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UK EE 468G - Designing With the TRF6900 Single-Chip RF Transceiver

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EE468G NOTES (5C) Reading assignment: Chapter 8 and Chapter 9 Contents: Magnetic material and inductance HW 5e: Chapter 8: 4, 6, 7 Due: 12:30pm, Thursday, October 9. HW 5f: Chapter 9: 1, 2, 3, 11, 12, 13 Due: 3:00pm, Monday, October 13. Objectives: (1) Magnetic material and permeability (2) Boundary condition of magnetic fields (3) Faraday’s Law of inductance (4) Calculation of self and mutual inductances TEST 2: 12:30pm – 1:45pm, Tuesday, October 14, 2003 Closed bookMagnetic material and boundary condition Rotation of electron around nucleus form magnetic dipole kM. The total magnetic dipole per unit volume is defined as the magnetization vector: Magnetization vector: 0,limkkvMMv∆→=∆∑G [A/m] The magnetic potential produced by the magnetization vector is ∫∫×+×∇=SnVmdSRaMdVRMA 'ˆ4''400GGGπµπµ Meanwhile, the magnetic potential produced by the imposed volume current and surface current is given by ∫∫+=SsVidSRJdVRJA '4'400GGGπµπµ We conclude that the effect of magnetization vector is equivalent to an induced volume and surface current: nSMMaMJMJˆ×=×∇=GGGG, is the outward normal direction of S. naˆUsing the induced currents in th Am e e’s la , we have GGGGGe p r w()MJJJJBMG×∇+=+==×∇000total0µµµµ GGGThis means: ()JMB =−×∇0/µ Comparing this equation with Maxwell equation: , we established that JHGG=×∇G ,/0MBHGG−=µ or MHBGGG+=0µ Simple magnetic material (isotropic), MG is parallel toHG, GGGGGG ()mmmHHHBHMχµµµχµµχµ+==+=⇒=10000 Where, we define relative permeability as: 1rmµχ=+ Example: A material has 2000rµ= and atom density of . If it is put in a uniform magnetizing field with [T], (1) calculate 27 -3210[m]×0.15mB =MGand the magnetic moments per atom. Solution: (1) From 0BHMµ=−GGG, we have 0000111rrBBBMHBµµµµµµ=−=− = −GGGGG G M is in the same direction of B, hence 67111 0.15 0.12 10 [A/m]410 2000Mπ−=−×=×× (2) Moment per atom is: 621270.12 100.06 10 [A-m]210m−×==××Magnetic field boundary condition is derived from , and 0CSHdl I BdS⋅= ⋅=∫∫GGGGvv The boundary conditions are: ()() ( )1212 1122ˆˆˆ0, or 0snHH JnB B n H Hµµ×− =⋅− = ⋅ − =GG GGG G G Example (8.3): At the material interface as shown, derive the relationship between the magnitudes and angles of BG on the two sides of the interface. Assume there is no surface current on the interface. 1µ2µ2B1θ2θ1BG Solution: The normal component of 1BG is 11cosθB GThe normal component of 2B is 22cosθB GBoundary condition for B requires that: 2211coscosθθBB= (1) GThe tangent component of 1H is ()111sin/θµB GThe tangent component of 2H is ()222sin/θµB No imposed electric surface current on the interface, the boundary condition for HG gives: ()()222111sin/sin/θµθµBB= (2) Solving (1) and (2), we have: 212222121111tan tan , sin cosBBµµ21θθθµµ− ==  θ+Numerical example: let 0111020.1 [T], 10 , 2 , 1000B0θµµµ====µ we find from the above expression that ()110221101222221 1 1122 21000tan tan tan tan102tan 88.1689.35 ,sin cos0.1 500 sin (10 ) cos (10 )0.1 7539.48.68 [T]oooBBµµθθµµµθθµ−−−=====+=× +=×=0Faraday’s Law of inductance: CSdEdl BdSdt⋅=− ⋅∫∫GGGGv (Recall: For magnetostatic case: no time varying B-field, the contoul integral evaluates to “ZERO”) The differential form of the Farady’s law is: GBGEG BEt∂∇× =−∂G Example (9.1): A high-permeability core that carries a uniform, time-varing B-field: ()0ˆcoszBB taω=G, calculate the electric field generated inside the core (the core is a circular cylinder that is infinitely long). Solution: From Farady Law (differential equation form): G ()0ˆsinzBEBtωω∂∇× =− =∂G taBy symmetry, E-field has φ-component only, and depends onρonly. Hence the curve operation give rise to: ()1ˆzEaφρρρ∂∂. This means: ()() ()0011sin , sin2EBtEBφφρtωωρωρρ∂=⇒=∂ωExample (9.2): Calculate the gap voltage in the rectangular loop formed by a conducting wire if the B-field passing through the loop is uniform and given by: ()0ˆsinzBBtω= aGG. gVy - + a x b Solution: For a coGGunterclockwise path around the loop, ˆz 00ˆˆ,sin()sin( )zzdS a dxdyBdS B ta adxdyB t dxdyωω=⋅=⋅=G Faraday’s law yields ()00sin( )() cosCS SddE dl B dS B t dxdydt dtab B tωωω⋅=− ⋅=−=−∫∫ ∫GGGGv On the other hand, the line integral of the electric field around the path evaluates to the voltage on the gap. Hence ()0cosgCE dl ab B tVωω=⋅=−∫GGv.Example (9.3): A magnetic field ()2ˆsin [Wb/m ]zBtaω=G links the circuit as shown. Calculate the voltage recorded by the voltage meter. 100V200VmV200V100Vi bi a a d Meter Meter c b d c (a) (b) Solution: Use Faraday’s law ()( )26(100 200) 2 10 cos 0.01CSdEdl i BdS tdtω⋅= + =− ⋅=×∫∫GGGGv Solve for the current, we have:()0.666 cos [A]itω= Hence the voltages on the two resistances are: ()()100200100 66.6 cos [ ]200 133.3 cos [ ]Vi tVi tωω=− =−==VV (1) For this case, we evaluate Faraday’s law on the path: abcd, since there is no magnetic flux passing this loop, we have GG100100066.6cos( ) [ ]mabcdmEdl V VVV tVω⋅=− =⇒==∫v (2) For this case, we evaluate the integral on the path of adcb, ()()200200200cos200cos 66.6cos( ) [ ]mabcd SmdEdl V V B dS tdtVV t tVωωω⋅= −=− ⋅=⇒=− =−∫∫GGGGvFlux linkage: total magnetic flux passing through a surface. It is defined as the following integral: Flux linkage (single turn): SBdSΦ=⋅∫GG [Web] BGS For N-turn wire wind on a cylinder, they form N-faces. The total flux linkage passing through this N-turn wire is then NΛ=Φ Using the flux linkage, the Faraday’s law can be written as: CdEdldtΛ⋅=−∫GGvSelf-inductance: 111111Flux linking C due to current in C [H]Current in CLI11Λ== 1C1SBG1ICurrent Loop Circuit model for self inductance: divLdt= v+−i To calculate self inductance: 1. Set current I on a closed loop C 2. Calculate the magnetic flux linkage 3. Take the ratio of the linkage with the current.Example: Calculate the self-inductance of the N-turn solenoid


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UK EE 468G - Designing With the TRF6900 Single-Chip RF Transceiver

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