Making a 1000 ohm ferrite bead - Page 1
Forum section: electronics / projects

author: Rickpercy87
author_rank: Contributor
timestamp: February 22, 2016, 12:52:41 pm
content: Hello, I have some ferrite beads knocking around and plenty of magnet wire in all sizes.I would like to make a 1000 ohm ferrite bead. I don't believe the frequency is too important. It will be to filter noise from our 50hz power supply.Can anybody point me in the right direction? Also how would I test this once built?Thanks

author: c4757p
author_rank: Super Contributor
timestamp: February 22, 2016, 02:51:51 pm
content: Frequency is absolutely important. A ferrite bead is equivalent to an inductor with a resistor in parallel - at low frequencies, the impedance is essentially zero. As frequency increases, the inductor starts to provide a bit of impedance (and then the voltage it drops appears across the resistor as well, giving the dissipative element). Any normal ferrite bead is essentially a short at 50 Hz, and in fact for a few orders of magnitude above.If you want to make a 1000 ohm ferrite bead effective at 50 Hz, you need an inductor whose impedance is significant (I'd say at least 1000 ohms...) at 50 Hz, which means athree henryinductor. Get yourself one of those, then put a 1k resistor across it...Unless you mean you want to filter noise going back the other way, in which case... specifics pls!If it turns out you really did want to filter much higher frequencies from going backintoyour mains, the usual way to 'make' your own "ferrite bead" with specified characteristics is to put R and L in parallel yourself. Much more effective. The performance of a true ferrite bead is highly dependent upon the exact material characteristics.

author: sarepairman2
author_rank: Frequent Contributor
timestamp: February 22, 2016, 07:43:14 pm
content: you want a high mu ferrite bead.

author: Rickpercy87
author_rank: Contributor
timestamp: February 23, 2016, 12:44:14 am
content: Cheers, I'm sure the schematic needs it to filter out power supply noise as the ferrite bead is needed right after the power supply. Also the circuit is an amplifier which amplifies a signal by 10,000X.Im still not too clear on what I need and I should probably look into the circuit a bit more.If my circuit was power by a switch mode power supply what would you recommend?Also how are you working this out???I can't remember the formula? I keep thinking or L/R which is the time constant of a inductor I think.

author: c4757p
author_rank: Super Contributor
timestamp: February 23, 2016, 12:46:37 am
content: Only equation I used up there is \$X_L = 2 \pi f L\$ to compute the equivalent reactance of an inductor at a given frequency.We need way more details to answer, "a switch mode power supply" isn't nearly enough.

author: c4757p
author_rank: Super Contributor
timestamp: February 23, 2016, 12:49:06 am
content: Quote from: sarepairman2 on February 22, 2016, 07:43:14 pmProbably - if you're working with quite low currents, or at least low differential current (that is, you're making a common-mode filter) -- neither of which has been suggested -- or otherwise if you like your filters to saturate and then do piss-all.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 23, 2016, 02:50:17 am
content: What are you working with? What is the source (power supply, I guess?)? What is the noise? Do you know what frequency range the noise consists of? (Can you identify it with an oscilloscope?) What is the load? Is it common mode or differential? Grounded, shielded, bypassed? What's the surrounding circuit?Depending on where you intend to put it, "surrounding circuit" may include everything from the mains outlet to other cables/devices attached to the box in question. Volunteer as much information as possible!Tim