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LM4900MM 数据表(PDF) 15 Page - National Semiconductor (TI) |
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LM4900MM 数据表(HTML) 15 Page - National Semiconductor (TI) |
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15 / 19 page ![]() Application Information (Continued) (5) R F/Ri =AVD/2 (6) From Equation 5, the minimum A VD is 1.55; use AVD =2. Since the desired input impedance was 20 k Ω, and with a A VD of 2, a ratio of 1:1 of RF to Ri results in an allocation of R i =RF =20 kΩ. The final design step is to address the bandwidth requirements which must be stated as a pair of −3 dB frequency points. Five times away from a pole gives 0.17 dB down from passband response which is better than the required ±0.25 dB specified. f L = 100Hz/5 = 20Hz f H = 20kHz x 5 = 100kHz As stated in the External Components section, R i in con- junction with C i create a highpass filter. C i ≥ 1/(2π*20 kΩ*20 Hz) = 0.397µF; use 0.39µF The high frequency pole is determined by the product of the desired high frequency pole, f H, and the differential gain, A VD. With a AVD = 2 and fH = 100kHz, the resulting GBWP = 100kHz which is much smaller than the LM4900 GBWP of 25MHz. This figure displays that if a designer has a need to design an amplifier with a higher differential gain, the LM4900 can still be used without running into bandwidth problems. www.national.com 15 |
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