# Series resonant circuit help

1. Jul 5, 2007

Hi I am new to the forum. I have been studying and recently doing experiments with (LC) series resonant circuits at home. I have been referencing numerous electrical engineering books for information on series resonance, but unfortunately I have been unable to find any books that break the phenomenon down to the most basic level as to what is really going on in a series resonant circuit to reduce impedance to to zero. I have seen the vectors and the math but that really doesn't explain to me how for example the capacitor is able to interefere with the inductor's ability to control current rise in itself. All I can do is theorize that the capacitor's voltage rises at the same rate as the inductor's reactive voltage which results in no voltage drop across the the LC network. If anyone has had it explained to them or understands it more deeply and is willing to share the information please respond.

Thanks

2. Jul 5, 2007

### Staff: Mentor

Welcome to the PF. It sounds like you have it mostly correct. I like to think about series and parallel resonant circuits like this:

** A parallel resonant LC circuit will sustain an oscillation on its own, once the oscillation gets started up. The energy flows back and forth between voltage stored on the capacitor (E-field energy storage in the cap) and current flowing in the inductor (B-field energy storage in the inductor). Since the voltage oscillation goes on with very little additional input current (assuming low loss in the L and C), that presents a high impedance to the drive circuit.

** A series LC circuit can have very low voltage across it when it is drawing maximum current, and hence presents a low impedance to the drive circuit. The complex impedances of the L and C are basically cancelling out at resonance, and all that is left is any real ESR in the cap and DCR in the inductor.

Does that help?

3. Jul 5, 2007

berkeman,

Thanks for responding. I guess I am trying more to understand the transient current through both the inductor and capacitor. As I understand it normally an inductor is able to induce voltage a voltage opposite of the current that is rising inside itself which controls current rise. I was trying to understand how having a capacitor in series with the inductor is able to interfere with the inductor's ability to control the current rise through itself.

4. Jul 5, 2007

### Staff: Mentor

In both series and parallel resonant circuits (assuming low loss for the purpose of this discussion), the energy stored does what I said before -- it oscillates back and forth between the inductor and capacitor. The difference between the series and parallel cases is just where you inject current and measure voltage. In the parallel case, picture the L and C sitting vertically side-by-side, with the bottom connection ground, and the top connection the point driven by the AC signal source. After energy is stored in this resonant circuit, you can disconnect the source and observe the behavior. When the current flowing is zero, you have maximum charge on the capacitor, and when the cap voltage is zero, you have maximum current flowing in the circuit (through the inductor). Draw a picture of these waveforms, and verify that they obey the differential equations for the voltages and currents in caps and inductors.

The series case is the same in terms of how the current changes and how the cap voltage changes, since it's the same circuit. Picture the L vertical above a vertical C, connected in the middle, and now with a separate wire that goes from the top of the L down to the bottom of the C. It's still just the same circuit, right? An inductor connected to the cap, forming the LC resonant circuit. But when you measure the input impedance, you now would be putting your test AC voltage at the top of the L and C, and measuring the current that flows. But you would measure (at resonance) that even a small delta-V gives a large current into the series combination, because their complex impedances cancel at resonance. Again, the differential equations for the currents and voltages for the L and C should help make it more obvious.

I get the feeling that what I've just typed may not help with the root of your question, though. Are we getting any closer? If not, maybe I can try to do it with the differential equations later this afternoon if I get time.

5. Jul 5, 2007

berkeman,

Once again thank you so much for your response. I have been studying parallel and series resonance for the last several months as well as experimenting with them in my garage for fun. I really appreciate you taking the time to answer my post. I guess the question I am asking is pretty unconventional. I am not really thinking of resonance in terms of AC because when you really think about it all a frequency translates to is a voltage applied for a certain time period. I have just finished studying parallel resonance and experimenting with it for now. I was able to get it to resonate with a pulsating square wave DC signal. The difference I noticed was that amplitude of negative side of the resulting AC sinewave was smaller than the positive. Parallel resonance is much more understandable to me than series resonance. Although, I am really trying to break it down to a ultra simple level. Here is the just of what puzzles me so much. Take for example a 1 henry inductor with 3 ohms of resistance and a 12 volt power pulse. Calculating the inductors time constant results in 0.3333 seconds per constant. The inductor should reach the full current of 4 amps after 5 time constants which should equal 1.66666 seconds. Now, when I add a 0.001 uf capacitor the inductor reaches it full current after 100 microseconds. The way I figured this out is by calculating the resonant frequency of that LC circuit resulting in approx 5028hz. I then devided 1 by 5028 which resulted in 198 microseconds. I then devided 198 microseconds by two because we know that current only goes in one direction for 50 percent of the time with single phase AC circuit which resulted in approx 99 microseconds, which I rounded up to 100 microseconds for simplification. Now we know that the current through the inductor reached the full 4 amps because we can check the power contained in the capacitor at its peak. I calculated this by calculating inductive reactance at 5028 hz which results in 31575 ohms of impedance. Now we know that at resonance the current of the circuit is limited by only the resistance of the circuit which I stated earlier in the example was 3 ohms which would mean 4 amps has to be flowing at resonance. 4 amps across 31575 ohms of reactance would produce 126303 volts across both the inductor and capacitor and when I check the energy in joules contained in both the inductor and capacitor at their peaks they both equal approx 16 joules. Provided there was no arcing in this high energy example. I included all my math so that you can point out any errors that I may have made in coming to the conclusion that the current in the inductor did indeed reach 4 amps in 100 microseconds when normally it should have taken approx 1.6 seconds. I am sorry this post is so long but my question; I guess which is not so simple is how did the capacitor make it so that current was able to rise so much faster through the inductor than it normally should have. Once again sorry for such a long post. Thanks again for your input.

6. Jul 5, 2007

### Staff: Mentor

Couple things (I'm not sure I understood all of what you were saying)....

3 Ohms is a moderate amount of loss, so you really have a damped oscillator.

The better way to experiment with the resonant circuit is to drive it with a fairly high source resistance. That is, if you have a 50 Ohm output signal generator, that 50 Ohms will interact with the tank circuit, and change its resonant behavior. You want to use a less invasive method of getting the tank circuit ringing, so connect something like a 1kOhm resistor between your signal generator and the parallel resonant tank circuit. You can also connect something like a 0.1 Ohm resistor (or 10 // 1 Ohms resistors) between the L and C, and measure the voltage drop across that resistor to see the current flowing in the circuit. Then you should be able to watch the sinusoidal current and voltage in the circuit, as you change the frequency of your signal generator from below resonance, to resonance, and above resonance.

I didn't get the part about the 1H inductor having 31575 Ohms of reactive impedance at 5kHz, giving a giant voltage for a 4A AC signal. You will not be pushing 4A AC through that inductance at 5kHz....

7. Jul 5, 2007