Today I tried connecting some 70-LED light strings to DC (using a single 1N4003 diode). Half the string would light. Changing the polarity made the other half light. This also applies to additional strings connected to the female connector at the end of the string. This could make an interesting multi-string flasher, automatically switching the polarity.
I have a few 70-LED icicle lights that work this way too.
I tried it on some others: rope lights, 25- and 35-LED strings, and even an incandescent (considering the 1A limit of the diodes). These would all light dimly on either polarity.
120VDC half-wave rectified, what did you expect? I wired a normal receptacle with the tab between the hot screws removed and a diode connected between then, and then repeated that with the diode wired the other way (to get the other polarity).
The 1N4003 diode can handle up to 200V (not the 1N4004 as I mistakenly posted earlier, that one's 400V). The diode has a current rating of
1A, more than sufficient for several of these light strings.
I had a few last year, but hadn't tried DC yet.
It ends with a period, therefore it is a statement. As for it being unclear, well, I guess it is since you say it is. To me it seems like a reasonable answer to your query... Q: "Why do you need..." A: "So [I] can tell..."
I understood you the first time. And as I said, by using the diode, he was able to tell that the LEDs WERE NOT "all pointing the same way," to wit:
Thus we see that using the diode to remove half of the A.C. wave from the string prevented half of the LEDs from lighting, which effect was reversed by reversing the polarity to the string. Therefore we know that the string of LEDs is actually wired as two anti-parallel strings.
Of course, if you were making a string of LEDs intended to run from A.C., you might also wish to use a series diode in addition to the LEDs themselves. Why? Because a normal rectifier diode has a much greater tolerance for inverse or reverse voltage than does a normal LED. If the peak-inverse-voltage (PIV) specification of a diode is exceeded, the diode will usually be destroyed. Some diodes are made to tolerate this effect, but the LEDs I know of are not intended for such operation.
I once thought so. It's not true. BTW, if it was, it would make it risky to test LEDs of unknown polarity.
Reverse biasing will not hurt a diode (LED) as long as the proper series resistor is present. It limits current to a safe value.
Note that exceeding the PIV causes "breakdown" which just means the diode starts conducting. It is not damaged unless the current becomes excessive. Zener diodes are operated this way (in breakdown) all the time. Right now, I have a device (part of a holiday light controller) that has 2 LEDs on RS232 outputs. these switch polarities (+/-), so the LEDs are reverse-biased half the time. They haven't been harmed yet (during about 60 hours of operation).
Some LEDs are intolerant of reverse breakdown. GaN and InGaN ones (most other than infrared, red, orange, yellow and yellow-green) can be significantly degraded, sometimes to the point of producing zero light at a significant fraction of normal current or even zero light at normal current, by reverse breakdown. It appears to me that some localized electrolysis process or microscopic localized heating causes a partial short. This damage occurs even from static electricity, and many of these LEDs are considered static-sensitive. Thankfully, in my experience at least, these LEDs tend not to get damaged by exceeding their typically 5V reverse voltage limit by small or moderate amounts. I can usually get away with testing one with a 9V battery (and appropriate resistor) with wrong polarity because in my experience they usually need more voltage than that to break down.
I suspect that at least some other LEDs can suffer localized heating effects within their chips during reverse breakdown, and in that case would not tolerate full rated power input during reverse breakdown. At least with LEDs other than ones with GaN or InGaN chemistry, I don't hear of this being so bad as to have an unnoticeable zap of static electricity causing actual damage.
470 ohm is generally OK. If you want, you could calculate an ideal value. It's (supplyvoltage-LEDvoltage)/LEDcurrent. Resistors also have a power rating, but you don't need to worry about that here.
BTW, 9V batteries can supply plenty of current for testing LEDs. I was using one when testing the SSRs (LED input) for Holiday light control.
Not in a simple series circuit. All resistor locations are electrically identical. Resistors are not polarity sensitive. Just make sure it's in series with the LED.
No. Failure to use a resistor will, but you don't need to do that.
Just make sure that resistor is in series with the LED in all cases. If the LED doesn't light, reverse its polarity. You're supposed to be able to tell by the length of the leads, but you may find it easier to try it.
BTW, I've done a lot of that when setting up LED circuits.
So if I wanted to test a LED with a 9v battery, how do I know which leads on the LED to connect to the POS and to the NEG on the battery, and what size resistor do I use? Does it matter if the resistor is on the POS or NEG side?
Also, reading what you said, does this mean that if I connect the LED backwards on the battery it will burn out?
If the LED has leads of unequal length, then the longer lead is positive.
If the LED has a round body with a flange and the flange has a flat spot on the same side as one lead, then that side is negative.
I have seen some exceptions - often through more-disparaged apparently-to-me surplus channels, indicating that they are rejects.
220 ohm, any tolerance, wattage 1/4 watt or any larger size is good for something like about 98-99% of LEDs with a 9V battery continuously, and probably something like 99.96% of LEDs for at least 3 seconds. And much closer to 100% if the LEDs have leads of style typical for soldering into pads with holes in a printed circuit board (or anything bigger, such as more than 2 leads for 1 chip ["high-flux'] or more overtly heatsinkable design).
No. There is tradition in favor of the resistor being on the POS side, but that gets into stepping towards the "tradition" side of the borderline between "tradition" and "general good practice".
With a 9V battery, usually not. By a factor of a large majority, usually not. But on the other hand blame only yourself if you get victimized by LEDs burning out on your watch if you power an LED backwards from a 9V battery.
- Don Klipstein (Jr) ( snipped-for-privacy@misty.com)
Largely good, despite my previous post recommending 220 ohms when battery voltage is 9V.
Usually not, good chance hardly ever or rarely and possibly so unlikely as to possibly next well-traceably happen halfway to a time when the USA does not have a "national debt", but LED manufacturers typically disclaim responsibility of LED failures being of anyone's fault other than yours if you can be held responsible for reverse voltage above 5V.
I figured I'd err on the high side, for testing.
Hmm. News to me. Said responsibility shall no doubt be limited to replacement cost of said device even in instances where said device can be shown to have manufacturing defect(s), even where said device was solely and directly responsible for burning the house down. Have I got that right? ;)
If the LED is defective, return it without abusing it. Return unused ones that can be shown to have a defect first becoming apparent when one was abused, but the abused one you have to eat. Yes, defective electronic components do only cost their manufacturer what they were sold for, or replacements thereof - and maybe lost future sales.
If a defective electronic component burns your house down, then the blame normally goes to the electronic component being in an inadequately fault-tolerant product/installation or being used in a risky manner. If you leave a homebrew electronic circuit running unattended on your bed and it starts a fire, I doubt your fire insurance company will go after the component manufacturers.
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