Parallel Resonant circuit with L with series R

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  • #1
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Main Question or Discussion Point

Most people in electronics know that the resonant frequency for a parallel LC network is given by w=1/sqrt(LC). But, few realize that when there is a resistor in series with the inductor, the resonant frequency is shifted downward.

When I solve the equation for the resulting impedance, I get:
Z=(R+jwL)/[(1-w2LC)+jwRC. I multiply the top and bottom by the complex conjugate and get:
Z=[R - j(wR2C-wL+w3LC)]/[(1-w2LC)+(wRC)2]

Since the phase at resonance is 0, I can equate the imaginary part to 0 and solve for w|w0:

[w0R2C-w0L+w03LC/]/[(1-w02LC)+(w0RC)2] = 0

=> w0=sqrt(1/LC -R2/L2).

So, here's my question....according to this equation, if R=0, we get the standard, w0=sqrt(1/LC). √. For any R, the resonant frequency will be shifted to a lower frequency. However, this is only true as long as the a 1/LC - R2/L2 is ≥0, which only occurs for R ≤ sqrt(L/C), since frequency must be real.
So, what would it mean physically, if R > than this limit allows? What should the expected behavior of the circuit be?

I suspect if I plug this negative frequency into the original equation for the total impedance I'd just get a different complex impedance at that frequency, instead of the resistance I'd get if I plugged in a real w that satisfies the equation for w0, but I"m not sure what that means.....no resonance at any value of R that results in w0 being imaginary?
 

Answers and Replies

  • #2
AlephZero
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It R is bigger than your limit, won't get "resonance" at any frequency. If you apply a steady voltage across the circuit and suddenly remove it, the current in the circut will fall to zero in a smooth curve instead of oscillating. If you apply a sine wave signal at different frequences, you won't get a peak in the response at any "resonant frequency".

Simple discussions of resonance in electrical circuits tend to ignore this situation, since in practice you usualy want R to be as small as possible. Google for tutorials about mechanincal vibration, and under-damping, over-damping and critical damping, for more on this.
 
  • #3
vk6kro
Science Advisor
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The actual effect you get depends on the surrounding circuit.

To see the resonance of this circuit you would put a resistor in series with it. Maybe 1000 ohms.

Now, when the circuit becomes resonant, the current it will draw through the resistor drops, so the resistor drops less voltage and the voltage across the tuned circuit rises to a nice peak, as you would expect.

Click on this:

parallel LC circuit.PNG


This diagram show the red trace from the circuit at the right, with no resistor in series with the coil and the green trace of the circuit at left with a resistor in series with the coil.

However, as you add resistance in series with the coil, the circuit gets more and more like a low pass filter.
The 1 K resistor and the capacitor form the low pass filter and the coil is unable to contribute much effect at all.

So, you get fairly constant output below the cut-off point and then a steady reduction in output above that point.
 
  • #4
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thank you both for your responses!
 
  • #5
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I had a chance to get back and see if the simulation matched the expected results.

Following the example from left hand side of the example provided by vk6kro, the transfer function I get for Vogreen is given by:

Vogreen=(s*L +R3)/(s2L*R2*C+s(R2+R3*C+L)+(R2+R3))

The roots of the transfer function should determine the poles, and also the damping of the circuit, based on whether the roots are 0, real, or complex.

When I equate the discriminant to 0, and solve for R3, I get that critical damping should occur when R3=210. R3 <210 results in under damping, and R3>210 should be damped. This matches the simulation (after applying a step response to transfer function in Laplace domain and then taking the inverse Laplace to get back to time domain)...

Vo(t)=5[(1/(R2*C)*(1/(beta-alpha)*(e(-alpha*t)-e(-beta*t))))+R3/(R2*L*C)*(1/(alpha*beta)+e(-alpha*t)/(alpha*(alpha-beta))+e(-beta*t)/(beta*(beta-alpha))),

where alpha and beta are the roots of the transfer function.

So this makes sense--when the total circuit is taken into account. However, when I'm just looking at the "tank" circuit (and the resistor has in series with the inductor), and the natural frequency is given by

w0 =sqrt(1/LC - R2/L2).

I still fail to relate this to the "complete circuit" solution or understand what it means then when the value of R makes w0 imaginary in the equation above. For example, solving for R for w0=0 results in R=100 (suggesting any R greater than that will result in an 'imaginary" resonant frequency --in which case the phase is not 0, and therefore, the circuit is no longer resonant. This 100 ohms, though, doesn't seem to show any distinctive attribute when I run the LTC Spice simulation, ie, it's not much different than 90ohms or 110 ohms, for that matter. Below is the circuit simulated....

V1 N001 0 PULSE(0 .5 0 10n 10 5M 10M) AC 1
L1 N003 0 100m
R3 N002 N003 {r}
C1 N002 0 10µ
R2 N002 N001 1k
.tran 0 50m 0 100n
.step param r list 50 90 100 110 150 200 210 250
.backanno
.end


I seem to be confusing 2 different things here. Can anyone shed some light?
 
  • #6
can i ask a quick question? how come the equation that i got from my lecturer is

Wo = sqrt ( 1/LC - (R/2L)^2 ) ?
is there any different? i tried using your equation in my homework, but i cant get the correct answer.
 
  • #7
psparky
Gold Member
884
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True or false?

At resonance....the source sees zero reactive power and the ciruit is at it's max power flow.
 

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