2
\$\begingroup\$

I'm trying to make a 1MegHz CM filter using a common mode choke and some capacitors but I have a very high gain at lower frequencies instead of attenuation i.e high Q factor. A simulation shows the issue:

enter image description here

The filter is meant to filter 1MegHz or higher freq. CM noise from a DC output of a supply. But my fear is to damage the load because of a possible high gain at lower frequencies. How can I model it more realistic? I tried to add 1 ohm series resistors but it didn't help much. How can the Q factor be reduced in reality?

For the choke I use two of this ferrite core on top of each other with tens of turns. And here is the the actual photo:

enter image description here

Solution to the issue from the answer of @Andy aka:

enter image description here

\$\endgroup\$
2
\$\begingroup\$

Without a load resistor this circuit will have a high Q value. However, if you apply a reasonable load resistor (say 100 ohms) you will find that the Q drops to a workable level and the potential for overshoot at the resonant frequency (circa 3.5 kHz) will be much diminished.

I estimate that with a 1000 ohm load the Q will be about 10 and the peak in the bode plot will be more like 20 dB. At 100 ohms the peak will be about 1 dB.

Of course if the load can vary across a large range, it would be better to to apply resistors directly in parallel with both inductors to reduce the Q. This method ensures that there is no "DC" attenuation due to using resistors in series with each inductor winding.

If you were trying series resistors, to get the Q down to a "reasonable" level you would need values around 100 ohms and this could severely restrict the power fed to the load.

\$\endgroup\$
5
  • \$\begingroup\$ Oh man you are the master! Parallel 100 Ohm resistors worked i.stack.imgur.com/beiyZ.png I have 5W 100 Ohm resistors. I guess I can use them since in parallel with winding they wound not pass much current right? Max load current is 3A in my case. \$\endgroup\$ – cm64 Dec 10 '18 at 12:51
  • \$\begingroup\$ If the load current is DC then all that load current passes through the coils and barely any through the resistors - those resistors are just there to snub the overshoot. \$\endgroup\$ – Andy aka Dec 10 '18 at 13:04
  • \$\begingroup\$ So as not to confuse folk, maybe you can add a few words before the solution picture you have just added. Something like: "Proposed solution from answer given". \$\endgroup\$ – Andy aka Dec 10 '18 at 13:06
  • 1
    \$\begingroup\$ @cm64 but be aware now that the roll-off slope isn't as steep as before and you may want to modify values to tweak things a little bit. For instance, you could see what things are like with 200 ohms and a lower value of inductance. In other words, a solution that stops one thing can usually have a knock-on effect that requires a couple of value changes to optimize performance. You have to ask yourself if -54 dB attenuation is good enough at 10 MHz bearing in mind that before you had over 120 dB. \$\endgroup\$ – Andy aka Dec 10 '18 at 13:58
  • \$\begingroup\$ Good reminder, Andy. \$\endgroup\$ – analogsystemsrf Dec 10 '18 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.