# Parallel LC filter

Discussion in 'Homework Help' started by anhnha, Oct 14, 2013.

1. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
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In parallel LC filter, what is filtered, voltage or current?
Assuming that all components and wires are ideal, with zero internal resistances.
In the picture, Vin = Vout at all times, right?
Now consider two extremes:
• If Vin is a DC constant voltage source, zero frequency. Vin = V = constant ≠ 0, L behaves as a short circuit and C as an open circuit. The equivalent reactance of LC will be zero and therefore, Vout= 0
And we have Vout = 0 ≠ Vin = V = constant ≠ 0
• If Vin is a sinusoidal voltage source with infinite frequency, and non-zero amplitude. L behaves as an open circuit and C as a short circuit . The equivalent reactance of LC will be zero and therefore, Vout= 0
And again we have Vout = 0 ≠ Vin ≠ 0

From two cases above, Vin is not always equal to Vout. I don't understand why.

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Jul 18, 2013
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Take a look at what the output will be at resonance?
Max.

3. ### #12 Expert

Nov 30, 2010
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It looks intentionally confusing to me. If you assume ideal for everything, the ideal generator (either AC or DC) will supply infinite energy into the ideal wires and the whole planet melts. For this example to work, you have to imagine that the source of energy has some impedance. In that case, either the inductor or capacitor will short circuit the generator and the voltage ends there.

In real circuits, nobody would try to see if a microfarad will short out a kilowatt at a gigahertz. It's always about supporting the voltage of a certain frequency and attenuating all the other frequencies, and there is always some impedance in the generator. That impedance forms a voltage divider with the impedance of the components, and that is how the output voltage can be different from the input voltage.

4. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
Yes, I know that at resonance the equivalent reactance of LC is infinite and therefore Vout = Vin. But what I am confused is about two cases above.

5. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
I know that those assumptions are not practical but I did it because I want to understand how it works in theory.
Yes, if we don't ignore the impedance of voltage generator, then all problems solved.
In my ideal model, now I see it looks like connect an ideal voltage source with an ideal wire.

6. ### WBahn Moderator

Mar 31, 2012
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This is another example of constructing a circuit that has no solution given the assumptions of an ideal model. Forget about the C and the R, just have the L there. Now apply a DC voltage Vin from an ideal source. Is the voltage across the inductor Vin because that is the voltage across an ideal source, or is it 0V because the inductor is acting like an ideal short? It can't be both, but the ideal models used don't allow for a resolution.

That doesn't mean that the exercise is pointless. What it does is give some insight into how, why, when, where, and what parasitic components can play an important role in circuit behavior.

Feb 17, 2009
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8. ### #12 Expert

Nov 30, 2010
18,076
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I think you found how it does not work in theory because you set up conditions that defeat its purpose. As long as every physical part is assumed to be "ideal", the only possible input that will not cause an explosion is the exact frequency of resonance. Only a practical version will act as a useful filter. The theory to be learned is about the impedance of the parts causing a slope in the graph of the output voltage vs frequency. Only then can you learn the theory of how the "Q" of the inductor affects the rate of change of the response and how certain types of capacitor fail at high frequencies. These things can not be studied in a circuit that only has one frequency that does not make it burn up.

9. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
Hi,
If we consider internal resistance of generator, what do you call the filter, voltage filter or current filter?
In this case, I see both of them are correct.

10. ### WBahn Moderator

Mar 31, 2012
23,556
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It's a current filter. If you make a filter that relies on the output impedance of the voltage source that supplies it having some minimum impedance, they you need to work on choosing a better circuit topology.

11. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
Sorry, I don't quite understand your point. Here is the circuit with the addition of Vs and Rs.

At zero and infinite frequencies, the equivalent reactance of LC are zero and therefore, the output voltage, Vout, is zero. All supply voltage drops in Rs.
At resonant frequency, the equivalent of LC is infinity and therefore, all voltage supply drops in LC part and therefore Vout = Vin.
And in this case, it seems that voltage is filtered here.
Similarly, the current is also filtered.
And so, can we say that it is both voltage and current filter?

Actually, I am reading about power amplifier and the filter in the circuit.

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12. ### LvW Well-Known Member

Jun 13, 2013
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anhnhna,

in general you are right. In reality, all frequency-dependent circuits (in particular, filters) exhibit a kind of filter effect - independent on the (real) source (voltage or current).
But the kind of filtering depends on the source resistance/impedance.
More than that, you must discriminate not only between the type of input sourc but also between the two cases: voltage-out or current-out.

Simple example: First order passive RC low pass (voltage-in and voltage-out):
* V-in and V-out: 1st order low pass
* V-in and I-out: 1st order high pass
* I-in and V-out: Ideal integrator
* I-in and I-out: Unity transfer

A short comment to your above example:
* Vout=Vin (as mentioned by you) cannot be true.

13. ### WBahn Moderator

Mar 31, 2012
23,556
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I see a circuit with the addition of some more circuitry. I don't see an identified Vs and Rs.

If the LC goes to infinite impedance, how does that result in all of the supply dropping across the LC part? Remember, you have an Rs and an Rload.

And, again, do you really think that it is a good circuit if it can't work if driven by a good (i.e., very low output impedance) voltage source (and you are wanting to drive it with a voltage signal)? Note that I am not talking about situations in which impedance matching is an issue.

14. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
Well, sorry, I forgot to attach it. I am going edit it now.

I didn't take the load impedance, RL, into account. Yes, if the equivalent of LC part are infinite, then the total impedance of L, C, and RL will be equal to RL.
And therefore, the voltage across RL:

$v_{ R_{L} } = \frac{ R_{L} }{R_{s}+ R_{L}} . V_{s}$

Now I am really confused. How do you classify low-pass, high-pass, band-pass filters?

I have just calculated frequency responses of them in two cases.
The first case: calculate frequency responses without resistance of the generator, Rs, and load resistance, RLoad.
The second one: calculate frequency responses with the present of Rs and RL.

In the first case, it is easy and intuitive to recognize the kind of filter( low-pass, high-pass, band-pass filters) from frequency responses.
However, in the second case, the frequency responses are totally different from the first case.
To me, they are not look like responses of filters.
No, that is not good.
I think we all usually want to have voltage source with very low output impedance. And a circuit has to work with the source. If output impedance of the voltage source are large then, maybe, losses by heat may be large.

15. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
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Here are the frequency responses of first order low pass filter I calculated in two cases.
Case 1: Not include output impedance of voltage source, Rs, and load impedance, RL.

$H(f)= \frac{ V_{c} }{V} = \frac{ \frac{1}{ \omega C} }{ \frac{1}{ \omega C} + R} = \frac{1}{1 + j \omega RC} = \frac{1}{ \sqrt{1 + ( \omega RC)^{2} } } \angle - tan^{-1} ( \omega RC)$

From this, we can see if ω is very high, Vc is almost zero and if ω = ∞ then Vc = 0.
As ω = 0 => Vc = V
Therefore, from the frequency response, I can see it is a low pass filter.

Case 2: Calculate frequency response at the present of both Rs and RL.

The impedance of RL and C part:

$Z_{1} = \frac{ R_{L} \frac{1}{j \omega C} }{ R_{L} + \frac{1}{j \omega C} } = \frac{ R_{L}}{1+ j \omega R_{L} C}$

The voltage across the capacitor:

$H(f)= \frac{ V_{c} }{V} = \frac{ Z_{1} }{ R_{s} + R_{L} + R + Z_{1} } = \frac{ \frac{ R_{L}}{1+ j \omega R_{L} C}}{R_{s} + R_{L} + R + \frac{ R_{L}}{1+ j \omega R_{L} C}}$

$
H(f)= \frac{R_{L}}{R_{s} + R_{L}+ R + j \omega R_{L}(R_{s}+ R)C }$

Now if input signal have zero frequency:

$V_{c} = \frac{R_{L} }{R_{s}+ R_{L}+ R }. V$

Output signal is attenuated significantly.
However, in this case, it is clearly an low pass filter. At high frequencies, output signal is also attenuated.

Well, that is not same as what I first calculated.
From this example, I see that the output signal is decreased as load is added and output impedance of voltage source is taken into account.

In reality, do you use any method to solve that problem?
I think if we can transform load impedance to infinity then the attenuation will decrease.

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16. ### Jony130 AAC Fanatic!

Feb 17, 2009
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In real world we almost always design the circuit in such a way that Rs<<RL.
But even if Rs = RL this LC circuit and RC circuit, still behave as band-pass/low-pass filters.

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17. ### LvW Well-Known Member

Jun 13, 2013
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Anhnha, did you realize that in your calculation you have assumed R=RL?
Or did you exchange R with RL in the second line of your calculation?

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18. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
We can ignore Rs but how about RL. It will have great effect to the circuit.

19. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
61
Well, thanks for spotting that. I am going to change it now.

20. ### anhnha Thread Starter Well-Known Member

Apr 19, 2012
866
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Can anyone explain that? I really get stuck.
Why we can ignore RL as it has a great impact in characteristic of the filter?