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I was watching this video on JTAG. It is mostly too difficult for me to follow, and I will likely learn more about UART first before returning to this.

At around 13:00 he uses a multi-meter to discover VCC, GND, and which pin does what. I was wondering if someone could explain to me what exactly he is doing?

My intention is to repair an old television with a broken panel (on/off, channel up, channel down, volume up, volume down) where I want to discover which pin does what. I think it will be a fun project to wire up some buttons to an Arduino, and send the correct signal to the correct pin.

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  • \$\begingroup\$ questioner - Hi, You seem to have a project to control a TV (with a broken front panel) using an Arduino. OK, got that. Then you link to a video about JTAG, but are asking about UART. There's a different video on that YouTube channel about finding a UART interface, so why not start with that one? Also, are you sure that you understand enough about that TV, to replicate the broken front panel functions via its UART interface (if you can find that)? I don't see the connection (no pun intended). This sounds like an XY-problem. I wouldn't use the UART in your case! How do you know UART is needed? \$\endgroup\$
    – SamGibson
    Feb 15, 2021 at 14:12
  • \$\begingroup\$ @SamGibson Admittedly, I was tired when I posted this question, however surely the technique of using a multi-meter in that video can also be applied to repairing my television as the connector for the panel has the same stuff like GND? \$\endgroup\$
    – questioner
    Feb 15, 2021 at 14:37
  • \$\begingroup\$ questioner - Hi, "surely the technique of using a multi-meter in that video can also be applied to repairing my television" What you seem to be asking about isn't a repair - replacing the functionality of the front panel via UART & Arduino is completely different! You haven't given photos and your reverse-engineered schematic of the broken front panel. Based on my guess of the missing info, no, the technique shown in that video won't help you, as neither JTAG nor UART interfaces are on the likely path to success. My approach wouldn't involve UART/JTAG hence IMHO it's an XY-problem. \$\endgroup\$
    – SamGibson
    Feb 15, 2021 at 14:47

2 Answers 2

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His initial measurements, when he mentioned resistance readings, were apparently done with the meter set to measure resistance, and no power applied to the board.

Later, he applied power to the board, and measured the voltages on the connector pins.

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  • \$\begingroup\$ I assume to do this he would have to already know which one was ground, or something? I was hoping someone would provide an answer with step by step instructions I could follow. \$\endgroup\$
    – questioner
    Feb 15, 2021 at 11:16
  • \$\begingroup\$ @questioner: I would start by trying to identify Power and Ground - electrolytic capacitors are generally connected between those points, so that gives a starting point. Where you go from there requires some knowledge of what you are looking for, and the ability to make educated guesses of where to look, and what to expect. I wouldn't expect an inexperienced person to make much progress attempting to reverse-engineer something by this, or any other technique. \$\endgroup\$ Feb 15, 2021 at 16:37
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VCC will have a higher voltage, txd will be almost as high, ground will be lowest, (and also be connected to other known grounds). and RXD will be somewhere in between.

tx is an output, idle uarts are high, but no transistors are perfect consuctors, so the voltage will be slightly less than the supply voltage.

rx is an input , it will have a high impedance to ground, possibly with a weak pull up, so, a voltage measurement will probably show a value between tx and ground.

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  • \$\begingroup\$ Why will tx be almost as high as Vcc? Why will RX be somewhere in between? \$\endgroup\$
    – Lorraine
    Feb 16, 2021 at 7:54
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    \$\begingroup\$ rx is an input , it will have a high impedance to ground, possibly with a weak pull ip \$\endgroup\$ Feb 16, 2021 at 8:24

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