I have put together a simple wake up light. It has an external powersupply, at the moment rated 7V and 600mA. Inside it has Arduino and bunch of LEDs controlled by a MOSFET.

It works fine at the moment but I would like to ensure that it wont burn my flat down while I am away. So far I have been thinking following features:

  • Fuse, rated somewhere below max output of the power supply
  • A decent heat sink for MOSFET. It gets warm without but doesn't burn my finger.

Anything else to consider?

Edit: The device is used in a bedroom. Not too much gas during night time.

  • \$\begingroup\$ do you have gas or other flammable things nearby?And other factors like LiIon batteries, chemicals etc etc. can't answer this question if so, because a little ESD charge in rubber would be able to burn down a gas station. \$\endgroup\$ – Standard Sandun Oct 12 '12 at 10:16
  • \$\begingroup\$ see : youtube.com/watch?v=j1dTSuwz0R8 \$\endgroup\$ – Standard Sandun Oct 12 '12 at 10:16

If this would be a professional design you would do an FMEA, for Failure Mode and Effect Analysis. Consultants in suit and necktie offer you expensive FMEA workshops, but it's all just common sense. Throw him out.

Organize a creative session with also other people than the designers present. You want to think of any possible thing which can go wrong with the product. The designer must be present to answer questions, but she's not the best person to do the assessment: every designer thinks her design is failproof, and issues overlooked during design will also be overlooked during the FMEA.

When you have listed up the things which can go wrong (that's the failure mode part) you add columns for Occurrence Rating (OR) and Severity Rating (SR). How likely is the failure to occur and how bad would it be if it happened. If the result of the failure is that the lights in the den go out that's a lower severity (1) than when the house would burn down (10). The product of OR and SR give you a Risk Priority Number (RPN). Sort the table by RPN, high to low, and you know which issues you have to attack first.

OK, that sounds like complicated, and no fun at all. For a hobby project you don't want to do all that, then you can better switch your hobby to knitting. But the principle remains: try to assess what can go wrong, how bad it would be if it would, and what you can do to prevent it.

The fuse is a simple solution to a lot of possible problems, and that's why you'll find one in most products. The fuse should be the first part viewed from the mains. Don't place it between the power supply and the circuit, because it won't protect the power supply itself (unless that already has a fuse).
If heating up is a risk, you can provide a heatsink (which you would probably need anyway to keep the FET within specs). If you want extra insurance add a thermistor, which you use as an overheating detector to switch off (part of) the circuit in case of overheating. Note that for instance voltage regulators often have thermal protection built-in, so for those you won't need the extra temperature sensor.

For more we'll need more details about the circuit, but overheating and overcurrent (short-circuit) protection often covers most of the critical failures.


When you design custom hardware with a safety approved low voltage supply that is isolated through a transformer that is only rated for < 5 Watts, the safety risk is pretty low for consumer use. More often a designer is concerned about protecting the components from failure, but it is good you are concerned about personal safety. A component failure could short the supply for which it must meet the flammability test requirements to get certified. This may consist of built-in protection such as a "Polyfuse" or resettable thermal resistor, a fuse or simply burn out the secondary winding after extended failure with no air flow.

In your case, I would be more concerned about choosing the right supply for your load and do not use more voltage than you need so that the series drop does not create excessive heat. DC wall supplies are usually unregulated, meaning they are higher voltage than specified until the load matches the rated current. This creates more V*I loss in your LED MOSFET switch. Rather than a 7V 600mA supply, I might consider a 5V 1A supply or more to drive more LED's in parallel with less drop. A low drop out or LDO regulator may or may not be necessary for the Arduino but run the drivers from the unregulated source with filtering as required.

If it is not too hot to touch, do no worry unless your LED wiring shorts out. So for protection of the MOSFET, add a resettable fuse in series rated for the LED current. These can be ganged together and cost two bits in small quantities.

That's my two cents worth.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.