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Help me understand a strange LC phenomenon

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billano786

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strange LC phenomenon

I designed a simple series tuned circuit comprising of L & C. The value of L=0.17mh and C= 10uF. The resonant frequency comes close to 3.3kHZ.
The dc resistance of the coil is around 5 milli ohms. I am tuning the circuit from a function generator + power amp connected to LC circuit via an impedance matching transformer (0.5 ohms secondary).

I am observing a strange phenomenon. The current at resonant frequency is through LC circuit is around 1A(RMS) and voltage developed across LC circuit is 552 mV (RMS) telling me that the impedance of LC circuit is around 555 milli ohms. The input voltage to primary of impedance matching transformer is 3.3V

Can some one explain the reason behind this and how to reduce this impedance?
 

flatulent

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parasitics

Your parts are large enough that the inductor also has shunt capacitance and the capacitor has series inductance and series and shunt resistance. This is especially true if your capacitor is polarized type. Another possible problem is magnetic core saturation. Scale the current down by 10 and see if the voltage drops by more than 10.

Try raising the inductance so that you can use a lower value capacitor which is not polarized.

If you make the inductor, use a pot core with three winding areas to reduce the capacitance. Also, use a large size core. If you buy the inductor, use one intended for power filtering which will have large wire and a core that does not saturate.
 

billano786

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I am already using Metalized Poly Propylene Capacitors (1uf*10) which are non polar. For inductor I am using an amorphous toroid with distributed gap (from honeywell). The core is certainly not going into saturation. Before opting for the core I throughly verified the power capacity. Also there is no heating of core. The core is meant for power filtering.

Any comments?
 

echo47

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Did you wind the inductor with one strand of heavy wire? Try five or ten strands of fine insulated wire. This dramatically reduces AC copper losses caused by the proximity effect.
 

flatulent

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more things to try

You can determine the Q by measuring the voltage across either part. This will give you another clue. The Q is the ratio of the center measurement to the whole measurement when measured at the exact resonance frequency. Divide the reactance of either part by the Q to get the efective series resistance. See if this matches your measurement.

Another thing that may be happening is the AC resistance of the wire is higher than the DC resistance.
 

ted

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As flatulent points out, the voltage at center (where L and C are connected together) gives an indication of the Q value. Be aware that in all resonant circuits the physical mechanism for obtaining the resonant impendance produces one or another extreme situation: In parallel resonance the current circulating in the circuit is multiplied by Q, while in series resonance the voltage is multiplied.

That high voltage can in extreme cases actually cause flashover, and thus shorts. As a result you might even have an internal short if the circuit first had very high Q (and low impedance you wanted), but subsequently something said "bang" due to the voltage stress, and now either the inductor or capacitor might be damaged. Such partial short would lower the Q value, rsulting that the resonant voltage is lowered proportionally. And you measure higher total impedance.

Also, if you think that the resonant impedance would ever be very close to the DC resistance, dream on. It will be substantially more, but should not necessarily be 100 times more as of now! A circuit has all kinds of additional losses: Coil copper AC losses, coil core losses, capacitor's internal losses.

The fact that you do not observe any substantial heating is maybe because you only dissipate about 1/2 watt total (1 A rms, 555 mV). If the components are bulky enough, it is hard to locate where it heats up.
 

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