I have to build the below series RLC circuit , and output its bode diagram using NI ELVIS Instrument Launcher. It is for a bandpass filter, so the output is taken across the resistor. Before proceeding I simulated the circuit on Multisim, and it worked perfectly, however the implementation didn't give me quite the expected results. I measured the two cut-off frequencies, and they had high percentage errors. Any reason as to why this could happen? Does the software assume certain things? enter image description here

  • \$\begingroup\$ The software assumes what you put in as parameters. Did you include ESR and ESL of the capacitor? Did it include DC Resistance of the inductor? Did it include inductance of the wire loop? Did it include the output impedance of the signal source? \$\endgroup\$ – Asmyldof Apr 15 '16 at 9:06
  • \$\begingroup\$ @Asmyldof Oh no I didn't, we never did that in our previous labs so I had no idea about such things, I thought they are negligible. Anyways it is a design lab, so maybe I need to find another combination of values that give me my required cut-off frequencies. \$\endgroup\$ – 32px Apr 15 '16 at 9:14
  • \$\begingroup\$ What means "high percentage errors" (5% or 20%)? More than that, because of the very small resistor (3.3 ohms), the bandwidth will be rather small. Did you take into account tolerances? \$\endgroup\$ – LvW Apr 15 '16 at 9:22
  • \$\begingroup\$ @LvW more like i am supposed to get lower frequency of 677Hz and upper frequency of 915Hz, but i got 549 and 1258, around 16% and 36% i guess. Yes I did, my theoretical resistor value is 3.47 ohm, simulation-wise it is still close. I tried using 1 ohm too for implementation to lower my bandwidth, but it didn't make that much of a difference. My problem statement requires me to build a bandpass series rlc filter with 677 and 915 as the cut-off frequencies, the choice of values was done by me. \$\endgroup\$ – 32px Apr 15 '16 at 9:30
  • \$\begingroup\$ The biggest difference between simulation and 'real life' components is tolerance. In simulation you can use values such as 18u6 (accuracy claimed <1%?) for the capacitor. In 'real life' this will probably be a 22uF with +/- 20%. (Don't believe the printed value is the actual value). Its not surprising that you get such different results. Large value capacitors are to be avoided in filters due to the large tolerance . \$\endgroup\$ – JIm Dearden Apr 15 '16 at 10:33

In theory, theory and practice are the same. In practice they're not.

A simulator will happily crunch away on unrealistic values that make real components far from ideal. Some non-ideal characteristics may be simulated for, if you specified them and entered their values, but the real world will always be more complicated.

In your case, your impedances are very low. Consider the current required to drive a 3.3 Ω load. It's just numbers in a simulator that assumes V1 is a ideal voltage source, but V1 certainly isn't. If V1 is a function generator, then it may have output impedance of 50 Ω. That's still a lot higher than the 3.3 Ω load that will be presented to it at the center frequency.

Simulators can be useful, but there is no substitute for actually thinking about the circuit. Use a calculator or let the simulator determine the details of the numbers, but you have to think about the broad picture first. In this case, 3.3 Ω loading on the voltage source should have been a obvious consideration.

Make the input impedance of your filter several times the output impedance of the signal source. If the signal source has 50 Ω output impedance, then R1 should be 500 Ω minimum. Using common values, so 1 kΩ it is. Now the ESR of the cap and the DC resistance of the inductor will be effectively irrelevant, which they weren't when acting against a 3.3 Ω load. At 1 kΩ output impedance, it will also be easier to find realizable values for the inductor and capacitor.

  • \$\begingroup\$ Thank you, I will try increasing the resistance and see how it goes. \$\endgroup\$ – 32px Apr 15 '16 at 11:00

32px - I think the problem with your circuit is the following: At resonance frequency the total load resistance will be approximately 3.5 ohms (neglecting copper wire resistance). This will require a current of app. 0.3 A. I think, this load will overload your signal generator. So - why not using a parallel RLC bandpass?

If you are required to use a series RLC circuit, use a larger ohmic resistor in aacordance with the required quality factor Q=midfrequency/bandwidth=(1/R)*SQRT(L/C).

Example: R=25 ohms, C=15.4µF, L=105mH.

  • \$\begingroup\$ Yeah, I have to use series! Q=3.31, giving me R=3.29 ohm which is almost 3.3 ohm, thanks for your effort though. I will ask the lab instructor to look over my circuit again, just in case. \$\endgroup\$ – 32px Apr 15 '16 at 10:46

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