I am working with a 6-axis robot (UR5) with a smart camera mounted on the end of arm, and I want to add a ring light to the camera lens which I will power using tool outputs from the robot interface. The tool outputs can supply either 12V or 24V, and the voltage is chosen programmatically using a touch screen or script.

The LED light rings have a constant current driver requiring 12-18V, so there is a potential for a programming mistake sending 24V which would damage them. I want to prevent that by adding a voltage protection mechanism in front of the ring light, so that if I or anyone else mistakenly sets the outputs at 24V instead of 12V the ringlight is protected.

As an aside, I also want my circuit to be very small, with the fuse fitting in a 150 series inline fuse holder from Littelfuse, catalog # 150274, And the zener diode fitting on this tiny PCB, which I can fit inside the case with existing electronics: 5050 LED breakout PCB, product ID 1762 from ADAFruit. I have located components already that fit these, so I don't need help with that part; I just wanted to point out my design requirement.

I'm a mechanical engineer, and not well-versed in electronics, so I want to make sure I am doing this overvoltage protection correctly before I implement it. Below is a diagram of what I've come up with after extensive internet research. My question is specifically about the fuse, but if anything else stands out that I am doing wrong, please feel free to comment and correct me.

I based my circuit off of this: (http://www.learningaboutelectronics.com/Articles/Overvoltage-protection-circuit.php) and I did a lot of reading on zener diodes, fuses, thermal runaway in LEDs, constant current vs. constant voltage drivers, etc. Here it is: enter image description here

Here is how I understand this: the Zener diode will prevent more than 12V from reaching the circuit. So if 12V is applied the Zener acts like an open circuit. If 24V is applied, the Zener dissipates 12V of that and the LEDs still get 12V...is that right? And am I correct in assuming that the Zener will draw the maximum current it can from the output, causing the fuse to blow? That is what I am hoping. I want the fuse to blow and alert the user to the programming error so the Zener doesn't just sit there and get hot.

I've chosen a zener diode that is 12Vz with a 5W power rating.

I couldn’t find a fast fuse in the size I want (more than 300mA, less than 600mA), but the slow blow fuse I spec’d out will take 3-20s to pop at 200% current rating. Is that good enough, or do I need a fast fuse? I've calculated 0.330Ax12V=3.96W across the Zener if the circuit has 24V applied to it, which is within its rating...Have I done this correctly?

If someone can clear this up for me, that would be great, and as mentioned above, if I have made other mistakes or taken the wrong approach let me know. Thanks a bunch!

  • \$\begingroup\$ As long as the zener impedance is low enough and the power supply does not fold back (programmed or parasitic), the fuse should blow at 24 V. Also,+1 for language and layout. \$\endgroup\$
    – winny
    Commented Jun 23, 2017 at 17:10
  • \$\begingroup\$ @Anna you would be better off using a SMPS ( and follow my advice) that accepts a wide range of input voltages 12 or 24 and maintains perfect CC regulation rather than follow the other advice. These are cheap online (few$) These boost the voltage to accommodate a wide range of LEDs >12V total \$\endgroup\$ Commented Jun 23, 2017 at 22:38
  • \$\begingroup\$ If using this circuit, I would use a 15 Volt Zener diode (or something a little over the normal "12 Volt" supply), otherwise the Zener may draw current under normal operation. Preferably, I would use a switch-mode regulator, as Andy aka suggests below, to bypass the problem and prevent blowing fuses. However, the RIGHT solution is to revise the software so users can't accidently program 24 Volts output to a device that requires 12 volts. \$\endgroup\$ Commented Jun 24, 2017 at 0:46

4 Answers 4


Why don't you just design (or buy) a small voltage regulator that feeds your bank of LEDs. Then there are no complex issues like fuses blowing and notifications etc. Think simple. Design a simple small buck regulator that can supply (say) 12 volts from the 24 volt power option.

It looks like this would be OK: -

enter image description here

But there are many other options from Linear technology and Texas Instruments.

if I or anyone else mistakenly sets the outputs at 24V instead of 12V the ringlight is protected

With this idea you set it at 24 volts and it regulates to 12 volts. If you set the input to be 12 volts then the LEDs will be a bit dimmer because the LT3970 won't be able to supply 12 volts from a 12 volt input (it'll be more like 11.75 volts as per graph on pg 6 of data sheet). But it won't fuse!

As an aside, I also want my circuit to be very small, with the fuse fitting in a 150 series inline fuse holder from Littelfuse, catalog # 150274, And the zener diode fitting on this tiny PCB, which I can fit inside the case with existing electronics

I expect the regulator circuit will be smaller.

So if 12V is applied the Zener acts like an open circuit

Of course if you do decide to go down the zener diode route and pick a 5% tolerance one (about as good as they come) it may only be regulating at 11.4 volts and this may blow the fuse under normal circumstances. Go for a 15 volt zener is my advice then there is never any possibility of it blowing the fuse at 12 volts.

  • \$\begingroup\$ Thanks..I took your advice and located a very compact and inexpensive voltage regulator. Thank you for the input, because that is a much better solution. For anyone interested, this is it: getfpv.com/… \$\endgroup\$ Commented Jun 28, 2017 at 16:49
  • \$\begingroup\$ At $2.29 it's worth a risk but be prepared for disappointment whenever buying from a source like this when there is no data sheet (general advice I give everyone in these circumstances when buying something with no apparent supporting documentation). \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2017 at 18:44
  • \$\begingroup\$ I got them in the mail today and stripped the heatshrink off one to take a look. They seem to be little buck regulators, just like what you suggested making. They contain this converter: monolithicpower.com/Products/Product-Detail/101/Step-down(Buck)/… Thanks for your help! \$\endgroup\$ Commented Jul 3, 2017 at 15:51

The current through the zener will be determined by the source impedance of your 12/24 volt supply and the associated wiring. Most supplies will have a low enough internal impedance that the zener diode will fail when it conducts.

A slow blow fuse is not applicable in this situation since by the time the fuse blows, it is likely that the damage has been done to the LEDs or their constant current supply. A fast blow fuse is required.

You may wish to investigate a technique called a crowbar circuit. This uses an SCR that fires when the maximum voltage is exceeded. The SCR can handle substantial current for a sufficiently long enough period to blow a fast blow fuse. These are commonly used in power supply circuits to protect the load in the event the voltage regulation mechanism in the supply fails. The fact that your current requirement for the LED ring is quite small makes it a good candidate for this type of circuit.


simulate this circuit – Schematic created using CircuitLab

If you have the room, you could simply add a voltage regulator ahead of your LED ring. In the event someone does switch on the 24 volt source, the regulator would step it back down to 12 volts.

  • \$\begingroup\$ although a good answer to the question, still a poor design solution overall. ( have a spare fuse?) \$\endgroup\$ Commented Jun 23, 2017 at 22:41
  • \$\begingroup\$ Understood but a $0.50 fuse is cheap insurance. I also listed a regulated power supply option. Personally, I would disable the operator's ability to change it. \$\endgroup\$
    – Glenn W9IQ
    Commented Jun 23, 2017 at 23:26
  • \$\begingroup\$ if the design permits multiple source voltages from 12 to 24 then the supply ought to be able to auto range with a simple SMPS as I have shown . Then the screen setting is not necessary. If indeed it is a linear CC, then it might be the problem with self heating and poor efficiency, not the LEDs. If it is the LEDs then again the design is flawed as the CC is a misnomer. \$\endgroup\$ Commented Jun 24, 2017 at 3:10

Don't overthink problems that don't exist. You have a CC LED driver that is obviously some form of Buck conversion. It works at a fixed pre-defined I(LED) current (adjusted by some control or a resistor value), and set at 270 mA in your case. The controller does not care what the LED voltage is since it will set create a voltage that supplies 270 mA.

From the input side your controller works on any voltage from 12-18 V, so it is clearly not rated for 24 V input. You are taking a risk if your input voltage is programmable (as you point out) ....but I imagine you actually have a fixed input for the rest of your CNC so you don't actually program the supply voltages.

If you supply the CC LED driver with 12 V then you are safe even if the 12 V is a bit high (you'd be safe up to 18 V obviously).

If you want to supply 24 V, then you should really get a CC LED driver that works at that voltage.

Your idea of the Zener is very crude and may be effective. Since you've chosen a 5 W Zener then the maximum rated current is about 0.4 A in conduction, it will probably have a pulse capability many times that value for a short time. Zeners are rarely accurately 12 V (especially power zeners) so I'd suggest that if you want to use the circuit you proposed you should use a 15-17 V Zener.
Here is the Datasheet for a 5 W 1N5349 device. Notice that the pulse current is about 7.5 A rated for 8.3 ms, this is many times the rated current of your fuse so should blow it rapidly.

If you simply want to blow the fuse when you inadvertently connect 24 V instead of 12 V, then the circuit you have will work (though I suggest a 15-18 V Zener). The downside is that it will draw an uncontrolled current (limited only by the wiring) until the fuse blows. Whether you use a slow or fast fuse, you may find your 24 V power supply shuts down (it's electronic overload detection) before the fuse would blow. This will depend almost entirely on the value of the output capacitors in your power supply. This loss of 24 V may mean you lose X,Y,Z positional integrity on your CNC ...not a nice result.

I think it would be a much better result overall if you sourced a CC LED driver that works up to 24 V, there are a plethora of these available online (an example here).


you said...

"The LED light rings have a constant current (CC) driver requiring 12-18V, so there is a potential for a programming mistake sending 24V which would damage them.""

The problem is the CC design spec: ( and resulting design)

If no inductor is used and just PWM then it is less efficient and heat loss rises with input voltage drop thus limiting the maximum input voltage as suggested 18V instead to > 3x the output voltage common in simple SMPS CC sources regulated by current sense Rs.

So the problem is in the spec. and the design of the CC.

Dont put a bandaid on a poor design spec with a poor inefficient solution such as a power clamp or inefficient zener or LDO (low drop out) linear regulator.

Design it correctly from the start with smart specs. and say it MUST function from Vin from 12 to 28V

TI and many other sites have DIY webdesign tools on their site or in datasheets.

e.g. of 11 LEDs (1W ea.) or 6 LEDs (3W ea.)

enter image description here

These are examples of a SMPS using highside CC source and boost voltage to Diode V+ (anode side) with a fixed current for any voltage 12 or 24V +/- ?? %


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