# AC Circuit + Parallel resonant band-stop filter

"The parallel LC components present a high impedance at resonant frequency, thereby blocking the signal from the load at that frequency. Conversely, it passes signals to the load at any other frequencies"

When the current is flowing from "1" to "2" it makes sense, but the thing i do not understand is that since the circuit is AC, the current will flow in the other direction eventually (from "1" to "0" to "0"). So how can the LC Circuit "get in the way" then ? (present impedance)

EDIT:

Ok let me rephrase my question: The LC acts like a kind of "resistor" when current passes through it. What i do not understand is if the current goes the other way, it does not have to pass through the LC so each "odd" frequency offersno impedance.

example: at Time0, the current goes "up" so the LC presents some impedance. But at Time1, the current goes "down" so it does not even "meet" with the LC.

EDIT 2:

I think there must be something VERY fundamental that i do not understand. Here is an example with how i "understand" it:

Let's say the tank circuit is at resonant frequency and R1 is a light bulb.

When the current goes into the red direction, it can't reach R1 ("stopped" by the impedance of the LC) but when it goes in the blue direction it can go through R1 without even having "to stop" at the LC circuit (so the light bulb will glow).

• No, the current is always going through the whole series circuit. Just because it's going back and forth doesn't mean it doesn't have to complete the circuit. – Daniel Dec 12 '15 at 10:47
• Yes but when the current goes in the other direction it can reach the load without stopping at the LC so whats the point of having an LC ? – d0bz Dec 12 '15 at 11:22
• You are right. There is something very fundamental that you don't understand. In a series circuit, all currents are equal. If you hang something on OUT, it's not a series circuit anymore. If OUT is disconnected as shown, nothing "gets" to R1 "first" or "later". Everything happens simultaneously. – Daniel Dec 12 '15 at 21:30
• And again, the exact signal you see has to do with where you pick your ground. If you have some current through R1, that develops a particular voltage across R1 (as I depicted in my answer). If your loop current is defined by R1+(L1||C1), the voltage across R1 is ALWAYS defined by that current through R1. If it's going the other direction, it's just negative. The direction of current changes, and the sign of the voltage changes...but the voltage across R1 is still defined by that current through R1...which is defined by $\frac{V1}{R1+(L1||C1)}$ – Daniel Dec 12 '15 at 21:41

Assuming you set node 0 as your "ground" or "reference" node, you only care about what goes in at node 1 ($v_{in}$) and what comes out at node 2 ($v_{out}$). The voltage at node 2 fully determines the voltage and current through $R_{load}$.

Maybe you should prove what's going on by turning the capacitor and inductor into complex impedances in parallel, then finding the voltage at node 2 via the resistor divider equation. You can plot this (the magnitude, not just the real part) graphically as a function of frequency and see the voltage fall off at a particular frequency.

Another thing to realize is that you can set your ground/reference node ANYWHERE, but looking at what we want to look at here, it's most convenient to place it at node 0. But everything still works out exactly the same if you've done it correctly.

Imagine we break it up into this pair of DC circuits. Just because the voltage is reversed, it doesn't skip the circuit. It just goes around the other way.

simulate this circuit – Schematic created using CircuitLab

• Sorry but i don't understand why i should only care about 1 and 2 ? – d0bz Dec 12 '15 at 9:44
• The magic of setting a reference node is that it is always 0V by definition! Current goes in, current comes out, but that node is always 0. – Daniel Dec 12 '15 at 9:59
• Im sorry i still don't understand. If i were to draw an arrow of the direction of the current, wouldn't that arrow change side every frequency? If so, there is one direction that does not go through the LC ... (what am i missing?) – d0bz Dec 12 '15 at 10:24
• No, we use the complex impedance transformation to make AC voltages and currents work like DC voltages and currents. That should have been covered at this point. The current though, will be the same in $R_{load}$ as it is in $V_1$. The current $i_{C_1}+i_{L_1}$ is also this same current, since that parallel is in series with the other two components. – Daniel Dec 12 '15 at 10:41
• To create an analogy: if the LC is a traffic light and the circuit is a road: a car (current) start at V1, makes (or not) a stop at the LC and goes to the load. But if the car goes in the OTHER direction, it can avoids completly the traffic lights – d0bz Dec 12 '15 at 10:51

So how can the LC Circuit "get in the way" then ? (present impedance)

The beauty about capacitors and inductors is that their impedances are truly opposite to each other. There is no simple component that has negative resistance so basic resistors don't have this property.

The reason L and C have opposing impedance values is down to the basic formulas that relate current and voltage for each. For instance, the current that flows through a capacitor is: -

$I = C\dfrac{dv}{dt}$ and for an inductor $V = L\dfrac{di}{dt}$

So if you mathematically integrated the inductor formula you'd have: -

$\int V\cdot dt= LI$ or

$I = \dfrac{1}{L}\int V$

It follows that if a sinewave is applied to both components, the current in the capacitor leads by 90 degrees while for the inductor it lags by 90 degrees: -

http://www.johnhearfield.com/RC/339Z.gif

Clearly there is 180 degrees of phase shift between both currents with the common voltage applied.

It's no great leap of faith to then recognize that their impedances are opposite.

So, if you apply a sinewave to a parallel L and C and you used the correct frequency (resonant) the net currents into and out of the pair will be zero.

Conversely, if L and C are in series and you applied a sinewave of the correct frequency the impedances cancel out each other and you get a short circuit at the resonant frequency.

For both cases, the resonant frequency = $\dfrac{1}{2\pi\sqrt{LC}}$

In case you want it put another way if you had an impedace Z in parallel with -Z the net impedance is product over sum AND clearly the denominator is zero due to the sum of Z and -Z.

• Sorry but i do not understand what this answer has to do with my question ? – d0bz Dec 12 '15 at 10:40
• I've explained that at resonance the LC parallel circuit blocks current completely. You asked : "So how can the LC Circuit "get in the way" then ? (present impedance) " – Andy aka Dec 12 '15 at 10:48
• I understand how a LC circuit works, what i dont understand is why it stills "block" the current when the current goes in the OTHER direction (counter clock wise) – d0bz Dec 12 '15 at 11:04
• I'm sorry but you may think you understand how they work individually but when brought together at resonance something extra happens for them to form an infinite impedance. Also continuing to try and analyse them in terms of each half cycle relies on understanding the leading and lagging nature of their currents which, I have tried to make as clear as I can. – Andy aka Dec 12 '15 at 11:10
• Do you have any issues with the current flow into a resistor when just an inductor is present? – Andy aka Dec 12 '15 at 11:11

example: at Time0, the current goes "up" so the LC presents some impedance. But at Time1, the current goes "down" so it does not even "meet" with the LC.

Imagine the circuit like a tube filled with Water. Some of the current might not "meet" the LC but others will. Current is not at only one place in the current (and it is VERY fast).

If you think of the "tank" (the LC circuit) as a variable resistor with a resistance (impedance, actually) that varies with frequency, and that at resonance its impedance becomes very high, then, at that frequency, the current through the load will be diminished because of the tank's high impedance. However, at frequencies above and below resonance, the tank's impedance will fall, allowing the current into the load to increase, and the voltage across it to rise.

Here:

EDIT

Hmmm... I just noticed I used 10 milliherys instead of 100. No big deal though, that'll just change the frequency of the suckout at resonance.

Here's the LTspice circuit list if you'd like to play with the circuit:

Version 4
SHEET 1 880 680
WIRE 128 16 80 16
WIRE 256 16 208 16
WIRE 80 64 80 16
WIRE 80 64 -64 64
WIRE 256 64 256 16
WIRE 368 64 256 64
WIRE 416 64 368 64
WIRE -64 112 -64 64
WIRE 80 112 80 64
WIRE 144 112 80 112
WIRE 256 112 256 64
WIRE 256 112 208 112
WIRE 368 112 368 64
WIRE -64 240 -64 192
WIRE 368 240 368 192
WIRE 368 240 -64 240
WIRE -64 288 -64 240
FLAG -64 288 0
FLAG 416 64 OUT
SYMBOL cap 208 96 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C1
SYMATTR Value 10µ
SYMBOL ind 112 32 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 5 56 VBottom 2
SYMATTR InstName L1
SYMATTR Value 100m
SYMBOL res 352 96 R0
SYMATTR InstName R1
SYMATTR Value 1000
SYMBOL voltage -64 96 R0
WINDOW 123 39 51 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value ""
SYMATTR Value2 AC 1
TEXT -48 264 Left 2 !.ac oct 1024 28 1000

• I think there is something very fundamental that i do not understand. I still do not understand with your answer how the current will meet with the tank before the R1 when it's going the other way (counter clockwise) – d0bz Dec 12 '15 at 18:44
• @ dobs: Yeah, this one's tough because it deals with "imaginary" quantities which are hard to grasp without understanding their mathematical underpinnings. What works for me when I run into something like this is to read as much as I can from as many sources as I can, and then read them over and over until I see the light. Try that and, eventually, things may make sense, the fog will clear and you'll be ready for your next foray into the unknown. :) – EM Fields Dec 12 '15 at 22:59