LED Fail on Magnifying Glass - Page 1
Forum section: electronics / beginners

author: Kagord
author_rank: Contributor
timestamp: May 30, 2019, 01:58:40 pm
content: Hi,I purchased this magnifying glass, 3 months later, an LED failed (flickered for awhile and went out), then within a week the other 2 LEDs failed in the same fashion.Product:https://www.amazon.com/gp/product/B015VGC15YOpening it up. The circuit is a 20 ohm resistor connected to 3 straw hat 5mm white LEDs in parallel. Vin is 6V (4 AAA batteries in series). Looking at datasheets for various LEDs, the LEDs seem to be 3V.Sample Datasheet:https://www.iled.com/class/INNOVAEditor/assets/YeniDatasheets/2718-2780(hiledmantar).pdf(part 2729).I bought these:https://www.aliexpress.com/item/wholesale-100-Pcs-5mm-White-Straw-Hat-Water-Clear-Super-bright-Wide-Angle-20000MCD-LED-free/1923509715.html(White, 3.2V minimum)It works again, but why, if min voltage is 3.2V and max is 3.4V per LED, how can 6V turn 3 LEDs on, isn't it too much voltage? How do you calculate the voltage after the resistor?

author: DDunfield
author_rank: Regular Contributor
timestamp: May 30, 2019, 02:51:11 pm
content: Min voltage 3V to max 3.4V .. lets guess 3.2Typical current 20maStarting with 6v and dropping 3.2v for the LEDS, that leaves us with 2.8v dropped across the resistor.By Ohms law, 2.8/20 = 0.14 amps, or 140maIn an ideal situation, the current would be being drawn equally by the three LEDs, which would mean 140/3 or about 47ma per LED.Seems a little higher than the typical from the datasheet.** Not an accurate calculation because we don't know the actual voltage drop on the LEDS, and if the LEDs are drawing more than typical, likely voltage is slightly higher than typical, but the calculation with typical values and finding current more than 2x typical is enough to determine that something is "not right" with the design. You could measure with a multimeter and determine more accurately what is going on.Also note that driving LEDs in parallel without series resistors on each LED is not good, as they will not be exactly balanced, one will draw more than another. Likely the first one to go was drawing the most current.It is common in cheap stuff to find LEDs overdriven, presumably in the hopes of getting a little more brightness from them.I bought a little nightlight to use when travelling - It used 3x1.5V (4.5v total) and a single white LED ... seemed OK for the first few trips, then the LED started flickering very noticeably.Upon taking it apart, I found a 5 (FIVE) Ohm resistor in series with the LED and batteries. This caused it to draw over 50ma which is more current than I deem suitable for a low power LED (I didn't try to find the data sheet).I replaced it with a 100 Ohm resistor which gave a current of about 15ma, and a hardly noticeable drop in brightness and no more flashing. Batteries last much longer too!Based on the datasheet and a lower limit of 3V drop on LED, leaving 3V on the resistor, solving for 20ma I would try a 150 Ohm resistor in series with EACH LED and see where it falls.Dave

author: Audioguru
author_rank: Super Contributor
timestamp: May 30, 2019, 03:43:17 pm
content: I agree that each LED should have its own series resistor and operate each LED at its 20mA rating. The resistor for each 3.2V white LED will be (6V - 3.2V)/20mA= 140 ohms which is not available, use 120 ohms or 150 ohms.

author: Kagord
author_rank: Contributor
timestamp: May 30, 2019, 04:17:53 pm
content: OK, please bear with me, I'm new at this, if you're supplying 6V to a 3.4V max input voltage component, is that OK? I guess I'm not getting what voltage drop means.

author: boffin
author_rank: Unknown
timestamp: May 30, 2019, 04:47:15 pm
content: Quote from: Kagord on May 30, 2019, 04:17:53 pmThat's why there's a resistor.The LED drops 3.4VThe resistor drops the rest 2.6VBy knowing both the value of the resistor 20R and the voltage across it 2.6, you can calculate total current: 2.6/20 = 130mA, a little over 40mA/LED (in a perfect world)The problem with the solution you have is that not all LEDs will drop EXACTLY the same voltage, so the electricity will tend to favour the one that drops the least voltage, burning it out; rather than evenly distributing through each LED Rinse, repeat until no LEDs remain.Take the advice here, and have a separate resistor for each LED, and drive the LED at about 25mA max, or about 100 Ohms (2.6V / 0.025mA) for the resistor.

author: Laszlo
author_rank: Contributor
timestamp: May 30, 2019, 05:15:33 pm
content: The Led is a diode, like all diodes it has a so-called forward voltage, which is necessary for it to turn in.If the Vin is below that value the diode will not turn on, and there will be no current flowing through your circuit. (It will act pretty much like an open switch.)This forward voltage equals the voltage drop across your LED. So if you use the datasheet in your example, the LED will drop around 3-3.4V and the rest of the voltage drop will be dropped across your 20 Ohms resistor which is in series with your load.Kirchoffs voltage law states that the sum of voltages in a closed loop equals to zero, therefore 6-3.2=2.8VAs others have stated, this circuit is not designed well, and the LEDs are being pushed over the limit when the batteries are fully charged. If you change the resistor to the recommended 150 Ohms, they will last longer but it won't be as bright as it was anymore.

author: DDunfield
author_rank: Regular Contributor
timestamp: May 30, 2019, 07:56:48 pm
content: Quote from: Kagord on May 30, 2019, 04:17:53 pmIf any current is flowing through the resistor, there will be voltage "dropped" across it. There will also be voltage dropped across the LED. The sum of all voltage drops in a circuit will equal the battery voltage applied. So how do we know how much will drop on the resistor?One of the most fundamental things you will use in electronic work is "Ohms Law", which is expressed as: E = IRE = Electromotive force (voltage) VOLTSI = Current (I don't remember why it's 'I') AMPSR = Resistance OHMSWe want 20ma (0.020 amps)From the data sheet, we know that this should occur at between 3v and 3.4v depending on the tolerances of the particular LED.To cover off worst case (3V) we need to drop 3V (from our original 6V) across the resistor.E = IR can also be expressed as E/I = R3(V) / 0.020(A) = 150(R)Which is how I arrived at 150 Ohms for the resistor (assuming one per LED).If you go to the other end of the data sheet range (3.4v) we need to drop 2.6V2.6 / 0.02 = 130 OhmsIf you take the middle road (3.2), we need to drop 2.8:2.8 / 0.02 = 140 OhmsHighest resistance is safest because lower will allow more than 20ma to flow for some LEDs which still conform to the datasheet.However the 20ma is listed as "typical", and it's quite likely that some amount more is OK, possibly a fair bit more - but to know for certain if it could damage the LED you would have to know the maximum allowed current, which didn't appear to be specified on the datasheet.Also note that the datasheet says there is a 10% tolerance on all current values, so adjust your boundaries accordingly.Dave

author: mariush
author_rank: Super Contributor
timestamp: May 30, 2019, 09:25:47 pm
content: Using a resistor is a very basic way of limiting current going through a led.The most common way is to have multiple leds in series and then have a single resistor to limit the current going through the whole chain.You have the classic formula Ohm's law : Voltage = Current X Resistance (V=IxR)So from this formula you can derive a formula which lets you determine resistance value or current value :Voltage input - ( number of leds in series x forward voltage of each LED) = Current x Resistance.In your case, the voltage of the batteries (4.5...6v depending on battery chemistry) is lower than the total voltage of the LEDs in series (~9v) so they resorted to connecting the 3 leds in parallel, and then have a single resistor in series with the leds to limit the current. This way, the current is in theory split between the LEDs.This is a problem, because the LEDs don't have the same forward voltage, though it's close enough. So in extreme cases, one led could have a 3v forward voltage, another could have 3.2v forward voltage, so one LED could have 25mA going through it, the other could have 20mA...Also, if one LED dies, you're left with two LEDs that now have more current flowing through them.If you're interested about efficiency and all that, it would be better to advance to using LED driver ICs, which allow you to not waste so much power in that resistor.For example, let's say you want 20mA through a 3.2v resistor and you power it with 6v battery :6v - 1 x 3.2 = 0.020 A x R => R = 2.8/0.020 = 140 ohm ... so you'll probably go with a 150 ohm resistor. Let's recalculate with the 150 :I = 2.8 / 150 = 0.0186 or 19mANow we can use same formula to calculate how much power is dissipate in the resistor : P = I x I x R = 0.0186 x 0.0186 x 150 = 0.052 watts.So basically, 3.2v x 0.0186A = 0.05952 watts go into the LED and 0.052 watts go into the resistor - over half of your battery is wasted as heat into a resistor.For a single LED or a chain of leds that have the sum of forward voltages less than input voltage, you can easily go with a buck (step-down) led driver IC.Here's a very cheap one, 0.45$ for one:https://www.digikey.com/product-detail/en/diodes-incorporated/AL8860WT-7/AL8860WT-7DICT-ND/6226981It's surface mount, but the pins are spaced apart so it's fairly easy to bend the pins and solder wires to them even by hand and the datasheet is very helpful and has example circuits:https://www.diodes.com/assets/Datasheets/AL8860.pdfSee page 2 in datasheet for example circuit - you only need a small inductor, a small resistor, a basic diode and a decoupling capacitor - the values for them are explained in datasheet (pages 11-13)If you want to put several leds in series and the total forward voltage will be higher than input voltage, you can use a boost (step-up) led driver ic.Here's a cheap example, 0.5$ each or 0.31$ if you buy 100 :https://www.digikey.com/product-detail/en/diodes-incorporated/AP3019AKTR-G1/AP3019AKTR-G1DICT-ND/7708820See page 2 of datasheet for example circuit, and pages 7-9 explain how to pick the other components required (inductor, resistor) :https://www.diodes.com/assets/Datasheets/AP3019A.pdfCompared to plain resistor to limit current, you get up to 84% efficiency with such a chip.Besides the actual chips, the inductor and everything else won't cost you more than 0.5$..1$ so for a total of 1$ plus shipping you could basically double your battery life, and have better current control (current won't vary depending on battery voltage as the led driver IC will dynamically adjust things to keep current constant)