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I need to match the input and output impedance of this part circled in red. In theory this stretch in red represents a coaxial cable with an impedance Zo=75Ω I need some power that can be activated by the 4071 IC's set and that supplies at least 300uA to the transistor base after it goes through the "cable". Does anyone have any ideas? The idea of using an amplifier has already been suggested to me, due to its low impedance, but I was still in doubt about the impedance matching.

The cable is 300 m long. There is no need to change quickly.

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I'd put 4.3 kohms between the 4071 output and the input of the cable, and 75 ohms between the output of the cable and the transistor base.

The 4.3 kohms limits the 4071 output current to ~1 mA, well within its drive capability, but still above the 300 uA you say you need.

The 75 ohms terminates the cable with a matched load (at least when the transistor be junction if forward biased) and minimizes reflections returning on the cable.

Likely the 4.3 kohm current limiting resistor would be enough on its own to slow down the signal edge and prevent any problematic reflections, but terminating the receiving end is good practice and doesn't cost much.

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  • \$\begingroup\$ But won't the resistors and cable circuit draw most of the current? I believe that 300uA will not make it to the other side. \$\endgroup\$
    – Aaa
    Sep 21 at 22:36
  • \$\begingroup\$ With what The Photon suggested, where do you think the 300 uA is going to go to? It's all just one series path. \$\endgroup\$
    – SteveSh
    Sep 21 at 23:12
  • \$\begingroup\$ @Aaa, the resistors are all in series. All the current that goes in to the left end of the resistor comes out the right end of the resistor. Anyway you have the circuit modeled in a simulation tool...so why not run the simulation and see what happens? \$\endgroup\$
    – The Photon
    Sep 22 at 0:02

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