Part of the designer's average 'EMI troubleshooting day' may involve taking a core from the shelf and just winding some turns on it. But what if he is really struggling to bring up the inductance and in the process winds several more turns on it. The core could now very well be saturating, making it quite ineffective. Self-defeating. But how can we tell? The simplest equation to check if a core is saturating or not is
where L is the measured inductance in H, N the number of turns, and Ae is the effective area of the core in m2. Ae is just the normal geometric cross-sectional area of the core (if it is an E core, we would take the area of the center limb, or twice the area of each side limb, whichever is smaller, though usually both ways we get the same result). So if we plug in the peak current we can calculate the peak B in the choke. If we know the material of the core, we would know its saturation flux density Bsat. Then we can easily check if we are saturating the core or not. Note that Bsat for powdered irons is usually around 10000 gauss (1000 mT or 1T) and for ferrites it is 3000 gauss (300mT, or 0.3T).
But what is the peak current? This is usually completely underestimated by most designers, and they don't realize that their DM choke is ineffective simply because it is saturating.
The temptation to wind a few more turns on the DM choke (if any) in an effort to boost its inductance, may just produce enough ampere-turns to saturate it. What happens after that depends a lot on the material. Powdered iron may be more forgiving, but then it has a lower initial permeability to start with. Ferrites can saturate sharply. Of course if the DM choke is just a leakage inductance created from the CM choke, then in effect it is air-cored, and we don't have to worry about saturating it. But we may have also wondered why the CM choke itself was running so hot? We did want to build a cost-effective filter, but not a poorly designed one!
In a typical off-line input stage (no Power Factor Correction), the input bridge conducts only for a part of the AC cycle as seen from Figure 1. The input (bulk) capacitor discharges slowly during the remaining time at a rate that depends on the power it is delivering to the converter (which is the PIN=POUT/Efficiency). If we put larger and larger bulk capacitance we can end up with extremely high peak and RMS currents through the input bridge and the filter chokes (along with a lot of low frequency harmonic content which is also regulated separately by various regulatory standards). This is attributable to the fact that the bridge conduction time becomes shorter as we increase the bulk capacitance. Since a certain amount of average input power is being demanded by the converter, the current bursts must increase in amplitude to compensate for the shorter gating interval.
Knowledge of the input RMS current is necessary to correctly estimate the copper losses in both the CM and DM chokes, whereas knowledge of the peak current is necessary to correctly estimate the DM core volume (its energy handling capability). In the CM choke, the input AC line current effectively cancels out so saturation is not normally a possibility.
The shape of the input current into the power supply is described as a 'haversine' which is simply a sine waveform offset on its vertical axis so as to make its minima coincide with the horizontal axis. The current waveform shown in the figure is a haversine (during the diode conduction time).
We can show that the following equations apply:
The time for which each diode conducts is
C is in Farads above.
The RMS and average values of the current waveform (calculated only over the conduction time) are
The peak value of the current is
Example: A power supply delivering 5A@14A at 70% efficiency has a 330uF input capacitor. of. What are the RMS and peak input currents at 265VAC/50Hz?
So A=0.978, and tC=0.67ms. We can calculate IAVG=4.05A, IPEAK=8.1A, IRMS=8.1*0.612=5.03A.
We have also provided a graphical method in Figure 1 for any general case at an AC input of 265VAC. The input capacitance per (input) watts is 330/100=3.3uF/W. We locate this value on the horizontal axis, and then we can see that this gives us about 0.05A for the RMS current on the vertical axis. But the vertical axis is the current per Watt of input power. So for our case, the RMS current is 100?0.05=5A. This agrees with our numerical calculation above.
Since we have two conduction intervals per AC time period, the input average and RMS currents now calculated over the whole cycle are (with T=1/fLINE)