# Finding a faulty chip that draws too much current

Note that this is a theoretical question - there is no schematic I can show. I will show some schematic, but it will be a very simplified version of an actual circuit, only for illustration purposes.

Assume I have a voltage converter that takes as an input my main voltage (from a power supply) and outputs a certain voltage, for example 1.8V. It would look something like this:

simulate this circuit – Schematic created using CircuitLab

When connecting my circuit to the P.S, I notice it draws too much current (the P.S shows that).

Since I have multiple voltage converters in my circuit (not shown here), I check resistance between each output of each converter to ground. I see that the resistance between 1.8V to ground is almost 0 Ohms. Now I know the fault is either in the voltage converter or one (or more) of the other components drawing power from that 1.8V.

I desolder the resistor shown in the image to disconnect the converter from the other components and see that the converter is fine, but checking resistance from the point connected to all those components still shows 0 Ohms.

My question is - how would you check which component is the faulty one, without desoldering each suspicious component? As you can see in the image, the 1.8V supply is connected directly to the components, without a resistor/bead.

For the sake of this question, assume I have access to whatever equipment needed (no matter how pricey). I wouldn't want solutions to be limited due to availability of equipment.

Thank you!

• I had a tech in my first job out of school that had his own special way of finding such shorts. Another part of the same division made electro-plating power supplies. These put out 5 V at 100s of A. He'd connect one of these between the offending net and ground. The resulting smoking hole gave you a pretty good idea where the short was. Jul 28, 2017 at 20:05
• Bad but fast solution, apply 1.8 V with no current limit and see what burns away. Jul 28, 2017 at 20:07
• @winny, no, you should set the voltage to 1.8V, and set a small limit on current. Then gradually increase the current limit until something started to heat up above the expected. Jul 28, 2017 at 21:31
• Borrow a thermal camera? Jul 29, 2017 at 2:00
• Do you have a non-populated board, to be sure that this is not a power plane short, a routing defect? Do you have any other board that works, and only this one has a fault? Jul 29, 2017 at 3:59

A thermal imager is very useful in this situation. They are not terribly expensive these days. If you don't have one, a bare finger can be substituted for a sensor.

ADDITION: There are also Thermochromic Paints for different temperature ranges that can be used to identify hot spots.

• Thank you, @Ali Chen. I actually use one, guess I should have mentioned it in the post. It does help a lot, but the point of this question was to see if there is any other way which I have not thought of. Jul 28, 2017 at 17:06
• I used to not see the usability of a thermal imager in my day-to-day work, until I got one. It is my second-most used piece of equipment, behind the o'scope. Jul 28, 2017 at 18:45
• be aware though, that thermal imaging on an unprepared board, can be misleading, as reflective parts will show a lot hotter then they actually are. Jul 28, 2017 at 20:13
• @Grebu, actually, the opposite is true - metal shiny parts look quite colder than they are, because their emissivity quite less than a typical black body. Jul 28, 2017 at 20:44
• @Eran, actually, if you have a solid short, I would vote for Spehro Pefhany method. You would need to apply a reasonable current into 1.8V plane; Voltage will be near zero at every point, but not completely zero. You would need a good DC millivoltmeter, and then plot a map of voltages with u-volt resolution across your board. Do the similar map for ground plane. In this case you might be able to locate your short, where the Vcc is lowest, and Gnd is highest. Jul 29, 2017 at 19:21

You could use a PCB current probe. A search showed up the following.

Figure 1. A TTi current probe.

The probe-head is held on the PCB trace under investigation and the output can be monitored on an oscilloscope and, presumably, in the case of DC on a multimeter.

I have never heard of a "Fluxgate Magnetometer" before and I doubt they'll give away too much detail. Good old Wikipedia says the following:

A fluxgate magnetometer consists of a small, magnetically susceptible core wrapped by two coils of wire. An alternating electric current is passed through one coil, driving the core through an alternating cycle of magnetic saturation; i.e., magnetised, unmagnetised, inversely magnetised, unmagnetised, magnetised, and so forth. This constantly changing field induces an electric current in the second coil, and this output current is measured by a detector. In a magnetically neutral background, the input and output currents match. However, when the core is exposed to a background field, it is more easily saturated in alignment with that field and less easily saturated in opposition to it. Hence the alternating magnetic field, and the induced output current, are out of step with the input current. The extent to which this is the case depends on the strength of the background magnetic field. Often, the current in the output coil is integrated, yielding an output analog voltage, proportional to the magnetic field. Source: Magnetometer.

• You can improvise a crude PCB current probe for AC with a small inductor connected to the tip of an ordinary 10:1 scope probe. Use a small SMT inductor on a ferrite core with an air gap, preferably a bobbon core. Some examples: digikey.de/product-detail/en/wurth-electronics-inc/74477420/… Jul 28, 2017 at 20:16
• Solder thin solid wires to the inductor and wrap them over the tip and GND sleeve of the scope probe. Feed a rectangle signal of a few Volts into the supply of the PCB and follow the current. Jul 28, 2017 at 20:25
• Wow. I didn't know this existed. Jul 29, 2017 at 7:03

Assuming the supply is putting out a large current (eg. hundreds of mA) you can follow the voltage gradient from the supply using a voltmeter on its most sensitive range. When you find the minima on the net (or plane) you've found the sink (Vcc) or the maxima on the ground net.

Kind of a manual implementation of a steepest descent optimization algorithm.

• I'm sorry, but other than the first sentence, I did not really understand what you were trying to say. Can you give a more thorough explanation? For example, which points to probe (if any)? Jul 28, 2017 at 17:41
• Say you put one probe near the regulator output (say at the output filter capacitor). The Vcc pin of the chip that is the lowest will be the culprit. If several tie, then the closest one of the group is the culprit. Jul 28, 2017 at 18:30
• So I actually measure the input voltage in reference to the regulator's output? Jul 28, 2017 at 18:43
• You should just see a small drop from the regulator output to the chip. Millivolts, usually, but on a good multimeter that's lots of counts. Jul 28, 2017 at 18:47
• @Eran One voltmeter probe might go to U1's 1.8V_out. The other voltmeter probe goes to U2's P1 input. If heavy current flows along that path, the voltmeter will display many mV. Then try probing U3's P1 input....then U4's P1 input...then U5's P1 input. On printed circuit boards, you can probe along a path with probes quite close to one another (perhaps separated a few centimeters) if DC current flow is fairly large. Jul 28, 2017 at 19:34

Ghetto FLIR:

Squirt some low boiling point liquid (like flux cleaner) on the board. See where it boils.

• Webcam with IR filter works good too. Jul 28, 2017 at 20:06
• Digital thermometer or multimeter temp probe anyone? Jul 28, 2017 at 20:30
• @winny, unless you've got a really unusual webcam, that will only pick up stuff that's on the verge of letting the magic smoke out.
– Mark
Jul 30, 2017 at 23:27
• @winny don't you mean with IR filter removed? Jul 31, 2017 at 4:49
• @AdamEberbach Yes, IR filter removed. If you have access to a piece of Woods glass to filter out the visible light, it's even better but I've had good results with just pitch black and setting the camera sensitivity to max. There was a thread back in the day with which brand and models of webcams where crappy enough to leave out the IR filter from the factory. At Mark, Your mileage may vary, but I've had great success using a webcam until we could afford a real IR camera. Aug 1, 2017 at 6:50

The fastest and cheapest way I learn from Youtube.

Power up your board and pour some alcohol. See which area dries up first.

• That's... amazing. No damage will be done (nothing will explode or something)? Can you provide a link of the YouTube video? Jul 29, 2017 at 4:58
• @eran search for Louis Rossmann videos. He uses this a lot. Jul 29, 2017 at 5:01
• I can vouch for this method. Just make sure that your alcohol is 99% or higher (i.e. pure). But, it depends. You might get away with 91%. Jul 29, 2017 at 11:14
• But the first order is a fingertip test, see youtube.com/watch?v=t5fICjcaJ3E at about 15:01'. Actually the youtube case is simple: having 0.6V indicates a fault in some semiconductor device, and not a hard solder bridge. Jul 30, 2017 at 21:17

There is a spray for that.

Google "cold spray electronics" and you will find many hits, like this one

Spray the stuff on and watch where it disappears most quickly. That is the point generating the heat - ergo drawing too much current.

This stuff has other troubleshooting uses - should be standard in any well equipped electronics lab.

I found a video on YouTube where this method is demonstrated. It is rather slow moving but gives the idea - the short is found around 4 minutes in. Incidentally they used a dusting spray with the can held upside down - even easier than buying freezing spray.

• +1, this is a very cost effective idea. The result might depend on power rail topology/routing, but it is worth to try. Jul 29, 2017 at 19:36

So you have a rail shorted hard to ground. In my experience this is usually a soldering problem.

My technique is to hook the rail in question up to a bench PSU. Set the voltage limit at the normal operating voltage of the rail and the current limit to about 1 amp. The current is something of a compromise, too low and the volt drops will be difficult to measure, too high and you risk burning things up. 1 amp seems a reasonable compromise for most boards.

I then use a multimeter on a sensitive voltage range to trace the flow of current around the board.

• So you advise to use @SpehroPefhany's method, but instead you suggest hooking the 1.8V from a power supply and not the voltage converter that provides it? Jul 29, 2017 at 5:02
• Yeah, bench PSUs will happilly drive a user-specified current into a short circuit indefinitely. Most on-board voltage converters won't. Jul 29, 2017 at 8:33
• +1. The title of OP question is misleading, which shifts the focus on heat dissipation side of the issue, and less on solder bridges or dead-short components, which might not dissipate much (this is a weak excuse for my likely incorrect answer :-() Jul 29, 2017 at 21:39

You haven't specifically mentioned that you can rule out the traces or visible solder points. So first thing I would do is take a microscope and check the traces (especially in homemade boards) and the solder points for shorts.

I have found many solder shorts (because I'm obviously bad at soldering) but also many copper shorts between traces on self made boards.

This method doesn't take long but won't help you find all the possible faults.

As you mentioned price is not an issue, I'd say this is another worthy method:

As another real high tech solution, you can use an X-Ray machine. With that you even have the possibility to see shorts under chips which is especially useful with BGA chips.

So that would look something like this:

By X-Ray_Circuit_Board_Zoom.jpg: SecretDiscderivative work: Emdee (X-Ray_Circuit_Board_Zoom.jpg) [CC BY-SA 3.0 or GFDL], via Wikimedia Commons

X-Ray images can be a bit misleading at times, but you get used to interpret what you see, much like a doctor does.

If the machines supports it, you can also look at different angles and do a full 3D-Scan, which is pretty impressive but often not necessary.

And it being X-Ray you have quite a bit of paperwork ahead of you getting it all set up.

Another method, which is related to the voltage drop method, might be using a Milli-Ohm-Meter and measuring all the Vcc to GND nodes near the chips.

While your normal meter might read 0 Ohm, a Milli-Ohm-Meter might show a value, the node with the least resistance would be the most interesting one.

• @Mast the question posed explicitly no limit on resources. Jul 31, 2017 at 17:13
• Right, quite explicit indeed. My mistake :-)
– Mast
Jul 31, 2017 at 18:35

Put some thermosensitive paper (like from a shopping receipt) on the circuit. Here is a Youtube video.

Power up. Wait. Check for discoloration. Of course, a really solid short circuit has a voltage of zero across and will not produce significant heat. But most faulty circuits with a large current draw will have enough resistance to be trackable with heat other than only at the voltage regulator.

Inject a square wave and scope the (tiny - obviously) ringing at the driven-end and then "walk" the scope's earth (and probe of course) along each path (toward each IC). The ringing will lessen until you arrive at the short itself (with both scope's earth return and probe-tip on either side of it).

Your problem is the result of management malpractice by the creators of the circuit board: they failed to design for testability. This is a common problem in automatic test engineering.

The answers above using thermal imaging or some other way to find the hot chip are your best bet. Note, however, that if the chip is an absolute short, it will not dissipate any power and will appear cool because ALL of the power is heating the internal resistance of the power supply. In that case, the current probe shown in the previous answer might work... if your circuit board traces are large enough and spaced far enough to isolate their magnetic fields.

Alas, if you have a modern circuit board with 17 layers and super tiny SMT chips, you are probably out of luck. Logistic support analysis generally designates such devices as disposable.

Welcome to the ATE world.

• +1 for failure to implement DFT, and for "disposable" designation. The OP will probably waste more engineering-hours than the cost of the entire board. Aug 9, 2017 at 15:52

This is just a thought experiment.

Using a current source pulsing at about 1 kHz DC square wave at about 0.9 µS rise or fall time: This will make an audible tone at the start of frequency range of a standart AM receiver. The ground plane junction of the fault path must be distinguishable at the most. You may adjust the antenna length to adjust sensitivity.