I should point out that the energy absorbed per bead is reduced by a factor of 4. The total energy absorbed by the beads is reduced by a factor of 2. Follow the current. Dave -----Original Message----- From: drcuthbert Sent: Friday, December 19, 2003 8:40 AM To: 'Ken Javor'; Price, Ed; 'Grasso, Charles'; '[email protected]' Subject: RE: Measuring a ferrite performance
By measuring the amplitude and angle of the conducted RF, thru the bead, you can determine the L and R of the bead. An oscilloscope can be used for this but a VNA is a better method. At any rate, the reduction in radiated energy is due to the reduction in antenna current flowing on the cables and such and not so much that the bead "absorbs" the RF. An example is when the number of beads is doubled. We can use the example of a zero-impedance RF source. The RF current is halved and the radiated energy is reduced by a factor of four. The losses in the beads is also reduced by a factor of four. So, we have a reduction in radiated energy yet the beads are "absorbing" less RF energy themselves. Dave Cuthbert Micron Technology This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. Visit our web site at: http://www.ieee-pses.org/ To cancel your subscription, send mail to: [email protected] with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: [email protected] Dave Heald: [email protected] For policy questions, send mail to: Richard Nute: [email protected] Jim Bacher: [email protected] All emc-pstc postings are archived and searchable on the web at: http://www.ieeecommunities.org/emc-pstc
