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So I'm making a board, and this is a recommended configuration for one of the chips I'm using (FT232H). I'm trying to set it in the Synchronous 245 mode. This resistor is to improve clock performance.

I'm just trying to figure out if this kind of diagram means I'm supposed to connect the pin to the resistor back to that pin.

enter image description here

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  • \$\begingroup\$ Ok, so apparently that pin is CLKOUT, and it says all signals should be syncronized to it. Does that mean I connect it to every signal? \$\endgroup\$
    – Relianthe
    Commented Apr 19, 2013 at 14:56
  • \$\begingroup\$ That's a new question. Please ask a new question for it, instead of commenting to this other question. \$\endgroup\$
    – user17592
    Commented Apr 19, 2013 at 14:58

2 Answers 2

Reset to default
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No, you are not right. AC5 is a pin of the chip. You have to place a 27 Ohms resistor near to that pin and connect the actual external circuitry to that resistor (on the diagram, this is called the AC5-Load).

AC5 and AC5-Load are different things: AC5 is the pin, AC5-Load is the circuitry that should be connected to that pin (apart from the resistor, which is for Synchronous 245 mode).

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The diagram indicates that the resistor is to be placed in series to the load, i.e. the signal from the AC5 pin will pass through the resistor to the AC5 load.

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