LED Blows Up - Simple Circuit FAIL! - Page 1
Forum section: electronics / beginners

author: EccentricRuss
author_rank: Newbie
timestamp: April 30, 2011, 04:39:50 pm
content: n00b to the forum, let's get that detail out of the way. My name is EccentricRuss and I am an electroholic...So the wife were baby stuff shopping (one on the way) and we bought a consumer electronics product that came with a wall wart, but charged $30.00 for the car lighter adapter accessory. No way I was going to pay $30.00 for a simple circuit like that. So, I decided to build it and it was all fairly straight forward until I decided I wanted a pretty blue LED to show that there was power at the lighter plug.The original design was straight off the datasheet. I chose a Rohm LM7809, placed a 0.33uF cap before the input pin and a 0.1 uF cap on the out pin which provided me with the exact voltage and current needed, 9v and 1A. Under breadboard design review and testing, I received consistent and anticipated results. I found a sacrificial cigarette lighter plug attachment at Ollie's (a discount clearance item store), gutted it and modded my board to fit in the enclosure. For under $5.00 in parts I had my adapter...and then I had to screw with it.I saw on one of Dave's blogs (live blog probably) that you could soak up some voltage with an LED. Since my enclosure was limited on room I wanted to avoid a heat sink. Under no load, the LM7809 is cool as the other side of the pillow, but under load it warmed up. So I wanted to suck up some of the front end voltage and provide just enough to allow the LM7809 to function and get my 9v. In hindsight, I should have considered a voltage divider, but instead chose to place the LED in series and see what happened. I could hear Dave Jones's voice in my head, "I hope your next...project...fails!" But hell with it, what was the worst that could happen.It worked, at least for a while. In reality the cigarette lighters put out over 13V and at that voltage the blue LED looked "purty" and I have 9V output. The next part is fuzzy, and now that I have laid out the background I need help figuring out what may have gone wrong in the order of likelihood. See there was a cat...my cat and she has an incontinence issue. Or she just hates me. So I heard her scratching and I got up to see what she was doing. When I came back, the smoke had been let out. So the test leads were off the table and in my memory I remember the sound of two alligator leads touching. The multimeter read "0.00" and the LED was no longer illuminated. My heart sank and all I could think about was that I fried the whole thing, but especially my Rohm LM7809 which was a Digi-Key purchase (and with their shipping rates you have to order a butt-ton of stuff $$$ before you come out ahead). After desolder, the damaged component ended up being the LED; I bypassed it in circuit and the board came back alive and 9V, 1A output.I hypothesize that short circuit is not only the obvious choice but also the only reason with no other underlying cause. However after looking for a datasheet in vain, I found that this LED burns up often under normal usage. Let me dissect that last sentence, I mean schmucks like me put in a circuit and overdrive it until it fails and then we blame the store where we got it and chalk it up to poor product design. I bought the LED at Radio Shack. Please spare me the lecture, its the only show in town and they have improved upon their selection of brands in stocking their parts drawers. Don't talk to the employees, they know cell phones and batteries only. The stock number is 276-0316. 5mm Blue LED. 3.7V 20mA 2600mcd. So after the blow up, I read the back of the package which was all I had since Radio Shack won't put up datasheets!!! Here are the details that put my hypothesis in question: "Foward  (supply) voltage 3.7 typ. 4.5 max", "Foward  (supply) current 20mA typ. 30mA max." I had "ass-u-me"d that since they misspelled forward, that they intended forward voltage drop, but I may just be stupid.So looking at it, was the short circuit alone the reason for failure? did I exceed the limits of the LED by subjecting it to higher voltage? should i place a fuse or another type of diode in the circuit to limit damage to components? its a simple circuit, but the voltage regulator and blue LED are the expensive elements. In my defense, the LED was powered at 13.8V for 20 minutes before I shorted it, so I still have a modicum of belief that without the injustice of a short circuit, it would have survived.I have attached a DaveCad of the circuit. The DaveCad has been hacked off its native OS of PostIt and ported to BackOfEnvelope. At least I hope the file stays attached. first post on this forum, you know...cheers,Russ

author: jahonen
author_rank: Super Contributor
timestamp: April 30, 2011, 04:50:13 pm
content: The problem is that the led would have to last 1 A current when the output is fully loaded. Input and output current of the linear regulator are just about the same, the input current is a bit higher due to regulator internal quiescent current. So 20 mA spec will be exceeded by quite a bit when you load the output, so it is not a wonder if the led burns out.Furthermore, the voltage drop might be too much if you want the regulator work with 12 V input. So just put the LED in parallel to the input, with an appropriate series resistor and it will work just fineRegards,Janne

author: Mechatrommer
author_rank: Super Contributor
timestamp: April 30, 2011, 05:13:52 pm
content: its not how much voltage across it, but how much current.

author: EccentricRuss
author_rank: Newbie
timestamp: April 30, 2011, 05:21:11 pm
content: Ok, so I have learned something about forward voltage and forward current. I really wish I had been able to locate a datasheet. The voltage regulator drop out is moot; though originally intended for 12V, both cars this adapter were used in put out between 13.7 and 13.8 volts.Jahonen: so following your explanation, if the LED will die after allowing the flow of 1A through a 20mA pipe, is the difference in voltage not a limiting factor as well? i understand that voltage and current are separate realms, but if the max voltage is expressed as 4.5v, does not the potential difference ~9V excess voltage across the diode negatively impact or break it down?I will put the LED in parallel though. When I bought it, it was more wishful thinking (absorb extra voltage and save on the heat sink space) than logical thought.Thank you Jahonen!Cheers,Russ

author: EccentricRuss
author_rank: Newbie
timestamp: April 30, 2011, 05:25:43 pm
content: Mechatrommer, I did not think the voltage was an issue either. but why then does the manufacturer have a max voltage? I think i have read into Jahonen's response and I should go back to the bible (art of electronics) and read up on diodes and focuse on LEDs. Although at this point, I may need to revisit Ch 1 to get my head straight on V vs A.thanks a bunch though.cheers!

author: sacherjj
author_rank: Frequent Contributor
timestamp: April 30, 2011, 05:27:29 pm
content: Yep, everyone else said it. The only way to get a current different from the load into the regulator is with a parallel connection. To get what you were thinking of by using the voltage drop of the regulator to light it, you would hook it in parallel with the regulator in and out. However, I have no idea how this would affect the regulation.Hooking in parallel with the input, you just need a resistor to consume the waste voltage at 20 mA. 14V - VfLED = 10.3V in this case. V = IR, so R = V/I or 10.3V/0.02A = 515 Ohms. Common value at 510 Ohms should be good.Now, make sure the power is good. P = IV = 10.3 V * 0.02A = 0.206W 1/4W will be fine, although close to max rating.I over voltaged a 3mm LED and it had an air bubble around the chip. When it got really hot, really fast, it blew apart and dented my ceiling. Scared the crap out of me.Edit: Just saw your response. Look at it this way:If you remove the LED from your schematic (replace it with a wire) and hook up a load to the 9V output that pulls 1 A. How much current needs to flow through the wire that replaced the LED? A little over 1 A, due to losses in the regulator. If you put anything in there, either it has to handle 1A or limit the output. Nothing else is possible.If the electrons flowing were people (coulombs of charge) walking down a hallway. They are walking at a certain speed (amperage), due to circumstances you can't control. Now you get out there in your crutches and can only walk 1/50th of the speed of them (1A vs 20mA). What happens? You get blown up! Slammed to the ground.Now if you have a side hallway that is narrower (current limit resistor) that runs in parallel with the high current hallway. The both start and end at the same place, but you can travel slower in the small hall without getting blown up. That is what hooking the LED in parallel does.The voltage is what is required to power the LED if nothing else affects it. What your circuit was doing is providing the voltage, while also forcing current through it. Both are important. What probably happened is the voltage across the LED was WAY over 3.7V. Because current needed to flow, so the only option was to power through the increasing load of the failing LED, until it blew.Take a look at Forward Voltage vs Current on page three of this data sheet (probably fairly similar to the LED you are using):http://www.vishay.com/docs/81260/vlww9900.pdfThere is a Max value of Forward Voltage. After this, so much current flows through the junction that it melts. Boom. You get what you got. The is nothing in an LED that will limit the voltage to a safe value. This is why you need current limiting resistors. There is a linear relationship between resistance and current for a given voltage. You can see in the Forward Voltage vs Current that this doesn't exist for an LED. After it turns on, the current increases MUCH faster than voltage, so it can't limit itself and will explode with too much voltage.

author: EccentricRuss
author_rank: Newbie
timestamp: April 30, 2011, 05:54:14 pm
content: Sacherjj: I have the current aspect down pat. My question was voltage and you answered it in your last post. If you refer back to the DaveCad, you will see that the LED was the first component on the V+ side. The voltage regulator was tied in to it in series, intentionally. It was poor implementation (LED maxed out at 30mA) and probably design (unless I found a ~ 1A LED).Just so i understand the concept, allow me to restate it with what was observed. So the LED had all 13+ volts going across it well in excess of 3.7V as you state. once there was a load (and shorting the test leads for 3 minutes is a remarkably efficient one), the excess voltage (~10V) carried with it an increased threshold for current?even if i don't understand the voltage angle right now, i get the gist of the problem. the easy fix is the parallel option.thank you all so much!russ

author: tecman
author_rank: Frequent Contributor
timestamp: April 30, 2011, 05:57:22 pm
content: Another big issue not mentioned is if you want 1 amp out, at 13.8 volts in that's nearly 5 watts of heat dissipation in the 7809. Tou need a pretty good heatsink on the regulator to keep it alive. A heatsink with 15 degC/watt is about the limit, including the Tcase-Tab thermals. So make sure you add a nice heatsink to the tab of the 7809.paul

author: sacherjj
author_rank: Frequent Contributor
timestamp: April 30, 2011, 08:00:55 pm
content: Quote from: EccentricRuss on April 30, 2011, 05:54:14 pmUnless there is a reason to need an LED with that much power, there is no reason to put the 1 A LED in series.First, do you need an indicator that consumes 4-5 W or power? Sounds like a waste.Second, voltage drop of an LED that power is 3-4V. Basic voltage regulators require a certain voltage drop to function and you are getting close to or past it. This doesn't just drop the output voltage, it can also add crazy noise.Third, what happens if someone decides to hook 1.5 A on it? Poof!The reason a parallel circuit is used is the voltage is the factor that is the most constant. The only advantage of doing it in series if your load was know and constant is that you don't have the loss in the series resistor.Quote from: EccentricRuss on April 30, 2011, 05:54:14 pmIt is hard to say how much the LED had across it with the 9V side without a load. It could have been just fine or it could have already been over. No idea. However, once you put a load in place that draws more across the LED than 20-30 mA, the voltage drop across the LED goes up. It probably failed before it had 13+V across it, but you were driving it so far out of spec that it would be hard to know for sure.Quote from: tecman on April 30, 2011, 05:57:22 pmYes, and here is the cost of some of the Auto voltage converters. Switching based DC to DC voltage converters. Much more efficient, but more costly.

author: Zero999
author_rank: Super Contributor
timestamp: April 30, 2011, 08:40:12 pm
content: All right, I've not read the whole long post, just looked at the schematic and noted that it's being established a typical LED is only rated to 20mA.Why do you need to reduce the voltage by 3.5V to power an LM7809 from a 13.8V supply?The LM7809 is designed to work from up to 25V, connect it directly to 13.8V

author: sigxcpu
author_rank: Regular Contributor
timestamp: April 30, 2011, 08:40:56 pm
content: sacherjj: amperage is intensity or count per section (if you refer to people). Your "speed" is voltage.if you put 1MOhm resistor in series, the speed would be the same, but the electron count, the electron quantity per time unit is much, much, less.

author: Unknown
author_rank: Unknown
timestamp: April 30, 2011, 09:51:30 pm
content: And just to piss everyone off, connect your LEDin series!(Note the 17V source). I am kidding, connect the LED as everyone already suggested.Welcome to the forum.Alex

author: Psi
author_rank: Super Contributor
timestamp: May 01, 2011, 12:38:02 am
content: If you really want to burn off some voltage before the regulator just use normal silicon diodes instead of the led.As long as the diode is rated over 1A (1.5A to be safe) it will drop the voltage by ~0.7V. (check the forward voltage vs current graph for exact value)And you can put multiple diodes in series to drop as much voltage as you need. Be aware though, that your regulator input voltage must be no lower than 11V because the regulator requires 2V to function.You may also want to put a 30V TVS diode across the input first, to absorb any high voltage spikes. The battery voltage in cars can have pretty large spikes at times.

author: Tony R
author_rank: Regular Contributor
timestamp: May 01, 2011, 03:07:44 am
content: yea you drew to much current though the LED.1) I am a little confused at why you choose a blue high intensity LED, for a power indicator.2) I also am not sure why you felt like you needed to drop some voltage, i mean to soak up some voltage but a resistor will do that too (however will vary based on the load.) Your voltage regulars are rated for up to 25V if I read the data sheet correctly. the drop in voltage you have wont drop it that much, If you are smart and use a metal case you can use the case as your heat sink. and depending on the current draw you may not need any at all3) The LED should be in parallel to the regulator with a 515 (510) Ohm resistor in series with the LED.4) <REMOVED>if you make these changes you should be good...

author: sacherjj
author_rank: Frequent Contributor
timestamp: May 01, 2011, 03:10:47 am
content: Quote from: sigxcpu on April 30, 2011, 08:40:56 pmAmperage is Charge (Coulombs) per Time. That is speed. If you put a 1 MOhm resistor in series, the speed drops. This is like narrowing the hallway. Less charge can flow per Time. Voltage is the potential or the number of people wanting to get down the hall.

author: sacherjj
author_rank: Frequent Contributor
timestamp: May 01, 2011, 03:13:44 am
content: Quote from: Tony R on May 01, 2011, 03:07:44 amI was with you till here. Are you talking for the LED or the 9V supply? Voltage dividers only work as a low current biased voltage. As soon as to start pulling current, the voltage tanks because the upper resistor voltage drop soars.

author: Zero999
author_rank: Super Contributor
timestamp: May 01, 2011, 10:35:03 am
content: Quote from: sacherjj on May 01, 2011, 03:10:47 amBoth sound like pretty poor analogies to me, my favourite one is water pressure being volts and flow current but that breaks down once you get past simple resistance, capacitance and inductance.

author: Tony R
author_rank: Regular Contributor
timestamp: May 01, 2011, 02:19:44 pm
content: Quote from: sacherjj on May 01, 2011, 03:13:44 amallow me to explain, if the issue was that the voltage regulator was getting to hot because it was dissipating so much power, you could create a voltage divider to cut the voltage down a little, if i remember correctly that was the point of the LED in the first pace to drop the voltage just a little bit. Yes you will not get very much current doing it this way, and it is not how i would do it period. but having designed a few projects in the past for classes and personal use, I can tell you at some point your willing to try just about anything to get it to work.For example he had basically a 14V (13.8 or something) input, if his regulator would get to hot taking the 14 down to 9 so you opt to take it from 11 to 9 smaller voltage drop across the regulator means you dissipate less power. Doing this is not without its faults as u said current dies off.As appose to just placing a resistor in series, your voltage drop would not change when your load does.I also posted that I would opt for this, that is a typo there should be a not in there somewhere...

author: sigxcpu
author_rank: Regular Contributor
timestamp: May 01, 2011, 03:49:15 pm
content: LE: ignore me, I'm an idiot

author: tecman
author_rank: Frequent Contributor
timestamp: May 01, 2011, 04:39:45 pm
content: Quote from: Tony R on May 01, 2011, 02:19:44 pmBottom line is if the load is 1 amp, with a linear regulator you will dissipate heat. If you drop voltage somewhere else, you are merely shifting the heat to somewhere else. The other risk is if you drop 3 volts before the LM7809, then you will fall out of regulation if the battery (input) voltage slumps for any reason (engine off for example). He needs a heatsink.paul

author: Zero999
author_rank: Super Contributor
timestamp: May 01, 2011, 04:49:15 pm
content: Quote from: Tony R on May 01, 2011, 02:19:44 pmAs mentioned above, a voltage divider won't work.Is the regulator shutting down?If it gets hot but remains functional then it's not too hot.If it's shutting down the power dissipation can be reduced by connecting a resistor in series with it to drop the input voltage but you have to ensure the input voltage to the regulator is high enough to maintain regulation at the highest output current and lowest input voltage.

author: allanw
author_rank: Frequent Contributor
timestamp: May 02, 2011, 03:55:54 am
content: If you're in the USA, use USPS first-class mail to ship from Digi-key. I get all my small parts with only $2.41 shipping and it takes two days.

author: EccentricRuss
author_rank: Newbie
timestamp: May 02, 2011, 07:27:01 pm
content: You all have been very helpful and its fantastic to see some veteran practicioners argue over what current really is and what metaphor should be used to explain the concept.Here are some answers or responses to some of the questions about my intentions/goals/failures.Hero999: Yup, the answer to why I am trying to reduce the voltage my regulator sees at input is in my long windup. I am trying to keep the size of the heat sink small. I am working with a hacked car lighter plug enclosure as this will be used when we travel in the car. I wanted to thread the needle and sap just enough voltage to reduce heat without starving the 7809. 12V was an approximate goal and the supposed output from a car lighter plug; in reality both of our cars put out ~13.7 volts. The LED did absorb the excess voltage and I had my 9 volts…until a load was introduced and my LED became a 20mA fuse.Alex: Thanks for the AlexCad schematic. It’s an alternative.Psi: I gave thought to the voltage spikes and although the 7809 is able to be used at a max 26v, it has an absolute maximum of 35v. I measured the voltage of my car cranking over and it was in the realm of 17-18V. I read some anecdotal accounts that “load dump” can generate spikes of 40-60V. This was so far out of my specs that under those conditions I preferred my $5.00 (and counting) charger to fail and arrest the spike before it touched the load. That was my thought process at the time, but then the LED blew up and my heart sank and I realized that I really liked my little circuit and maybe I should look into protecting it. And a TVS diode should be easy to introduce. And if my car has a true “load dump,” more than my car charger is going to bite the dust.Tony R: (1) I like blue LEDs and wanted one, that easy. (2) Sorry, metal case is not an option. I want it to fit inside a car lighter plug.Tecman: “If you drop voltage somewhere else, you are merely shifting the heat to somewhere else.” That was the point. I want to reduce the amount of thermal dissipation at the regulator. I have a heat sink. I want to keep it small to fit the car lighter plug case.Allanw: Good to know on Digi-Key and USPS. I used the iPhone app to order the parts and FedEx was what was sent. And I did not get it in 2 days!!Cheers,Russ

author: Zero999
author_rank: Super Contributor
timestamp: May 02, 2011, 07:54:20 pm
content: You're going to have a problem with heat, as long as you use an inefficient linear regulator.The heat sink can be eliminated completely by using a switching regulator.A simple switching regulator can be built using a couple of transistors, an example is linked below. As you want 9V at 1A you'll need to use higher current transistors and a 9V zener and change some of the other component values.http://www.romanblack.com/smps/smps.htm