# Placement of filter components to reduce EMI

An electro-magnetic interference (EMI) source is connected to a filter and to a LISN.

The EMI source is a common-mode source: it generates an undesirable signal which flows from the generator $V_g$, equally divides itself and flows through the branches HOT and NEUTRAL and comes back to the generator through ground (notice that LISN too is connected to the same ground).

The filter is placed between the EMI source and the LISN in order to deviate to ground the EMI signal before it reaches the AC mains. Input voltage in the LISN is a measure of the EMI signal which will exit into the AC mains.

There are two available topologies for the filter.

First configuration:

simulate this circuit – Schematic created using CircuitLab

Second configuration:

simulate this circuit

$L_1$ and $L_2$ are (in both circuits) mutually-coupled inductors, with $L_1 = L_2 = L \simeq M$, so both have a total inductance of $L + M \simeq 2L$.

Notice that the voltages $V_L$ are the same because the two branches are equal and the current produced by $V_g$ equally splits between the two branches. The first configuration gives a higher $V_L$ than the second, so the first configuration is the worst.

Why?

I drawn the equivalent circuits for the whole signal: the returning path to which I am interested in is through ground. So, in the first configuration the equivalent circuit is

simulate this circuit

and in the second one it is as follows:

simulate this circuit

The left resistance is $R_{LISN} / 2$ because it is the parallel of the input resistances of the LISN seen from HOT to ground and from NEUTRAL to ground: if they are both $R_{LISN}$, $R_{LISN} || R_{LISN} = R_{LISN} / 2$.

Then I obtained the $V_L$ voltages: in the first case it is

$$V_L = V_g \displaystyle \frac{(R_{LISN} / 2) || (2C_y)}{R_g + sL + (R_{LISN} / 2) || (2C_y)}$$

and in the second one it is much more complicated. If I did not make mistakes it should be

$$V_L = V_g (R_{LISN} / 2) \displaystyle \frac{sL + (R_{LISN} / 2)}{R_g(2C_y + sL + (R_{LISN} / 2)) + 2C_y(sL + R_{LISN})}$$

I know that $R_g \gg R_{LISN}$; moreover, in such cases the inductors should be placed after a small impedance and the capacitor after a great impedance. This is confirmed by the fact that the second configuration is better than the first. But, again: why?

I can't immediately see this from the expressions I obtained: both have $R_g$ in the denominator. Even if it is not immediately visible from the maths, which could be the physical reason for this behaviour?

• I think the impedances in your simplified circuits are supposed to be Rg, Cy/2 and 2L (because it is a common-mode choke). Although it has been a while since I have done this kind of thing. – mkeith Sep 9 '15 at 16:55
• Two coupled inductors gives 4L – Andy aka Sep 9 '15 at 16:59
• @Andyaka why? If the two coupled inductors have the same current, they have a total inductance of $L + M = 2L$ and $2L || 2L = L$. It is like resistance. – BowPark Sep 9 '15 at 17:35
• @mkeith $R_g$ is correct; the capacitor impedance in the equivalent circuits is $1/(s2C_y)$ and yes, it is half the beginning impedance $1 / (sC_y)$. For $L$, please see the comment above. – BowPark Sep 9 '15 at 17:37
• Actually regarding the capacitors, I was wrong, but unclear, and you found a way to interpret what I said as something correct. I was thinking of differential impedance. For common mode impedance, it really is 2C. Did you try simulating the circuit to confirm your calculations? Regarding the CM choke, I am not too sure at the moment so I will say nothing. – mkeith Sep 9 '15 at 18:13

## 1 Answer

Sparing the maths, an easy way to think about EMI is a high frequency voltage will take the path of lowest impedance back to its source. In short, the filter direction will make a difference, in the real world but not so much in the ideal case.

If the filter is placed with caps to the noise source, the preferred current path will be through the caps. You could imagine L as a large impedance placed after a small impedance, the high frequency currents preferred path back to the source is through the smaller resistor, of course you have to make sure the inductance of the filter is significantly higher than the inductance of the cap (which is going to be in the nH range).

simulate this circuit – Schematic created using CircuitLab

The problem with the second configuration is there is no place for the HF current to go after it is blocked from the filter inductor. Any voltage RF that gets through the inductor will be seen by both the load and the caps equally. It may also be attenuated, however, there may be another preferred path an that is through mutual inductance between cables and capacitive coupling through the air to other physical things to get back to the source. It is extremely difficult to come up with a circuit model of exactly what would happen in a situation like this. There are too many parasitics and cables and terminals start functioning like antennas. If you have ever studied antenna theory, it is more of an art. If the inductor could do the blocking on its own, then why would you need caps in the first place? Ideally they look similar, once you add the parasitics in and account for other effects they are not.

simulate this circuit

How do I know this? I actually encountered this very problem with a product this week. I had an issue with a switching load. The load was producing RF, which was causing interference on the PCB, even to the extent of causing a micro controller to 'drop out' and have a communications failure. I placed a ferrite on the cable which would be just like adding a filter with only HF inductance\impedance. The problem was attenuated. Then I added capacitors the problem was gone, and it worked better when the caps were pointed toward the load.

Remember the rule: HF will take the lowest impedance pathway, if you provide a better low inductance pathway, you can attenuate the problem. Oh, if your building or selecting a filter, make sure it has low inductance between the caps by placing components close together.