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I have a signal generator outputting 60 hz AC. I've connected the output to a capacitor (922uf) and inductor (0.0076H) which are in parallel with each other creating a tank circuit. These two values will have resonance at approximately 60hz. However by the loud squeel my signal generator is making, I can tell it's shorting out. I also connected two parallel diodes in series to the tank to indicate if there is current going through and it lit up like Christmas. Isn't it supposed to have a net current of zero??

Here are my sources: http://www.deephaven.co.uk/lc.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-6/parallel-tank-circuit-resonance/

Can someone explain how to make the tank circuit work? I want to achieve a net current of zero by blocking out the 60hz signsl from my signal generator.

Thanks.

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  • \$\begingroup\$ What is the DC resistance of your inductor? \$\endgroup\$
    – user1582568
    Commented Jan 27, 2016 at 4:20
  • \$\begingroup\$ It's about 1.6 ohms. I didn't think of that but isn't it too small to do anything? \$\endgroup\$
    – user29150
    Commented Jan 27, 2016 at 4:33
  • \$\begingroup\$ why you don't use a notch filter? \$\endgroup\$
    – HOPE
    Commented Jan 27, 2016 at 4:34
  • \$\begingroup\$ I'm doing a tank circuit for the purpose of understanding how it works. So far I feel betrayed by allaboutcircuits.com \$\endgroup\$
    – user29150
    Commented Jan 27, 2016 at 4:38
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    \$\begingroup\$ AllAboutCircuits showed you a SPICE simulation of the circuit with an Rbogus = 1p\$\Omega\$ 'because of SPICE limitations'. However, that R there is not so much Rbogus as Runavoidable. In your case, it's 1.6\$\Omega\$. Run their SPICE simulation with that value, and see what you get. If you don't have SPICE, use the circuit simulator available on this site when you edit your question. \$\endgroup\$
    – Neil_UK
    Commented Jan 27, 2016 at 6:19

2 Answers 2

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A 60 Hz tank circuit will need a much bigger inductance and a much lower capacitance in order to achieve a decent Q. With the value you have, it resonates at 60 Hz but the overall impedance of the circuit is very low.

Feeding the tank from a 100 ohm source and sweeping the frequency I see a 25 dB attenuation at the peak of resonance and this is equivalent to an impedance of about 6 ohms i.e. 6/106 = 0.057 = -24.9 dB.

If you had an inductor that was 76 mH (3.16x the turns and 5 ohms) and a capacitor that was 92.2 uF, the attenuation at 60 Hz is now only 4 dB or a reduction in signal to 63%. This is more like an impedance of 170 ohms at resonance.

Just take note that the quality factor of a tank circuit is proportional to \$\sqrt{\frac{L}{C}}\$ so making L ten times bigger whilst reducing C by ten times increases Q by 3.16 times (all other things remaining fixed).

Also, I'm concerned that you may be using an electrolytic capacitor - this is a polarized capacitor and may have really poor performance in a tank circuit.

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  • \$\begingroup\$ Yes I am using electrolytic. Is there a way to increase the Q without changing the windings of my coil? I can't increase my inductance with the coil I have without making a new one. Thanks \$\endgroup\$
    – user29150
    Commented Jan 27, 2016 at 14:20
  • \$\begingroup\$ You are aware that electrolytics are polarized and that they will conduct current in one direction (reverse polarized) - this will make a mess of anything you are trying to achieve. Even with a perfect 922 uF capacitor you will not achieve a parallel impedance better than 6 ohms and no, without winding a new inductor you won't be able to increase inductance in any significant way. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2016 at 14:32
  • \$\begingroup\$ This is a good suggestion, you will still need to keep losses low. The components will now have around 28ohns reactance at resonance, so the inductor resistance must be much lower than this and core loss low. Aldo see my post above about saturation. Plastic capacitors will be s goof choice with values around the ones that @Andyaka suggests. \$\endgroup\$
    – user1582568
    Commented Jan 27, 2016 at 14:35
  • \$\begingroup\$ If you got hold of a power transformer, it might have a primary inductance of (say) 10 henries, then you could put a non-electrolytic capacitor in parallel (about 1uF ballpark) to achieve maybe a few kohms impedance. \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2016 at 14:36
  • \$\begingroup\$ @user29150 I don't know if you were aware but there is an "answer acceptance scheme" for the answer that has proven most useful when raising a question. For instance on this question (electronics.stackexchange.com/questions/130877/…) and this (electronics.stackexchange.com/questions/82528/…), you might want to consider formal acceptance (symbol under up/down arrows on relevant answer). \$\endgroup\$
    – Andy aka
    Commented Jan 27, 2016 at 14:45
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At resonance your inductor and capacitor have a reactance of about 2.85 ohms. In order to have close to infinite impedance, the Q must be very high, i.e very low loss. The resistance of the inductor must be very low w.r.t the reactance. Also other losses ( inductor core loss, capacitor ESR etc) must be low.

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  • \$\begingroup\$ Shouldn't there still be some noticeable drop? I got a short circuit instead which from what I understand is suppose to happen only when you put the capacitor and inductor in series. LCs are simple filters that you can make with regular inductors and capacitors. So the Q will never be high unless you use small inductance and high capacitance so I don't buy that answer as the cause of my circuit's failure to block 60hz. Thanks for the help though. \$\endgroup\$
    – user29150
    Commented Jan 27, 2016 at 4:56
  • \$\begingroup\$ There will be a drop in impedance at resonance, however the impedance will always be low as the Q is low. Your inductor has nearly as much resistance as inductance so it will never give a high Q at 60Hz. Why do you not accept this answer, it is the correct explanation of the problem that you posted? \$\endgroup\$
    – user1582568
    Commented Jan 27, 2016 at 5:08
  • \$\begingroup\$ I looked into it some more. Yes you are right the drop in signal is insignificant. Thank you \$\endgroup\$
    – user29150
    Commented Jan 27, 2016 at 14:18
  • \$\begingroup\$ Selecting the components is very important I'm this type of circuit. Losses must be low. If you use in inductor with a core you also need to take account of core loss and saturation. The saturation current will limit the voltage that you can put across the tank circuit. Also, as Andyaka points out, an electrolytic capacitor is not a good choice. Plastic capacitors will be best for this application as they are fairly low loss and available in reasonable values (10s of if) \$\endgroup\$
    – user1582568
    Commented Jan 27, 2016 at 14:30

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