I have a CMC that has is rated at about 1kR at 100Mhz at 3A
However my circuit uses only 1A. Do I need to derate my inductance since there is less current flowing through it ?
common-mode-chokederatinginductance
I have a CMC that has is rated at about 1kR at 100Mhz at 3A
However my circuit uses only 1A. Do I need to derate my inductance since there is less current flowing through it ?
Even though this questions looks like very specific, it can be treated indeed as a much more general case filtering question: "How can one filter out electrical noise coming from power electric motors?".
The first information data we need to gather in advance is the type of noise our circuit is exposed to. Sometimes it is really difficult to get this data in advance, sometimes it is even harder to measure the noise without prior experience and high-end laboratory equipment.
In general, we can assess our noise sources in terms of:
The above is a partial list, incomplete, which may serve as a starting point only.
Then, there are a lot of techniques, I mean literally hundreds of tricks and broader approaches depending on the case.
Delving into the specifics of the original question, this is my best guess on the sort of noise that may be originated by the system,
Some practical hints and techniques for dealing with the noise in the system above:
Returning to the circuit you posted, my initial approach would be:
Last but not least, devise a simple way to measure your circuit at critical points, in order to verify the effectiveness of the different approaches. Do, please, try to test under similar circumstances as the real device will operate under.
If neeeded, I can provide more references (books, articles) to the approaches aboves. If you can specify in greater detail some parts of your system, additional filtering techniques will surely apply.
For the circuit you have drawn, yes, the voltage spike has nothing to do with the value of the inductor. It is given entirely by the value of the current in the inductor, and the resistive load R2 across it.
What does that mean if R2 is a higher value, or even absent? In theory, with what you have drawn, if R2 was open circuit, then the voltage spike would be infinite. As you can guess, that doesn't happen in real life.
In practice, there are two things omitted from your drawing.
a) the stray capacitance across the inductor, and due to any wires from the inductor terminal to ground
b) any breakdown mechanism for your switch
capacitance
As the switch opens, the current will start to charge the stray capacitance, which will limit the rate of rise of the voltage. For large value inductors, with many turns in close proximity, this capacitance can be surprisingly large.
Sometimes an external capacitor is added to the inductor deliberately to reduce the rate of voltage rise.
No matter whether your switch is mechanical one with opening contacts, or a semi-conductor one like a MOSFET, it will not support an infinite voltage.
the switch
Mechanical switches are especially poor at breaking the current flow, as at the first break, the contact separation is very small, and an arc needs very little voltage to form. This arc will keep the current flowing, and damage the contacts. It is responsible for switch and relay failure, unless controlled.
In the old-style contact breaker car ignition system, the 'points' that connected and disconnected the coil to the battery could be subject to excess erosion from arcing. Often, the first sign that your 'condenser' (capacitor) had failed would be excessive wear at the points. The capacitor, fitted across the points, slows the rate of voltage rise so that the points are sufficiently far apart before the voltage gets high enough to create an arc.
The specifications for a MOSFET will typically give a breakdown voltage figure. Good ones will also give an energy they can withstand when the breakdown voltage is exceeded. As long as the stored energy in the inductor is less than that figure, a MOSFET can switch current off to a coil, limit the open circuit voltage to its breakdown voltage figure, and survive.
Best Answer
If your common mode choke is passing a regular circuit current of 1 amp then that will be a differential current i.e. it flows into one terminal on one side and returns in the opposite direction through the equivalent terminal on the other winding: -
Under these circumstances the currents are in opposite directions and the fluxes cancel out hence there is no need to derate the inductance. In any case, inductance will only be reduced (usually) when the common mode current reaches a significant level: -
Picture source.