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Why is bandwidth of series RLC circuit independent of the value of C ? Reason without derivation

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  • \$\begingroup\$ Any equations, links where did you get this? \$\endgroup\$ Aug 25, 2017 at 6:53
  • \$\begingroup\$ @MarkoBuršič RLC series bandwidth = R/L for any C \$\endgroup\$
    – TVV
    Aug 25, 2017 at 6:54
  • \$\begingroup\$ I didn't know it and it sound weird, can you post the article or the source of this claim. \$\endgroup\$ Aug 25, 2017 at 6:56
  • \$\begingroup\$ @MarkoBuršič google.co.in/url?sa=t&source=web&rct=j&url=https://… equation 1.19 \$\endgroup\$
    – TVV
    Aug 25, 2017 at 7:00
  • \$\begingroup\$ * Sorry B = L / R \$\endgroup\$
    – TVV
    Aug 25, 2017 at 7:00

3 Answers 3

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There is no 'reason' per se, it's just the way the mathematics works out.

The tutorial paper you mentioned in your comments (add link to your OP please) is pretty clear and thorough, it leads you by the nose from first principles to equation 1.19, which is $$B=\frac{L}{R}=\frac{\omega_0}{Q}$$

The useful part of that equation is the \$\omega\$ and \$Q\$ bit (which you omitted from your question). Both of those have L and C in them to half powers. It just happens that when when you expand them and combine, for the series circuit, you get a whole L and no C. You might as well ask why sqrt(2) times sqrt(2) is 2.

For the parallel circuit, they combine differently to give you a whole C and no L.

Most of the time, we're concerned with fractional bandwidth, which is why Q was invented. Dealing with absolute bandwidth while varying the C of a series circuit is not something that people ever tend to do, or would want to understand quantitively and intuitively. But if you did, then that's the formula that you would use.

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Why is bandwidth of series RLC circuit independent of the value of C ?

It isn't. You are mistaken.

If you construct a series tuned RLC circuit with a fixed value of resistor you can tune it to resonance with L and C (for instance) or, you could tune it (to the same resonant frequency) with 2L and C/2. In other words: -

\$\dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{2L\cdot\frac{C}{2}}}\$

When you tune it with progressively higher inductance (which results in progressively lower capactance) the bandwidth gets smaller progressively for a fixed value of R.

And all of this relates to the Q of the tuned circuit where Q = \$\dfrac{1}{R}\sqrt{\dfrac{L}{C}}\$

Because Q = \$\dfrac{F_R}{BW}\$, as Q gets larger (L/C gets bigger), the bandwidth must get smaller for a fixed value of resonant frequency.

You can easily work this out for yourself if you recognized that at resonance, the inductive and capacitive reactances cancel out - if both impedance are (say) ten times bigger then they still cancel out but, at a fraction away from pure resonance, the net impedance must be ten times higher.

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For a series \$\small RLC\$ with the output taken across \$\small R\$, the amplitude ratio is:

$$\small \lvert H(j\omega)\rvert=\frac{\omega RC}{\sqrt {(1-\omega^2LC)^2+\omega^2R^2C^2}} \:\:\:...\:(1)$$

At resonance, \$\small \omega=\omega_0\$, we have \$\small \omega_0^2LC=1\$, hence, $$\small \lvert H(j\omega_0)\rvert=1 $$ Therefore, the bandwidth, \$\small\omega_B\$, is the difference between the two frequencies where: $$\small \lvert H(j\omega)\rvert=\frac{1}{\sqrt{2}}\:\:\:...\:(2)$$

Equating \$\small (1)\$ and \$\small (2)\$ and solving the resultant quadratic equation for the upper and lower bandwidth frequencies gives: $$\small \omega_B=\frac{RC+\sqrt{R^2C^2+4LC}}{2LC}\:-\: \frac{-RC+\sqrt{R^2C^2+4LC}}{2LC}$$ Hence: $$\small \omega_B=\frac{R}{L} $$

But note that the resonant frequency is contained implicitly in this equation, since \$\omega_0^2=\frac{1}{LC}\$, and so we may also express \$\small\omega_B\$ in the form: $$\small \omega_B=\omega_0^2RC $$

where the relationship to \$\small \omega_0\$ is now explicit.

Interesting to note the relationship of \$\small\omega_B\$ to the time constants of 1st order \$\small RL\$ and \$\small RC\$ circuits.

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