I am trying to simulate LC circuit using external source that generates square wave of 503Hz of amplitude 2V as below: enter image description here

Natural frequency of circuit above is 503Hz.So it must pass this much frequency.When I run the simulation of above circuit,capacitor gets damaged or shorted (don't know exactly) and it appears in circuit like :

enter image description here

and output becomes zero in oscilloscope.

But when I pass square wave of frequency 5kHz,sinusoid output is obtained.(as shown in image below)

enter image description here

Please guide me why such outputs are observed ? If it acts as band-pass filter,then it should pass simply frequency around natural frequency and same output as input i.e square wave should be observed.At higher frequencies (5kHz),output should simply be zero but why sinusoid ?

PS :

  1. I read from wikipedia that series LC circuit acts as band-pass filter having zero impedence at natural frequency.
  2. I am using Multisim 11.
  • \$\begingroup\$ The simulated driver presumably has a "zero" output impedence; put a resistor in series with it, such as 8ohm or 50ohm. \$\endgroup\$ – pjc50 Feb 8 '15 at 18:46
  • \$\begingroup\$ It's been decades since I looked at this stuff, but it looks to me like you've wired the components into a low-pass filter. \$\endgroup\$ – Hot Licks Feb 9 '15 at 1:41

That's the problem with simulating with "ideal" components — you see behaviors that you would never see in the real world.

Your circuit has no resistance in it anywhere. The function generator is an ideal voltage source with zero output resistance. The oscilloscope has infinite input resistance (open circuit). And the components have no parasitic series or parallel resistance, either.

Therefore, the behavior that the simulator is showing you is correct. The 503 Hz sine wave is the L-C circuit continuing to "ring" from the start-up transient. This ringing will never die out. And you see none of the 5 kHz square wave at the output because your filter has infinite "Q" (quality factor), which means it blocks other frequencies perfectly.

When simulating a circuit that has only ideal components, you need to remember to model the parasitic effects of real components. Depending on the accuracy that you need, you might include the series resistance of both inductors and capacitors, and maybe some parallel capacitance on inductors and parallel resistance on capacitors as well. Usually, when simulating more complex circuits — which almost always contain resistors anyway — the effects of these parasitic components are insignificant.

  • \$\begingroup\$ First,thanks a lot.I understood that square wave of 5kHz can be represented as sum of sinusoids and it is only passing the 503Hz component and blocking the rest.But when I pass 503Hz sinusoid or square wave,why capacitor blast ? Filter must simply pass input as output.Isn't it ? It even blast when I connect resistor of 0.1 ohm in series with inductor to make circuit more realistic as you mentioned. \$\endgroup\$ – Abhinav Feb 8 '15 at 12:48
  • \$\begingroup\$ I'm sorry, but I don't know what you mean by "blast" in this context. I assume you mean that the capacitor is damaged in some way, but do you mean shorted or open? \$\endgroup\$ – Dave Tweed Feb 8 '15 at 13:10
  • \$\begingroup\$ For clarification,look at capacitor in second image of this question.Isn't it distorted from normal representation ? By "blast",I mean capacitor is damaged in some way.When this happens,output simply goes to zero. \$\endgroup\$ – Abhinav Feb 8 '15 at 13:18
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    \$\begingroup\$ Are you saying that the simulator changed the drawing as a result of running the simulation? I don't know what that means, but if the output went to zero, it must mean that the capacitor is shorted. \$\endgroup\$ – Dave Tweed Feb 8 '15 at 13:27
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    \$\begingroup\$ Ah, I didn't understand that you were talking about two different simulation scenarios. Again, ideal components. When driving the circuit at its resonant frequency, the energy supplied by the function generator builds up in the L-C circuit because there's no resistance to dissipate it. Eventually, the voltage (and current) levels get so large that numerical overflow occurs in the simulator software, and they must be indicating that by "shorting" the capacitor. \$\endgroup\$ – Dave Tweed Feb 8 '15 at 13:46

Your circuit would be better if you added a third component in series - a (resistive) load.

One channel of the oscilloscope would be connected to the input source, and the second chanel connected across the load.

Also note: many circuit simulators can't handle ideal inductors, which have an infinite voltage in response to a change in current and zero ohms resistance at DC. Real inductors have a "Q" factor which you can emulate by adding another small resistor (0.1 ohm) in series with the inductor.

  • \$\begingroup\$ Thanks for answering.As you mentioned,I connected a resistor of 0.1 ohm in series with inductor but still capacitor blast in all cases when I pass square wave or sinusoid of frequency 503-510 Hz.Can you explain why ? \$\endgroup\$ – Abhinav Feb 8 '15 at 12:28
  • \$\begingroup\$ There's a difference between an "ideal" inductor and an "infinite" inductor. \$\endgroup\$ – Dave Tweed Feb 8 '15 at 13:11
  • \$\begingroup\$ The voltage across the inductor is: inductance, times rate of change of current. Without resistance, the current is huge - and the voltage is even bigger.The 0.1 ohm resistor HELPS model a more "realistic" inductor but its not enough. Put another, 1000 ohm resistor between the capacitor and ground, to act as the "load". \$\endgroup\$ – Alan Campbell Feb 9 '15 at 5:17

I really don`t know how a capacitor could be "damaged" during simulation - nevertheless, your circuit does not work as a bandpass but as a LOWPASS because you pick up the signal BETWEEN both parts. Use in addition a grounded resistor and measure the voltage across this resitor.

UPDATE: A series resonant bandpass consists of a frequency-dependent voltage divider consisting (a) of a LC-series combination and (b) of a resistor R. The output signal is available across this resistor R. The bandpass has a very sharp resonant response (small bandwidth) for small R values (1..10 Ohm). You should use a resistor of at least 50...100 Ohms.

I repeat (although somebody does not agree): At present (measuring the output between L and C) you have a second order lowpass with a very high Q value (large amplitude peaking at the pole frequency). In the vicinity of the resonant point it looks like a bandpass - but it is a lowpass output

  • \$\begingroup\$ When I run simulation,representation of capacitor changes(look at capacitor in second image).It's my interpretation that capacitor must have damaged.But now I guess that it is shorted when sinusoid of frequency 503Hz is passed through above LC circuit because output also goes to zero when this happens.Sorry for confusion caused. \$\endgroup\$ – Abhinav Feb 8 '15 at 13:44
  • \$\begingroup\$ At resonance, the inductive and capacitive reactances of the LC circuit will be equal and 180 degrees out of phase, they will cancel and the current through them will only be limited by the resistance in the circuit. At resonance, the voltage at the LC junction will also be at its maximum, and at any other frequency the circuit will drop out of resonance and the voltage at the LC junction will fall, making the LC a bandbass filter, like this: \$\endgroup\$ – EM Fields Feb 8 '15 at 20:56
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    \$\begingroup\$ @EMfields - seriously, will you really deny that a L-C-R series connection does NOT provide a lowpass output between L and C? Watch the phase response (starting at ZERO!) of your own simulation and/or extend the magnitude simulation to lower frequencies down to 1 Hz. It`s a high-Q second-order LOWPASS! Have you ever seen a bandpass with a phase between 0 and -180deg? Perhaps you reconsider the down voting? \$\endgroup\$ – LvW Feb 8 '15 at 21:33
  • \$\begingroup\$ From your comment, it appears that you're considering the spectrum from DC to light, which doesn't seem to be quite what the OP had in mind. And, while it's true that below resonance the voltage at the LC junction will be attenuated as frequency increases because of the choke's increase of reactance with frequency, that voltage will also be attenuated as frequency increases because of the capacitor's decrease in reactance as frequency rises. \$\endgroup\$ – EM Fields Feb 11 '15 at 22:18
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    \$\begingroup\$ Maybe I am a bit "sophistic" - however, a series connection of R-L-C gives a lowpass if the output is across the capacitor. Of course, it is a high-Q lowpass with a resonant peak at the pole frequency (neither an "anomalous" nor a "weak" lowpass, what is that?) - but it remains a lowpass. As far as the phase is concerned - I didnt speak about a "180 degree change" but about a phase that STARTS at 0 deg. And thats NOT the case for a bandpass. Agreed? \$\endgroup\$ – LvW Feb 12 '15 at 8:14

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