# Can I increase current using AC resonance?

In a resonance, specially if we resonant at the natural frequency, the generated waves' intensity will add up thus resulting in a wave of ever increasing intensity.

This phenomena is responsible for taking bridges down, when the wind reaches the resonant frequency of the bridge material, and the wind force starts adding up to the already resonating waves and the bridge material fractures and collapses. I wonder if we could do the same for electric current.

Given an AC voltage source (not a very high voltage), with constant upper and lower boundaries for the source AC voltage, could I feed it to a "resonant circuit" such as the current measured by the amperemeter over time would be like: And use this as an artificial way to increase the current in a load? (knowing that this would probably be the recipe to burn down everything for excess of current)

• What do you expect to gain? Power in = power out - losses in any circuit. Also, you can adjust to any voltage you want with a simple transformer. Nov 14, 2016 at 11:37
• I expect to gain more current. For example, if after the resonant circuit I have a coil, I expect it to get a stronger magnetic field over time until it heats and burns. All of this with a very limited current and voltage at the AC source. Hence the resonant circuit used to "build up" such large scale currents from a limited weak source. Nov 14, 2016 at 11:39
• For steady state, you still need to satisfy power in >= power out. "More current" is vague. Nov 14, 2016 at 11:46
• What you mean "steady state"? That graphic I draw certainly isn't steady, it shows current building up as time goes on. Moreover, it does not violate power in >= power out law, it just stores some of that power to add up with even more power for the next cycles. Nov 14, 2016 at 11:49
• The Pi filter of power supplies avoid line frequency resonance for that reason. That high 'Q' resonance is used by a "Tesla Coil" to achieve high frequency voltages. But there is no free power gained. Nov 14, 2016 at 14:41

If the passive resonant circuit has a high 'Q' then energy can build up in the circuit over many cycles. It will similarly die down naturally over many cycles.

The intensity is not "ever increasing", however. As the intensity increases, so do the losses and at some point the losses equals the input power, so you have an equilibrium.

Imagine a tuning fork. If you keep exciting it, the vibrations will build up (as will the losses) and at some point the metal shape might change if the excitation is powerful enough, but it's unlikely. Most real circuits have a rather lower Q than the mechanical Q of a tuning fork.

In a real LC resonant circuit, the losses are usually due to inductor resistance, to core losses (if a core is used) and to electromagnetic radiation, especially at higher frequencies. Capacitor dielectric losses contribute too. Superconducting circuits can have huge Q's (in the thousands), resonant circuits made with parts from your favorite distributor, much more disappointing.

• Exactly! And yes, it would have a limit to the maximum current it could achieve. Knowing that circuits have a lower Q, is it still possible to achieve such a result? Nov 14, 2016 at 11:52
• Lower Q means the losses per cycle are higher so the equilibrium amplitude is commensurately lower. The Q is roughly the number of cycles it takes for the resonance to die down from losses. Nov 14, 2016 at 12:09

Yes, resonance works with voltage or current too. Look up something called a tank circuit. This is a inductor and capacitor in parallel. With ideal components, the circuit stores energy as a sinusoidal voltage and current together. It constantly sloshes the energy between the cap and the inductor. This happens at the resonant frequency, which is the frequency where the impedance magnitude of the two parts are equal.

Again, for ideal components, this energy has no place to go once put into the tank. If you keep adding more energy, then the total energy in the tank increases, and the amplitudes of the voltage and current have to too. The total energy is proportional to the square of the voltage or current.

The magnitudes of the impedances of a capacitor and inductor are:

Zcap = 1 / ωC

Zind = ωL

By setting these equal, we find that

ω = 1 / sqrt(LC)

f = 1 / 2Π sqrt(LC)

When L is in Henries and C in Farads, then f is in Hz.

So far this has been with ideal components, which are unfortunately difficult to obtain. Real components have real losses. These are mostly due to the resistance of the wire in the inductor, and losses in the ceramic of the capacitor.

One way to quantify the lossiness of a tank circuit is with the Q factor, which stands for quality factor. Higher values mean less loss. The ideal tank has a infinite Q factor. The reciprocal of the Q factor is related to the fraction of the energy lost each cycle. Real tank circuits can have Q factors in the several 100 range. You can build something yourself rather easily with a Q factor in the 10s range.

You can make the current in a coil very high compared to the input current if the coil is part of a low-loss tuned circuit: - Here we have a 1 uH coil resonated at about 300 kHz. The 1V RMS voltage source feeds through a 10 nF capacitor (C1) and the peak voltage on the inductor is 36.5 dBV (66.8 V RMS) at resonance. The voltage across the 1 uH coil implies a current of 35.4 amps. This current must also flow proportionally through C1 and C2 but, because C2 is 27.2 times bigger than C1, the current supplied by the voltage source is only about 1.25 amps.

At the end of the day, the AC voltage source is stil supplying power to warm up the resistor that is inevitably in series with the coil. Also, if you tried to extract power from the coil's magnetic field, that power would also need to be supplied by the AC voltage source and you would find that the resonance peak would start to drop because of the Q of the tuned circuit falling.

• What happens if after the resistor I insert a diode bridge rectifier (converting that high AC current to DC) and after the rectifier, I put another coil. Would I be able to enjoy of a very large magnetic field in this coil without disrupting the tuned circuit? Nov 14, 2016 at 12:13
• The diodes have a forward voltage drop and forward resistance that is many times greater than the ohms in my diagram and they would kill the Q of the circuit and make the "current gain" significantly lower. Just get hold of a free sim and try these things out yourself. LTSpice is free and although there is a learning curve you won't regret it. Nov 14, 2016 at 12:16
• Are those simulators reliable to the real world? Or are usually too optimistic by always considering optimal scenarios? Nov 14, 2016 at 12:20
• Well, the results I put in my answer are from a simulator and the trick is, as always with a sim, is to make sure your circuit represents the parasitic components that can affect performance. In my sim above, the 10 milli ohms resistance is very relevant because if it were not present, I would get a voltage magnification factor of infinity and this of course doesn't make sense. So, in short, sims are very useful providing you don't try and be too silly with values and you understand that a perfect inductor is never a real practical part. Nov 14, 2016 at 12:38

I found something: http://www.richieburnett.co.uk/resonant.html

The author says that this is possible but, it would require some engineering to prevent voltage burst in milliseconds, and to achieve stable voltage gain.

I'm not a scientist, but to say it is impossible to achieve free voltage gain by resonance it's limiting. We should accept this as an idea, and try to prove it right, rather than wrong (it's easy to say anything is impossible and wrong). This would be huge discovery, so saying that "it cannot be done" should be replaced by, "we don't know how it could be done", and scientists should keep experimenting.

• Stable voltage gain is absolutely possible. Happens all the time; I just measured it in a school lab experiment the other day. Just put an inductor in series with a capacitor and stick some AC into it. Apr 24, 2017 at 13:17