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The problem is that I have an increased current in my whole circuit than should be, the required current is 0.26 amp, but I have 0.45 amp.

After conducting continuity tests, I found that one of the capacitors is shorted, then, when I traced what other components are connected to this capacitors, there were 3 more capacitors connected to it in parallel, as well as the main microcontroller pin Vddfl, fl here stand for flash.

How can I know which components of those are causing the real short? do I have to remove every one of them including the microcontroller and test the circuit? or there's another way?

Update

A voltage is present in those capacitors of 3.3V, the resistance across the capacitors is 18.6 ohm, I didn't measure the resistance of all of them yet, only a sample of the shorted ones.

Also, just found out there're other 20 capacitors and 5 resistors that are also shorted to ground(not necessarily faulty) and also another IC that is responsible for power management that has one pin shorted to ground.

Also, the MCU has about 10 pins that are shorted, not only the Vdd.

The name of the shorted MCU is : Infineon Tricore TC1762

The name of the power management IC is : 40049 Bosch

Both the MCU and the power management IC is producing heat. My board is an ECU from a car.

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  • \$\begingroup\$ John Sall - Hi, Re: "found out there're other 20 capacitors and 5 resistors that are also shorted" Just to be clear, you do realise that it's very likely not all (perhaps not any) of those components are actually faulty, right? That sentence in your question isn't clear to me how much you are posting bare measurement results and how much you are posting a conclusion about what is faulty. Can you explain how you determining where you have a short? Are you using a DMM on continuity test? Thanks. \$\endgroup\$
    – SamGibson
    Jun 28, 2023 at 15:17
  • \$\begingroup\$ @SamGibson Yeah, initially, there was only one capacitor that I made a continuity test for it and it was shorted, then I found out there're many other components connected to this capacitor in parallel e.g. 20 more capacitor, 5 resistors and 2 ICs. I made a continuity test from the shorted end of the initial capacitor to those other components and it was positive. The real challenge as you have mentioned is I don't know which one of those components is shorted. Because they're connected in parallel, then if only is shorted, then all of them will have a ground connection \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 15:55
  • \$\begingroup\$ @SamGibson A voltage is present in those capacitors of 3.3V \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 16:08
  • \$\begingroup\$ @SamGibson the resistance across the capacitors that I measured so far are 18.6 ohm, but when I do a continuity test across a capacitor it beeps \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 16:33
  • \$\begingroup\$ John Sall - Thanks for the reply. Unfortunately I find terminology like "the shorted end" and "if only {one} is shorted, then all of them will have a ground connection" too ambiguous for me to be able to add any more in the time I have available (too much to-and-fro would be needed to clarify everything you are saying) so I have to stop here. I will leave you with one recommendation: If you haven't done so already, be sure to understand the effect of ESD diodes within MCUs, when using a DMM continuity test (on an unpowered PCB, obviously). A "beep" from a continuity test does not mean a short. \$\endgroup\$
    – SamGibson
    Jun 28, 2023 at 16:36

5 Answers 5

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There used to be a cheap and cheerful impedance measurement instrument called a ToneOhm, e.g. the Toneohm 700. Having used one at a previous place of work, I snapped up a second hand one Toneohm 550 for myself when I found it on (auction site).

enter image description here

(picture source and partial schematic here)

It's basically a relaxation oscillator with Kelvin probes; the slightest change in resistance produces a clearly audible pitch change. You just move the probes around until you find the highest frequency, while annoying the hell out of anyone else within hearing range.

I can't express how brilliantly simple it is to use. It doesn't get a whole lot of use but whenever it does, it pays for itself.

The current version, the Polar Electronics ToneOhm 950 is far from cheap; I don't know how much added value it gives in practice.

Thermal imaging is a pretty good alternative, though it does require the failed equipment to be powered up until you find the fault, while the ToneOhm should be used on unpowered equipment.

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You don’t, but the best way I’ve found to find it is a thermal camera and see what heats up. If several things does, you may need a reference board to tell designed/intended heating from offending components.

Before they became available, I recall my professor’s favorite method to find hard shorts was to just inject DC with a lab supply and increase the current limit until you could see which part started to smoke.

An intermediate version of the above two is to either freeze the board so that you have ice crystals all over and see which melts first when you apply power or just use freeze spray.

Happy hunting!

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  • \$\begingroup\$ Okay so I tested my reference board and there was no heat on the microcontroller when I touched it, however, the original one had high temp on the microcontroller. Also, in the original board, there's a program installed on the microcontroller, do you think because the microcontroller in the reference board lacked any software, that's why it wasn't heating up? \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 13:37
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    \$\begingroup\$ Unless this is a pretty high power MCU ( like > 200 MHz) , this is pointing to damage (possibly ESD) to the MCU on your suspect board. If the Cs are easy to remove, there is little to be lost by lifting them and verifying the short is still there. That's usually easier than lifting the MCU, but it's likely you'll have to do that anyway. \$\endgroup\$
    – user16324
    Jun 28, 2023 at 13:54
  • \$\begingroup\$ @user_1818839 the MCU is from an ECU, the name of the MCU is "Infineon Tricore TC1762", also, I just found out there're about 20 Cs and 5 resistors that are shorted \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 14:52
  • \$\begingroup\$ Depending on the rest of the circuit, it would be possible to have some unfavorable condition before the GPIO pins are configured to have some current sink going on, but it should not be the MCU which is heating. How’s your ESD condition? Any other damage like voltage into pins of an unpowered MCU? \$\endgroup\$
    – winny
    Jun 28, 2023 at 15:52
  • \$\begingroup\$ @winny I'm not sure if I have caused an ESD unintentionally to the MCU or other components. \$\endgroup\$
    – John Sall
    Jun 28, 2023 at 16:10
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Usually, you will need to remove components and check. But, if you have a suitable layout and suitable tools, you may be able to measure which component is causing the short.

The trick is to measure the current in tracks connecting the suspected components. This is done by either measuring the mV-level voltage drop across the track, or the magnetic field around the track with a Hall probe.

If your tracks are very short, or worse still part of a plane, tough, you're back at removing components.

Some companies make 'short detecting' tools that inject current pulses into the suspect tracks. It's easier to detect small AC signals than DC ones.

If there's enough voltage across the component, it may get hot enough for you to feel the temperature rise with a finger, or to spot the smoke rising.

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If you can measure the resistance of the "short" with a DMM (0.3'ish Ohms an bigger) the short will produce a fair amount of heat when pumping enough current through it.

If you don't have access to a thermal camera, just set a lab supply at or below the working voltage of the circuit (so you don't fry other components) and slightly increase the current limit. First thing you can try is to find the short with alcohol: Just look out for spots where the alcohol evaporates very quickly.

The more brutal method is to increase the current of your supply until the faulty component calls your attention by turning black and/or emitting smoke.

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Years ago I came across a piece of equipment that was blowing its fuse when the battery was connected.

A multimeter check revealed a dead short between positive and negative supply rails.

A study of the schematic showed a number of capacitors on a number of boards directly connected across the supply rails.

Every such capacitor had to be de-soldered from every board and checked. They tested 'OK' one after another and I was keeping my fingers crossed till the last one, a 1 μF 25 V tantalum capacitor, tested 'NOK'! It was replaced with a 1 μF 25 V electrolytic capacitor.

It may be worth your while to remove and check those four capacitors with a hope that there's nothing wrong with the microcontroller.

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