Why don’t manufacturers put reverse protection in LED lights ?

[Jay R]

All the single cell ones have to boost the voltage up anyway so why not boost it an extra half volt and stick a diode in ? It shouldn't affect the runtime much. Chances are that I'm not seeing something obvious so feel free to chip in and tell me what I've missed.

 

[Ragiska]

many lights do have a mechanical reverse polarity protection. an electronic reverse polarity protection does significantly affect runtime and brightness.

 

[Th232]

The further the output voltage is from the input, the less efficient the converter becomes, and the higher the current drain the lower the effective mAh of the battery. Also, it seems like you're thinking about putting it between the converter and LED? The converter is also a polarised component, and so it should be between the battery and converter otherwise a reversed battery will blow the converter unless it's already got inbuilt protection, at which point a diode becomes a moot point anyway.

Note that it'll also dissipate a lot of power, especially in higher applications. Since you say that "all the single cell ones have to boost the voltage up anyway" I'll assume you're talking about AAs and lithium primaries (compared with Li-Ions at a max 4.2V, at which point you'll need to step down the voltage).

So let's take an AA and assume that it's at 1.3 V and we desire to draw 500 mA from it. Without a diode, we simply draw the 500 mA, the converter is getting 0.65 W.

With a diode that has a Vf of 0.5 V, this leaves the converter with 0.8 V. To get the same 0.65 W, it'll have to draw 0.8125 A. Note that the current is also flowing through the diode, which is now dissipating (as heat) 0.5*0.8125 = ~0.41 W, so we're now drawing a total of 1.06 W from the battery instead of 0.65 like before.

I think that's a bit of a worst case scenario though.

Also hope my logic and maths is correct, I'm not fully awake yet.

 

[kramer5150]

There is a Vdrop through a diode. Some say its negligible, others will say its significant enough. The better designs I have seen use a mechanical solution at the B+ terminal. Fenix and eagletac are two off the top if my head.

 

[65535]

I've thought about this, bottom line for me personally is, how hard is it to put the battery in the right way? Honestly, I feel like you'd have to be pretty stupid to put it in the wrong way.

 

[lolzertank]

All the single cell ones have to boost the voltage up anyway so why not boost it an extra half volt and stick a diode in ? It shouldn't affect the runtime much. Chances are that I'm not seeing something obvious so feel free to chip in and tell me what I've missed.

Sticking a diode on the input doesn't work, since you're left with too little voltage for the boost converter. If it's on the output, then it does nothing to protect the boost converter's chips.

The best way is to connect a P-MOSFET in series with the battery's + terminal and the load's input, with the gate going to the battery's negative terminal. This way, there's very little loss and there's no chance that an oddly shaped battery will manage to bypass the protection.

 

[hoongern]

I've thought about this, bottom line for me personally is, how hard is it to put the battery in the right way? Honestly, I feel like you'd have to be pretty stupid to put it in the wrong way.

Mistakes can and do happen, unfortunately. Especially when you're very tired, or if you're doing it the dark, etc.

I guess it could be that I'm stupid, but just the other day I accidentally put in a battery into the charger the wrong way. Luckily I noticed in a second.

Now I color code the positive end red and match it with the end of the charger which I have also color coded red.

I definitely think that battery manufacturers should improve things by, say, coloring a bright red ring (or some other color) around the positive end of the battery :shrug:. At least with flat top batteries, sometimes differentiating the ends can be hard.

 

[Kuryakin]

The problem is even the best silicon power diodes have about 1/2 volt Vf. This is a full 1/3 of the battery voltage in a single cell alkaline, more if it's a NiCd or NiMH! That means 1/3 of the delivered energy is wasted from the cell, probably more since you're wasting the energy from the fresh part of the cell's life.

All the single cell ones have to boost the voltage up anyway so why not boost it an extra half volt and stick a diode in ? It shouldn't affect the runtime much. Chances are that I'm not seeing something obvious so feel free to chip in and tell me what I've missed.

 

[Jay R]

The boost on most of the cheap lights I've taken apart has just been a tiny wire wound, ferrite core, transformer type. Should be no problem putting the diode after that and then you'll only be knocking 4.2v down to 3.7 rather than 1.2 to 0.7. A much better ratio.
Still, I guess it's still a bit on the high side and I suppose that the cheaper brands I'm thinking of such as U***fire or S***fire don't really worry too much if you burn out the LED after you bought it.

 

[uk_caver]

The best way is to connect a P-MOSFET in series with the battery's + terminal and the load's input, with the gate going to the battery's negative terminal. This way, there's very little loss and there's no chance that an oddly shaped battery will manage to bypass the protection.

Would that work for a single-cell boost? It would need a FET that turned fairly fully on at ~1V.

For drivers running from a nominal 2.4V, I've used reversed N-channel MOSFETS for reverse polarity protection (drain connected to battery-, source to circuit-, gate to battery +), and they seem to work OK, but they do have to be the 'wrong' way round, otherwise the inherent diode in the FET will conduct if the battery is misconnected.
Presumably you'd be doing something similar with P-channel FETS?

If extra components weren't a problem, with a switch using an N-channel FET, would it be possible to do a little bootstrapping if using a single AA cell supply?
As long as there was enough current getting through at startup to turn the boost circuit on to *some* extent, some of the boosted output could be fed back to the FET gate, (or the gate of a parallel FET) which could then be driven fully on.

 

[HKJ]

The boost on most of the cheap lights I've taken apart has just been a tiny wire wound, ferrite core, transformer type.

These inductors has some electronic controlling them and it is the electronic that is damage when reversing polarity. Led has no problem with reverse polarity as long as the voltage is below 5 volt.

 

[uk_caver]

Led has no problem with reverse polarity as long as the voltage is below 5 volt.

That does depend on the LED.
Crees XR-E and old Luxeon I/III were fine.
Seoul P4s have an integral reverse-connected diode which will start conducting at low voltages (~1V)