magnetic field and self - Page 1
Forum section: electronics / beginners

author: texane
author_rank: Regular Contributor
timestamp: February 28, 2010, 01:16:37 pm
content: Hi,I am trying to understand how the following schema works:fromhttp://www.gae.ucm.es/~padilla/extrawork/magamp.htmlesp. the preamplifier part.This is a mag card reader but I am interested in how the magfield is used to encode information, not the reader itself.As far as I understand, the data are encoded by same polaritymagnetic dipoles being interfaced on the card stripes, creatinga force (since they are the same polarity), thus allowing to encodedata.Regarding the schema, what I guess is that by creating a force,the characteristics of the self are modified (by the flux created bythe same polarity dipole interface on the stripe). I guess that this isthe reactance of the self that is modified, but I do not know muchabout that, could you point me to some links about that? If I amright, the equivalent resistance C1 L1 has to be computed, thenthe equivalent resistance with R2, then we have the base voltagefor the transistor base when mag field is present, and when magfield is not present. Is my understanding correct?I noted there is no value for the self in the schema, but it is nota problem, I am interested in "how" such a schema should beinterpreted, in order to make mine later. If you have any linkswith the related maths, from a practical point of view...Thanks for helping,Fabien.

author: jimmc
author_rank: Frequent Contributor
timestamp: February 28, 2010, 11:13:35 pm
content: Sorry, I find it a little difficult to follow your reasoning.Remember that you have to swipe the card past the magnetic head in order to read the card.This means that, in effect, you are moving a series of small magnets past a coil of wire.Of course this will induce a corresponding series of voltage pulses in the coil.These are then amplified by the circuit you gave.The magnetic head looks as if it came from an old tape recorder, I don't think it's very critical.Jim

author: texane
author_rank: Regular Contributor
timestamp: February 28, 2010, 11:26:04 pm
content: sorry for not being clear in my reasoning. The question is basically how to compute the induced voltage in L1. Which characteristics from both the self and the stripe are to be taken into account?Regards,Fabien.

author: mkissin
author_rank: Regular Contributor
timestamp: March 01, 2010, 08:00:05 am
content: The voltage induced in L1 by the magnetic stripe moving will be given by:V = N * d(phi)/dtwhere d(phi)/dt is the rate of change of the magnetic flux produced by the stripe with time. N is the number of turns in L1.To increase the induced voltage you'd need to increase the number of turns in L1, or increase the strength of the magnetic flux. Calculating phi is going to be very tricky (to the point of being impossible). You'd need either a very good datasheet for the card you're using, or to simply measure it using a known coil.Bear in mind that d(phi)/dt is also time-dependant, so moving the card more quickly will increase the induced voltage, but will obviously also decrease the pulse widths that you get.In terms of links, check wikipedia for Faraday's law, or any introductory electromagnetics textbook.

author: texane
author_rank: Regular Contributor
timestamp: March 01, 2010, 07:18:04 pm
content: Hi,Thanks for your reply, that is exactly whatI was looking for. I read the book of Crowellabout elec and magnetism, along with someother text courses, but it is a bit difficult todigest all in once, however very passionating!Fabien.