# Cost effective way to protect circuit from overvoltage

Which is the most cost effective way to protect a DC circuit that requires 5V input (and draw up to 1A) from external power supply?

Suppose that a careless user can accidentally attach an external power supply and power up with 6-9-12-24V, this situation can literally blow up my circuit!

This can appear to be a silly question but i don't have any ideas!

• Will your circuit really blow up if it is fed 6 volts (or even 24 volts)? Aug 13, 2020 at 10:54
• With 7 volts it will blow up. absolutely Aug 13, 2020 at 14:27

Probably the most common method is to use a fuse with any (and sometimes all) of the following: -

• A regular power zener diode
• A power shunt regulator (like a zener but programmable)
• A crow-bar circuit (on over-voltage, it clamps to close to 0 volts)

So, if the voltage exceeds a certain level, too much current will flow into the aforementioned devices/circuits and then the fuse blows.

You can use resettable fuses too (polyfuses they are sometimes called). Once power is removed, the polyfuse resets.

An alternative is a series-connected over-voltage circuit that "disconnects" when the input voltage exceeds a certain critical level. Not too tricky to design but, they should also be used in conjunction with a zener to ensure that fast voltage rising events are dealt with effectively.

I would consider something like this ciruit:

simulate this circuit – Schematic created using CircuitLab

• The fuse will blow when the input voltage exceeds 5.1 V
• If replacing a blown fuse is an issue, consider a polyfuse which self-resets
• Instead of the zener diode many other solutions can work, a zener + Thysristor solution like this:

Borrowed from here, there are more solutions in that article.

What solution is most suitable in your situation depends on many things:

• is replacing a fuse OK?
• what is the current consumption of your circuit?
• Cost of the components

Also don't forget physical solutions to protect your circuit, like using a USB connector for powering. You will still need to cater for cheap USB chargers that output up to 7 V though.

The fuse and zener is a good and commonly used solution ... Others include, marking clearly the DC voltage intended (or AC otherwise) to be used, close to the power supply input. This would normally be on the product case, next to the supply input. Also making it difficult to otherwise plug anything else but your intended voltage, such as a mini-USB or USB connector if you're running of 5V DC only ... Usually its a combination of all of these ways.

The most common solution is to have a voltage regulator in your device.

Your circuit itself runs on 5V, but the regulator can accept a much higher voltage (say, up to 40VDC) and make 5V from whatever input it gets.

That's what most consumer devices do.

If you want to protect against more extreme mistakes, then a circuit such Andy aka describes can handle much higher voltages.

Regardless of what solution you choose, be aware that they will all fail if the user tries hard enough.

I knew a kid who attached a 120VAC power cable to the barrel plug and connected it to an electronic game just to see what happened. The (expensive at the time) game smoked and burned out - of course.

The folks who designed it didn't include "protect 9VDC input from 120VAC line voltage" in the game because that would cost too much.

Even if it weren't too expensive, you have to draw the line somewhere. 120VAC? Why not 240VAC, or maybe the local distribution transformer shorts and you get 3000VAC? At some point you just have to figure the "protection" is more trouble than it is worth, and the user's device is going to get damaged from misuse.

An economical solution is Toshiba 5.5V (5.1 Vth) TVS (\$0.036 @ 1k) with desired 1A Polyfuse

Use a micro USB socket for the power connection.

All USB shaped power connectors provide 5V. So the 9V plug doesn't fit.

• That's a very 2007 - 2014 answer.
– towe
Aug 13, 2020 at 4:01
• We've had USB C for that long, I didn't realise, but yeah that would work too. USB 3.1 muddies the waters, but not an any way that matters. Aug 13, 2020 at 9:54
• USB C still provides 5V per default, unless the device communicates to the power source that it wants more (see also Wikipedia: USB Hardware/USB Power Delivery) Aug 13, 2020 at 13:46

There are a surprisingly large amount of things to tackle for input protection: over voltage (constant), over voltage (ESD) and reverse voltage protection.

One of the easier solutions is to use a fuse or polyfuse, and a TVS or zener shunting to ground in the case of overvoltage. The tricky part is sizing the fuse to allow proper operation, but prevent damage of downstream components in the time it takes to trip. The diode will heat up considerably and may be destroyed, depending on the power supply.

To tackle reverse voltage protection, you can use a diode in series with the supply (if you can handle the dissipation/voltage drop), or a P-Channel MOSFET circuit if you need lower loss, in the milliohms region.

If your voltage input source is not variable, the easiest solution would just be to use a voltage regulator with some transient input protection. If it is variable (such as in the case of a wide input buck converter), you may want to explore the above options.

Finally you can use many commercial e-fuse or hot swap controller ICs if you want to reduce your BOM or design effort - These ICs can (but not always) combine ALL of the above functions into a tiny package.

• You have some commercial part reference to advise? Aug 13, 2020 at 6:41
• I recently used a TPS25944 to protect a FPGA module - it only handles up to 20 volts overvoltage, but handles undervoltage, overvoltage, current inrush, current limiting and reverse polarity protection. Generally anything over 20 ish volts requires an external FET with the eFuse IC. They are more expensive than a fully discrete or simpler protection scheme though.