# LC resonance with AC and DC

Discussion in 'General Electronics Chat' started by wes, Jun 8, 2012.

1. ### wes Thread Starter Active Member

Aug 24, 2007
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2
I was wondering if it is possible Set up a circuit where you use a AC sine wave at the resonant frequency of the circuit and limit the current from the AC source to maybe the low mA range. Then you use a DC source across the inductor to provide the high current. So the whole idea is to use the resonance of the circuit so as to lower the opposition from the inductor to the DC source current. The DC source would be at the same voltage as the AC source.

The whole reason for this idea is if you wanted to pulse a high inductance coil to a specific current level quickly, you would be met with a huge reactance limiting the current flow in a certain amount time ,ex... 1A-1us on a 100uH coil requires 100V. A more extreme example is 1A-1ns on a 1 H coil requires 1GV( 1,000,000,000 V).

The reason for DC is so the current is always only going in one direction unlike in AC resonance where the current is alternating. That's why you would limit the AC current

The reason for this is I know with resonance you can get a inductor to rise in current and thus the magnetic field to rise much faster than when it is not. The problem though is the current is always going back and forth and so the magnetic field is going north, south, north and so on. But how do you get the benefit of faster rise times but current in only one direction?

If someone knows how then that would be great. I have been wondering about this for a while now.

Last edited: Jun 8, 2012
2. ### WBahn Moderator

Mar 31, 2012
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Assume and ideal inductor and connect it to an ideal DC source. What is the current in the inductor?

3. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
Infinite I think, because an ideal ideal inductor would have no resistance or any parasitic effects. An ideal voltage source as far as I know is one that can supply infinite current. So the current in the inductor would just constantly keep rising to infinity (doesn't really exist, lol but works here). It would not matter what the inductance was, the inductor would eventually reach infinity, it would just take longer for larger inductors to reach it or never reach it, lol.

or am I mistaken and it is zero, if so why.

also what am I supposed to take away from this?

4. ### #12 Expert

Nov 30, 2010
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The DC results are entirely divorced from the AC behavior...unless you get into saturation (discontinuity) or have to consider resistance ohms.

5. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
umm what do mean?

Do you mean the DC is seperate from the AC , regardless if the AC is at the resonate frequency or not?

6. ### #12 Expert

Nov 30, 2010
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How to say that simpler?

The characteristics of an inductor will act just the same when suddenly connected to a DC voltage regardless of what some amount of AC current is doing at the same time, as long as you don't violate the rules about excessive current through a real inductor..give or take the effects of resistance that is always present in a real inductor.

Your AC can be dancing back and forth at any amount of current and voltage and the answers to a DC math about the inductor will not change, as long as you don't overload the real inductor. As long as you stay in theoretical ideal inductors, there is no limit to the AC or DC current.

7. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
but doesn't the dc voltage effect the ac voltage? Since if you have AC source, then it will at certain times be positive say 40 volts and if the DC source is in series with the AC then the net voltage will be a positive 80 volts and -0 volts. or were you talking parallel since I said in my post the DC would be across the inductor, meaning parallel. In which case, I think I see what you mean.

Now that I think of it, I don't think using a a dc source to try and drive the high current would work because like you said,
means the DC would still encounter the reactance of the inductor as if it was just the inductor and the DC and no capacitor or AC voltage.

Well is there a way to get a inductor to operate at resonace in one polarity so you can have the high rise times and high current but in only one polarity.

Last edited: Jun 8, 2012
8. ### #12 Expert

Nov 30, 2010
16,705
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Just examine the math. (Do some pencil work.) If you add a positive DC voltage to increase current through the inductor while the AC stimulated current is rising, the resulting change in current will be the change caused by the AC component plus the change cause by the DC stimulus.

I hope this confirms your theory, as much as the laws of physics allow.
Meanwhile, the idea of placing a DC voltage in parallel with an inductor, thus not going through the inductor, confuses me. If you want to carry this conversation farther, we will need to be talking about schematics that you have posted in this thread.

9. ### WBahn Moderator

Mar 31, 2012
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As long as the inductor is behaving as a linear component, then the response to a combined AC and DC signal is exactly the same as the sum of the responses to the AC signal only and the DC signal only.

Your problem is in trying to think that the polarity has anything to do with it. It's also apparent that you aren't being careful to distinguish the difference between a change in polarity of the current and a change in polarity of the voltage.

For an ideal inductor v(t) = L di(t)/dt. Notice that there is no i(t) and so neither the polarity or the magnitude of the current matters. So if you apply a constant 2V to a 1H inductor, the current will change at 2A/s and it doesn't matter what the magnitude or polarity of the current is. If it starts out at 5A then after a second it will be 7A. If is started at -10A then it will be -8A, if is started started at -1A then it will be +1A.

Now, you could bias the inductor with a DC current of, say, 10A. Once established, this will result in no voltage across the inductor. If you then applied the AC voltages you mentioned you would get exactly the same results. To CHANGE the current in a 1H inductor by 1A in 1ns will require a voltage of 1GV. It doesn't matter if you are going from 10A to 11A or -10A to -9A or -500mA to +500mA. To decrease the current back to the initial level in the same amount of time will require a negative voltage of the same magnitude.

#12 likes this.
10. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
Well I have a new idea that doesn't use a DC source parallel at all. The idea for getting a Inductor to operate at higher frequencys and high current rise times is to use transitors with some resistors and diodes.

I have a image of one circuit I built in NI Multisim

I'm not totally sure if I am reading the circuit right?

Across the Inductor the current is 16.4A DC and 16.4 RMS (why are these the same, isn't rms for alternating currents?)
The P-P is only 1.16A.

From this I gather there is very little AC current and High DC current in one direction.

Basically when the Pulse source pulses to +40V, the transistors turn on and the circuit acts as a normal resonant circuit. Once the Pulse source goes to 0V, the transistors turn off and the capacitor and inductor are now separate circuits. The inductor discharges through the diode and so does the Capacitor. They are now ready for the next cycle. So the end result is a current in one direction.

I know this circuit is really weird, but it seems to work, this circuit is just for testing, doesn't need to be exact. I just want to get the basics of the how it works down.

So does this circuit look about right for generating a high current and fast rise times using resonant frequency's. As you can probably tell, I am not exactly an expert on circuits, lol.

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11. ### WBahn Moderator

Mar 31, 2012
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RMS values are intended to give DC equivalent values of a waveform in terms of telling you what DC value would result in the same power being dissipated in a resistive load as the waveform being measured. Not surprisingly, the DC equivalent value of a DC current is the value of the DC current.

As for the rest, we've already been through why having the current going always in one direction makes no difference and gains you nothing.

12. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
Ok I think I have it now. What you are saying I think is that regardless of the resonant frequency, the current through a certain inductance will always take a certain amount of time, that time depends on the inductance of the coil and the voltage applied. The reason a inductor looks like a short circuit or low resistance at the resonant frequency is not because the inductors actual inductance has decreased, it is because the capacitor counter acts it to create what looks like a low resistance resistor or short. So the inductor and capacitor taken as whole look like a short but individually, they still have the same properties they normally would which means the inductor still takes x amount time to reach 1 A. This means that no matter how you wire it up, physics states that the inductor can not rise faster then a certain amount of time regardless of whether it is at resonate or not, no fancy transistor circuit will make it rise faster either, lol.

So I think this is right and is what you both were getting at.

Is it correct?

13. ### WBahn Moderator

Mar 31, 2012
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I'd say you pretty much got it. It's likely that some of the fine points are still not quite there, but I think you've described the big picture pretty well.

14. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
Yeah it makes a lot more sense now, but I still have alot to learn, lol.

Thanks for the help WBahn and #12

15. ### crutschow Expert

Mar 14, 2008
13,509
3,385
One technique to increase the rise time of inductive circuit is to apply a large pulse of voltage. You then reduce the voltage when you reach the desired inductor current.

Last edited: Jun 9, 2012
16. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
hmm interesting, I'll have look more into that.

Thanks crutschow

17. ### Ron H AAC Fanatic!

Apr 14, 2005
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I don't think anyone mentioned this basic relationship:

i=(1/L)∫vdt

If v is constant and independent of t (no series resistance), then this simplifies to
i=v*t/L.

This makes it clear that, if you apply a step of voltage, a ramp of current will occur in the inductor. The higher the voltage, the steeper the ramp.

18. ### WBahn Moderator

Mar 31, 2012
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While that form of the equation was not mentioned, the exact same point was made more directly.

Looking at it from both views is good practice for someone just learning this stuff. The more directions you come at the problem and see that they all end up in the same place, the more path that are available to you to solve other problems in the future.

19. ### Ron H AAC Fanatic!

Apr 14, 2005
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I saw that, but I thought that your statement "there is no i(t)" was immediately contradicted by "current will change at 2A/s". I know that you know what you meant, but it wasn't clear to me.

20. ### WBahn Moderator

Mar 31, 2012
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It doesn't contradict it at all. The expression involves v(t), L, and di(t)/dt. It does NOT involve i(t) at all. But I supposed that a slight rewording, or even just using a semi-colon between the two sentences, would have made the meaning more obvious/explicit.

If I were sitting down with someone or lecturing at a whiteboard, I would almost certainly have waved my hand over the expression as I pointed out that there was no i(t). In writing it in text, you mind tends to infer those same clues without realizing the reader won't have them.