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Is there any rule of the thumb at which power dissipated in an IC one should be able to feel that it is hotter than the other components?

More specifically, I have a "signal" that should consume nearly no power (<2mA @ 4.7V.) After identifying that the signal was always low, I removed the IC driving it, and supplied it with a lab supply. The track (with components attached to it) now behaves approximately like a resistor (resistance in the 10-30 ohms range) to ground.

I suppose one component is bad, and is leaking to ground (and therefore dissipating all this power.) At how much power dissipated in a single component should I be able to feel that it is hot with my finger?

Of course, there is no precise number (it depends on the exact component, the layout, the sensitivity of my finger, etc.)

Do you know at what power I'm sure to feel it, provided it's a "normal" IC?

EDIT: the components connected to the signal:

Note that I cut the track going to the LT4294IDD#TRPBF as it was my main suspect at that time.

Also note that everything worked fine until a few days ago, so it's not a component placed wrongly.

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    \$\begingroup\$ Whatever power is necessary to get it to somewhere in the vicinity of 40 to 50 °C, or noticeably hotter than ambient, which depends on way too many factors to answer. A thermal camera is better than your finger for things like this; you can get a FLIR One for only a couple hundred dollars. \$\endgroup\$
    – Hearth
    Dec 7, 2023 at 14:30
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    \$\begingroup\$ This doesn't make any sense without a specific component and package in mind. Obviously a TO-220 part with built-in heatsink will tolerate heat better than some SOT23... \$\endgroup\$
    – Lundin
    Dec 7, 2023 at 14:30
  • \$\begingroup\$ The issue is that I don't know what component dissipates the heat. But among the "suspects", there is nothing particularly big. I would say the 2 with the biggest thermal dissipation should be LT4321IUF#PBF or LT4294IDD#TRPBF (excepted of course if the problem is somewhere completely unsuspected). That's why I'm asking a a rule of thumbs instead of a specific value. \$\endgroup\$
    – Sandro
    Dec 7, 2023 at 14:41
  • \$\begingroup\$ @Hearth : I agree that a thermal camera would be helpfull (I'm currently trying to put together the caracteristics needed for a decent one, and will try to convince management to invest, but it will be at least a week before I get one) \$\endgroup\$
    – Sandro
    Dec 7, 2023 at 14:46
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    \$\begingroup\$ Squirt some isopropyl alcohol on it and see where it evaporates first. \$\endgroup\$
    – bobflux
    Dec 8, 2023 at 11:16

2 Answers 2

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You can find thermal resistance junction-to-ambient and junction-to-case in most datasheets.

You should have some 'feel' for what temperature in °C (or °F if you must) feels warm. A human body is 37°C. Lukewarm is around 40°C. Somewhere around 60°C is too hot to hold your finger on for long, depending on how hard you press, calluses etc.

Your finger is imprecise- an object with a lot of thermal conductivity (eg. copper or diamond) will feel hotter or colder than one that has less thermal conductivity (eg. plastic or cloth). It may be difficult to sense the temperature of a small object as easily as one that allows more of a fingertip to contact. Perhaps to the point that you could get some skin damage before noticing how hot a small object is.

Anyway, to pick something big like the top epoxy of a TO-252, if we assume the 25°C ambient, to heat to 45°C (warm) would require about 400mW . Smaller parts could require much less power to get equally hot, but as noted that may be hard to sense with a finger (and also hard to sense without a non-contact temperature sensor since even a tiny sensor with thin wires will tend to affect the temperature).

In general if you can't directly measure an object you may do better to measure the power dissipation and calculate the temperature rise (but exercise caution with SMT parts which are very dependent on the PCB particulars- often you can find the footprint and materials they assume specified somewhere other than the datasheet). There are sometimes tricks that you can use to measure the internal temperature of a component by using some characteristic that can be measured quickly after power is removed.

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    \$\begingroup\$ For smaller devices, using a smaller finger helps. Your little finger also tends to have thinner skin than the ones that work harder. And while fingertips are sensitive, the back of the hand (or the back of the finger) can be more sensitive to heat. \$\endgroup\$
    – Chris H
    Dec 8, 2023 at 9:24
  • \$\begingroup\$ I've just tested on a coffee cup I'd used a few minutes ago (ceramic is a decent model material for us). With the back of my little finger I can clearly feel the 3°C difference between the handle (17°C) and the side (20°C). With the tip of my little finger, maybe, but this is a big area, and the tip of my index finger can feel it, just about, and more slowly than the back of my finger. Checked with an IR thermometer \$\endgroup\$
    – Chris H
    Dec 8, 2023 at 9:26
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Generally, check the component's thermal impedance, which very very strongly depends on its package, but also on its internals. Specifically, the value you are looking for is often called junction-to-ambient impedance, \$\theta_\text{j,a}\$. (Actually the difference between the junction-to-ambient and the junction-to-case impedance, but the latter is usually very small.)

For something to feel hot, it should warm by perhaps 25 K over room temperature. E.g. for an impedance value of 100 K/W, you would cause such heating at a dissipation of 0.25 W.

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    \$\begingroup\$ Yes, 1/4 Watt is about the threshold to feel heat for a typical IC. Maybe less for a very small IC. \$\endgroup\$
    – Mattman944
    Dec 7, 2023 at 15:59
  • \$\begingroup\$ 25K is more than you need to feel a difference, and may or may not be enough for "too hot" - regulators often run hotter for example, to the point where you want to take your hand away before long - more like RT+40K. Even those are too hot for long term use if you take it away instantly with an "ouch". I tested with a couple of TO-220 diodes (on the plastic face of the package), one at room temp (18°C) and one warmed by body heat (to 25°C), and the difference was clear with a cool fingertip. But the calculation is the right way to go, & "too hot" for signal components isn't all that hot \$\endgroup\$
    – Chris H
    Dec 8, 2023 at 9:34
  • \$\begingroup\$ @ChrisH as far as I understood it is about being "hot to the touch". I went with that being ~45°C because a small component like an SOT23 also cools down considerably when you touch it. \$\endgroup\$
    – tobalt
    Dec 8, 2023 at 18:05
  • \$\begingroup\$ The OP used both "feel hot" and "hotter than other components" so there's room for interpretation. But you've already got my upvote for teaching the OP how to work it out \$\endgroup\$
    – Chris H
    Dec 8, 2023 at 21:03

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