Efficiency of power supply circuit (rectifier + filter section with load)

Hello All,

I am calculating an efficiency of full wave bridge rectifier after step-down transformer with a filter capacitor used.

I am looking a circuit as simple as this:

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There, I understand that Pin = Vr.m.s (secondary) x Irms (secondary), and P out = VLoad x Iload.

Efficiency is a ratio of Pout to Pin.

It is easy to get Vrms, but as the current wave from is not sinusoid, I don't know hot to calculate Irms.

Can anyone please suggest me an approach?

Regards

Reply to
Myauk
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I've also been looking for some of the derivations of the "rules of thumb" for output voltage, output current, efficiency of rectification of various rectifier topologies and have discovered it's not so easy to calculate...:) The current in the rectifier is going to be governed by a differential equation, namely I = Ec/Rl + C*dEc/dt, where ib is the rectifier current, Rl is the load resistor, Ec is the voltage across the capacitor, and C is the filter capacitor. A solution to that differential equation is ib = Ep*sin(wt)/Rl + wC*Ep*cos(wt). You could find the RMS of that but it would involve an integration probably best left to a computer program. The DC voltage on the capacitor is going to be the peak voltage minus (ripple/2). Finding the ripple voltage is hard though because you have to calculate the cut-out time of the rectifier - the ripple voltage is given by Ep*sin(wt) before the cutout time and the above differential equation with ib set to 0, it's hard because there's a transcendental equation involved in solving for the cut-out time.

I remember to simplify things one of my textbooks makes the assumption that the conduction angle is a small part of the cycle and the capacitor discharges into the load linearly. That gives the ripple voltage as Idc/2fC and the Dc output voltage as Edc = Ep - Idc/4fC.

Reply to
Bitrex

I meant to say "the differential equation above with ib set to 0 after the cut-out time when the recifier is not conducting"

Reply to
Bitrex

"Myauk"

** Fraid that is 100% wrong.

The power input to the rectifier is simply the DC power output PLUS the power lost in the diodes of the bridge.

No easy way exists to calculate the latter accurately as it depends on the characteristics of the transformer - mostly primary and secondary resistances.

..... Phil

Reply to
Phil Allison

To make the problem tractable to any kind of first order numerical analysis I think one has to assume a 100% efficient transformer and diodes with no diode drop. In that case the Vrms of the secondary times Irms of the secondary would be the input power (since the transformer for all intents and purposes doesn't exist). If one considers the capacitor ripple current to be part of the load current then in this simplified case as well the input power would also equal the DC power output plus the power lost in the didoes of the bridge - well yuh, all the other sources of loss have been accounted for and since there's loss inherent in full-wave rectification it has to be taking place in the rectifiers.

Reply to
Bitrex

"Bitrex is off his Face "

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** In that case, efficiency is automatically 100%.

The OP asked only about the efficiency of a rectifier and cap filter circuit with resistive load - where the ONLY loss of energy is in the diodes.

** You need to look up the term " VA" and discover for yourself that Power is not the same as VA.
** This dude likes to think out loud while he is writing his mindless bollocks on usenet.

How very quaint.

..... Phil

Reply to
Phil Allison

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It's not, because the Irms thing is a horrible, non-sinusoidal waveform. And you'll have losses in the diodes and the filter caps.

If (IF!) you can fully characterize the transformer and capacitor parasitics, and have a good diode model, you can Spice it to get close to actual performance. If not, test it with real parts.

When I buy custom transformers, I let the transformer company do the work!

John

Reply to
John Larkin

OK...

In this case it's easier just to directly integrate V*I over a period and average rather than compliating the matter with RMS voltages and currents, IMO. The problem is that, as you're aware, the current is quite messy, since it comes in bursts to "top off" the filter cap every cycle and the turn-on/off points are a function of not only the real power delivered to the load but also the imaginary (reactive) power sloshing around... and of course also the output voltage relative to the diode drop voltage. Hence, while this is certainly a good mathematical exercise, in practice most of us "cheat" and either use measurements on the actual circuit, SPICE or rough approximations.

National Semicondcutor's Audio/Radio Handbook (from 1980) has a nice appendix on power supply design which summarizes many of these approximations and has various graphs for helping to determine reasonable capacitor sizes for desired amounts of ripple; it's well worth reading. They suggest that your transformer needs to be able to handle ~25% more VA than your intended DC output and mention that in nearly all cases the total losses in the an unregulated linear power supply will be dominated by transformer loss (at least if you're trying to keep the price reasonable, I suppose). If you then follow such a linear power supply with a linear regulator, often the total efficiency is only around 50%-75% after you consider the margin you need for low AC supply input and (particularly for low voltage outputs, e.g, 5V) the diode drops.

---Joel

Reply to
Joel Koltner

It can't be 100%, even in the impossible hypothetical case with a perfect transformer and rectifiers that have zero resistance. A full wave rectifier has a theoretical maximum efficiency which comes from the fact that you're converting the input sine wave into a rectified wave that has non-zero higher order Fourier components after the DC term. When that DC term is all that is being applied to the load then the other terms are lost. The resistance of the transformer or the rectifiers doesn't even enter into the equation.

Reply to
Bitrex

There are plenty of data sheets application notes explainning how to calculate or measure efficency.

Here is one. I suggest you start with "Techniques for Measuring Efficiency"

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Reply to
Hammy

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No, the transformer needs to have a VA rating of about 1.5 times the output watts.

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Reply to
MooseFET

I was more pleased to discover this:

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The author claims that a full closed-form solution of the voltages and currents in a capacitor input filter may have never been computed before.

Reply to
Bitrex

Looking at the results, this might be because doing so appears to be more of an exercise in math/algebraic manipulation than anything particularly useful, especially in that he ignored capacitor ESR and a proper transformer model? :-)

Although I suppose I mean "useful" from the perspective of someone doing design with paper and pencil; perhaps there might be some good opportunities to speed up simulators given the analytical results. So kudos to him for a doing the work -- apriori it's never 100% certain that something useful won't emerge.

Reply to
Joel Koltner

(secondary),

the=20

the=20

=20

=20

Still NO. You need the integral (summation) of the (point by point) = vector=20 product of V(sec) and I(sec). There _IS_ no other way.

Reply to
JosephKK

If there are no losses in the transformer, diodes, or caps, output power must equal input power. Conservation of Energy.

The "theoretical maximum efficiency" of an ideal bridge rectifier is

100%.

John

Reply to
John Larkin

Or, in the frequency domain, you can compute the harmonics of the output current, In, and compute PL=sum(abs(In)²)*RL, or something similar with the output voltage harmonics.

Pere

Reply to
oopere

y),

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> >>> To make the problem tractable to any kind of first order numerical
 Click to see the full signature
Reply to
MooseFET

(secondary),

the=20

on the=20

times=20

=20

=20

all=20

loss=20

vector=20

=20

with=20

OK, you could add the Fourier transforms to the math to do and do it in = the=20 frequency domain, but is adds a lot of work for no better answer.

Reply to
JosephKK

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