From: Bobhelenmcclure@.......

Date: Thu, 5 Oct 2006 16:11:35 EDT

I submitting this note in order to give others the benefits of my own sometimes trial-and-error efforts to build a seismic sensor. I am sure that Chris Chapman will continue to contribute, as well. Both Chris Chapman and I favor the use of Neodymium Iron Boron magnets arranged in a four-pole structure for generating the magnetic field needed for velocity sensing and damping in seismic sensors. The magnets are powerful, relatively cheap, and easily obtained. The magnets I favor are block magnets, 6 mm thick, 18 mm wide, and 50 mm long. They are magnetized in the thickness direction and are of the composition known as N38. They have a coercive force of about 12,000 Oersteds (CGS units), corresponding to 1.2 Teslas in the SI system. Construction of a four-pole magnet assembly is very simple, but also very hazardous. These magnets are attracted to steel and to each other with a force of about 57 pounds per square inch, and any skin caught between them can be badly nipped. You must keep loose magnets very far away from steel or each other and wear heavy leather work gloves when handling them. The magnets are brittle, and will break if they crash together or with steel. The pole plates are mild steel, thick enough to carry the flux of the magnetic circuit. More on that later. One of the plates must have clearance holes for 3/16" bolts, one hole in each corner. The plates, for the above magnet dimensions, should measure two inches by at least three inches. Slide two magnets onto each plate, with opposite poling on each plate. When you have one magnet in position on a plate, carefully hold the second magnet above the first. You will feel either an attraction or a repulsion of the second magnet toward the one on the plate. Make sure the force is repulsive, then slide the second magnet into place beside the first. When the magnets are positioned on each plate, make a wooden or plastic shim somewhat thinner than the final magnet gap you intend to use. I would use a shim about six inches long and the same width as the paired magnets, which would be about 1.5 inches for the example. With the shim covering the magnets of one plate, hold the second plate over it, getting no closer than enough to determine if the force between them is attractive or repulsive. The correct alignment is indicated by an attractive force. You now place the upper plate (magnets down) on the shim at one end of the lower plate and slide the upper plate and magnets up over the lower magnets. You now have a sandwich of upper and lower assemblies tightly squeezing the shim in between. Place four bolts in the plate with the holes. Note than these bolts are used to hold the plates apart, not together. Use the bolts to jack the plates apart to get the desired gap spacing, and then you can withdraw the shim. If each of your plates have clearance holes drilled in them, you will need two nuts for each screw. Now it is time to discuss what field strength you get, and what thickness of steel is required. Suppose for my example using 6 mm thick magnets, I use a gap of 6 mm. The resulting distance between plates is 18 mm, 12 mm filled with magnet, and 6 mm with air. I call the ratio of total magnet thickness to total plate separation the filling factor. To a first approximation, the gap field the coercive force of the magnetic material multiplied by the filling factor. In this example, the coercive force is 12 KOe and the filling factor is 2/3, so the gap field is 8 KOe. The accuracy of this estimate depends on the magnet width compared to the gap size. If the gap becomes appreciable compared to the width, you will get more fringing field and less-than-expected gap field. I have written a program that solves LaPlace's equation by iteration which lets me estimate more exactly what the penalty is for any geometry. For my example magnet structure, I use 1/4 inch thick steel plates. I find that there is a very slight saturation of the plates in the region between magnet pairs. When I use 24 mm wide magnets, there is a lot of saturation and a lot of external stray field around the whole assembly. I have to put steel side plates on the assembly to eliminate the saturation and the stray field. Steel can carry a flux density of about 20 to 24 KOe without saturation. Without saturation, the magnetic circuit of a 4-pole structure consists of a closed rectangular flux path, with lines traveling through magnets, gaps, and steel. The steel must carry the same total number of lines of flux as the gap and magnets. The number of lines is proportional to the gap flux density multiplied by the magnet width. In the example of 8 KOe and 18 mm width, this corresponds to 8,000 times 1.8, or 14,000 lines per lineal centimeter of magnet structure. The quarter-inch-thick steel (0.625 cm) must carry a flux density of 22,000 KOe. It does so, barely. The next topic is coil design. By following the above magnet assembly design principles, you know the approximate gap field, so by also knowing the magnet length you will be able to estimate the number of coil turns you will need to achieve a given output sensitivity in volts per meter per second. If the end loops of the coil extend beyond the ends of the magnet, each turn is immersed in a total field length of twice the magnet length. In the example design, the magnet length is 50 mm, so L per turn is 0.1 m. The output voltage generated by the moving coil is Volts = B * N * L * Vel, where B is field strength in Teslas, N is number of turns, L is length per turn in the field, and Vel is velocity in meters per second. One Tesla corresponds to 10,000 Oersteds. Suppose n = 1100, B = 0.8 Tesla, and L = 0.1 meter. Then Volts/Vel = 0.8 * 1100 * 0.1 = 88 v-s/m, which is a good number to strive for. A wire size of #38 or less will allow this number of turns to fit comfortably within the gap field cross-section. The coil will have a resistance low enough to permit resistive shunt damping of a pendulum weighing up to a kilogram, in my opinion. My sensors have a pendulum mass of about 0.1Kg, and critical damping is achieved at about 30 kOhms. Since the coil resistance is only 340 Ohms, the shunt damping imposes negligible loss on output sensitivity. Even a kilogram mass would require only about 10% loss of output using shunt damping. There is one significant complication to this type of magnet and coil design, having to do with the fact that pure copper is diamagnetic, and coil and its pendulum are subject to forces other than the desired restoring force of a garden gate pendulum. In the example design this diamagnetic force is a decentering force, making it very difficult to adjust the sensor to a state of stable long period equilibrium. I reduce this effect by essentially potting the coil in acrylic plastic, which is also diamagnetic, and having the boundaries of the plastic extend well outside the edges of the magnetic field. Using a heavier pendulum would be of great benefit, as well. Bob McClureI submitting this note in order to give others the benefits of m= y=20 own sometimes trial-and-error efforts to build a seismic sensor. I am sure t= hat=20 Chris Chapman will continue to contribute, as well.Both Chris Chapman and I favor the use of Neodymium Iron Boron=20 magnets arranged in a four-pole structure for generating the magnetic field=20 needed for velocity sensing and damping in seismic sensors. The magnets are=20 powerful, relatively cheap, and easily obtained. The magnets I favor ar= e=20 block magnets, 6 mm thick, 18 mm wide, and 50 mm long. They are magnetized i= n=20 the thickness direction and are of the composition known as N38. They have a= =20 coercive force of about 12,000 Oersteds (CGS units), corresponding to 1.2 Te= slas=20 in the SI system.Construction of a four-pole magnet assembly is very simple, but=20= also=20 very hazardous. These magnets are attracted to steel and to each other with=20= a=20 force of about 57 pounds per square inch, and any skin caught between them c= an=20 be badly nipped. You must keep loose magnets very far away from steel or eac= h=20 other and wear heavy leather work gloves when handling them. The magnets are= =20 brittle, and will break if they crash together or with steel. The pole plate= s=20 are mild steel, thick enough to carry the flux of the magnetic circuit. More= on=20 that later. One of the plates must have clearance holes for 3/16" bolts, one= =20 hole in each corner. The plates, for the above magnet dimensions, should mea= sure=20 two inches by at least three inches.Slide two magnets onto each plate, with opposite poling on each=20 plate. When you have one magnet in position on a plate, carefully hold the=20 second magnet above the first. You will feel either an attraction or a repul= sion=20 of the second magnet toward the one on the plate. Make sure the force is=20 repulsive, then slide the second magnet into place beside the first. When th= e=20 magnets are positioned on each plate, make a wooden or plastic shim somewhat= =20 thinner than the final magnet gap you intend to use. I would use a shim abou= t=20 six inches long and the same width as the paired magnets, which would be abo= ut=20 1.5 inches for the example. With the shim covering the magnets of one plate,= =20 hold the second plate over it, getting no closer than enough to determine if= the=20 force between them is attractive or repulsive. The correct alignment is=20 indicated by an attractive force. You now place the upper plate (magnets dow= n)=20 on the shim at one end of the lower plate and slide the upper plate and magn= ets=20 up over the lower magnets. You now have a sandwich of upper and lower assemb= lies=20 tightly squeezing the shim in between.Place four bolts in the plate with the holes. Note than these bo= lts=20 are used to hold the plates apart, not together. Use the bolts to jack the=20 plates apart to get the desired gap spacing, and then you can withdraw the s= him.=20 If each of your plates have clearance holes drilled in them, you will need t= wo=20 nuts for each screw.Now it is time to discuss what field strength you get, and what=20 thickness of steel is required. Suppose for my example using 6 mm thick magn= ets,=20 I use a gap of 6 mm. The resulting distance between plates is 18 mm, 12 mm=20 filled with magnet, and 6 mm with air. I call the ratio of total magnet=20 thickness to total plate separation the filling factor. To a first=20 approximation, the gap field the coercive force of the magnetic material=20 multiplied by the filling factor. In this example, the coercive force is 12=20= KOe=20 and the filling factor is 2/3, so the gap field is 8 KOe. The accuracy of th= is=20 estimate depends on the magnet width compared to the gap size. If the gap=20 becomes appreciable compared to the width, you will get more fringing field=20= and=20 less-than-expected gap field. I have written a program that solves LaPlace's= =20 equation by iteration which lets me estimate more exactly what the penalty i= s=20 for any geometry.For my example magnet structure, I use 1/4 inch thick steel plat= es.=20 I find that there is a very slight saturation of the plates in the region=20 between magnet pairs. When I use 24 mm wide magnets, there is a lot of=20 saturation and a lot of external stray field around the whole assembly. I ha= ve=20 to put steel side plates on the assembly to eliminate the saturation and the= =20 stray field. Steel can carry a flux density of about 20 to 24 KOe without=20 saturation. Without saturation, the magnetic circuit of a 4-pole structure=20 consists of a closed rectangular flux path, with lines traveling through=20 magnets, gaps, and steel. The steel must carry the same total number of line= s of=20 flux as the gap and magnets. The number of lines is proportional to the gap=20= flux=20 density multiplied by the magnet width. In the example of 8 KOe and 18 mm wi= dth,=20 this corresponds to 8,000 times 1.8, or 14,000 lines per lineal centime= ter=20 of magnet structure. The quarter-inch-thick steel (0.625 cm) must carry a fl= ux=20 density of 22,000 KOe. It does so, barely.The next topic is coil design. By following the above magnet=20 assembly design principles, you know the approximate gap field, so by also=20 knowing the magnet length you will be able to estimate the number of coil tu= rns=20 you will need to achieve a given output sensitivity in volts per meter per=20 second. If the end loops of the coil extend beyond the ends of the magnet, e= ach=20 turn is immersed in a total field length of twice the magnet length. In the=20 example design, the magnet length is 50 mm, so L per turn is 0.1 m.The output voltage generated by the moving coil is Volts =3D B *= N * L=20 * Vel, where B is field strength in Teslas, N is number of turns, L is lengt= h=20 per turn in the field, and Vel is velocity in meters per second. One Tesla=20 corresponds to 10,000 Oersteds. Suppose n =3D 1100, B =3D 0.8 Tesla, and L=20= =3D 0.1=20 meter. Then Volts/Vel =3D 0.8 * 1100 * 0.1 =3D 88 v-s/m, which is a good num= ber to=20 strive for.A wire size of #38 or less will allow this number of turns to fi= t=20 comfortably within the gap field cross-section. The coil will have a resista= nce=20 low enough to permit resistive shunt damping of a pendulum weighing up to a=20 kilogram, in my opinion. My sensors have a pendulum mass of about 0.1Kg, and= =20 critical damping is achieved at about 30 kOhms. Since the coil resistance is= =20 only 340 Ohms, the shunt damping imposes negligible loss on output sensitivi= ty.=20 Even a kilogram mass would require only about 10% loss of output using shunt= =20 damping.There is one significant complication to this type of magnet and= =20 coil design, having to do with the fact that pure copper is diamagnetic, and= =20 coil and its pendulum are subject to forces other than the desired restoring= =20 force of a garden gate pendulum. In the example design this diamagnetic forc= e is=20 a decentering force, making it very difficult to adjust the sensor to a stat= e of=20 stable long period equilibrium. I reduce this effect by essentially potting=20= the=20 coil in acrylic plastic, which is also diamagnetic, and having the boundarie= s of=20 the plastic extend well outside the edges of the magnetic field. Using a hea= vier=20 pendulum would be of great benefit, as well.Bob McClure