# Resonance Frequency

Discussion in 'General Electronics Discussion' started by MikaM, Jun 21, 2015.

1. ### MikaM

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Jun 21, 2015
hi
some one can help me find the resonance frequency of this circuit?
(i have the answer but could not find the way)
w = sqrt( (Rs+Rc)/RcLC) )

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2. ### LvW

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Apr 12, 2014
How do you define "resonance" for this circuit? The point (frequency) of maximum current is not identical with the point of max. output voltage.

3. ### ramussons

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Jun 10, 2014
Resonance will be dependent ONLY on L and C.

Rc and Rs will only affect the "Q" of the Resonant Circuit.

Minder likes this.
4. ### LvW

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Apr 12, 2014
I don`t think that this is true for the shown circuit. The (different) peaks of the current and output voltage depend on the R values. Therefore my question how the resonance point is defined for this circuit.

5. ### ramussons

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Jun 10, 2014
I don't see how there can be more than 1 resonant frequency. Let's see how the discussion progresses.......

6. ### Minder

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Apr 24, 2015
I agree, Rs and L + Rc and C appear representative of two single circuit components.
Rs and Rc would affect the Q.
M.

7. ### LvW

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Apr 12, 2014
As I have mentioned - it depends on the definition (max. current or max. output voltage).
The current (magnitude) through the whole circuit has a maximum at a certain the frequency f1, but this does NOT mean that the magnitude of the voltage drop across Rs+L has a maximum at the same frequency. You shouldn`t forget the influence of the various phase shifts.

The calculation (finding of the magnitudes) is a bit involved because of the complex expressions - thus, the most simple method is circuit simulation.

8. ### Minder

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Apr 24, 2015
What I got from the OP is the resonant freq required only?
M.

9. ### LvW

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Apr 12, 2014
Yes - but how do you define it?

10. ### Minder

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Apr 24, 2015
I go along with post #3.
M.

11. ### LvW

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Apr 12, 2014
It`s not as simple.
As an example, the resonant frequency of a lossy L in parallel to a capacitor C is (see, for example, wikipedia):

12. ### Laplace

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Apr 4, 2010
It should be a good assumption that resonance is that point where the current flowing through the RLC network has no reactive component.

Since I=V/Z, take the reciprocal of the RLC impedance, multiply the numerator by the complex conjugate of the denominator, isolate the imaginary component of the numerator and set it equal to zero, then solve for the frequency. However, I am unable to find any expression that includes Rs in determining resonance. Note that in the simple series RLC circuit, the series resistor has no effect on resonance, so one must question whether the given 'answer' is actually correct.

13. ### MikaM

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Jun 21, 2015
thanks everybody

14. ### LvW

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Apr 12, 2014
Yes, that is another possible - perhaps the best(?) - criterion for defining resonance. It is intersting to note (prooved by simulation) that the peak of the total current through the network is NOT identical with the frequency where the current phase is zero (no reactive component).

15. ### LvW

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Apr 12, 2014
MikaM - what is your conclusion? Which definition will you use - and why?

16. ### LvW

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Apr 12, 2014
Perhaps it is interesting to numerically compare the various alternatives (definitions):
Example circuit:
Rc||C=1kohm||1nF
Rs=250 Ohms; L=10uH.
Results (Simulation):
|Imax| at 1.76 MHz
IMG(I)=0 at 1.583 MHz
|Vout,max| at 2.78 MHz.

17. ### Ratch

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Mar 10, 2013
This is a series resonant circuit. There are no reactive elements in parallel. Therefore, there is only one resonant frequency.

Ask if you have any questions.

Ratch

18. ### LvW

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Apr 12, 2014
Hi Ratch, I didn`t check your calculation - and I was too lazy for calculating by myself.
However, apparently there is a conflict beween your result (C*Rc²>1) and my simulation.
As I have mentioned above, according to simulation (symbolic analyzer as well as PSpice) the imaginary part of the total input impedance is zero at f=1.583 MHz.
This is true (my example) for C*Rc²=1E-9*1E6=1E-3 [Vs/A]
More than that, there is a conflict in units (dimensions): C*Rc² is given in [Vs/A]. Hence, it cannot be combined with "-1".

Here is the input impedance (symbolic analyzer):

Zin=[Rc+Rs + s(L+RcRsC) + s²(RcLC)]/[1+s(RcC)]

Last edited: Jun 24, 2015
19. ### Ratch

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Mar 10, 2013
There is no conflict between your answer and mine. If you plug in the numbers for my solution of omega=sqrt(-1+C Rc^2)/(C sqrt(L) Rc), you will find our answers are identical. Your equation for Zin is identical to mine if you expand it and substitute j w for s. C*Rc²>1 is a condition for resonance, not frequency calculation, so the units do not have to match. If that condition is not met, then the term under the square root sign of my solution will be negative, and omega will not be real. So, my calculation matches your simulation.

EDIT: I have to point out a mistake on my part. Because 1 and l look so much alike on my screen, I mistakenly used 1 instead of l. Therefore, the condition of resonance is C*Rc² > L, not C*Rc²>1. Sorry for the confusion.

Ratch

Last edited: Jun 24, 2015
20. ### LvW

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Apr 12, 2014
OK - problem solved because C*Rc²=1E-9*1E6=1E-3 > 1E-6.