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I want to achieve electrically HIGH state in this receiver (Vo tag), when there is HIGH state in the fiber. What I get now is 5 V when there is no pulse in the fiber and 0.48 V when there is (RXVCC=5 V). I can't understand why the manufacturer decide to make the electrical signal as inverse function of the optical signal. I made a PCB and I can't figure out how to do this without adding components.

https://docs.broadcom.com/doc/AV02-0176EN

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winny
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Jesús Álvarez
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    Add a simple output inverter or choose an optical interface that does what you want. – Andy aka Nov 02 '21 at 11:05
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    There is no way to do it without adding components. Use an inverter. Many things with open collector outputs are active low. – Justme Nov 02 '21 at 11:13
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    _"I can't understand why the manufacturer decide to make the electrical signal as inverse function"_ It's cheaper to make it with open collector output = inverted output. – winny Nov 02 '21 at 11:18
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    Just put an inverter. – zewill Nov 02 '21 at 11:32

1 Answers1

3

Why inverted output?

1.Convenience for HV CMOS > 5V

The datasheet says "The absence of an internal pull-up resistor allows the open-collector output to be used with logic families such as CMOS requiring voltage excursions much higher than VCC"

But if using a long cable or load C greater than specified for Tr, Tf the negative leading edge will be faster from lower Rce.

2.Because the input is inverted

Now you must learn how to implement a TTL, CMOS or discrete inverter with cuts and jumpers, very tiny and neat.

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Falstad SIM

Tony Stewart EE75
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