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schematic

simulate this circuit – Schematic created using CircuitLab

I have a LC low pass filter, L=39 mH, C=0.45 uF. How to calculate the voltage across the capacitor? The voltage can be magnified at resonant frequency. What about low frequency? If I want to know the voltage with 300 Hz input, how to calculate it in math. If I want to prevent the voltage magnification in low frequency, I need a series resistor with the inductor or a parallel resistor with the capacitor, how to calculate the resistance?

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    \$\begingroup\$ Since you appear interested in frequency questions, the usual route is to use Laplace. But I'm not fully sure about all the question sin mind. You want a voltage number for a specific circuit and you want a resistor number for general circuits of this nature where the voltage number is given? What math have you tried? What math are you familiar with? \$\endgroup\$ Commented May 26, 2023 at 3:13
  • \$\begingroup\$ XL=jwL, XC=1/jwC. Z=XC/(XL+XC), VC=VINZ. w=2*pif, can I just use 300Hz as the f to calculate the voltage? If I know the output voltage, I can do a simulateion without the resistor, it's around 320Vrms, how to calculate the resistor number to prevent voltage magnification \$\endgroup\$
    – CYJ
    Commented May 26, 2023 at 3:25
  • \$\begingroup\$ Do some simple research into series RLC circuits and bandwidth. Perhaps start here: electronics-tutorials.ws/accircuits/series-circuit.html \$\endgroup\$
    – Nedd
    Commented May 26, 2023 at 3:28
  • \$\begingroup\$ en.wikipedia.org/wiki/RLC_circuit For your values, f0 = 1.2 kHz. For a series circuit, if you don't want any peaking, you want a damping factor (zeta) of 1, R about 600 ohms. 300 Hz is well below the natural freq, magnification at 300 Hz will be minimum at any zeta. \$\endgroup\$
    – Mattman944
    Commented May 26, 2023 at 9:25
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    \$\begingroup\$ @Mattman944: "...want a damping factor (zeta) of 1...". For maximally flat, \$\zeta=Q=\frac{1}{\sqrt{2}}\$ \$\endgroup\$
    – RussellH
    Commented May 26, 2023 at 17:32

2 Answers 2

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Since you seem to understand complex impedance, and how they will divide, you can use Excel for the calculations. This is a modification of a spreadsheet that I made a long time ago. Thanks to RussellH for pointing out my mistake in the comments.

To determine the resistor value, you need to know how much gain is tolerable at 300 Hz. If no gain is allowed, then the resistor value should be 416 ohms (determined by trial and error).

Note that the gain at 300 Hz may not be the most important factor. The gain at the resonant frequency may be more important.

enter image description here

And with the formulas shown (ctrl~)

enter image description here

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The problem you pose is 2nd order. Which is convenient and well-analyzed.

In one way, the problem is fairly simple. But before you can say it is so simple, you need to be a little bit comfortable with a few ideas:

  • The magnitude (vector length) of complex values plotted on the complex plane.
  • The concept of \$s=\sigma+j\,\omega\$ and multiplication of complex numbers as a combination of both rotation and scaling of the vector length.

In general, a low-pass filter has the transfer function form of:

$$A\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1{Q}\left(\frac{s}{\omega_{_0}}\right)+1}$$

(\$A\$ is the gain, \$\omega_{_0}\$ is the pole angular frequency, and \$Q=\frac1{2\,\zeta}\$ is the shape, where either \$Q\$ or \$\zeta\$ are just two different ways of thinking about the shape.)

You can set \$\omega_{_0}=1\$ and \$A=1\$ to eliminate those distractions and then just realize that everything really has the even simpler form of:

$$\frac{1}{s^2+\frac1{Q}s+1}=\frac{1}{s^2+2\zeta\,s+1}$$

If you study that, then you can always come back later and shift the gain by making \$A\ne1\$ (usually just a movement of the plot up or down in the \$y\$-axis gain value) and shift the angular frequency making \$\omega_{_0}\ne 1\$ (moves the plot left or right along the \$x\$-axis angular frequency value.)

In your schematic, \$Q=\frac{L}{R}\$ and \$\zeta=\frac12\frac{R}{L}\$. Note that the value of \$C\$ isn't present there. So you can adjust the shape by focusing only on \$R\$ and \$L\$, leaving \$C\$ to help set the angular frequency when combined with \$L\$.

Setting \$\sigma=0\$ (no scaling, focused only on frequency for now) and \$A=1\$, find that the magnitude at any particular \$\omega\$ is:

$$\frac1{\sqrt{1+4\cdot\left[\left(\frac{\omega}{\omega_{_0}}\right)^2-\left(\frac{\omega}{\omega_{_0}}\right)+\zeta^2\right]}}$$

From this, you should be able to find an appropriate value for \$R\$ given the magnitude you'd prefer at some angular frequency ratio. Or, alternatively, find the magnitude at any given angular frequency ratio for some given \$R\$ value.

Use the above as a guide of sorts. See if you come up with similar results, or not. If you have questions, feel free to ask.

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