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The schematic below is the input circuit of a signalling PCB, which we buy from one of our suppliers of fire detection systems. Said PCB has to be built into a so called geographic evacuation panel which allows firemen to see in which zone of a building a fire has started, and as such is part of a safety system.

schematic

simulate this circuit – Schematic created using CircuitLab

The LED shown is actually the IR LED of an optocoupler (needed for reasons of common mode rejection). Each fire detection zone has such an input. The optocoupler's outputs are fed into an Atmel MCU where they are processed in order to light certain LEDs on a floor plan of the building. In the absence of any input signals, the MCU will reset all LEDs on the panel.

The 820 ohm resistor is an SMD type, and from its dimensions I estimate it to be package 0805 and as such it is rated for 125mW. Documentation from our supplier claims the input voltage range is from 2.2 up to 24V. This is by design, in order to support many brands of fire detection computer. Not all, but a number of systems do actually output 24V. By my own calculations the resistor dissipates about 600mW at 24V input, assuming a total forward voltage of 1.9V for both diode and LED. Actually applying 24V at the input for as short as 5 seconds makes the resistor heat up so much that you can't touch it. At this point the input current is about 26 mA. As I am not very experienced with SMD components, being out of electronics for many years, I need to know if there is any risk that the resistor will burn out, and the panel will not do what is is designed for, potentially resulting in the loss of human life.

The moment the fire service can eyeball the panel is on average first detection + 15 minutes. This means that the resistors in the activated inputs will be subjected to those conditions for at least 15 minutes, in densely populated areas. In rural areas with less firefighting personnel this can be even longer.

Authoritative answers, or links to them, are highly appreciated.

Picture of a geographic panel:

enter image description here

Picture of board with input circuit:

enter image description here

There are eight identical input circuits. I added the text "820 ohm" below one of the resistors. To the left of this resistor is the diode, above and to the left is the opto-isolator. It is a 4 pin device with SMD code 824.

Very close-up view of the resistor in question:

enter image description here

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    \$\begingroup\$ Raise your concerns with the supplier - they sound well-founded. \$\endgroup\$ – Andy aka Oct 3 '16 at 8:16
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    \$\begingroup\$ I forgot to include this detail, the current measured with 24V input was 26 mA. \$\endgroup\$ – Bart Oct 3 '16 at 10:01
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    \$\begingroup\$ If all else fails, you can still request written confirmation that the parts in question (SMD resistor) are operated within their specifications even with 24V input. \$\endgroup\$ – JimmyB Oct 3 '16 at 10:46
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    \$\begingroup\$ Why don't you measure the voltage across the resistor as Olin suggests? This will eliminate the guesswork about power dissipation. Another thing to check for is the exact brand and type of the resistor, although this could be difficult to determine without some help from the manufacturer. There are certain types of 0805 resistors rated for higher power (see, for example, this datasheet for Vishay Dale PCAN series, which lists a 1W power rating for the 0805 types), and it's possible, though unlikely, that the manufacturer specified some of these. \$\endgroup\$ – uzde Oct 3 '16 at 12:40
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    \$\begingroup\$ Are you sure that the diode is in series? It is common to have a diode antiparallel to the IR LED to protect it against reverse voltage from ESD. \$\endgroup\$ – CL. Oct 3 '16 at 14:07
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From the data you provide, this indeed seems like a bad design. I also get about 600 mW dissipation in R1 in the circuit you show.

The fact that the resistor is getting really hot is direct evidence that it is dissipating significant power for its size, but not necessarily too much. Resistors can run indefinitely without harm at temperatures that would burn your finger. A finger test doesn't really tell you whether it is dissipating just within the limit, or over it.

One possibility is that the circuit isn't as you show. Perhaps there is something else going on that isn't easily visible from the outside of the board. A good test would be to measure the actual voltage across the resistor. That together with the label on the resistor will give you a definitive answer to how much power it is dissipating.

Note that 0805 resistors are labeled with 3 or 4 digits. This is a floating point format with the last digit being the exponent of 10 and the previous digits the mantissa. A 5% 820 Ω resistor will be labeled "821", which means 82 x 101 = 820.

The power dissipated by a resistor is the square of the voltage across it divided by the resistance. In common units,

$$\mathrm{W}={\mathrm{V}^2\over\mathrm{\Omega}}$$

Therefore, the voltage that causes a particular dissipation is

$$\mathrm{V} = \sqrt{\mathrm{W}\cdot\mathrm{\Omega}}$$

At 125 mW, a 820 Ω resistor will have

$$V = \sqrt{125\mathrm{mW}\cdot820\mathrm{\Omega}} = 10.12\mathrm{V}$$

across it.

If the resistor is really 820 Ω, is really only good for 125 mW, and has more than 10 V on it, then yes, this is a flawed design. From the data you've given us, these premises seem to be true.

If it turns out the resistor really is overloaded, then probably what happened is that the unit was originally designed for a lower voltage. Somebody realized they were missing too much of the market by not supporting higher voltage. Whoever was supposed to check this in engineering either didn't, was generally incompetent, or just missed this one.

Of course why it is like this doesn't matter to you. You absolutely need to reject this system. Currently, it's just some other company putting a bad product in the field. If you incorporate that into your system, you are putting a bad product in the field, and own the resulting liability, and it will by your reputation that gets damaged.

While you definitely don't want to use this product (again, assuming things really are as you say), The device is very unlikely to catch on fire as a result. Such overloaded resistors will usually just burn out and fail open. There isn't enough flammable stuff around to cause a fire. However, the resistor could burn out and open before the fire fighters arrive, giving them wrong information as to where the fire is. That's the real danger of this system. Or, the system could latch the information until manually reset, so there are no symptoms during the first incident. However, now that channel is broken and won't respond to future fires in that zone. That's obviously really bad too.

Do the voltage measurement and point out your concern to the manufacturer. It might be worth hearing what they have to say, but it would have to be something really good for me to ever trust their products again. Remember that with electrical engineers, just like with any large group of people, there are really good ones at the top end, the decent-enough majority in the middle, and incompetents at the bottom. There are certainly incompetently designed products out there. You may have found one.

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    \$\begingroup\$ The SMD resistor has text 8200. Researching this information yields a value of 820 ohms. Direct measurement with a multimeter confirms this. Current measurements at different input voltages are consistent with an 820 ohms resistor in series with an optocoupler input and diode. There seem to be no components that can shunt away some current. The component measures about 2x1 mm, which suggests it is a 0805 package. I was not able to find info on the opto isolator. It has text 624 on top, which may point to ISP624, though I am not sure about this. \$\endgroup\$ – Bart Oct 3 '16 at 12:16
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    \$\begingroup\$ This is a product from a certified fire detection company in the Netherlands. \$\endgroup\$ – Bart Oct 3 '16 at 12:17
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    \$\begingroup\$ Measuring the voltage across the resistor is certainly a good idea. The reason I did not yet do that is it is not easy to get a good connection on those small components. In the follow-up test I will certainly have this done. \$\endgroup\$ – Bart Oct 3 '16 at 12:21
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    \$\begingroup\$ The designation 8200 suggests a 1% resistor (whereas the 821 designation mentioned above would suggest a 5% resistor). Generally-speaking, 1% resistors have a lower power dissipation specifications because they are designed to provide precise resistance rather than power dissipation. \$\endgroup\$ – EBlake Oct 4 '16 at 5:59
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Well, you're kind of guessing on a lot of things, yet you seem very sure that an 0805 resistor, which you think these are, is rated for 125mW.

There are 1W-rated (at 70°C) 0805 resistors. Of course they will run very hot, but they are designed to do that. In the value you have, it's more likely to be 500mW maximum at 70°C. Or perhaps lower rated, but there would be no visible difference.

I would not personally feel comfortable in this particular situation running parts even near their printed specification, but in fact surface mount parts are very sensitive to the PCB details- from tests a very small part can dissipate a lot of power (similar to a much larger part) if mounted over a ground plane. A very large part on a single sided board with thin traces might run hotter than an 0603 part with fat leads, a ground plane etc.

I don't see any redundancy in this circuit so any kind of single point failure- the opto, the wires to the unit, the resistor, the diode could cause failure to recognize the signalling so this is not being treated as a safety-critical device design in the slightest.


(Edit: I do have one suggestion- that you confirm that the input is actually rated at 24V DC. The power dissipation with 24VAC would be about half what it is with 24VDC input.- okay you covered this in a comment)


On the other side of the equation, if the voltage in question is coming from a backup battery bank the '24VDC' might be more like 28VDC which would increase the power dissipation considerably- to more than 850mW. The resistors are close to each other so they will heat each other up.

Do raise this with the supplier.

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  • \$\begingroup\$ The 8200 designation suggests a 1% tolerance resistor which is very unlikely to have an enhanced power dissipation rating. \$\endgroup\$ – EBlake Oct 4 '16 at 6:00
  • \$\begingroup\$ Are these systems using voltage or absence of voltage to signal an alarm - in one case, failure would at least be immediately evident by causing a literal false alarm (unless the burning resistor actually sets something on fire)... \$\endgroup\$ – rackandboneman Oct 4 '16 at 10:16
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    \$\begingroup\$ Heh, that's how alarm multiplexing works with positive signalling too: If a fire is detected, a current is sent to the panel, setting the panel on fire, so you only need one fire detector near the panel which then causes the actual alarm. \$\endgroup\$ – rackandboneman Oct 4 '16 at 10:19
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Typical general purpose SMD resistors are usually of the thick film type.

This type of resistor is not designed for long term overload (indeed, none are), but the effect of overstressing the part thermally (via too much power) is to change the resistance down in the initial phase:

Delta R vs Time on overload

(Source)

In this type of circuit, where the forward voltage of the LED and diode will not change significantly for changes in current (for a silicon diode it is 60mV per decade of current), that will increase the current in the circuit with close to constant voltage across the resistor during this period leading to more heat in the part. This could conceivably cause a thermal runaway.

Whether it will burn out or not is unknown (but highly likely if it is subjected to this sort of overload continuously), but it will definitely have a shorter than stated life (normally stated at 25C although some endurance ratings are at rated temperature); indeed, raising the temperature of a device to deliberately induce failures is a common test for manufacturers as the failure rate increases exponentially with increasing temperature.

This process is used by manufacturers to predict the useful life of components using the Arrhenius equation in many cases by deliberately causing early failures at elevated temperatures. This leads to a predictable life of the component at more benign conditions.

I thoroughly agree with Olin that you should reject these units as the reliability of the unit is guaranteed to be low at the extremes of voltage specified by the supplier, even if they survive the overload.

A proper design will never permit overstress of a part although there are parts deliberately designed to withstand short term pulse events and are often found in ESD and lightning protection circuits.

[Update] It is possible, as Supercat comments, that this is a PTC, such as this series

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    \$\begingroup\$ Would there be any kinds of resistors that might look similar to what's pictured but would have a moderate positive temperature coefficient (reaching 2x-3x nominal resistance at a temperature below that required to cause damage)? I would think such a resistor might be ideal for this sort of application. \$\endgroup\$ – supercat Oct 3 '16 at 15:54
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The only way the circuit makes sense is that the 820 ohm resistor is a current limiting resistor. It works together with the collector resistor of the previous driving stage. The collector voltage of the driver might be 24v, but if its collector resistor is 1 k ohm, then the current through the 820 resistor will only be about 12ma and the power loss will be about 118mW.

This shows that this circuit must not be used with open collector input drivers!

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    \$\begingroup\$ There is no "previous driving stage". These are inputs that takes a voltage from 2.2 up to 24V. \$\endgroup\$ – pipe Oct 6 '16 at 20:53
  • \$\begingroup\$ @pe There are inputs with what source impedance? Is it really just a battery? \$\endgroup\$ – user207421 Oct 7 '16 at 1:12
  • \$\begingroup\$ @pipe: there has to be a "previous" driver. Even if it is just a battery in series with some sensor. The sensor's impedance will work the same as the collector resistor mentioned. \$\endgroup\$ – Guill Oct 7 '16 at 4:35
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    \$\begingroup\$ Yes, but still, this is designed to take a voltage, up to 24 volts, connected to its inputs. Whatever is on the other side is irrelevant, and if it's designed with some unknown source resistance to drop the voltage down, it's no longer 24 volt. The question isn't about designing drivers for this unit - he already has those. \$\endgroup\$ – pipe Oct 7 '16 at 4:40
  • \$\begingroup\$ It is a "white lie," the input is not designed to take 24V. Who ever made the claim, conveniently forgot to mention it limitations. It probably was designed for 12V, but to have it used on more applications, they just changed the spec. \$\endgroup\$ – Guill Apr 15 '17 at 5:05

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