How to Calculate L, C, and R Values for Resonance Circuit at 100 Hz Frequency?

171 12

#1 21661069 04 Feb 2012 07:00

Kobi Aflalo Kobi Aflalo

Anonymous
Post #1
21661069 04 Feb 2012 07:00

i need to find values of capacitors coils and resistors in order to create resonance mixed circuit on the frequency of 100 hrz (as shown in the following screenshot)
ω0=1/√(L*C)
ω0=2∏f0

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#2 21661070 04 Feb 2012 08:54

Mike Burr Mike Burr

Anonymous

Post #2
21661070 04 Feb 2012 08:54

Do you need help understanding the formula or just working through it to find close arbitrary values for the capacitor and inductor?
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#3 21661071 04 Feb 2012 10:17

Kevin Parmenter Kevin Parmenter

Anonymous

Post #3
21661071 04 Feb 2012 10:17

if you have an andriod phone there is a very cool program - free, called electrodroid - if you download that and open it you will find some very nice tools for calculating reactance. You don't list other constraints on the circuit so I assume you just want to make R=xl=cl and you are not limited on the resistance values? I would round those values off to real world capacitors you can buy as standards if you plan to build it vs a theory exercise. in any case this will get you there I believe.
#4 21661072 04 Feb 2012 11:03

Kobi Aflalo Kobi Aflalo

Anonymous

Post #4
21661072 04 Feb 2012 11:03

i know how to calculate those values manualy but only for simple series or paralar circuits, but in this example i have a mixed circuit and i'm not sure if i shuld calculate the total XC and make it equel to the XL in order to make resonance on the frequency 100 hrz

i don't think the resistance values makes any different
and i don't need real world capacitors
i just make this circuit in Orcad and simulate
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#5 21661073 04 Feb 2012 11:04

Kobi Aflalo Kobi Aflalo

Anonymous

Post #5
21661073 04 Feb 2012 11:04

i understand this formula, i just need to find values for the capacitor and inductor to get resonance on 100 mhz.
#6 21661074 04 Feb 2012 11:07

Mike Burr Mike Burr

Anonymous

Post #6
21661074 04 Feb 2012 11:07

when L and C impedances match, (i.e. resonance), then the resistance becomes noting more than series resistance. The best thing to do will be to pick an arbitrary value and work the formula backwards solving for the missing component value. For example, setting a 10mH inductor would use a 10mF capacitor.
#7 21661075 04 Feb 2012 11:47

Kobi Aflalo Kobi Aflalo

Anonymous

Post #7
21661075 04 Feb 2012 11:47

as you said i picked an arbitrary value for the inductor (L=10mH) and found the value of the capacitor to recive a resonance frequency at 100 hrz and i found that C=253.302uF

my calculations:
http://img593.imageshack.us/img593/8049/photo030212173055hdr.jpg

and i tried to place those value on the simulation program orcad pcspice
http://img94.imageshack.us/img94/286/42599051.jpg
only that i placed this values on both of the parts c1 and c2 l1 and l2

and i recived this:
http://img580.imageshack.us/img580/8971/63874598.jpg

is this the correct answer?
and is it possible to get two different resonance frequancy in one circuit such as this?
that the wave will be something like this?
http://img823.imageshack.us/img823/5848/11132918.jpg
#8 21661076 04 Feb 2012 12:29

Mike Burr Mike Burr

Anonymous

Post #8
21661076 04 Feb 2012 12:29

O.k. formula is correct. and numbers work out as expected. the first formula is a little easier to work with though. 1/(2pi*(sqr(L*C))= resonant frequency. which can then be worked out to L*C=((1/(2*pi*100)^2) or L*C=2.53X10^-6 Then divide the number by L or C to find the opposite one. 2.53*10^-6/10mH = 2.53uF. It will make it easier to find the common value combinations without having to have special capacitor/inductor values.

The Resistance at that point does not affect the resonance.
#9 21661077 04 Feb 2012 12:43

Kobi Aflalo Kobi Aflalo

Anonymous

Post #9
21661077 04 Feb 2012 12:43

thank you mike

and what about the second question?
is it possible to get two different resonance frequancy in one circuit such as this?
that the wave will be something like this?
http://img823.imageshack.us/img823/5848/11132918.jpg
#10 21661078 04 Feb 2012 12:55

Mike Burr Mike Burr

Anonymous

Post #10
21661078 04 Feb 2012 12:55

What that looks like is a band-stop or band-reject filter. with a high and low pass filter to taper the outer edges.

An LC resonant filter is best for a notch filter, to allow a specific range through, because as frequency increase the inductor impedance increases. For frequency decrease, then the capacitor impedance increases. You might be able to work them to where the roll off of each section can provide something similar, but you will get a greatly reduced output from a given input.
#11 21661079 04 Feb 2012 14:54

Kobi Aflalo Kobi Aflalo

Anonymous

Post #11
21661079 04 Feb 2012 14:54

one more thing
when i calculate the Q-factor i use the total resistence or from a specific resistor?
Q=(w0*L)/R
#12 21661080 04 Feb 2012 17:52

Mike Burr Mike Burr

Anonymous

Post #12
21661080 04 Feb 2012 17:52

because each of the filters is "parallel", in that the capacitor and inductor are across from each other. the formula is a little different than the series formula. Parallel Q is

Q=R*SQR(C/L)

Series

Q=(1/R)*SQR(C/L)
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#13 21661081 05 Feb 2012 09:51

Kobi Aflalo Kobi Aflalo

Anonymous

Post #13
21661081 05 Feb 2012 09:51

what is SQR?
and shuld i use the total resistence or any specific one?
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Topic summary

✨ The discussion focuses on calculating inductance (L), capacitance (C), and resistance (R) values for achieving resonance in a mixed LC circuit at 100 Hz. The fundamental resonance formula ω₀ = 1/√(LC) and ω₀ = 2πf₀ is used to determine component values. For a chosen inductor value (e.g., L = 10 mH), the corresponding capacitor value is calculated (e.g., C ≈ 253 µF) to achieve resonance at 100 Hz. Resistance is noted to have minimal effect on resonance frequency but influences the quality factor (Q). The Q-factor formulas differ for series and parallel configurations: Q = (ω₀L)/R for series and Q = R√(C/L) for parallel circuits. The discussion also addresses the possibility of multiple resonance frequencies in one circuit, suggesting such behavior resembles band-stop or band-reject filters formed by combining LC sections. Simulation using OrCAD PSpice confirms the theoretical calculations. The use of practical tools like the Electrodroid app for reactance calculations is recommended for real-world component selection.

FAQ

TL;DR: For 100 Hz resonance, the LC product is ≈2.53×10^-6; “The Resistance at that point does not affect the resonance.” [Elektroda, Mike Burr, post #21661076]

Why it matters: This FAQ helps students and hobbyists quickly size L, C, and R for 100 Hz LC resonance and simulate it correctly.

Quick Facts

Targeting 100 Hz requires L·C ≈ 2.53×10^-6 (H·F). [Elektroda, Mike Burr, post #21661076]
Example picks: 10 mH needs about 2.53 µF; 100 mH needs about 25.3 µF. [Elektroda, Mike Burr, post #21661076]
At resonance, reactive impedances cancel; R acts as series resistance. [Elektroda, Mike Burr, post #21661076]
Parallel vs series Q differ: Q_parallel = R·√(C/L); Q_series = (1/R)·√(C/L). [Elektroda, Mike Burr, post #21661080]
Handy tool: ElectroDroid app provides quick reactance calculators. [Elektroda, Kevin Parmenter, post #21661071]

How do I choose L and C to hit 100 Hz exactly?

Use f0 = 1/(2π√(LC)). Rearrange to LC = 1/(2πf0)^2 ≈ 2.53×10^-6 for 100 Hz. Pick one value, solve the other. Example: choose L = 10 mH, then C ≈ 2.53 µF. This keeps resonance on target and avoids overfitting specific parts early. “1/(2π·√(LC)) = resonant frequency.” [Elektroda, Mike Burr, post #21661076]

Do resistor values change the resonant frequency at 100 Hz?

At resonance, the inductor and capacitor impedances cancel, so resistance behaves like series resistance and does not shift the resonant frequency. It still influences losses and peak amplitude, but the frequency stays set by L and C. “The Resistance at that point does not affect the resonance.” [Elektroda, Mike Burr, post #21661076]

Can a mixed LC network show two resonant frequencies?

Yes. Combining sections can produce a band-stop (notch) response with two skirt corners and a dip between them. The shape depends on how LC sections interact and where you probe. Expect lower output between the peaks and tapered edges from the high- and low-pass behavior. [Elektroda, Mike Burr, post #21661078]

What does SQR mean in the Q formulas, and which R should I use?

SQR denotes the square root. Use the resistance in the configuration of interest: the R in the parallel expression for parallel-tuned sections, and the series resistance for series-tuned ones, per the stated formulas with SQR(C/L). [Elektroda, Mike Burr, post #21661080]

What are the Q-factor formulas for series and parallel LC?

For parallel resonance: Q = R·√(C/L). For series resonance: Q = (1/R)·√(C/L). Higher R raises Q in parallel tanks and lowers Q in series tanks. These relations guide bandwidth and attenuation choices around 100 Hz design targets. [Elektroda, Mike Burr, post #21661080]

How do I compute LC if I only know the target frequency?

Start from LC = 1/(2πf0)^2. For f0 = 100 Hz, LC ≈ 2.53×10^-6. Any L and C pair whose product equals this value will resonate at 100 Hz in the ideal model. Then select practical values near those results. [Elektroda, Mike Burr, post #21661076]

If I pick 10 mH at 100 Hz, what capacitor do I need?

C ≈ (2.53×10^-6)/L. With L = 10 mH (0.01 H), C ≈ 2.53×10^-4 F = 2.53 µF. You can similarly scale: L = 100 mH needs ≈25.3 µF, a useful rule of thumb for low-frequency tanks. [Elektroda, Mike Burr, post #21661076]

My simulation changed when I duplicated L and C into two branches—why?

Duplicating L1/L2 and C1/C2 changes the network topology and effective reactances. The circuit no longer behaves like a single LC tank, so the response can split or shift. Model one resonant branch first, then add sections intentionally and re-solve LC per branch. [Elektroda, Kobi Aflalo, post #21661075]

How do I design a simple 100 Hz notch (band-stop) with LC?

Use an LC resonant section to create a notch near 100 Hz, then add gentle high-pass and low-pass shaping if needed. The resonant tank provides the rejection; outer sections taper the skirts. Expect reduced output in the stop band compared with the pass band. [Elektroda, Mike Burr, post #21661078]

Quick 3-step: set 100 Hz resonance in SPICE (OrCAD, LTspice).

Compute LC = 2.53×10^-6 for 100 Hz; choose L, then C = 2.53×10^-6/L.
Build a single LC tank and verify the AC sweep peak/dip at ~100 Hz.
Add R to study Q using Q = R·√(C/L) or Q = (1/R)·√(C/L), per topology. [Elektroda, Mike Burr, post #21661076]

Is there an Android app to speed up reactance and LC math?

Yes. ElectroDroid offers convenient calculators for reactance and many other quick RF/analog utilities. It’s a free, popular choice to explore LC values before you simulate or prototype. [Elektroda, Kevin Parmenter, post #21661071]

Can I round L or C to common values without breaking 100 Hz?

Yes. Pick a nearby standard value and recompute the partner to keep LC ≈ 2.53×10^-6. This aids parts availability and speeds simulation. “Round those values off to real world capacitors you can buy as standards.” [Elektroda, Kevin Parmenter, post #21661071]

Does resistance ever spoil my filter even if f0 stays put?

Yes—edge case to watch. In series tanks, higher R lowers Q and flattens the peak; in parallel tanks, too little R lowers Q and weakens the notch. Use Q = R·√(C/L) or Q = (1/R)·√(C/L) to predict bandwidth and depth. [Elektroda, Mike Burr, post #21661080]