) so during the next flyback pulse not as much energy is transferred from the transformer (coupled inductor really) to the load. At the next turn-on, Imin is therefore a bit larger, but dIp/dt is the same so Ipeak gets larger by the same amount (as the duty cycle is constant). pretty soon the output voltage is back where it chould be, and both Ipeak and Imin have increased b the same amount.

the converse applies when Rl increases (decreasing load).

at some point, the load reduces (Rl increases) below the CCM/DCM boundary, at which point Imin = 0, and the load resistance enters the voltage transfer function.

but I wont go into that; the purpose has been to conclusively disprove your erroneous mathematics that P = 0.5*L*(Ipeak - Imin)^2*F, rather than detail the exact workings of a flyback converter

what, no eng >Any average intelligence engineer could write an entire book on a >flyback transformers(without basic errors)

its pretty funny really.

(that which needed to be done)

Cheers Terry

LMAO Must be that crappy Windows 98 you are using .Its time to upgrade. Was that the distinguished medal the Army awarded you for your service?

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

You are welcome. Its one of my favorite links!

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

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

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

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

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

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? 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.

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

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

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

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? 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?

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

snipped >

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

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)

duh. you went on about integration (which although technically correct is trivial with DC)

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

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

yes of course I do. its a trapezoid

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

I never wrote this. therefore you must have.

I did write this, and its true.

you wrote the rubbish about average current

yeah, thats right.

I never said that. I dont know where you got that line of text from, but

off on a tangent

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.

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.

I went about it ...???? Check any of my posts ...I never mentioned it!

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.

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.

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.

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.

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

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

Yes you have to decide that when designing it.

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.

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.

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

Till later

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

which is of course integration.

I can. I have doubts about your literacy though.

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

true, but your design equation is WRONG.

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

Estart = 0.5*L*Imin^2

Efinish = 0.5*L*Ipeak^2

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

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).

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 bullshit 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:

simplifying:

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.

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

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.

what, no duty cycle?

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

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

yep, and its right

yep, and its wrong.

really?

LOL

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:

Estart = 0.5*L*Imin^2

Efinish = 0.5*L*Ipeak^2

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

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

Cheers Terry

---> snipped

ummmm

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

Oh come of it ... The per capita income of a nation is an average value but you will never be able to calculate that using an integration. The case above is a different but IMO Ohms law is sufficient.Well if you like playing around with mathematical terminology/techniques that is another story.

You must admit that more then 90% of your rambling is about a case that was never raised by the Op ... and that I was not talking about. Either you misunderstood the case because of lack of clarity or as it seems you wanted to launch a personal attack on me. Whatever the case Mr instead of round abouting the most important issue just try like me to talk/argue about the main issue ..

And as you notice above I also agreed on the possibility of that.

which examples? Sorry maybe I missed them. I think I thoroughly answered them.

My expression is for power ... The above equations at first looks, seem to be valid but are unrelated and so they do not disaprove in anyway my formula.

As you and any engineer is well aware of Power = average voltage * average current To give you an example your 100W lamp implies that at 240V ac you have a current of 0.416A. Both the 240V and the 0.416A are rms values or in other words averages ... square roots of the average of square values because we working with negative sinusoids.

So my approach is to calculate power you have to work with average currents and voltages and forget all the other nice looking formulae.

As you said the input voltage is normally a constant value ... even if directly operated from the ac mains we would be working with a rms value in the expression for power.

Vin = Ldi/dt (1)

di/dt is the gradient of the inductor's waveform which as I showed in my previous post is very simply seen to be equal to Imax - Imin /ton substitutuing in eqn (1) Vin = L * (Imax - Imin)/ton (2)

The voltage change across the inductor at the primary is caused by a current change of Imax-Imin. The average value of this current change is Iavg =( Imax-Imin)/2 (3) Power = Vin * Iavg (average current causing voltage change at primary inductor) (4) substituting eqn (2 ) and (3) in eqn (4) you get

Power = 0.5 * L* (Imax - Imin) ^2 /ton

The rest is self explanatory.

You can derive the same formula using an alternative approach by applying differential analysis methods.

Yes it is related to the duty cycle ... I imagine with all the equations you've written its kids play for you to see how the above is related to the duty cycle.

OK ... agreed. I was considering more then 1 cycle but even then with an unrectified input , low frequency etc Vin can change .

Yes. But even with no load a current will flow in the primary ... it could even be designed to be larger then the limiting current.

well ... its a free world

disagree on that

Well I feel I have sufficiently disproved your equations / techniques but as I said its a free world. I would add basic physics will tell you that dc components don't contribute to inductor voltage obviously ignoring resistive effects. Your equations above contain dc components.

Frankly I'm getting bored with this topic. I think we've exhausted this argument ... whether you agree or disagree with me :) Maybe its time to pass on to something else . What do you think?

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

Well, I'm glad I wasn't in on this one.

Looks like lemonstains has been either been squirming since he got it wrong in the first place in an effort to maneuver himself into a position where he can claim Terry is some sort of thick for not realising what lemonstains was saying in the first place rather than admit to his faux pas. Or lemonstains is just stupid and can't admit to it.......

?I think I just said the same thing twice?

Anyway, pity he's given up on the argument.

I think he may have realised that pushing his last resort 'Ahaaaa but my (incorrect)equation gives the same answer as your (correct) equation in the case of discontinuous operation and I have been talking about discontinuous operation all along.' argument was likely to be laughed at.

DNA

Yeah I can bet ... It wasn't for kids.

nice dreaming hey hey

Hey hey ... some men will rob you with a 9mm others with a pen

Me stupid hey hey ... great ... women love ruthless and stupid men

Given up ... hey hey... the tide always turns so never never never never never give up Genome

Dare you say I wrote an incorrect equation! hey hey

Discontinous operation? Hey hey ...

Nice anecdote it may have backfired... but that is lemonjuice ... sour, bitter and pungent but always to be tasted. Keep up the funny work ... Your posts are among my favorites

lemonjuice

lemonjuice

ALL INCOMING MESSAGES CONTAINING BARTOLI ARE DELETED BY SYSTEM

lemonjuice wrote: [snip]

thats because its discrete, not continuous. Otherwise, it is an integration (or a summation, which is the discrete form)(

The case above is a different but IMO Ohms law is

no personal attacks dude, just poking a large hole in your incorrect equation.

personally, I prefer the variable INGWELLRIGHT. its just an arbitrary placeholder....

your equation is indeed "correct" for DCM, it just falls to pieces in CCM.

LOL

I just wrote that lot down OTTOMH. thats not the approach I use either, but is fairly similar.

not yet. but a little bit further on....

ROTFLMAO!

So, for a 50Hz ac power system driving a resistive load:

Vaverage = 0

Iaverage = 0

==> Paverage = 0

by your reasoning. which (given that my electric oven heats things up) is patently untrue, and is of course (in the case of AC) why we use RMS.

then they are NOT averages. by definition. RMS = Root Mean Square.

I dont use any averages at all, except in deriving the equations, and they are really integrals, just with "easy" waveforms.

yep.

nope. impressing a voltage across an inductor causes the current to change; a constant voltage gives rise to a linear change in current (unless of course the inductor saturates)

yep. just not very useful.

indeed, but the explanation is that you were wrong.

you just stated:

" Power = 0.5 * L* (Imax - Imin) ^2 /ton "

yet previously you stated that:

"Energy entering the primary winding during each cycle of the core is from the definition of inductance E= 1/2* L * i2 or more like 1/2 *L* (i peak - i min)^2 "

We all agree that Power = delta_E*F

but *YOU* said delta_E = 0.5*L*(Ipeak - Imin)^2

ergo P = 0.5*L*(Ipeak - Imin)^2*F

yet above you just wrote Power = 0.5 * L* (Imax - Imin) ^2 /ton

where ton = D*T = D/F

ergo Power = 0.5 * L* (Imax - Imin) ^2 *F/D

one of these things is not like the other.

oops, you just contradicted yourself!

they cant both be right.

this is semantics, but we were talking about changing the load current....

there is an F missing: P = 0.5*L*(Ipeak^2 - Imin^2)*F

CCM or DCM

?!

I think you made a mistake and dont want to admit it.

well, its appropriate, seeing as it was a losing argument (and, of course, mathematically incorrect).

OTOH it was amusing to derive a complete set of CCM equations OTTOMH.

plus, of course, the quip re. power = average voltage * average current was PIMP funny :)

Cheers Terry

Terry Given ha scritto:

Root mean square is a statistical mean. Mean is an average. His reasoning on this point is intact implying that Vaverage or Vmean, I averge and P average are not necessarily zero.

RMS are by definition statiscal averages

Cheers

Charles

Terry that woman or man is right there.

He calculated the energy expenditure in t= 0 to t= t on as L* (Imax - Imin) ^2 /(2*ton)

His energy expenditure then in 1 second which defines Power is L* (Imax - Imin) ^2 /(2*T) = 0.5 * L* (Imax - Imin) ^2 *f

I think he doesn't want to be very clear so as to confuse so that he comes back and say you are wrong.

Best regards

Jonas

look at the units of this equation. L*I^2 = J, L*I^2/t = W

I havent checked, but I think his latest version of the power equation, P = 0.5*L*(Ipeak - Imin)^2/ton is actually the right answer.

IOW P = 0.5*L*(Ipeak - Imin)^2/ton = 0.5*L*(Ipeak^2 - Imin^2)/T

he just defined the energy wrong.

this is still wrong.

its really very, very simple:

when the switch turns on, I = Imin

inductor energy = 0.5*L*Imin^2

when the switch turns off, I = Ipeak

inductor energy = 0.5*L*Ipeak^2

thus the change in energy in the inductor is:

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

with constant input voltage, output voltage and load, Imin and Ipeak are the same for every cycle. the total power is then P = delta_E*F

yes, I think so. or he genuinely misunderstands it.

Cheers Terry

yes, thats true.

the term In mathematics, an average or central tendency of a set (list) of data refers to a measure of the "middle" of the data set. There are many different descriptive statistics that can be chosen as a measurement of the central tendency. The most common method, and the one generally referred to simply as the average, is the arithmetic mean.

one could rightly say that:

average power = RMS voltage * RMS current

and:

instantaneous power = voltage * current

but LJ's statement "Power = average voltage * average current" is just plain wrong, given that an unqualified "average" refers to the arithmetic mean

indeed. which are *NOT* the same as averages. turns out there is a reason we have all those different words

Cheers Terry

Terry Given ha scritto:

Sorry Terry but you are pla In mathematics, an average or central tendency of a set (list) of data

refers to a measure of the "middle" of the data set. There are ---many

different descriptive statistics--- that can be chosen as a measurement of the central tendency. The most common method, and the one generally referred to simply as the average, is the arithmetic mean.

There are ---many different descriptive statistics--- that can be chosen as a measurement of the central tendency.

The rms is one of those descriptive statistics.

Terry I'm starting to think you are being too sure of youself on too many things.

Cheers

Charles