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I'm following this guide for battery powering some Adafruit NeoPixels. It states that I should put a 1N4001 diode in my circuit to lower the voltage to avoid having too high a voltage and destroying the LED strips.

I tried connecting the +ve terminal of 4 x 1.5V alkaline batteries to one end of the diode and the other end of the diode to the -ve terminal of the batteries to measure the voltage drop and confirm it was about 0.7V (bringing the voltage to around the desired 5.3V). When I did, the diode became very hot very quickly and even started smoking a little. Is this normal?

Also, I'd like to use Lithium batteries for longevity in my final project. The lithium batteries are reading about 1.8V each for a total of about 7.2V across 4. So, in order to get to a reasonable voltage for the LED strips, should I use three diodes connected together in series for a total of a 2.1V drop to bring it down to 5.1V?

// Edit For anyone reading this question later, the solution to the second problem is to use 3 Lithium batteries instead of 4 for 5.2 volts total. No diode even required for a voltage drop since 5.2V is in the perfect range for the NeoPixel strips.

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    \$\begingroup\$ I won't go through that TLDR guide, so add a schematic. You've probably put the diodes in series with the battery without any means (e.g. resistor) to limit the current. \$\endgroup\$
    – the gods from engineering
    Oct 25, 2015 at 17:20
  • \$\begingroup\$ What kind of diode? What current in the LED strips? \$\endgroup\$
    – user16324
    Oct 25, 2015 at 20:40
  • \$\begingroup\$ Added the diode type. Current in strip is a lot harder to estimate. It's approx. 60 mA per LED at full brightness. 1m of 60 LED per metre means 3.6A per metre if all LEDs are on at full brightness but, in practice, not all LEDs are on at full brightness all of the time \$\endgroup\$
    – the_new_mr
    Oct 25, 2015 at 22:02
  • \$\begingroup\$ Why the down votes? \$\endgroup\$
    – the_new_mr
    Apr 28, 2016 at 15:10

4 Answers 4

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Your diode is smoking because the current through it is excessive; you are damaging it. It is not a good test either since voltage drop over a diode depends on the current and when the current is higher the drop is larger (with real Neopixel load the drop over a diode will be smaller and you'll overload the LED). For a good test put an appropriate resistor in series with the diode to produce the current desired and measure that.

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  • \$\begingroup\$ Thanks for you answer. Right, I was thinking that perhaps it was overheating because there was nothing else loading the circuit. How big a resistor should I go for? The biggest I have available at home at the moment are 390 ohm which I'm guessing are not big enough? \$\endgroup\$
    – the_new_mr
    Oct 25, 2015 at 17:32
  • \$\begingroup\$ Also, do you think the diodes are damaged and should not be trusted now? They were in there for a few seconds? \$\endgroup\$
    – the_new_mr
    Oct 25, 2015 at 17:34
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    \$\begingroup\$ Ohm's law is your friend :-), the current in question is one for which a Neopixel is rated. You can add resistors in series, or you can make a strip out of diodes, like 4 say, plus an LED, and measure current through it. Remove diodes one by one until the desired current is reached. \$\endgroup\$
    – Oleg Mazurov
    Oct 25, 2015 at 17:39
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    \$\begingroup\$ Diode could be damaged, don't use it in final circuit. \$\endgroup\$
    – Oleg Mazurov
    Oct 25, 2015 at 17:52
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    \$\begingroup\$ It's not about the risk; to me it's small in either case. If you're experimenting then just keep using the diode until it breaks (and it could be even mechanical since overheated packages become brittle). If you want to build something then don't use parts that you over stressed during experiments so that your creation will function longer. \$\endgroup\$
    – Oleg Mazurov
    Oct 25, 2015 at 18:12
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You have badly misunderstood your source. You need to put the diode in series with your load, not parallel. In other words,

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ I think perhaps I wasn't clear enough. I don't have anything else in my circuit so there is no load. I was performing a test to check the voltage drop. According to other answers, I guess this was my problem. \$\endgroup\$
    – the_new_mr
    Oct 25, 2015 at 17:34
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    \$\begingroup\$ No, you were entirely clear. But if you don't understand the right-hand diagram, you don't understand how diodes conduct current, and you need to do some studying. \$\endgroup\$
    – WhatRoughBeast
    Oct 25, 2015 at 17:44
  • \$\begingroup\$ I obviously get the difference between the left and right hand diagrams. And I understand it too. But I didn't realise that a diode would have issues carrying all of the current in a test circuit. That's all. I actually studied some electronics over 10 years ago and was quite good at it (even if I say so myself) but I've since gone down the software route and forgotten all the electronics stuff. \$\endgroup\$
    – the_new_mr
    Oct 25, 2015 at 18:00
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Connecting a diode straight across the battery by itself? Yes, that'll destroy it quite quickly by putting the full current of the batteries through it.

This is what the "diode" setting on multimeters is designed for: to tell you the forward voltage and polarity of diodes.

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Zener diode is heating due to power dissipation. More the power dissipation more the heat, similar to resistor, for example: 2 ohms resistor carrying current of 1A is dissipation 2W of power. Smoking is due to "excess power dissipation", i.e you are crossing the limit of maximum power withstanding capacity of Zener.

In above example, if resistor with maximum capacity 1W is used to dissipate the 2 Watts then it can cause the smoking. However in same circuit resistor maximum power dissipation capacity increased to more than 2W then it won't smoke but it still heats-up.

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    \$\begingroup\$ Does this add anything compared to the other answers? \$\endgroup\$
    – Solar Mike
    Jul 30, 2018 at 9:29

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