0
\$\begingroup\$

If a parallel capacitor is placed in a circuit before a series inductor or ferrite bead (CL instead of an LC), will this combination be capable of any self-resonance? Or does the inductor have to be first in order? This will be in a DC circuit but dealing with potential unwanted self-resonance from noise and 60Hz wall power / switching PS / data transmission line frequencies and harmonics. Thanks in advance.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 2
    \$\begingroup\$ In your schematic, C1 serves no useful purpose: V1 dominates it entirely. Very large alternating currents flow around the V1, C1 loop. There is no resonance in this circuit. In practice, true voltage sources like this invariably include series resistance and inductance. Then resonances are possible. \$\endgroup\$ – glen_geek Aug 18 at 13:59
  • \$\begingroup\$ I'm trying to avoid resonance, the sine wave power source is just supposed to represent an alternating frequency capable of causing resonance with an LC combo. My question is, under any circumstances, will a resonant circuit be able to be produced from having the capacitor in front of the inductor? Or does the inductor have to come first in order to be able to cause self resonance from the components? \$\endgroup\$ – wdbwbd1 Aug 18 at 16:41
1
\$\begingroup\$

It will still have self resonance, but the Q will generally be much lower due to the low resistance of the power source 'shorting out' the capacitor (though if the source resistance is high then significant resonance could still occur). Putting the capacitor after the inductor has the opposite effect. Since the source resistance is now in series with the capacitor, a lower resistance will increase Q.

The simplified circuits below show the difference. In the first case the resistor is in parallel with the resonant circuit, so lower resistance reduces Q because the resistor draws more current from it. In the second case it is in series with the resonant circuit, so lower resistance increases Q because it drops less voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

But this doesn't mean putting the capacitor in front of the coil is good idea. The main reason for having it after the coil is to attenuate frequencies above the resonant frequency, which it cannot do when placed before the coil.

If you have a very low resistance source that causes significant resonance then you should decrease Q by other means, eg. by adding resistance in series or parallel with the coil. What method is best will depend on the particular parameters of your design. In your example, simply using an inductor with internal resistance of 1Ω is enough to damp the resonance.

\$\endgroup\$
1
\$\begingroup\$

Self-resonance usually refers to a resonance within a single element (L or C) caused by the parasitic inductance/capacitance in combination with the primary capacitance/inductance.

The effect is independent of the position in the circuit, but can vary a bit if the element is in series or parallel orientation due to the impact of the ground plane.


With regard to the resonance of the LC combination as stated in the question, the sequence also doesn't matter.

For example the cut-off frequency:

\$f_c = \frac{1}{2 \pi \sqrt{LC}} \$

is independent of the L/C sequence.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.