tailieunhanh - Current Trends and Challenges in RFID Part 2

Tham khảo tài liệu 'current trends and challenges in rfid part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 20 Current Trends and Challenges in RFID Fig. 6. Common-source stage with RLC-load. The load impedance for this case becomes 1 _ R Ls Zt R Ls - - L Cs R Ls Cs 1 And substituting this value in 2 one can find that A _ gm R Ls _ gmR s L I R 1 v R Ls Cs 1 s2LC sRC 1 3 4 Observe that the inductor added a zero which always increases the bandwidth and also two poles. These poles can be complex conjugate and this also can increase bandwidth yet they introduce peaking hence the name of the method. On the other side the difference between the number of finite poles and finite zeros is still one. This means that the asymptotic decrease of gain is the same as in the previous circuit -20 dB dec. Thus the inductor allows modifying the gain locally in the vicinity of the frequency 1 and the designer should use this possibility to his her advantage. Consider the amplitude of the frequency response for this circuit given as K ia gmR 1 L IR 2 1 1 _ 2LC 2 RC 2 5 To facilitate subsequent derivations it is introduced a factor m defined as the ratio of the RC and T L R time constants RC R2 m LI R L IC R2 p2 6 Here p yỊLIC is the wave resistance of the L2 R2 useful relationships namely r2m 2 LI C LC relationships 5 can be written as load. and This allows writing two more . L R2 rm RC . Using these RLIC Main RF Structures 21 Av ja gmR í ữĩ 2 1 1 - 2T2m rm 2 7 The right side of 7 is considered the normalized gain. First the bandwidth will be maximized without any consideration regarding the behavior to the gain in the bandwidth. The frequency where the right side equals 1 V2 is denoted as a -3dB. Considering a new parameter defined as x T then one has the equation 2 x2 1 1 - x2m xm 2 or x4m2 x2 m2 - m - 2 -1 0 From this equation one can find that 8 9 10 .2 2 x m m2 _ - m 1 2 But A2 xm a-3dB m a-3dB RC P-3dB I 11 k a1 And maximizing the right side of 10 by proper choice of m one can find the maximum available bandwidth given as f-3dB m Ạm 2 - 2m- 2 4m2 - m2 2m 12 Differentiating and .