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Natural Frequency

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kaosad

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How to compute the natural frequency of a series-parallel combination of inductors and capacitors? I have the formulae for series only and parallel only but not for mixture of both.

For example:

.....o
.....|
.....L
.....|
..+-+-+
..|.....|
..L.....C
..|.....|
..+-+-+
.....|
.....o
 
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Please draw the circuit properly.

Going from your rough description, you're talking about a circuit with more than one mode of resonance, series and parallel. The formula remains the same, it just means that it has more than one resonant frequency so calculating the response becomes harder.
 
Like I say, there are two modes of resonance:

Series, C1 and L1.

Parallel, L1 and C1.

There's nothing complicated about that.
 
Hero999 said:
Like I say, there are two modes of resonance:

Series, C1 and L1.

Parallel, L1 and C1.

There's nothing complicated about that.
If you work out the impedance, it is parallel resonant at L2, C1.
It is series resonant at (L2 || L1), C1.
In other words, if L2=1m and L1=3m, it will be series resonant at 0.75m, C1.
Of course, we know that Fr=1/(2pi*sqrt(LC))
 
speakerguy79 said:
Just take the Laplace transform and sub in jw for sigma.

Input that into a 555, output into a constant velocity joint, and read with a left handed monkey wrench.
 
Roff said:
It is series resonant at (L2 || L1), C1.

I thought this too. It all depends if the circuit is driven by a low impedance voltage source, then L2 appears to be in parallel with L1 and the circuit has one resonance at the frequency where the sum of the inductive reactances minus the capacitive reactance = 0.

Or use that other equation for frequency that you cited with the f = 1/(2*pi*(L*C)^0.5)

1/L = 1/L1 + 1/L2.
 
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Bob Scott said:
I thought this too. It all depends if the circuit is driven by a low impedance voltage source, then L2 appears to be in parallel with L1 and the circuit has one resonance at the frequency where the sum of the inductive reactances minus the capacitive reactance = 0.

Or use that other equation for frequency that you cited with the f = 1/(2*pi*(L*C)^0.5)

1/L = 1/L1 + 1/L2.
With a voltage source, the node where the 3 components join will indeed be parallel resonant at the single frequency determined by the parallel combination of L1, L2, and C1, but the source current will still have a series resonant frequency (current limited only by parasitic resistance), and a parallel resonant frequency (zero current).
 
The final form of the equation that I derived before my first post is in the graphic below. This has a zero at the origin, complex conjugate zeros (series resonance) at w=+/- 1/sqrt((L1||L2)C1), and complex conjugate poles (parallel resonance) at w=+/- 1/sqrt(L2C1).

 

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  • eqn.PNG
    eqn.PNG
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Hi everyone,

Thanks for all the feedback. I did some simulation by hooking up the circuit to a 72V (peak-to-peak) ideal AC source. The parameters were L1 = 318mH, L2 = 100mH, and C2 = 100uF. The simulation duration was 100s.

Here are the results:
  1. Using source frequency = 57.70Hz (obtained form Roff's series resonance formula), the voltage across the parallel LC tank oscillated at the source frequency but modulated by a lower frequency at around 0.374Hz. The modulation amplitude decreased and ceased at around 20s where thereafter the voltage stabilized at around 1260V (peak-to-peak). During the first 20s transient period, the maximum peak-to-peak voltage was 2700V.
  2. Using source frequency = 50Hz (obtained form Roff's parallel resonance formula), the voltage across the parallel LC tank oscillated at 55Hz but modulated by a lower frequency at around 7.16Hz. Over the period of 100s, the voltage across the parallel LC tank increased slowly from 132V to 240V (peak-to-peak) and the modulation did not cease.

I played with other values and found that 57.70Hz gave the highest peak-to-peak voltage across the parallel LC tank, whereas 50Hz seemed to be ordinary like any other frequencies around 57.70Hz. Is this what you expect to see?
 
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