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This is an ECU PCB and the measurement of resistance between Vcc and GND for the TC1762 Tricore MCU in the PCB is 18.7 Ω.

I have a reference board, the resistance there is 400 Ω between Vcc and GND for the same IC. When I power up the PCB in the first board, the MCU gets hot, while it doesn't in the reference board, although the reference board lacks a firmware in the MCU.

The voltage reading on the Vcc is 3.3 V. I'm suspicious that this resistance between Vcc and GND is the reason why the PCB's current is higher in the first board, it has 0.41 A, while in the reference board is 0.26 A.

Is 18.7 Ω between Vcc and GND in this PCB the possible cause for this problem? Given that the there's a problem in the first board, and I was debugging it. Is such resistance can be considered as a short between Vcc and GND?

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  • \$\begingroup\$ @Velvel that makes sense, it's almost the difference between the overall current between the two boards \$\endgroup\$
    – John Sall
    Commented Jul 7, 2023 at 7:15
  • \$\begingroup\$ What's the resistance if you just short your multimeter probes? Depending on length and quality, you'll have some "base resistance" there and that base resistance adds to the measurement. \$\endgroup\$
    – Lundin
    Commented Jul 7, 2023 at 8:21
  • \$\begingroup\$ @Lundin it's 0.0 ~ 0.1 Ohm \$\endgroup\$
    – John Sall
    Commented Jul 7, 2023 at 9:02

1 Answer 1

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Yes, but it's unlikely that you actually have a resistance, it's just that your meter says 18.7 ohms because the MCU and all other components connected to VCC and GND consume current when voltage is applied, and the multimeter will apply some voltage and measures how much current flows (or applies a current and measures voltage, it doesn't matter) and based on voltage and current it presents you the apparent resistance.

The point is that your load is not a resistance, your load is everything between VCC and GND, not only the MCU, but also the regulator that makes the VCC, the capacitors on supply pins, other chips, etc.

As the measured resistance of two boards is so different, it may show that the boards are different, but you don't know which one is normal and which one isn't.

But if the MCU heats up, it's a sign that the MCU could be damaged. Or some other part is damaged and current flows through MCU pins to the damaged part, which may short MCU output pins to GND or VCC.

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  • \$\begingroup\$ I'm certain the damaged one of them is not the reference board, I bought it new, the original board wasn't providing the required functionality and that's why I was debugging it. So is such apparent resistance is considered a short between Vcc and GND? \$\endgroup\$
    – John Sall
    Commented Jul 7, 2023 at 7:07
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    \$\begingroup\$ Depends on context. It is possible something inside a chip is damaged somehow and excess currents flow. It's not a direct short though, obviously, you would be reading the same ohms than shorting the multimeter leads together. \$\endgroup\$
    – Justme
    Commented Jul 7, 2023 at 7:16
  • \$\begingroup\$ I have 11 ICs and 20 capacitors that are using the same Vcc pin in that PCB, and all those 11 ICs have the same resistance value of 18.7 Ohm to ground, I assume one of those paths is the cause of the short, and the voltage is taking the least resistance path. \$\endgroup\$
    – John Sall
    Commented Jul 7, 2023 at 8:53
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    \$\begingroup\$ No, that's incorrect, there is one voltage on everywhere on VCC node, and current takes all paths, not one. So it does not matter if you measure resistance between VCC and GND at chip A or B because you are mesuring on the same wires and all chips are on the same wires. \$\endgroup\$
    – Justme
    Commented Jul 7, 2023 at 9:04
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    \$\begingroup\$ @JohnSall No you confirmed the MCU is damaged. The firmware may not be readable due to damage and manufacturer may have disabled the firmware readout anyway so it may not be possible to read the firmware anyway. Unless the chip is hackable in some way, but it's like a lock, it keeps common people out but does not prevent determinate criminals from breaking the lock. \$\endgroup\$
    – Justme
    Commented Jul 7, 2023 at 15:37

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