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CLC412 数据表(PDF) 8 Page - National Semiconductor (TI) |
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CLC412 数据表(HTML) 8 Page - National Semiconductor (TI) |
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8 / 12 page ![]() CMRR over frequency can be achieved through the placement of an RC network between the outputs (A and B) of the two amplifiers of the CLC412. Non-Inverting Current-Feedback Integrator. The circuit of Figure 7 achieves its high-speed integration by placing one of the CLC412's amplifiers in the feedback loop of the second amplifier configured as shown. Figure 7 Low-Noise Wide-Bandwidth Transimpedance Amplifier. Figure 8 implements a low-noise transimpedance amplifier using both channels of the CLC412. This circuit takes advantage of the lower input- bias-current noise of the non-inverting input and achieves negative feedback through the second CLC412 channel. The output voltage is set by the value of Rf while frequency compensation is achieved through the adjustment of RT. Figure 8 Buffered 2nd-Order Sallen-Key Low-Pass Filter. Figure 9 shows one implementation of a 2nd order Sallen-Key low pass filter buffered by one of the CLC412's channels. The CLC412 enables greater precision since it provides the advantage of very low output impedance and very linear phase throughout the pass-band. ½CLC412 ½CLC412 Av 1 = -0.5V/V Av 2 = 1.5V/V Vin Vout Rf1 Rf2 R Rg 2 Rg 1 R1 R2 Ro1 Ro2 + + - - V in 3 Figure 4 ½CLC412 ½CLC412 (Av 1 = -1V/V) (Av 2 = -1V/V) -Vin +Vin Vout1 Rf Rf Rg RR R1 R1 Ro ++ -- Figure 6 R Av = +1 Av = -10 R1 Rf R2 Rg Vout R Rf R1=Rf Rg=Rf/10 Ro +Vin -Vin ½CLC412 ½CLC412 Differential Line Receiver. Figures 5 and 5a show two different implementations of an instrumentation amplifier which convert differential signals to single-ended. Figure 5a allows CMRR adjustment through R2. Figure 5 Figure 5a High-Speed Instrumentation Amplifier. For applications requiring higher CMRR the composite circuit of Figure 6 uses the two amplifiers of the CLC412 to create balanced inputs for the CLC420 voltage- feedback op amp. The DC CMRR can be fine tuned through the adjustment of Rb. Further improvement of CLC420 ½CLC412 ½CLC412 Rf 2 Rf Rf 1 Rin2 Rb R Rin 1 Vin 2 Vout Vin 1 Rin 1 = Rin2 Ra A B Ro Rg + + + - - - ½CLC412 ½CLC412 Vin Vout Rg R R2 R1 Rb C + + + - - VV R R sRC oin = 2 1 A R R f g 2 2 2 =− V IR R Z(s A s o sf T p p = + + F H GG GG I K JJ JJ 1 1 2 ) ω ω ½CLC412 ½CLC412 Rin + + - - Vout Vin R2 Rf C2 C1 Rf R1 Rg Figure 9 ½CLC412 ½CLC412 + - Is Rg2 + - Rf2 Ro Vo R RT Rf Cf Cin V V K RR C C ss RC R C K RC R R C C out in o = ++ + − + 12 1 2 2 11 2 2 9 22 1 2 1 2 11 1 1 http://www.national.com 8 |
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