Question about "Inductors and Calculus"

Thread Starter

p75213

Joined May 24, 2011
70
I've been going through some of the electronics education on this site. In chapter 15 of Direct Current there is a heading "Inductors and Calculus" - https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and-calculus/
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? If the switch was designed to open at the point where current was at its maximum you would get maximum voltage across the capacitor. I done a bit of googling and found that the formula for energy of a capacitor is .5CV^2.
I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.
 

DickCappels

Joined Aug 21, 2008
5,952
Accepting the idea that the neon bulb is only there to illustrate that with high di/dt you get high voltage, you can replace the bulb with a capacitor to collect the energy, but what happens to the voltage across the capacitor once all of the energy that was stored in the inductor has been transferred?
 

Thread Starter

p75213

Joined May 24, 2011
70
Some extra circuitry would be required to siphon of the energy. Some back to take the place of the battery and some to drive the load. I have no experience in building electronics. However I believe diodes allow current to travel in one direction only. That should stop the capacitor from emptying back into the inductor.
 

crutschow

Joined Mar 14, 2008
23,544
You realize, of course, that the energy you capture can be no more than the energy required to generate the current in the inductor?
 

Thread Starter

p75213

Joined May 24, 2011
70
Would you want it to resonate? How would you invoke the switching within the tank circuit? The sudden switching off of the current is how you get the high voltage.
 

Thread Starter

p75213

Joined May 24, 2011
70
You realize, of course, that the energy you capture can be no more than the energy required to generate the current in the inductor?
That makes sense, But how does that fit with the formula for energy in a capacitor -> w = .5CV^2? Because the formula is telling me the higher the voltage the greater the energy.
 

Tonyr1084

Joined Sep 24, 2015
3,680
The higher the voltage the greater the energy for shorter the time. Over all, the energy out can not exceed the energy in just like Crutschow said. You can turn 1.5 volts into 1500 volts, where the 1.5 volts may be a steady constant source but the 1500 volts may be 0.001 of a second (If my math is correct - I'm good at screwing up numbers). Where the 1.5 volts (assume one amp) is constant, the 1500 volts would likely contain 1 milli-amp. Same amount of power (in a perfect world).

Bottom line, nobody has figured out how to make 1 + 1 = 2.001 or more. And much greater minds have tried.
 

DickCappels

Joined Aug 21, 2008
5,952
In post #4 dl324 tried to steer you in the same direction I thought this thread would go -toward discussing resonance.

Some extra circuitry would be required to siphon of the energy. Some back to take the place of the battery and some to drive the load. I have no experience in building electronics. However I believe diodes allow current to travel in one direction only. That should stop the capacitor from emptying back into the inductor.
I hope you realized that you just invented the boost or flyback power supply!
https://en.wikipedia.org/wiki/Flyback_converter
You have the stuff of an inventor. There is a large difference between those who copy designs and debug them and those who can jump to a new idea on their own.
 

Thread Starter

p75213

Joined May 24, 2011
70
The capacitor current formula is i = c dv/dt. On that basis there would be plenty of current as dv is large and dt is small.
 

ebeowulf17

Joined Aug 12, 2014
2,946
The capacitor current formula is i = c dv/dt. On that basis there would be plenty of current as dv is large and dt is small.
Although if your goal is boosted voltage, then the rapid drop in voltage in your "dv is large and dt is small" scenario means you better be switching really, really fast or else your boosted voltage won't be very boosted anymore!
 

Thread Starter

p75213

Joined May 24, 2011
70
In post #4 dl324 tried to steer you in the same direction I thought this thread would go -toward discussing resonance.



I hope you realized that you just invented the boost or flyback power supply!
https://en.wikipedia.org/wiki/Flyback_converter
You have the stuff of an inventor. There is a large difference between those who copy designs and debug them and those who can jump to a new idea on their own.
Flattery will get you everywhere - ha ha. I'm not so keen on reinventing the wheel though.
 

MrAl

Joined Jun 17, 2014
6,649
I've been going through some of the electronics education on this site. In chapter 15 of Direct Current there is a heading "Inductors and Calculus" - https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-and-calculus/
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? If the switch was designed to open at the point where current was at its maximum you would get maximum voltage across the capacitor. I done a bit of googling and found that the formula for energy of a capacitor is .5CV^2.
I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.
Hi,

As others have said, the energy stored in the cap can not be more than what you put into the inductor. This is a matter of energy and energy transfer and not about voltage or current alone. Remember that voltage alone does not represent any kind of energy level, it takes both voltage and current to have anything called energy.
What does this mean. It means that in order to charge the inductor we would have had to pump it up with energy which would take both current and voltage. There would also be some resistance which would dissipate power which would take some of the energy and convert it into heat which would be considered lost from the system.

This has actually been done, and it actually is being done right now as we speak right in our own computer power supplies. Energy is constantly being forced into an inductor and the inductor is discharging into a capacitor in three or more circuits inside the power supply. This is where we get 12v, 5v, 3.3v, and -12 volts to run the computer.
A standard buck circuit does this 10 thousand or more times per secoind. Many computer power supplies do it 20 thousand times per second.

With each cycle though there are various losses that come into play. The simplest is resistance. The current has to flow through a resistance for both the charge and discharge of BOTH the inductor AND the capacitor, because for one thing there are no ideal elements in normal circuitry. The resistor(s) dissipate(s) power P=i^2*R so that gets lost as heat.

Now if we could somehow create an ideal inductor and ideal capacitor, we might be able to transfer the energy back and forth. The problem then becomes, if we try to take any of that energy to use it for anything useful, we loose the energy that way and so it is gone again.

I think i understand your reason for asking this though. If you pump an ideal inductor with current there's no power there because it's only current and no voltage. IF you then transfer that energy to a capacitor the capacitor then has voltage that can be used for something useful like running a light bulb.
However, there's no way to pump an inductor with current without also having some voltage there to overcome the resistance because there's also no ideal current source and as the inductor charges up there is also a buildup of voltage v=L*di/dt.

It's an interesting question though, because if we could pump an inductor with pure current then when discharged into a capacitor the current would cause a buildup of voltage which then can be used with a resistance which constitutes energy. Since there was no voltage when charging the inductor there was no energy, but then what about the behavior of the inductor itself:
v=L*di/dt

this tells us that there is even energy as the inductor charges because the longer we apply current the more the voltage in the inductor builds up too, and so this will require the current source to possess a higher and higher voltage as time progresses, and so there we see the energy expenditure from the current source.

So the short answer is that the current source expends energy into the inductor and that energy has to come from somewhere to begin with, and since one of the first principles of physics is conservation of energy whatever energy we put into the system we get our of the system at some point, but there's never an increase in that energy that we got from seemingly nowhere. IF we seem to get energy from nowhere, we just missed where it actually came from.

When i was in grammar school and first heard about a photo voltaic cell, i told the professor why cant we just shine a light bulb on the cell and then have the cell power the light bulb. It would run forever. He was quick to knock it down and made me find the reason it would not work (ha ha).
 
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Thread Starter

p75213

Joined May 24, 2011
70
I've been giving this some thought. It all comes down to the formula for energy in a capacitor: w = .5CV^2. If we have 2 capacitors both of the same capacitance. One we charge from a 6 volt voltage source at 1 amp. The other one we charge from a 24 volt voltage source at 0.1 amps. In the first case the power is 6 watts and in the second case 2.4 watts. However the second capacitor would have more energy -> 0.5C24^2 joules as opposed to 0.5C6^2 joules. Power is defined as the rate of using energy. Therefore the second one would also have more power provided the energy was used at the same rate in both cases.
 
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ebeowulf17

Joined Aug 12, 2014
2,946
I've been giving this some thought. It all comes down to the formula for energy in a capacitor: w = .5CV^2. If we have 2 capacitors both of the same capacitance. One we charge from a 6 volt voltage source at 1 amp. The other one we charge from a 24 volt voltage source at 0.1 amps. In the first case the power is 6 watts and in the second case 2.4 watts. However the second capacitor would have more energy -> 0.5C24^2 joules as opposed to 0.5C6^2 joules. Power is defined as the rate of using energy. Therefore the second one would also have more power provided the energy was used at the same rate in both cases.
I think you're ignoring the time factor involved. It will take much longer to reach 24V at 0.1A charging rate than it will to reach 6V at 1A charging rate. Comparing the power you use to charge them is meaningless if you ignore how long you have to provide that power in order to reach the same state of charge.
 

Thread Starter

p75213

Joined May 24, 2011
70
The thing I was trying to point out is that the formula for energy in a capacitor is not concerned with amps - only volts. But your right - I hadn't thought about the time difference to charge the capacitors. Thanks for pointing that out.
 

Thread Starter

p75213

Joined May 24, 2011
70
I think you're ignoring the time factor involved. It will take much longer to reach 24V at 0.1A charging rate than it will to reach 6V at 1A charging rate. Comparing the power you use to charge them is meaningless if you ignore how long you have to provide that power in order to reach the same state of charge.
I'm flying by the seat of my pants here. Just done some googling and found out that the formula for time taken to charge a capacitor is: t = 5RC.
R = resistance
C = capacitance
On that basis it should take the same amount of time in both cases?
 

ebeowulf17

Joined Aug 12, 2014
2,946
I'm certainly no expert, so I'm guessing a little here too, but I think the distinction is your current limit. I think that formula assumes you apply the target voltage through the resistance R into the capacitance C and it takes 5 RC time constants to reach a full charge.

In you earlier proposed scenario, you're charging the cap with a fixed current until it reaches the target voltage. The choice of lower current when charging to a higher voltage means you're using a higher resistance to limit the current into the capacitor. Therefore, once you adjust the relative values of R for the two current-based charging schemes, the formula will predict larger charge time for lower current (higher resistance.)
 

Thread Starter

p75213

Joined May 24, 2011
70
I'm certainly no expert, so I'm guessing a little here too, but I think the distinction is your current limit. I think that formula assumes you apply the target voltage through the resistance R into the capacitance C and it takes 5 RC time constants to reach a full charge.

In you earlier proposed scenario, you're charging the cap with a fixed current until it reaches the target voltage. The choice of lower current when charging to a higher voltage means you're using a higher resistance to limit the current into the capacitor. Therefore, once you adjust the relative values of R for the two current-based charging schemes, the formula will predict larger charge time for lower current (higher resistance.)
Your right. The example I gave was probably not the best. I had some misgivings about writing it at the time. However I think it's important to remember that a capacitor can store a lot of energy using a high voltage low current. In my proposed circuit the time factor would come into play. Particularly at the instant when the circuit is open. V is at its maximum and I is at a minimum -> R = V/I.
 

nsaspook

Joined Aug 27, 2009
6,510
If a coil has been charged with 5 volts and has reached a steady state with 0 volts and 50mA. The power source is switched of and the voltage goes to x volts (greater than 5). What does the current go to? Is it greater than 50mA?
Your question is not very clear but the thing to think about is energy as others have said. Electromagnetic Energy is stored in the 'charged' magnetic field of the coil. How this stored energy from a collapsing magnetic field is dissipated by a possible circuit element depends on the properties of that circuit element and the circuit in whole. If the element resistance is low the current will be higher, if the element resistance is higher the current will be lower but the same amount of energy will ultimately be dissipated from the magnetic field of the coil.
 
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