Reguarding physics and boost converters

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So, I'm a programmer getting into electronics and micro controllers. I'm studying everything myself, with nobody to ask or tutor me in any real availability. I've developed images of how different circuits and components work in my head, but in this particular situation I'd like to verify the validity of my ideas as to how a boost converter works.

In a very basic boost converter, you have an inductor, diode, capacitor, switch of some sort (transistor), voltage source, and some sort of load ala Fig 1 here: http://en.wikipedia.org/wiki/Boo... . I know it works as I've simulated and built a working model, but as for how it works I'm worried I may be making incorrect assumptions.

I have read:
1. Inductors gain higher current charges with higher voltages.
2. Capacitors gain higher voltages with higher currents.
I'm also to understand that:
3. Inductors, once "fully" charged (drawing 0V or close to 0V while maintaining a current) will let out voltage when a load is placed on them/the supply is cut. In the case of a fully charged inductor, suddenly putting a load on it will cause a sudden release of a voltage much higher than the source originally used to feed the inductor.

So, in the case of a boost converter, we charge up the inductor a whole bunch while the "switch" is on. Then when we turn the switch off, the inductor pops up this incredible voltage and current which gets sucked up by the capacitor as it reacts to the change. Is it the higher voltage being shown to the capacitor from the sudden release from the inductor that makes it charge to more than the voltage source (see 3 above), or the current bursting out of the inductor pushing up the voltage in the capacitor (see 2 above) that is giving us this boost in voltage? Or, since current is simply a factor of voltage when employing Ohms law with impedance/resistance, are both factors actually just part of the same thing?

Sorry if I've expressed my question poorly, I'm still working through all this so it's difficult to even know how to ask something like this. If nobody understands I'll try and reword it (and make it shorter?).

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I covered this subject in the build a nixie clock tutorial. Third post down. That Wikipedia article is a bit mathematical - hopefully you'll find my explanation easier to understand. Note, there is more than one kind of boost converter. This kind is properly called a "flyback" converter, so called because it's the way the EHT is generated in a television set, by catching and rectifying the energy in the line output transformer during the "flyback" period between one line and the next.

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peret, that was fantastic! I completely neglected to consider how capacitors have an exponential ramp of impedance as they approach source voltage. So the physics of the collapsing field dictates the current has to go somewhere, and since there is nowhere to go but the capacitor the voltage just naturally ramps up because it has to in order to release the energy resultant of the collapsing field.

That brings up one question for me: If we were to keep the capacitor charged (assuming it was perfect with no loss of charge, and accumulate the same ammount of field in the inductor and release it over and over again, the voltage would rise in smaller and smaller increments? I'm assuming we are sacrificing current to increase the voltage each time, so say we were incrementing our voltage in equal time slices it may go something like 6V-9V-10.5V etc. ?

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The charge on a capacitor is Q=CV and Q=It. On a perfect capacitor, that doesn't (e.g.) blow up when its voltage tolerance is exceeded, the voltage will linearly with a constant current flow.

You see exponential charging because the differential between the voltage on the capacitor and the charging voltage decreases as the capacitor charges. You can't keep adding charge if the capacitor already has a voltage higher than you're trying to charge from; the charge will flow the other way.

However, the instantaneous voltage on an inductor is (theoretically) infinite at the moment the circuit is made or broken. In a boost switch-mode PSU, what you're doing is connecting an inductor to a low-impedance voltage supply and shorting the other end to ground. When you open that short, the voltage at that point will rise to a level limited largely by resistance and capacitance losses, and is connected through a diode to your capacitor. For the time that the inductor voltage is higher than that on the capacitor, the capacitor will be charged; once the voltage decays below the capacitor voltage, the diode blocks the charge from leaving.

In the real world, you take current from the capacitor which reduces its voltage; it's topped up at the next cycle. You adjust either the speed of the cycle (or more commonly, the mark/space ratio) to keep the capacitor working with your desired minumum ripple.

Neil

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Quote:
the voltage would rise in smaller and smaller increments?
This is something I've never thought about. Let's analyze it. We're transferring energy here. The energy in an inductor is 1/2.L.I^2, and the energy in a capacitor is 1/2.C.V^2, so if the energy from the inductor is constant, the incremental rise of energy in the capacitor is also constant. As capacitor voltage V increases, the additional energy must cause smaller and smaller voltage increments to satisfy the square-law relationship. Without doing a full mathematical analysis, I don't believe it's an exponential curve that approaches some limit - I think it's parabolic and will continue rising (slowly) to infinity.

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barnacle: You've made me very glad I continued to ask questions. It almost seems too simple now. The comment "However, the instantaneous voltage on an inductor is (theoretically) infinite at the moment the circuit is made or broken." took a second to rationalize in my head. Physically, I'm assuming the collapsing of the field forces current, and that current can't (easily?) be physically stopped. So when it encounters something, this unstoppable force is going to put as much pressure (voltage) as it needs to get through the weakest accessible obstacle. The perfect capacitor when fully charged to some voltage will resist infinitely all charges up to that voltage, so this unstoppable force just has to push harder to get through, and it does so by raising the voltage. But because there is a defined current from some voltage greater than the current charge voltage of the capacitor should allow charging at the same rate as an infinite voltage would?

I guess I'm trying to ask: is the voltage just innately always infinite to start out with, or does it have the potential to be infinite when faced with an infinite resistance?

EDIT: I went out today without submitting this comment to mill it over. I just got back and hit submit and found peret's comment which implys a non-linear charge rate?

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Thinking about it, if a capacitor were to be like a cylinder, its radius the size of the amount of columbs of charge it could take and the height to which the tube is filled the amount of voltage. We fill it with a constant current from a certain height (voltage) and the cylinder fills up, electrons stacking up onto each other like balls in a ball pit. Once the cylinder is filled to the source height (voltage) no more electrons can be pushed in from our tube feeding into this cylinder. So if we had an infinitely high voltage and a constant rate of flow the balls will just continue to fill up in the cylinder at a linear constant rate as barnacle says. Even if we were just moving the tube feeding this cylinder up as we filled it such that we have all the current free to flow out uninhibited this model would work. That is: not always infinite voltage, but rather a potential to go up as much as needed with no upper limit. Since we have a predefined current, measured in amps, and amps has an inherent amount of columbs of charge associated with it if we charge the inductor to the same amount each time and discharge it fully each time we will always have the same amount of charge being pushed up into the capacitor: a linear rate of rise in voltage when only plotting at the end of each cycle (eg: +3, +3, +3...). Is this a flawed analogy?

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Quote:
Since we have a predefined current, measured in amps, and amps has an inherent amount of columbs of charge associated with it if we charge the inductor to the same amount each time and discharge it fully each time we will always have the same amount of charge being pushed up into the capacitor ... Is this a flawed analogy?

Yes, it's a flawed analogy. You forget the transformer action. 1 amp at 12V will translate to 100mA at 120V or 10mA at 1200V, so the current flow into the capacitor will get less and less the more the voltage has to be stepped up to drive it in. Consider the energy expression. Since the energy in the coil is always the same, the energy delivered to the cap is always the same, but because of the square-law relationship between energy and voltage, successive voltage steps will be smaller to achieve the same energy gain.

For example, suppose you are delivering 1 Joule per cycle to a 1 Farad capacitor. Since energy=1/2.C.V2,

after 1 impulse   1 = 1/2.V^2,  V=root(2) (1.414)  rise = 1.414
after 2 impulses  2 = 1/2.V^2,  V=root(4) (2.000)  rise = 0.707
      3                         V=root(6) (2.449)  rise = 0.449
      4                         V=root(8) (2.828)  rise = 0.379

Like I said, a parabolic function. If you were to work out the charge injection from the current.time, you would get the same curve, because the current falls (by transformer action) as the voltage rises.

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peret: thank you for that. I had originally suspected that's how it would have worked, but when I read the line, "the voltage will linearly with a constant current flow" in barnacles' post above I started getting confused. I came up with that cylinder model in my mind to try and explain to myself how something like that could work, but you've quite clearly pointed out why it was false (and help me confirm my understanding of the physics of voltage/current).

I believe you've also indirectly answered my question about weather or not an inductor supplys instantaneous infinite voltage or simply if it doesn't have an upper limit. In fact the statement "the instantaneous voltage on an inductor is (theoretically) infinite at the moment the circuit is made or broken" is an oversimplification in that there has to be a limit to how slow current can move before the motive force is less than that required to pop them out of the valence bands. So while the upper limit is no doubt extremely immense, there has to be a limit. And because we have measurable current coming off of the inductor each time we blow it out to the rest of the circuit, no more voltage than what is required to allow further charge of the capacitor would be presented each step.

Assuming all I just said was correct I feel I have a concrete understanding of how it all works now. I can't thank you enough peret for your excellent explanations, as they have confirmed many of my suspicions and provided quite a bit of information necessary to solidly connect some of the fragments of electrical physics ideas that were floating around in my head. If you ever write a physics book please tell me as I found your methods of explanation very effective.