So many gadgets for your smartphone, but it's not cheap for you buy it. So now i will show you how to make unique gadget for your smartphone for cheap and under 30 minutes to make it. I will make Smartphone notification lamp because its so useful when i have new notification or any new message and this gadgets can use as flashlight.

Let's Make It!!

Step 1: Tools and Materials You Will Need


  • Old audio jacks (Use 3.5 mm jack)
  • 1 Low voltage LED (Use the low voltage led for best result)


  • Soldering iron
  • Hot glue gun / any glue for make the case

Step 2: Open the Case

Take the audio jack and open the case slowly, Don't broken the inside.

Step 3: Search the Output Voltage

  • Take your smartphone
  • Plug in the audio jack
  • Listen any music on your smartphone and set the volume full.
  • Paste the led and search the positive and negative output
  • And its look like image 4 and 5

Step 4: Solder It!

Take your soldering iron and solder it slowly. After that its look like image.

Step 5: Make the Case

Make the case for looking best. I use hot glue to make the case like image.But you can use anything for make the case. And cover it using heat shrink tubing or anything. I use white tape to cover it.

Step 6: Finish

  • Plug in this gadget to your smartphone
  • Set the audio volume on your smartphone full
  • And if you get new Notification or new message, the led will on
  • To see how it works, watch the video

Step 7: Thanks

All finish, I hope you enjoy this. Don't forget to like, subscribe, and vote me :)

Sorry for my bad english.


<p>not a bad idea. I wonder if it could be done with an rgb led</p>
<p>Does this light work on Iphones, I noticed you have a samsung and when I assembled mine, the LED never lights(polarity is correct and wiring is correct) And when tested the jack should drive enought power to light the LED (I tested a 1W 8Ohm speaker which worked on the jack.) But the LED never lights even when I know the jack is working. I also tried multiple LEDS.</p><p>I have an iPhone SE.</p>
<p>Nice thinking! If you use a headphone splitter you can still use your headphones while getting LED notifications.</p><p>This would be great to bring to the library!</p>
<p>Cool idea, but why no series resistor? This looks like a good way to damage your phones audio output</p>
<p>Modern audio out ICs have protection and were low current anyway. IF the output were high enough to cause damage it would tend to also be high enough to damage the LED because it's AC output instead of DC.</p>
<p>Again &amp; again &amp; again... the audio is NOT AC.. it's AV, (Alternating DC-VOLTAGE) The amplifiers of audio uses +v and -v for source voltage just to put the &quot;resting signal = NO audio present&quot; in to the middle of the source Voltage e.g. if your system has a 12V power source then the amplifer puts it to ~6V with no variations voltage from the input, only the inner end of the output capacitor has the 6V... THEREBY the Audiosignal variouses between 0V and 12V thus giving you an audiosignal in form of an electromagnetical variation to your &quot;membrane&quot; in your loudspeaker thus resulting the air to move thus giving an audiable &quot;sound&quot;. If you put just DCVolt to your louspeaker the membrane would be &quot;ON&quot; and it gets warm/smoke... right?</p>
<p>Not all audio amplifiers need a decoupling capacitor at it's output. I will mention three designs: first: using a split power supply (v+ and v- voltages relative to GND) if the output use complementary transistors one of the speaker lead is attached to the two emitter junction of the output transistors and the other speaker's lead is attached to GND, At rest, with no input signal, the two emitter junction is at 0 V relative to GND. As each half amplifier is designed for class B operation (or better class AB operation for less crossover distortion) one transistor manages the positive semicycle of the signal an the other transistor manages the negative semicycle. Even though you are feeding DC voltages to the load (the speaker) it sees an acts as if it is (and REALLY IS) AC. This design is used in high power audio amps, not probable in smartphones. The second design, uses a similar thinking as the first, but with a trick:, as the amplifier is powered by a single power supply ( v+ and GND) as you mention earlier, with no input signal, at the output you will have v/2. The trick here is that if you build and active voltage divider with an output at exactly v/2 you can attach one speaker lead here and the other speaker lead to the amp's output, as the two speaker terminals are at V/2, you got 0 V betwen speaker terminals at no signal input. At last but not the least, the for the third design, you can use a BTL (Brigde thru load) design. This design uses two amplifiers to drive one speaker, each speaker lead is connected at each one of the amplifier's output. The input signal is feed to both amplifiers, but one is in-phase and the other is 180 degrees out of phase. With no input signal both amplifier outputs are at v/2 but the speaker see 0 V thru it's terminals; with input signal one amplifier output swings toward GND and the other amplifier output swings toward V+, these swings alternate between amplifiers, so the speaker sees effetively, AC. This design, though, has a drawback to be used in smartphones, Can you tell which one it is?</p>
<p>Why is this so difficault? A audio output is NOT AC, (Alternating Current, that is when the current goes from one pole to another and vice versa, NOT the Voltage)... geee... it instead is ALTERNATING DC-Voltage look at the schema hereby, ONLY alternating voltage, and again... the output GOES through a capacitor, in this case 2000uF. </p>
<p>Oh probably, but the whole point of my comment is that without a resistor your technically shorting the output. An LED is only a diode, so any voltage above the diodes voltage drop is actually being fed back at a near infinite current (or at least at the max current the driver can supply). All due to ohms law.</p><p>AC has nothing to due with it, the reverse voltage threshold will likely be higher than any typical audio device can provide. </p>
<p>I guess the LED is the one to get fried?.In pictures he uses a blue led, that would be that usually the blue leds have an initial voltage of about ~3,2V , so, the voltage to be detroyed, (depending on your phone battey voltage), is 0,7volt. Surely the LED fries first, (it's kind of a fuse here)</p>
<p>It's maybe the LED, maybe the output stage of the audio port.. Who knows? For sure I'll never put a LED on an output I don't know without protecting it (it = the output port) with a resistor or other current-limiting protections</p>
<p>Please don't use a led as a &quot;fuse&quot; ewen it can't stand for more than say 50mA. It want stand for them other requierements for a fuse</p>
<p>Sorry, I couldn't understand what you meant, since it appears to me the opposite as what you were saying before. What I wrote is that you can't use it as a fuse, or using it saying &quot;well, this will fry before the output&quot;, because it is not guaranteed. Maybe the led fries, maybe the output. For sure I will always use a resistor in series, even a very low value one if the source voltage is too low, but I'd never put an LED directly to a constant voltage source</p>
<p>I DO put a LED directly to power source if the voltage is right, (the resistor only LIMITS the current with TOO high voltage hmmmmmm........). The resistor ONLY EAT POWER, use appropriate voltage PERIOD</p>
<p>Well, since an LED (and diodes in general) are highly non-linear components, a small change in the voltage is translated into a high change in current. For instance <a href="http://electronicdesign.com/site-files/electronicdesign.com/files/uploads/2013/01/IFD2527_FIG1.gif" rel="nofollow">this image</a> shows the first graph I found searching on google. Passing from 3V to 3.1V will change the LED current from 1A to approx 1.5A, and at 3.2V it will be over 2A. And keep in mind that this varies with the temperature.</p><p>Now, if you add a 0.1Ohm resistor in series, you will use approx 0.85A with 3V and 1A at 3.1V, going to around 1.2A at 3.2V.</p><p>As you can see, without resistor small changes in the voltage will result in high changes in the current. If you have a 3V +- 5% source, you can get half the nominal current or double....</p><p>So, according to my experience, there is no such thing as appropriate voltage. You can either choose to protect it, using at least a resistor to limit the current to avoid overstressing both the LED and the driver (or better by using a current source instead of a voltage one), or decide you don't care about shortening the LED and the driver life, like they do in cheap chinese-like products.</p><p>Period.</p><p>PS: maybe you are saying this because in some circuits you made you didn't use it with some voltage outputs. Please note, however, that if you didn't get twice the current the reason is that every driver has a so-called output impedance. This is the reason why you didn't blow up your circuit nor the LED, and I totally agree it works. But this is exactly my second case: you are using the internal resistance to mitigate the effect of non-linearity, making the driver dissipate the power internally. This will definitely shorten its life; I'd prefer using a dedicated resistor (of which I know the power rating) rather than dealing with an unknown value/unknown power rating parasitic resistance to keep the circuit working...</p>
<p>Yes You are right about allmost ewerything You said...but I've learned decades ago to NOT use the initial voltage &amp; current for them led's, because you are at the limits of the from manufact. given tolerances, (they wanna give the led a high lumens values to be the brighist one). I usually go for 10---15mA for a standard led rated @2,0volts, it will lit good enough. The outcome will be : I surely &quot;allways&quot; use resistors to limit the current. The point is: You really don't &quot;need&quot; the resistor if the voltage is correct, (e.g. a flashlight with a 1,5Volt battery &amp; a LED-light rated to 1,8Volt) hmm........put a resitor in series there and what?</p>
<p>The problem is that you can't &quot;go for 10--15mA&quot;, because you can't set a current without a) a current supply or b) a resistor to try to set the current. The 1.5V battery circuit you propose is perfectly legit, but falls in my &quot;cheap chinese-like circuits&quot; class: good if you don't care about performances (your LED will not light completely, maybe your battery will not last its maximum life). And of course in this kind of circuits you will always have to use a lower voltage for the LED (1.5V for 2V leds, 2.5V for 3V leds), otherwise you will not be able to guarantee that it will not blow.<br>In the end, I'm not stating that you can't power an LED without protections. I just want to point out that this, in my opinion, is a bad practice and should be avoided unless you are trying to use as few components as possible, as long as you are aware that in this case you will get very poor performances (for instance low light, shorter life or higher consumption with respect to results)</p>
<p>You'r, both right, (&amp; not so.... ecxatly right), at all your commetnt's. Howewer.. if you go and buy a lightbulb to your car.. do they ask it to be with or without a serial resistor? humor ;) ;)</p>
<p>I'd like to do some precisions: first it's not voltage who would fried either the phone's output or LED, it's current. Second, I doubt it's the battery full batery voltage at the audio output even at full power. Try a smartphone's earphone at an audio system and feed an audio signal 3.2 V amplitude and you'll see what happen to the earphones. Third: taking as example a red LED (one of the less electrically exigent LED) for a good illumination level it need an voltage of about 1.8 Volts at 20 mA. My question will be if the smartphone's output its capable of supply such 20 mA?</p>
<p>1. IT IS the Voltage that distroys things. The Ampereage comes along wit the Voltage, that is: &quot;how long time can you produce a certain amount of voltage consumed by something that draws ???? Amps?&quot;</p><p>2. The &quot;audio-output&quot; is in 99,99% of cases &quot;after&quot; a capacitor, thus blocking the DC-voltage. The analogue signal you get out of your headphones/speakers is actually a &quot;variable DC-signal&quot;, (there are exceptions, you use a phase shift of 180 degrees on them output poles, then you &quot;virtualy&quot; achive double the Volts applyed to your heaphones, rearly used,). </p><p>2.1 Try to apply &quot;raw&quot; power, not audio like in abowe. </p><p>2.1.1 Still. Apply your phone voltage, say ~3,7Volts ower a standard LED say with a initial voltage of 1,8V &amp; watch which goes out first? Both go warm, but....?</p><p>3. Your battery on your phone may be something like ~1000 ----2000mA, that is: 1/10 to 2/10 of an Ampere. The power, (W=Watt), you use is the Volt*Ampere, (that is: Volt multiplied with Amperes equals to Watt)</p><p>4. It's them Amps,(your battery), it can delievery at a given voltage, (say 3,7Volt). </p><p>How hard can this be? I mean the elementary with electricity?</p>
<p>Point 1 is relative: Example: a CRT anode voltage can be as high as 30 kV (kilovolt) but only few microamperes, if you short this voltage to earth nothing happen at the high voltage side of the flyback, instead you blown fuses at the low voltage side. By the way, fuses blown by amperage not voltage (indeed the voltge drop trhu a fuse before blow can be less than .1 volt). </p><p>Point 2 its far less than 99.9 % off cases an audio output is after an capacitor.</p><p>Point 2.1 You oversigth that when you apply an audio voltage to the earphones, the amplifier see an impedance, which can be tenfold the DC resistance of the earphones, obviusly, if you apply a DC voltage to an earphone you blown it</p><p>Point 3 is wrong, 2000 mA is 2 Amp.</p><p>In solid state design you must take into account current first, voltage second.</p>
<p>Right.</p><p>Point 1, The resistance ower a certain &quot;surface&quot; like human skin is of a certain ohm's, (wet/dry/distance), and there will be a &quot;flash ower&quot; when the potential of Voltage ower-rides the resistance. e.g. the CRT voltage (about 30kV, got that a couple of times), that still remains only to a potential differense of the poles...</p><p>Point 2 Check again them outputs of Audio. They &quot;really&quot; are decoupled through a capacitively source. Otherwise =&gt; Your loudspeaker would drain DC-current with no noice present, and that would &quot;fry&quot;/ burn your speaker.</p><p>Point 3. Just like in my statement 2000mA=2A, (1mA is a 1/1000 of an Amp)</p><p>Point 3.1 NO. The Voltage comes first. The risc for &quot;flash-ower&quot; comes with the high voltage, not the current. THE CURRENT IS ACTUALLY JUST THE SUM OF VOLTAGE RUNNING THROUGH A CERTAIN PLACE in a certain time, (and them multiplied with each other gives you the amount of work W, energy J done Geeeeee.... why is this so hard to tell</p>
<p>Point 1 is relative: Example: a CRT anode voltage can be as high as 30 kV (kilovolt) but only few microamperes, if you short this voltage to earth nothing happen at the high voltage side of the flyback, instead you blown fuses at the low voltage side. By the way, fuses blown by amperage not voltage (indeed the voltge drop trhu a fuse before blow can be less than .1 volt). </p><p>Point 2 its far less than 99.9 % off cases an audio output is after an capacitor.</p><p>Point 2.1 You oversigth that when you apply an audio voltage to the earphones, the amplifier see an impedance, which can be tenfold the DC resistance of the earphones, obviusly, if you apply a DC voltage to an earphone you blown it</p><p>Point 3 is wrong, 2000 mA is 2 Amp.</p><p>In solid state design you must take into account current first, voltage second.</p>
Its low voltage output, if you add resistor, the led is not turn on. but this gadget is safe for my audio output. Btw Thanks for look my works. Happy making :)
<p>It doesn't work. Can't even strip the plug !</p>
<p>mantap gan...</p>
Terima kasih!
<p>Maybe a stupid question. Do you plug out when you answer the call? I think speaker and microphone will be off when it is plugged in.</p>
No. you can loudspeaker your call
<p>I suspect this will only work if you have some kind of notification sound enabled with enough energy to light the LED. For instance the Pop on my droid probably won't work.</p>
<p>So, dumb question: the power for the LED comes from the audio signal traveling through the audio port, right? </p>
<p>This is really neat. I like it. It would be cool to use RGB LEDs and watch how the LEDs mix colors with the notification. I think I'll try it!</p>
<p>Cool enough.. to be a 'ible Yeah. but...... where do you put this gadget when you wanna plug in your headphones? :) By the way.. I'm irritated of there to be these plugs, (jack's), of several dimensions like 1,4mm 1,8mm 2,1mm 2,4mm 2,8mm........3,1mm ...6,0mm. hmm......</p>
<p>I wish to correct the author:</p><p>&quot;Old audio jacks (Use 3.5 mm jack)&quot;</p><p>That is the wrong item or description. The correct item is a plug since the jack is already in the smartphone. I recommend correcting this error soon because I may be the first of many comments about this error.</p>
<p>Wow! this so clever! NIce job bro I'm gonna try this!</p>
thank you. yes, you have too :)
Excellent. My iphone 6 s plus has this feature under accessibilty. I love it and it is very useful, especially when I am in a noisy setting and wouldn't hear the notification. I can easily see this in my pockets and purse's outer pocket.
Thanks. i hope you enjoy this. Happy making
<p>This is so cool yet so simple! I bet it can even activate other types of things haha</p>
thanks for look my works. happy making!

About This Instructable




Bio: I love writing, I make video in MY UPLOADED VIDEOS channel in Youtube for my instruction, and i love making something. www.instagram.com/balsuryana
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