Maximizing Efficiency: Back EMF and Transformer Behavior in an Ideal Circuit

[Crazymechanic]

Hello , I have a question , assuming ideal devices here without losses for the sake of simplicity.
Imagine I have a circuit like the one attached.
Now if I would manage to flip either one of the switches fast enough that the switch is cut off right when the magnetic saturation of the transformer core reaches it's highest point , would the capacitors (either one of them ) get charged?

I ask this because as long as I know theoretically assuming no real life losses the back emf induced for a given turn ratio and core material and size transformer should 100% oppose the incoming current so if the current would be turned off when the mag field reaches it's highest point in every cycle (no secondary winding load condition) then shouldn't the transformer impose no load on whatever supplies the primary winding ? And also should the cap's stay uncharged as the field reaching it's maximum and then falling back forming a current in the opposite direction and then the same again.

 

[NascentOxygen]

The best model when applying DC to a transformer with open circuit secondary is, I think, an inductance.

However, if you want to pretend it really is an ideal transformer, then the primary side looks like being an open circuit since di/dt in the secondary is 0.

 

[Crazymechanic]

well you can imagine the transformer with both load and no load on the secondary only with no load i can better express the meaning of my question.As with load the situation changes a bit.

Well the point here is that if you would do it slowly enough , flippin each of the switches closed and open it would make each of the capacitors to charge up , while they would charge , a current would rush through the primary at each end towards the capacitor, bigger at start and getting less and less when the capacitor gets charged , well you know this.
So next time when I would close the switch nothing or almoust nothing would happen as the capacitors would be already charged and no more current would go to charge them.

So my question is what happens if i close each of the switch (one switch at a time) fast enough so that each of the switches cuts off right before or at the peak of the induced magnetic field in each cycle , would then the caps be able to charge?

This kinda puzzles me because the applied potential to the primary and through the closed switch is almoust instantenous (speed of electric field) but so is the back EMF ,

so does the back EMF completely opposes the inrushing current in an ideal transformer on each cycle until the core saturation peak (for a given transformer)?

 

[Crazymechanic]

okay maybe my long and sometimes not specified explaining of the problem is why i am getting no responses again. So i'll try it simpler.

Can the back EMF in any case both theoretically and practically, in a transformer , inductor or electric motor can be so strong as to fully oppose the incoming current , so that the current flow would be stopped for a while (depending on each case specifics) until the field or maybe at the very moment, when the magnetic field that opposes it, reaches it's highest value?

 

[Drakkith]

Back EMF is generated by a changing magnetic field. Right before the field reaches its highest value the change is very very small, so very little back EMF is generated. In contrast, when a voltage is first applied the change in the magnetic field of an inductor is greatest, and the back EMF fully opposes the EMF for an "instant", decaying gradually over time as the magnetic field builds up.

However, it's the reverse when you already have a current flowing through an inductor. The instant you cut off the applied voltage the current wants to stop flowing and the change in the magnetic field is the greatest, applying an EMF and causing the current to continue to flow for a while and gradually decaying over time just like the previous example.

I don't know what would happen in a transformer though, as it's a little more complicated than a single inductor.

 

[Crazymechanic]

Well that's what I'm thinking, if you managed to apply a very very short pulse when the EMF is the greatest at the very start , also when capacitor has the highest inrush current also at the very start , would then the current get to the capacitor , as the capacitor would demand a very high inrush current being empty but the transformer would block that because of that sudden change ... ?

 

[Drakkith]

Well that's what I'm thinking, if you managed to apply a very very short pulse when the EMF is the greatest at the very start , also when capacitor has the highest inrush current also at the very start , would then the current get to the capacitor , as the capacitor would demand a very high inrush current being empty but the transformer would block that because of that sudden change ... ?

Let's simplify this to a circuit with just a single capacitor and inductor in series. When you first apply a voltage, the capacitor acts like a short but the inductor acts like an open. Since they are in series no current initially flows through the circuit thanks to the back EMF from the inductor. However, as the rate of change of the magnetic field decreases, the current increases until the limiting device starts to be the capacitor instead.

Make sense?

 

[Crazymechanic]

yes it does , that is what I thought , just wanted to ,make sure.
Yes as the EMF decreases the current starts to flow more and more , so to get the same effect you have to reverse the current flow and it will start over once again.
Although the thing I'm curious here is if we cut the current fast enough while the back EMF hasn't decreases much and we do this in each cycle (constantly reversing polarity) what would the effect be on the capacitor , would it charge up, stay empty or charge up slowly ?

The situation looks kinda similar to a hungry man and a guy offering him a pie , although every time the man wants to get to that pie somebody pulls him down on the ground (back EMF) and he has to get up slowly , while he's doing that the pie is already behind him. :D

 

[Drakkith]

The cap would never charge more than a minuscule amount unless it has a tiny amount of capacitance.

 

[Crazymechanic]

now the " heavier" the load will be on the secondary the less back EMF will be produced ?

 

[Drakkith]

now the " heavier" the load will be on the secondary the less back EMF will be produced ?

That I don't know. I'm not too familiar with transformers.

 

[Crazymechanic]

thanks so far drakkith :) Hope to see someone else commenting here too.

 

[nsaspook]

now the " heavier" the load will be on the secondary the less back EMF will be produced ?

Yes, with no load you have maximum flux density in the core and the CEMF almost cancels out the applied voltage. As we draw real current from the secondary. copper losses decrease the total flux density and the CEMF decreases as the electrical impedance turns more resistive as power in transferred from the flux instead of just stored while idle.

I think the term here would be counter EMF, I would normally use 'back' EMF when talking about generator/motor circuits where there is physical motion. It's not right or wrong, just the way I was taught.

http://www.tpub.com/neets/book2/2a.htm
http://www.allaboutcircuits.com/vol_2/chpt_9/1.html

 

[cabraham]

Yes, with no load you have maximum flux density in the core and the CEMF almost cancels out the applied voltage. As we draw real current from the secondary. copper losses decrease the total flux density and the CEMF decreases as the electrical impedance turns more resistive as power in transferred from the flux instead of just stored while idle.

I think the term here would be counter EMF, I would normally use 'back' EMF when talking about generator/motor circuits where there is physical motion. It's not right or wrong, just the way I was taught.

http://www.tpub.com/neets/book2/2a.htm
http://www.allaboutcircuits.com/vol_2/chpt_9/1.html

The IR drop you discuss is a small part of it. When secondary current is drawn, the flux due to this load current cancels the original flux. Since primary is connected across constant voltage source, primary current increases since flux and emf decreased from load current cancelling original flux. The increased primary current almost restores original flux, small difference due to IR drop.

 

[Crazymechanic]

so from that point of view if the current restores the flux (assuming the source is capable of doing this) then the back emf is back again to it's or close to it's original strength so in overall nothing much changes can we say ?

 

[cabraham]

so from that point of view if the current restores the flux (assuming the source is capable of doing this) then the back emf is back again to it's or close to it's original strength so in overall nothing much changes can we say ?

Yes I would agree. The cemf (counter-emf) is almost restored. The slight difference is due to IR drop you described. The constant voltage source at the primary provides emf which sets the flux. Time changing flux generates cemf and magnetizing current is small because difference between emf and cmf is small. Whn load is connected at sec, sec current produces counter-counter-emf, ccemf which cancels cemf and pri current increases until equilibrium. Pri current produces cccemf and flux is restored except for a few pct due to losses.

 

[Crazymechanic]

if the primary voltage wouldn't restore the back emf then basically the primary current could go up even though the secondary load should keep it down , and the transformer wouldn't be coupled anymore , so the back emf has to be there no matter what , I guess.