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Flyback + Cap charging time, help

Nikša

May 7, 2018
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Can you help me figure out how long it takes for a let's say 100W flyback outputting full wave rectified 5000V pulsed DC at 20mA to fully charge a 1uF cap?

For simplicity of things we can assume a pure DC 5kV source that can output 20mA max.

I am using 20mA cause that is approximate output current of the flyback according to my calculation.

We can't just use a RC constant to calculate the charging time here. If we did we would multiply capacitance and resistance of the secondary as so 0.000001F * 100Ohm =0.0001s.

According to time constant cap would fully charge in 5T, that is, 0.0005s, but this assumes current of 50A is available!

Yet, only few dozen milliamps is available. So i roughly estimate it would take about 1 second for it to fully charge.

Am i making a mistake somewhere, i hope i do, cause i would love it to charge faster.

Maybe some of you can simulate this.
 

Nikša

May 7, 2018
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I asked this at stackexchange, they told me to use I = C dv/dt (current through a cap formula) rearranged to solve for t so t = C * V/I. So t = 0.000001F * 5000V/0.02A = 0.25s.
 

Nikša

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Yes, i'm aware of that, i used a steady current to simplify the calculation. I did not write it here but i did write it on stackexchage, that high duty cycle is to be used, so figures should not be far.

Of course not too high duty cycle cause we sure don't want to feed pure DC to the primary. Maybe 80% at say 50KHz.
 
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Nikša

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I got a question about flybacks. We all know the principle of the flyback, exactly the same principle used in bucking converters and ignition coils, feed, break, collapse, repeat.. the thing is i'd like to convey max power from the supply and i don't like the fact that power is conveyed only half of the time. So i would like flyback without internal diodes, to use the AC output (and full wave rectify it), but ferrite transformers for sale are either too small or too big. I know i could tear FB apart and take the diodes out but that gets messy and winding can get damaged. I asked them on alibaba if they could sell just the core and secondary winding, we'll see.

I know i can also use bigger PS and two flybacks in parallel in normal flyback mode (with internal diode) to convey the same power, that sure is an option.

But the real question is if using it as normal transformer (pulsed DC or AC input and AC output) vs flyback mode (pulsed DC input, pulsed DC output) conveys more energy at all.

Consider first the normal transformer behaviour. Take 220V tranformer, it can be plugged into the wall 24/7, primary is a short circuit, yet it dissipates very little energy, of course this is due to inductance of the primary making the V-I 90° out of phase. Once load is put on the secondary to the power source driving the primary it appears as if the core suddenly got more lossy and this brings V-I more or less in phase, making the circuit resistive, dissipating energy.

Now for flyback, current is fed into the primary and is then cut off, but if it's let to flow even slightest bit beyond what's needed to establish max current, that is already an energy waste, so frequency/duty cycle must be optimal. Then, as source is cut off, field collapses and this induces a voltage spike in the secondary. This voltage takes a shape of a falling ramp, more less a sawtooth. Great article about this Avian’s Blog: On the output voltage of a real flyback converter (tablix.org)

This field collapse sure is not constant, constant change of flux would produce a square, not a sawtooth. Clearly, flux first falls off quickly and then tapers off. But this is beside the point, just my little observation.

And so the energy that was stored in the primary core is transferred to the secondary.

But still question is, if we use it as normal transformer with secondary conducting during both halfcycles, current will surely not rise to same height in the primary as it would were the secondary open during the first halfcycle. Overall, it is not impossible that flyback stores more energy in the core during first halfcyle while secondary is open, then it does for both halfcycles combined in normal transformer operation mode.

But i am not sure, i am speculating and i want to know to chose which way to go with this, goal is like i said, to convery max power, hopefully avoiding doubling power supply and flybacks. So question is, what gives more power, flyback operated in flyback mode with half rectified output or normal transformer mode with full rectified output.

I would go with the latter with about 90% certainty but would like your opinion.
 

crutschow

May 7, 2021
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A normal AC transformer uses the magnetic core to transfer the energy from primary to secondary with the turns-ratio determining the primary voltage to the secondary voltage..
It never stores any significant energy as the inductance is just to determine the value of the small magnetizing current.

A flyback stores significant energy in the transformer inductance when the primary conducts current, and transfers that energy to the secondary when the primary is open-circuited.
The secondary voltage step-up is determined by the primary peak voltage when the primary is open-circuited times the turns ratio.
This primary peak voltage can be much higher the the primary applied voltage.
The transformer is basically a multi-winding inductor.

So the two modes of operation are entirely different and can't be readily compared as you are attempting.
So i would like flyback without internal diodes, to use the AC output (and full wave rectify it),
That makes little sense to me.
You cant have a flyback without a diode to rectify the output.
There is no full-wave output from a flyback.
 

Nikša

May 7, 2018
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A normal AC transformer uses the magnetic core to transfer the energy from primary to secondary with the turns-ratio determining the primary voltage to the secondary voltage..

That is assumed, basics.

It never stores any significant energy

It does when secondary is open. It stores significant energy and then returns it, this is reactive power meaning voltage and current are 90° out of phase or near. For energy transfer to occur voltage and current must be in phase (except in antennas). Also, even when secondary is loaded, technically speaking, energy is still "stored" in the core and "released" almost at the same time (that is with slight delay cause it takes a short amount of time for magnetizing field to orient the domains etc).

as the inductance is just to determine the value of the small magnetizing current.

Who says magnetizing current is small, it can be any. At lower frequencies greater inductance is needed to convey the same energy, thus "normal" transformers are bulkier.

A flyback stores significant energy in the transformer inductance when the primary conducts current, and transfers that energy to the secondary when the primary is open-circuited.

That is what i said.

The secondary voltage step-up is determined by the primary peak voltage when the primary is open-circuited times the turns ratio.
This primary peak voltage can be much higher the the primary applied voltage.
The transformer is basically a multi-winding inductor.

That's what i said, i just didn't go into obvious details. Of course, like in ignition coil, after 12V is cut off from the primary voltage across primary can jump up to 150V (thus a cap and a resistor in parallel with a transistor) which is then multiplied by the turns ratio to 30-40kV. This is why they say flyback is "not a transformer" (input-output voltage is not directly proportional to turns ratio) but a coupled inductor, which is of course utter nonsense. It is a transformer with delayed, compressed induction.

So the two modes of operation are entirely different and can't be readily compared as you are attempting.

Not true, the only difference is that in normal operation induction happens in phase with the primary current while in flyback mode it is delayed an instant and then released more suddenly to somewhat higher potential. But essentially the same transformer effect.

That makes little sense to me.
You cant have a flyback without a diode to rectify the output.
There is no full-wave output from a flyback.

It is common to refer to these ferrite transfomers as flybacks even when they are not used literally as flybacks. If you care about exact terminology then call it ferrite core high frequency transformer instead.

So all you did with your post was to try to prove how flyback is supposedly something totally different from "normal" transformer, popular malpractice, while they are actually absolutely one and the same, with a slight difference in operation.
 
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crutschow

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So all you did with your post was to try to prove how flyback is supposedly something totally different from "normal" transformer, popular malpractice, while they are actually absolutely one and the same, with a slight difference in operation.
That is not true, but you are are condescending know it all, so I don't see why you posted here. o_O
 

Nikša

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That is not true, but you are are condescending know it all, so I don't see why you posted here. o_O

Ok, you noted some basics, some of them not true ("normal transformer never stores significant energy in the core"..) and then you concluded that they are two totally different things and thus cannot be compared (again, not true). I'm not a condescending know it all, would i be asking for help if i was, i am just expecting better reply than "they cannot be compared". They can and need to be compared, so question remains, which mode of operation transfers energy better, flyback or "normal".
 

crutschow

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Okay, let's try again.
normal transformer never stores significant energy in the core
A normal transformers core energy is the core inductance times 1/2 the magnetizing current squared.
If you want to call that significant, that's fine, but that's not directly related to the much larger energy that can be transferred by the transformer from the primary to the secondary by the load current.
which mode of operation transfers energy better, flyback or "normal"
In the "normal" mode the transformer energy transferred is determined by its maximum current rating due to winding resistance (as long as the primary voltage and frequency does not cause core saturation, of course).

In the flyback mode the energy transferred is stored in the core inductance when the primary is active, and transferred to the secondary when the primary is open-circuited.
The maximum transferred energy is thus limited by the core inductance (again the core should be kept below saturation at the maximum primary current).
So typically, for the same core size, a flyback can transfer less energy than a normal-mode transformer.

A flyback is thus typically used when a simple circuit to generate a high voltage output is needed, since the primary voltage, when the energy is transferred to the secondary, as limited the by the primary switch rating, can be much higher than the primary source voltage.
This reduces the turns-ratio required to get a specific output voltage.
 

Nikša

May 7, 2018
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Okay, let's try again.

A normal transformers core energy is the core inductance times 1/2 the magnetizing current squared.

That formula is energy stored in an inductor, inductor is coil + the core. Many factors affect inductance of the coil, number of turns, solenoid length, coil radius and finally permeability of the core material. Core inductance is one of many factors affecting the coil's inductance.

If you want to call that significant, that's fine, but that's not directly related to the much larger energy that can be transferred by the transformer from the primary to the secondary by the load current.

You seem to be under a strange conviction that in "normal" transformer most of energy is not transfered via the core. And how did you imagine it is transfered, by magic?

Core contains all the flux of the coil, all the energy is transfered via the core. Like i said, that is the reason low frequency transformers are bulkier, they need bigger inductance (more copper and bigger core) to transfer the same energy as smaller, high frequency ones.

In the "normal" mode the transformer energy transferred is determined by its maximum current rating due to winding resistance (as long as the primary voltage and frequency does not cause core saturation, of course).

Current is important in both cases but energy transfer capacity of a transformer comes down to inductance * frequency. Again, at higher frequency, smaller inductance conveys same energy as bigger inductance at lower frequency.

Of course saturation must not happen, both normal and flyback are predicted for their corresponding frequencies, 50Hz-1Khz and 15-150KHz. Inductive reactance XL= 2πfL being taken into account.

In the flyback mode the energy transferred is stored in the core inductance when the primary is active, and transferred to the secondary when the primary is open-circuited.
The maximum transferred energy is thus limited by the core inductance (again the core should be kept below saturation at the maximum primary current).
So typically, for the same core size, a flyback can transfer less energy than a normal-mode transformer.

Core inductance is a determining factor for both normal and flyback mode, if core inductance for the normal mode was not sufficient, you may as well have an air core transformer which would be useless. Likewise, if you operated a flyback at low frequency, due to low inductance it would be useless.

As for the comparison of the flyback vs normal mode, the key point is, like i said before, that in a flyback, source is feeding the primary only one halfcycle. Logically if source is feeding both halfcycles, approximately twice the energy is conveyed.

Again, saturation can hardly happen in normal operation, saturation flux density is known for electrical steel (~1.5T), ferrites (0.2-0.5T) and everything is designed in accordance.

A flyback is thus typically used when a simple circuit to generate a high voltage output is needed, since the primary voltage, when the energy is transferred to the secondary, as limited the by the primary switch rating, can be much higher than the primary source voltage.
This reduces the turns-ratio required to get a specific output voltage.

You are again repeating what i said. In the very start i said operation of the flyback is EXACTLY the same principle used in boost converter and ignition coil, fast collapsing magnetic field steps up source voltage which is further stepped up by the turns ratio.
 

crutschow

May 7, 2021
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Core inductance is one of many factors affecting the coil's inductance.
That makes no sense to me.
Core inductance is core inductance.
You seem to be under a strange conviction that in "normal" transformer most of energy is not transfered via the core. And how did you imagine it is transfered, by magic?
No I'm not under that strange conviction and I don't believe in magic.
.And lets eliminate the sarcastic, snarky comments, OK?

I stated the energy stored in the core is due to the magnetizing current, which is different from the energy transferred by the core. Energy stored and energy transferred by the transformer are two different and unrelated things.
Like i said, that is the reason low frequency transformers are bulkier, they need bigger inductance (more copper and bigger core) to transfer the same energy as smaller, high frequency ones.
It has nothing to do with amount of energy transferred.
The reason that low frequency cores are larger is that they need a higher inductance to keep the magnetizing current below core saturation, since magnetizing current is proportional to inductance and inversely proportional to frequency.
in a flyback, source is feeding the primary only one halfcycle. Logically if source is feeding both halfcycles, approximately twice the energy is conveyed.
You logic is flawed.
A flyback doesn't transfer energy from the input to the output, it stores the energy in the inductance on the first half-cycle and then transfers that energy to the secondary on the last half-cycle.
So how do you propose to store the energy and transfer the energy on the same-half cycle? :confused:
 

Nikša

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That makes no sense to me.
Core inductance is core inductance.

Core inductance is a term that doesn't really make sense. Inductance is a property of conductors. Permeability of the core material is just one of factors affecting the coil's inductance.

No I'm not under that strange conviction and I don't believe in magic.
.And lets eliminate the sarcastic, snarky comments, OK?

Yea you are and you asked for it.

I stated the energy stored in the core is due to the magnetizing current, which is different from the energy transferred by the core. Energy stored and energy transferred by the transformer are two different and unrelated things. It has nothing to do with amount of energy transferred.

It has everything to do with amount of energy transferred. They are both the same phenomena mediated by the same flux in the core, only difference is one acts immediately while another is built up and released like a spring.

Also remember what i wrote about normal trasnformer operation, while secondary is open, primary is like a short circuit, but IV are 90° out of phase thus energy is stored and released from the core... When secondary is loaded this makes the core suddenly appear more lossy which brings IV into phase (Lenz) and energy transfer happens.

I could also get into what magnetic field really is, Magnetic vector potential (aka A field) and other subtler things like two kind of inductions discovered by Henry, difference between flux cutting (Bvl) and flux linking law (dphi/dt), or that Lenz's law is not really a law... but i am keeping things mainstream for your sake. Let's just say there is far more to magnetic fields than usually assumed.

The reason that low frequency cores are larger is that they need a higher inductance to keep the magnetizing current below core saturation, since magnetizing current is proportional to inductance and inversely proportional to frequency.

Higher the frequency of light or any wave for that matter higher it's energy, so it is with oscillating magnetic fields. Basic Maxwell–Faraday's equation, faster the oscillation of the magnetic field, greater the induced electric potential, and thus with higher frequency smaller inductance can be used for the same power. This is the clearest and most valid way of explaining why high frequency transformers are smaller.

You logic is flawed.
A flyback doesn't transfer energy from the input to the output, it stores the energy in the inductance on the first half-cycle and then transfers that energy to the secondary on the last half-cycle.
So how do you propose to store the energy and transfer the energy on the same-half cycle? :confused:

Oh the irony, says you with such flawed logic and misunderstanding. First flyback does ineed transfer energy, what else would it do, it just does it with bit of delay. Also, i never said or implied "how do you propose..", i very clearly said if "flyback" (ferrite transformer with no diodes) is used with AC input and AC output, then power is delivered on both halfcyles and thus the average power is greater.
 

crutschow

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So it's quite clear you can't resist piling on the condescending, snarky comments, while misstating (or misunderstanding) what I said.
And it's obvious your superior knowledge can't be improved (or you are a troll).
So I'm not wasting any more of my time with you, since that's beating a dead horse.
Good luck with your flawed understanding of how transformers work. :p
 

Nikša

May 7, 2018
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Eh. Accusing me of snarky comments while using those emojis all the time, my my.

If you understood even remotely how transformers work you would be appalled by the amount of idiotic statements you made in this thread, to remember just a few.

Energy stored and energy transferred by the transformer are two different and unrelated things. It has nothing to do with amount of energy transferred.

Just this statement deserves a medal of shame. Transformers ability to to transfer and store energy is DIRECTLY proportional to it's inductance, core PERMEABILITY being a major factor determining it.

The maximum transferred energy is thus limited by the core inductance

It is core PERMEABILITY, not core inductance, you clown, and COIL inductance.

These is nothing to be misunderstood in your statements, they are just shamelessly wrong.

Needless to say, i had to explain all the essentials of the transformer operation, all you did was parrot how energy transfer and energy storing are unrelated (#facepalm).

1) It was me who explained exactly how transformer works, why it can be plugged into the wall 24/7 with primary being a short circuit yet it uses no energy, how loading the secondary brings IV in phase etc.

2) It was me who mentioned the key word of this subject, Lenz's law (not really a law but nevermind that) which makes the primary "see" the opposing flux being developed by the secondary which makes the power supply work harder to overcome this reluctance. Etc.

And it's obvious your superior knowledge can't be improved

Hypocrisy is so sweet, isn't it.

Taken all this into account who is wasting time on who beating a dead horse... but don't you worry, enjoy your "correct" understanding of how transformers work. ; )
 

Nikša

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Looking back, i'd like to say there was no need for somewhat heated exchange of words. Let's call it mutual misunderstanding.

To recap on the subject, for "normal" (non flyback) transformer, i'd summarize it like this to cover all the angles.

When transformer is unloaded (secondary open) only magnetizing current flows through the primary which is tiny compared to load current and is always (almost) 90° outta phase with the driving voltage, almost but not 90° due to various losses, namely, eddy currents, hysteresis, magnetostriction and copper losses (P = I²R). Average flux in the core is maximum in no-load state and slightly smaller in the full-load state.

When secondary is loaded, counterflux developed by the secondary demagnetizes the core and this makes the voltage across the primary to drop since this voltage is directly proportional to rate of change of flux, we all know Faraday's (or should i say Henry's) law V=--N*dΦ/dt.

Voltage across an inductor can also be expressed as V = L(di/dt)

And current through an inductor I = (V-E)/Z where V is voltage of the source driving the primary and E is voltage drop across primary's inductance. Clearly, when secondary demagnetizes the core and flux through the primary drops, so does it's inductive reactance Z and voltage across it E, V remaining the same means current must rise and so it does trying to bring the flux back to the original value but it never fully manages to do so, so, as said before, max load flux is slightly less than no load flux.

To the circuit driving the primary, it appears as if a resistor appears in parallel with the inductance of the primary, bigger the load smaller the resistor appears, obviously.

As said above magnetizing current is always almost 90° out of phase with the driving voltage while the load current is always in-phase. At least when the load is purely resistive.

If load is inductive then picture is not so clear, but extending what happens with the resistive load, we can assume this inductive load will also appear as an inductor and resistor in parallel and larger the work done larger the virtual resistor in parallel will appear again bringing IV in phase.

I guess we can extend the last paragraph to capacitive loads too, of course, electric field just like magnetic field can also be used to do work and as you all probably know there are various electrostatic motors, some newer ones of significant power (some even speculate about replacing magnetic ones).

As for flyback, the only difference is, as said before in the thread, induction in the secondary happens with a delay (due to internal diodes blocking the current in one direction) when the primary flux collapses. Ignition coil uses the same principle, store, collapse, get 10x (or more) voltage in the primary and x turns ratio in the secondary.

Talking of ignition coil, the reason for cap across the switch (be it mechanical relay or a MOSFET) is to limit the peak voltage, cap appearing as a short for an instant and lower the resistance in the circuit of the collapsing flux, lower will be the peak, obviously, higher the resistance higher the peak and faster will it burn out.

ignition-coil.jpg
 
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