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I have 2 exact electronic circuits, with one malfunctioning while the other is functioning. The total current consumed by the malfunctioning circuit is 0.48 A, while the functioning PCB has 0.271 A. My question is, how can I know which parts in the PCB is eating up this current difference?

What I tried so far is, to compare the resistance of the ICs in both of them to notice if there is any difference. Because:

V = IR

And so if resistance is low, current will be high. But this process is taking too long, I still have so many electronic parts to compare e.g. ICs, capacitors, transistors...etc. There are more than 300 parts in that PCB.

Is there an efficient way to find out which part to replace that is causing the high current without taking too much time?

Also: The problem isn't in powering up the PCB, it runs up just fine, but the problem is after it runs, in one of the driver ICs perhaps. I can add more details but I doubt it would be helpful.

More details

enter image description here

The above picture is the backside of the PCB, in the right-up, there are 2 similar driver ICs which I think might be causing the problem. They don't heat up though.

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    \$\begingroup\$ Find out what’s heating up in the the defective one. A thermal camera is what I find to be the fastest. \$\endgroup\$
    – winny
    Jan 12 at 12:22
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    \$\begingroup\$ You are starting from the wrong end. Start from the loss of function and work out what components that might cause it. Compare good circuit voltages with bad circuit voltages and don't worry about the current (much harder to measure than volts). \$\endgroup\$
    – Andy aka
    Jan 12 at 12:23
  • \$\begingroup\$ yes, please add much more detail, circuits, photos, application information. Without all this, any advice can only be guessing. \$\endgroup\$
    – Marcus Müller
    Jan 12 at 12:25
  • \$\begingroup\$ A thermographic camera is great for finding problematic parts, though the good ones are a bit expensive. \$\endgroup\$
    – Lundin
    Jan 12 at 13:02
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    \$\begingroup\$ "What I tried so far is, to compare the resistance of the ICs in both of them to notice if there is any difference." Huh... do you have any idea how complex most ICs are internally? They are not resistors. You can't measure their "resistance". \$\endgroup\$
    – Lundin
    Jan 12 at 13:07

6 Answers 6

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The difference in power consumption is about a watt (assuming 5V power). That's enough to get a reasonable size component noticeably warm (or perhaps remove a bit of skin if it's a small part such as an SOT-23 or SOIC-8). So look for charring or smell with the power off. If safe and feasible you might be able to probe the circuit with a finger you don't need too much.

Aside from some kind of software related issue (you have not disclosed the nature of the circuit so ignore if that's not a relevant concern) one possible problem is a CMOS chip that has been damaged and draws a lot of Vdd current. It will get very hot and (of course) wont function properly. It's even possible to have hairline shorts in the PCB that open and close with heating. If this is a production unit you will want to find the root cause whatever it takes, but sometimes it costs a heck of a lot more than the unit itself is worth.

You can also use traces as kind of a current shunt if you solder thin wire-wrap wires to instrument the board. A current change of 100mA is easily measured with a multimeter set to the lowest voltage range.

Edit: Aside from the (stressed) driver chips, also look for any tantalum cap (those orange parts) that has a charred or discolored look to it. Failures of those are not uncommon.

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  • \$\begingroup\$ I wouldn't really recommend "is it the voltage regulator?" place finger on it "Ouch!!! it was the voltage regulator!" trouble-shooting :) Done that way too many times. \$\endgroup\$
    – Lundin
    Jan 12 at 13:04
  • \$\begingroup\$ @Lundin Licking your finger will provide a bit of protection against high temperature by the Leidenfrost effect (but would make any electrical shock hazard worse). \$\endgroup\$
    – Spehro Pefhany
    Jan 12 at 13:18
  • \$\begingroup\$ I was actually taught to do that by a senior EE at some point :) But... I vastly prefer the thermal camera nowadays, as do my fingers :) Licking on fingers that have just been handling flux, fiberglass and other icky things encountered on the PCB is probably not a good idea either. Plus my fingers taste like disinfection alcohol these days... \$\endgroup\$
    – Lundin
    Jan 12 at 13:26
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    \$\begingroup\$ Upvote just for "probe the circuit with a finger you don't need too much". \$\endgroup\$
    – linuxfan says Reinstate Monica
    Jan 12 at 14:48
  • \$\begingroup\$ Just to follow up, you see the voltage regulator in the lower right part of the PCB? That's the only chip that is getting very hot. Is it normal for a voltage regulator to get this hot? Also, the driver chip aren't getting hot at all, but they are consuming more voltage than the driver chip in the functioning PCB. \$\endgroup\$
    – John Sall
    Jan 15 at 19:25
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I do not know how you are measuring "resistance," but if doing so with a multimeter (while circuit powered off), then you cannot measure the resistance of an IC (or rather, it has nothing to do with how it behaves once powered.)

In your place, I would look for any hot components in your bad circuit. Be careful, depending on the faulty component and the voltage, it can be quite hot. I would recommand touching first all components with a thin glove (to be sure there isn't one that is really hot(100°C+,)) and if you don't find it, then it is safe to start again with bare fingers.

If you find a hot component (that isn't on the good circuit,) then you found the problem (it might be the hot component itself or one "driving" it.)

If you happen to have a thermal camera, then it's even easier to find the hot component.

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Best thing? Thermal camera if you can find one, the components that are using up the most power (and usually the most current) will heat up the most. It's especially useful for finding shorts or other issues with one singular component that is bringing down the whole power rail.

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What I tried so far is, to compare the resistance of the ICs in both of them to notice if there is any difference.

Won't work. Putting a resistor in front of your broken device will then highlight the resistor as the broken thing.

Also, you're assuming that your measurement device is even capable of correctly assessing the voltage across a point. That is usually not the case for electronic circuitry including ICs: they can have very dynamic current draw.

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Often distributed high currents come from oscillations in CMOS. Using a short 10:1 probe as a loop antenna, you can sniff around to verify if this is the case, then determine is supply ripple is the cause then what controls that ripple , the unexpected load from ESD damage or the source caps. Comparing with a KGB (known good board ) helps alot.

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Lots of answers and comments contain "thermal camera". A cheaper alternative is pasting thermosensitive paper from a supermarket receipt over your circuit board and powering it for a while. If there are spots consuming more power than they are readily prepared to dissipate, they should become apparent.

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