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I have a board which has a CON1 connector being currently used for a battery. Are all CON1 ports serial? Why would a battery need a CON1 port?

My guess is, this port can be used to interface and debug the board. Where can I buy a suitable cable?

photo

EDIT: Sorry, the port is called CON1. The print was really small.

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What is the name and make of the board? – Turbo J Jan 10 at 15:37
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That image shows a small 4-pin Molex connector, which looks like it carries power (guessing based on wire colour). I was expecting a 9-pin D-subminature connector when you said "COM1"! – pjc50 Jan 10 at 15:44
basic serial link only needs rx and tx signals, when generic connection also add vcc and gnd. COM ports could drive modems which need more signals. If this cable is used to supply power, then i guess only red and black wire are used. – Arcadien Jan 10 at 16:11
The board is a proprietary board from Seagate called Satellite_V2R4. And I'm sorry I misread the fine-print. It says CON1 on the board where this wire fits in. So this isn't the way to go then? – Vegetto Jan 10 at 16:31
There are no standard for such things. The question is not answerable without more information, such as documentation, personal experience, or tracing the signals back to source chips. – Chris Stratton Jan 10 at 16:42
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closed as too localized by Brian Carlton, Olin Lathrop, Dave Tweed, zebonaut, Nick Alexeev Jan 10 at 20:57

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.

1 Answer

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CON1 undoubtedly stands for Connector 1, which is an arbitrary label on the board used to refer to the given connector. It has no specific meaning outside of "This is the first connector", in the same way R1 is used for the first Resistor, Q1 for the first Transistor, TP1 for the first Test Point, etc.

Seeing the picture that the OP provided in the comments, this is the most obvious answer.

User's Board

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