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TS615 数据表(PDF) 25 Page - STMicroelectronics |
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TS615 数据表(HTML) 25 Page - STMicroelectronics |
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25 / 27 page ![]() TS615 25/27 INCREASING THE LINE LEVEL BY USING AN ACTIVE IMPEDANCE MATCHING With a passive matching, the output signal ampli- tude of the driver must be twice the amplitude on the load. To go beyond this limitation an active matching impedance can be used. With this tech- nique, it is possible to keep a good impedance matching with an amplitude on the load higher than the half of the output driver amplitude. This concept is shown in figure 74 for a differential line. Figure 74 : TS615 as a differential line driver with an active impedance matching Component Calculation Let us consider the equivalent circuit for a single ended configuration, Figure75. Figure 75 : Single ended equivalent circuit Let us consider the unloaded system. Assuming the currents through R1, R2 and R3 as respectively: As Vo° equals Vo without load, the gain in this case becomes : The gain, for the loaded system will be (eq1): As shown in figure76, this system is an ideal gen- erator with a synthesized impedance as the inter- nal impedance of the system. From this, the out- put voltage becomes: with Ro the synthesized impedance and Iout the output current. On the other hand Vo can be ex- pressed as: By identification of both equations (eq2) and (eq3), the synthesized impedance is, with Rs1=Rs2=Rs: Figure 76 : Equivalent schematic. Ro is the synthesized impedance R4 R2 Vi Vi Vo Vo RL 100 Ω 1:n Hybrid & Transformer GND Vcc+ 10 µ 100n 100n 100n 1k 1k Rs1 Rs2 10n 1 µ R3 R5 Vo° Vo° GND Vcc+ Vcc+ + _ + _ GND 1/2 R1 1/2 R1 Vcc/2 1/2RL 1/2R1 R2 R3 + _ Vi Vo Rs1 -1 Vo° 1/2RL 1/2R1 R2 R3 + _ Vi Vo Rs1 -1 Vo° 2Vi R1 --------- Vi Vo ° – () R2 --------------------------- and Vi Vo + () R3 ------------------------ , G Vo noload () Vi --------------------------------- 1 2R2 R1 ----------- R2 R3 -------- ++ 1 R2 R3 -------- – ------------------------------------ == GL Vo with load () Vi -------------------------------------- 1 2 --- 1 2R2 R1 ----------- R2 R3 -------- ++ 1 R2 R3 -------- – ------------------------------------ eq1 () , == Vo ViG () RoIout () – = eq2 () , Vo Vi 1 2R2 R1 ----------- R2 R3 -------- ++ 1 R2 R3 -------- – ------------------------------------------------ Rs1Iout 1 R2 R3 -------- – ----------------------- eq3 () , – = Ro Rs 1 R2 R3 -------- – ----------------- eq4 () , = Ro Vi.Gi Iout 1/2 RL |
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