If the source voltage is the same as LED voltage, is a resistor necessary? - Page 1
Forum section: electronics / projects

author: geratheg
author_rank: Regular Contributor
timestamp: July 27, 2014, 05:39:10 am
content: If the source voltage is the same as the voltage drop across an single LED, is a resistor in series necessary? If so, why?

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 05:47:21 am
content: If you connect anything directly to a voltage source, then the voltage drop will always be equal to the supply voltage. It has to be so if you assume there is no voltage drop in the connecting wires.But this is not the question, however. The question is: what current will flow through the LED? Will it be too much? Will it damage the LED?

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 05:53:17 am
content: To elaborate on that a little:The current through an LED is very sensitive to the supply voltage. Increase the voltage slightly and the current can increase a lot. The current can also increase if the LED is warms up. And one sample of an LED may be slightly different from another.So you can never reliably know exactly what voltage is right to deliver, say, 10 mA through a given LED.If you put a resistor in series with the LED it provides some automatic current regulation and stabilizes the circuit. The resistor automatically adjusts the voltage seen by the LED to keep the current the same.

author: the1cyrus
author_rank: Contributor
timestamp: July 27, 2014, 06:01:53 am
content: If you know the led source voltage is exactly the same as the forward voltage of the led and you live in a perfect world then you probably wouldn't need a resistor. But if there is even 0.1v difference then you have problems and for that reason you always should include a resistor. 0.1v/ 20mA = 5 ohms.You have to have a higher source voltage and put a resistor or have a lower source voltage than the led needs and hope that led never changes as it ages or hope the source voltage never goes higher than the forward voltage.Constant current source is a better solution.

author: PedroDaGr8
author_rank: Super Contributor
timestamp: July 27, 2014, 06:04:28 am
content: IanB is correct kinda. The biggest thing to remember is for LEDs you should care about the current through it not the voltage. It's a weird concept for a beginner but LEDs are current controlled devices. small changes in voltage can cause dramatic shifts in current through the LED. For example, 3.3V might be 10mA but 3.4 might be 20mA and 3.5 might 50mA.Resistors aren't always necessary but are almost always a good idea because the rated voltage is usually the point of roughly max current. So without a current regulator you might want to run the LEDs at a lower voltage to give you some headroom for instability.

author: KedasProbe
author_rank: Frequent Contributor
timestamp: July 27, 2014, 08:57:41 am
content: http://en.wikipedia.org/wiki/Load_line_%28electronics%29See graph on the right. (the intersection is the current and voltage on the diode/led)A change of supply voltage moves the straight line parallel.(If R=0 then that straight line is going to be vertical, in that case it will be easy to get too much current hence destroying the led.)

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 11:18:06 am
content: QuoteNo.QuoteBecause the source voltage, by your own question, would vary with the LED's forward voltage. Thus, the same current flows through the LED.Such a source is effectively a current source (aka with infinite output impedance).

author: German_EE
author_rank: Super Contributor
timestamp: July 27, 2014, 01:24:57 pm
content: I once built a frequency counter and, eager to avoid soldering 70 resistors, I fed the display with a 1,8V supply and connected the segments directly to the decoders. It worked, but, as the regulator warmed up the displays got brighter and brighter until the decoders began to fail.I then had to desolder ten decoder chips and fit current limiting resistorsSorry, but sometimes we cannot cheat.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 01:41:02 pm
content: QuoteI have two frequency counters in front of me that have no current limiting resistors on its display, both running at 3.3v.At 1.8v, your leds are quite dim, if lit at all.

author: SeanB
author_rank: Super Contributor
timestamp: July 27, 2014, 02:01:12 pm
content: CMOS drivers at 3V will have an output resistance when on of around 100-300R, plenty enough to limit current in the displays. The actual value will vary from batch to batch but if all are from the same batch or you are multiplexing the displays with another CMOS sink then it will not need resistors. You do however get LED display drivers that have built in current sources that are quite capable of driving LED displays directly.I did a similar mod with changing from filament displays to LED, using series resistors of 180R. Unfortunately the LED's were too bright so a simple fix ( other than removing and resoldering around 500 resistors from a sandwich of boards) was to add 2 diodes in series with the common cathode lead of the whole display. The 1V2 drop was just right to get the brightness down. 1 was too bright and 3 was too dim. I used 1N5404 diodes as the total current there was around 2A at the test display of all 8's.

author: rob77
author_rank: Super Contributor
timestamp: July 27, 2014, 02:06:38 pm
content: Quote from: dannyf on July 27, 2014, 01:41:02 pmare you trying to tell us that 7segments with forward voltage 1.8V doesn't exist ? just the ones you have for 3.3V without the need of current limiting resistors ?

author: free_electron
author_rank: Super Contributor
timestamp: July 27, 2014, 02:15:34 pm
content: If the supply voltage is perfectly equal to the led no voltage then no current will flow.To have current you need a potential difference.So there, you have the answer.

author: hamdi.tn
author_rank: Frequent Contributor
timestamp: July 27, 2014, 02:23:32 pm
content: yes you can , i saw a video once where he make some experience with 100W led directly with the battery here the link

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 02:55:09 pm
content: QuoteNot sure what laws of physics you are relying on there.when the supply voltage is perfectly equal to the led's forward voltage, the current going through the led is exactly the forward current specified in the V-I curve that generates the forward voltage.For "no current" to flow through the led, the supply voltage (thus the led's forward voltage) has to be zero.No other way around it.

author: Tinkerer
author_rank: Frequent Contributor
timestamp: July 27, 2014, 03:10:07 pm
content: Quote from: PedroDaGr8 on July 27, 2014, 06:04:28 amI wouldnt call it weird. You can adjust the voltage within a range and the LED gets brighter or dimmer. The thing is that its typically decribed that LEDs will take as much current as they possibly can(regardless what the voltage is) and without something to regulate that, they burn themselves up.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 03:13:22 pm
content: It is really an issue of thermal stability: Vfwd of a diode / LED goes down as it heats up. So if you apply a constant voltage to the led, its current goes up as the led heats up, which causes the led to dissipate more power -> heating up more. In no time the LED is destroyed.The current limit resistor is really there to provide a negative (current) feedback.

author: hamdi.tn
author_rank: Frequent Contributor
timestamp: July 27, 2014, 03:30:14 pm
content: Quote from: free_electron on July 27, 2014, 02:15:34 pmwell this will not apply to a circuit with a voltage source and a led since the forward voltage of the led does not create a potential it self , it's the voltage that can be measured when applying a current with a current source.In this case with a voltage source almost near the forward voltage i think you only count on voltage drop that will be equal to led forward voltage and your voltage source will be practically a current source as dannyf explained before.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 06:38:39 pm
content: Quote from: hamdi.tn on July 27, 2014, 03:30:14 pmFree_electron was making a funny.If you consider an idealized system where there is no potential difference between two points there will be no potential gradient to produce a current between those points. So if the source voltage and the destination voltage are equal, then physics says that no current will flow. In the real world if you connect a battery to an LED and make the LED light up there will be slight voltage difference between the ends of the hookup wires proportional to the current flowing in the wires.

author: free_electron
author_rank: Super Contributor
timestamp: July 27, 2014, 07:16:04 pm
content: Quote from: dannyf on July 27, 2014, 02:55:09 pmit's a joke ... a joke ! got it ?an LED is a current driven device. it is the current that creates the forward drop. just like a normal diode.ever so often these questions pop up about 'can i use led's without series resistors... yes, if you drive them from a current source, not from a voltage source

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 07:34:04 pm
content: Quotenot so sure.QuoteTrue, if the two (destination + source) are voltage sources. In this case, the source is a voltage source and the destination is a load (!= voltage source).However small of a voltage is applied to the diode, even if it is less than the Vfwd spec for the diode / LED, there is always going to be a current flowing through the diode / LED.That is indisputable.QuoteIn this case, the voltage from the source and the voltage on the diode will be (nearly) identical, with the qualifier there being the wire resistance - which is irrelevant to this discussion.

author: retrolefty
author_rank: Super Contributor
timestamp: July 27, 2014, 07:38:54 pm
content: Quote from: free_electron on July 27, 2014, 07:16:04 pmYour luck here, at the Arduino forum it's a all too common questions, or maybe they are trolls that knows it will kick off a tired long repeat that we see way too often.

author: geratheg
author_rank: Regular Contributor
timestamp: July 27, 2014, 09:07:44 pm
content: Thanks for all the replies. I don't mean to annoy you all with a simple question as this. I'm new to electronics.This is how I understand it:An LED rated at 3.6V and 20mA will draw 20mA if supplied with 3.6V. Correct me if I am wrong.However, as it heats the current draw may vary. And in the real world, exact voltage is not the case.So adding a resistor basically acts as a "safety" device in that it limits the current draw of an LED to safe levels, so it does not burn out. In addition, it gives the LED flexibility in case of a varying voltage, and thus protects it.Another thing I realized is that an LED isn't a resistor like a regular light bulb (which I assume is basically a resistor), correct me if I am wrong about light bulbs being resistors.Instead an LED is like a diode. Diodes have an exponential looking increase in current due to a slight increase in voltage once it passes a certain point.If I said anything wrong or slightly off in this post, please correct me. If all is right I think my question has been answered.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 09:19:43 pm
content: I think you have it basically right.Quote from: geratheg on July 27, 2014, 09:07:44 pmYou could make this even better if you turned it around:"An LED rated at 3.6 V and 20 mA will drop 3.6 V if supplied with a current of 20 mA."An LED has a certain current (or range of currents) that it is designed for, and the voltage then follows the current.

author: mikerj
author_rank: Super Contributor
timestamp: July 27, 2014, 09:27:57 pm
content: Quote from: geratheg on July 27, 2014, 09:07:44 pmIt might do, but probably not. The Vf (forward voltage) of an LED is not rigidly defined, the datasheet will only give a nominal value. Even if you picked up a handful of otherwise identical LEDs you can expect to see some variation in the Vf for a given current. This is why you shouldn't run LEDs in parallel unless they each have their own current limiting resistor.The second issue is that of the diodesdynamic resistance. Because a diode is a semiconductor junction, it doesn't behave like a resistor. Once the LED is forward biased, increasing the applied voltage will cause an exponential increase in current because it's resistance actually reduces with increasing current.The real killer is that the LED's forward voltage reduces as the LED heats up, so if it's being driven from a constant voltage then the current through the LED will increase. The extra current will make the LED junction run even hotter, which lowers the Vf even more and causes even more current flow and so on... This thermal runaway only stops when the LED has burned out (or possibly your voltage source can't supply any more current).

author: free_electron
author_rank: Super Contributor
timestamp: July 27, 2014, 09:33:38 pm
content: Quote from: geratheg on July 27, 2014, 09:07:44 pmNO . that is where you got it wrong.We need to delve a bit into material physics here to explain this component . anything else will just run us in circles.An Led, or any diode for that matter, is a semiconductor. Semiconductor as in 'a kind of device', not as in "semiconducting material".The device is constructed from two semi-conducting materials: one positively doped, one negatively doped. ( no need for purists to go screaming 'what about schottky diode , that is a metal-semiconductor diode , but irellevant fo this explanation.)So , we have two materials : one is short a few electrons to be happy, one has a bit of extra electrons circling around in it.if i fuse these two materials together an effect called 'recombination' happens at the point of contact. the material that had excess electrons will donate them to the material that had a shortage. This process goes on until the supply of 'extra electrons' are depleted.So now you have a zone of material that has no more free electrons and a zone where still a shortage exists.Note the zone that had excess electrons is called N because electrons carry a negative charge. The zone that has a shortage is called P for positiveNow, if you apply 'pressure' in the form of voltage two things can happen :a) You cram in more electrons into the zone that originally had an excess : the applied electrons will now fill the remaining area that has a shortage and current will start to flow. The amount of pressure you need to get this to work is called the forward voltage. in case of a traditional silicon diode this is about 0.5 to 0.6 volts. in case of an led this is 1.4 to 3.1 volts depending on the materials used ( and the materials determine the vibration speed of emitted photons , thus the color of the emitted light)b) you pull electrons out of the excess zone : the area that had recombined is now being depleted completely. so nothing can pass and the entire structure has no more free electrons. so no current flows at all. If yo keep pulling hard enough eventually the structure will break down. We call that 'the breakdown voltage' . if you keep pulling eventually something will give : total destruction.So, if you apply a pressure in the right way the element will start to let electrons pass at a certain point.you have to take care not to apply too much pressure though. that pathway is only so wide. try to send a massive amount of electrons across and it will burn out.so we need to limit the flow ( the flow , or number of electrons per second is the current. Ampere ) What limits current ? a resistor !so if you don't make sure the current is limited to a safe flow you rip the diode or led apart. a power supply doesn;t limit current by itself. it either needs a control mechanism in the form of a current limit or you do the protection work by inserting a resistor.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 09:50:08 pm
content: QuoteGenerally true for diodes and signal LEDs (small current leds), and less true for power leds - if you look at their datasheet, you will see that they are far less exponential and more linear -> ie power leds are actually a lot closer to resistors than you think.The issue is with their temperature coefficients: Vfwd drops as the led heats up. The drop per degree of temperature rise varies based on the color, connection (serial vs. parallel), etc. But typically -2 to -10mv / C.For a 50c degree temperature rise, that means a Vfwd dropo of 500mv. Most of those LEDs have a resistance of only a few ohm (3ohm being fairly typical). That means a current increase of 170ma, assuming no resistor.That current increase, by itself, will increase power dissipation of the led which pushes down Vfwd further and pushes up Ifwd, thus more power dissipation.To put it another way, powering a power led with a voltage source is perfectly viable if you can manage the thermal situation - which often you cannot.

author: geratheg
author_rank: Regular Contributor
timestamp: July 27, 2014, 10:59:39 pm
content: Thanks for all the technical explanations.Makes more sense as to why an LED is a current device.If I had a batch of 100 LEDs of the same type, manufacturer, size, model, etc. rated at 20mA and 3.5 V I bet supplying 20mA to each one will produce a different forward voltage for each one?But if I were to supply a constant 3.5 voltage to each one I bet some would get too much current and blow, others will work with the wrong current and lose life, others may get too little current and be dim, while the rest will get the appropriate current and work fine. Is this correct?

author: georges80
author_rank: Frequent Contributor
timestamp: July 27, 2014, 11:16:37 pm
content: Vf varies LED to LED (within bins and worse yet across bins). Vf varies with temp (consider also the changes in junction temp due to changes in ambient temp). Vf varies with LED 'aging'.You need to either use a resistor which works fine for low power LEDs or a current regulated supply for high power LEDs. A resistor is ok for low power LEDs (20mA - 50mA etc), but not a good solution for high power LEDs.Constant voltage is a bad idea due to the Vf variations mentioned above.High power LEDs do have a non-linear Vf curve. Constant voltage is a really bad idea. Just look at a Cree XML datasheet. NOTE it states 25C junction temp and is non linear. Now factor in the junction temp will increase and cause the Vf to drop - compound that with a constant voltage supply and you have a bad situation, especially *IF* you are trying to get up near the 2 - 3A range where a small applied Vf variation changes the current by a large amount.cheers,george.

author: dannyf
author_rank: Super Contributor
timestamp: July 27, 2014, 11:23:46 pm
content: QuoteThat's usually not how people talk about LEDs. Just think of them as a resistor whose resistance is current (or voltage) dependent.QuoteYes and yes, aside from the "wrong current" part.

author: IanB
author_rank: Super Contributor
timestamp: July 27, 2014, 11:30:19 pm
content: Quote from: dannyf on July 27, 2014, 11:23:46 pmAbsolutely it is. The datasheets even describe the situation that way, giving a range of suitable drive currents and showing the expected light output and other parameters for each.

author: geratheg
author_rank: Regular Contributor
timestamp: July 27, 2014, 11:50:33 pm
content: Thanks guys!As for the "wrong current" part, I meant higher current, which will reduce LED life but not instantly blow it, though eventually it would.

author: dannyf
author_rank: Super Contributor
timestamp: July 28, 2014, 12:24:47 am
content: Take the XML part for example - see the picture attached.Assuming that you are driving it at 3v (1100ma per the datasheet, at 25c). The datasheet specifies that it has a tempco of -2.1mv/c. Say that our operating temperature is 75c (for a temperture increase of 50c). The entire V-I curve is shifted to the left by 2.1mv/c * 50c = 100mv, marked as 75c curve on the chart attached below.If the led is driven by the same 3v source, you will see that the current going through the led is now 1500ma, or an increase of 400ma (=1500ma - 1100ma, or an increase of 40%, at a base current of 1100ma.The dissipation has gone from 3.3w (=3v * 1100ma) to 4.5w (=3v * 1500ma).So while the "resistance" of the led isn't that bad - just shy of 3ohm at 3v@1100ma, the negative tempco can cause a deadly situation, if not killing the led outright, it will greatly reduce its life by running at much higher temperature.So unless you are sure that you have the thermal conditions under control, be careful driving it with a constant voltage source.BTW, you can easily work out the case when the chip is driven by a constant current source -> the power dissipation will decrease when the led hits up, making it inherently stable.

author: IanB
author_rank: Super Contributor
timestamp: July 28, 2014, 01:00:40 am
content: Quote from: retrolefty on July 28, 2014, 12:53:44 amOK, which part of "or range of currents" was not clear to you?Furthermore, can you show me a datasheet for an LED that will emit usable light when supplied with one million amps? (I don't consider an exploding flash to comprise usable light)

author: dannyf
author_rank: Super Contributor
timestamp: July 28, 2014, 01:00:52 am
content: QuoteWow!With "experts" like you, no wonder that we are not a country of makers,

author: retrolefty
author_rank: Super Contributor
timestamp: July 28, 2014, 01:04:14 am
content: Quote from: dannyf on July 28, 2014, 01:00:52 amIn that case I will delete the posting, didn't mean to offend you.

author: dannyf
author_rank: Super Contributor
timestamp: July 28, 2014, 01:12:15 am
content: Don't worry about it. I saved it for your and others' enjoyment,

author: sleemanj
author_rank: Super Contributor
timestamp: July 28, 2014, 01:44:39 am
content: Given that it is a practical impossibility to match the LED forward voltage with a supply voltage EXACTLY, and that you don't have superconducting wires, solder, or led legs... the whole thing is moot.So just dispense with the theoretical "exactly the same" nonsense that doesn't exist in the real world, we can make a much better example by just using numbers so imperceptibly higher than zero that it's still silly to think about, but far more easily visualised.LED Forward Voltage = 1vSupply Voltage = 1.00001vInherent Resistance (ESR of supply, resistance of wires, connections etc) = 0.000001 OhmsCurrent = (1.00001-1)/0.000001 = 10 AmpsYou can see that in practical terms, the inherent resistance is the only thing limiting your current if you place the LED directly across the supply with the "same" forward voltage, if that inherent resistance is low (eg a lab supply without current limiting), bang goes your LED, if it's high enough (eg a CR2032 battery or some such), no problem.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 28, 2014, 05:47:48 am
content: Quote from: sleemanj on July 28, 2014, 01:44:39 amthat is a seriously impractical practical example. the LED forward voltage is related to the current. It needs a graph it is not a number.Hate to say it but I agree with Dannyf.

author: sleemanj
author_rank: Super Contributor
timestamp: July 28, 2014, 06:20:09 am
content: Quote from: HackedFridgeMagnet on July 28, 2014, 05:47:48 amYes the forward voltage will change. Like I said, it's silly to think about.The original question "do you need a resistor if the supply and led forward voltage is equal" is impossible to achieve in the practical world and makes for the poster's confusion.By simply shifting the goal posts a minute fraction it makes something impossible into something merely ridiculously impractical, but looking at that, the answer to the question becomes quite obvious to the poster I hope, without getting into anything more than good old Ohm's law.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 28, 2014, 09:08:26 am
content: QuoteOk I can see where you are coming from now.Still it would be nice to have a single post answer to this question that can be understood by hobbyists.

author: rob77
author_rank: Super Contributor
timestamp: July 28, 2014, 09:32:04 am
content: the single and understandable answer for everyone is:use a constant current LED driver, or for small power use a current limiting resistor

author: Rufus
author_rank: Super Contributor
timestamp: July 28, 2014, 01:51:46 pm
content: Quote from: rob77 on July 28, 2014, 09:32:04 amThen the question becomes what compliance voltage does the driver need or how many LEDs can I chain with a given compliance voltage...

author: Richard Crowley
author_rank: Super Contributor
timestamp: July 28, 2014, 02:06:45 pm
content: There are several handy resistor calculators available online. You input the number of LEDS, the LED specs (current/voltage), the available supply voltage, etc. And the calculators give you the most efficient circuit (series, parallel, etc.) and the value of the current-limiting resistors. I find them quite handy.For example:http://led.linear1.org/led.wiz

author: UrQuan
author_rank: Newbie
timestamp: July 28, 2014, 04:24:28 pm
content: Quote from: mikerj on July 27, 2014, 09:27:57 pmOk, I think you just answered a question that's been plaguing me for two decades. When I was a teen, I had a paintball gun with a red-dot sight. This sight took a 3V button-cell battery which, back in the early '90s, was both expensive and hard to find. So I rewired it to take 2 AA batteries instead. Popped the diode as soon as I turned it on.I now work at an MEP engineering firm, and none of my EEs have been able to explain how that could've happened (though most of them can't remember a thing about DC).If I'm reading the explanation correctly, the LED in my sight probably wasn't rated on its spec sheet for 3V at all. It was probably rated for a current level that the dinky little button cell could provide at 3V, and that a pair of AAs sailed right past at 3V.Is my understanding reasonable?

author: Richard Crowley
author_rank: Super Contributor
timestamp: July 28, 2014, 04:37:14 pm
content: I suspect you are correct. Because of the nature of those small batteries, they have higher internal resistance which would have limited the current. There is a whole class of inexpensive and ultra-simple devices that operate LEDs directly from small coin cells, without a current-limiting resistor. Everything from "throwies" to coin-ring "flashlights" ("torches" in BrEnglish).

author: IanB
author_rank: Super Contributor
timestamp: July 28, 2014, 04:38:45 pm
content: Quote from: UrQuan on July 28, 2014, 04:24:28 pmIt's because of the internal resistance of the button cell. All batteries have an internal resistance to current flow inside them and small button cells have a much higher resistance than big AA batteries. So you could think of your 3 V button cell as a 3 V supply with a series resistor to limit the current (except the series resistor is inside the cell rather than outside it).As soon as you connect the AA cells their internal resistance is much less, it's like connecting 3 V to the LED without much of a current limiting resistance. So a large current flows and the LED goes poof.QuoteI don't think you have any real EE's at that firm

author: SeanB
author_rank: Super Contributor
timestamp: July 28, 2014, 05:55:27 pm
content: If you had used regular Zinc carbon cells ( not alkaline ones) the high internal resistance of the cells would have limited the current to something the LED would withstand for a while. The older and more used the cells the better.

author: free_electron
author_rank: Super Contributor
timestamp: July 28, 2014, 06:42:12 pm
content: Quote from: UrQuan on July 28, 2014, 04:24:28 pmFire them all, immediately. They don't understand the concept of battery internal resistance...those button cells cannot deliver more than 10 to 15 milliampere , some 40 or 50. pull more and the voltage collapses. so this becomes a self regulating system.throw in AA that can deliver amperes and you pop the diode in a millisecond.

author: UrQuan
author_rank: Newbie
timestamp: July 28, 2014, 07:24:42 pm
content: Quote from: IanB on July 28, 2014, 04:38:45 pmWell I'm not in a position to judge, but I will say I'm probably the first person to have asked any of them about DC circuits in at least 20 or 30 years. We do large commercial buildings and campuses. Some of them probably got licensed before LEDs were even in production :-)

author: German_EE
author_rank: Super Contributor
timestamp: July 30, 2014, 03:20:29 pm
content: I just thought of a way to prove the original problem. The original post talks about LEDs that have a nominal 2V to 3V voltage drop but if we increase the voltage there is an experiment we can try.Take a 5,1V zener diode and wire it directly across the output of a variable power supply. Supply 5,0V and you will probably find that the diode is still there, increase to 5,1V and it might survive if you are lucky and the PSU meter is accurate. Increase the output to 5,2V and that diode will die.Resistors in series are therefore required

author: free_electron
author_rank: Super Contributor
timestamp: July 30, 2014, 04:21:54 pm
content: Quote from: German_EE on July 30, 2014, 03:20:29 pmThou shallt not harm zener diode's in stupid experiments !

author: jlmoon
author_rank: Unknown
timestamp: July 30, 2014, 05:05:39 pm
content: QuoteWhich I have seen in some really cheap designs involving mpu multiplexed LED displays and not so well regulated power supplys. I love it when you can see the various segments intensity sag due to the number of segments being lit at one given time. Either using current limiting on the common anode/cath line of led or no current limiting at all.

author: Unknown
author_rank: Unknown
timestamp: July 30, 2014, 10:05:52 pm
content: Quote from: UrQuan on July 28, 2014, 07:24:42 pmJust curious, but are these EE's Electrical Engineers (or equivalent) or Electricians?

author: UrQuan
author_rank: Newbie
timestamp: July 31, 2014, 01:50:05 am
content: Quote from: steverino on July 30, 2014, 10:05:52 pmElectrical Engineers.

author: IanB
author_rank: Super Contributor
timestamp: July 31, 2014, 02:24:19 am
content: Quote from: UrQuan on July 31, 2014, 01:50:05 amIf they are licensed engineers (PE) then they should be able to explain about the internal resistance of a battery.Either they have forgotten everything they ever knew, or they were pulling your leg.If they seriously couldn't answer your question you should fire them, as was said by someone else above.

author: SeanB
author_rank: Super Contributor
timestamp: July 31, 2014, 04:50:22 am
content: They probably have forgotten a lot about stuff not concerned with AC power distribution. They likely will know without calculation what cable will do for a circuit of any arbitrary load at any arbitrary distance with any variety of mounting methods and locations, along with what protective devices and connection methods at each end and at any joint in the middle without even stopping to think. They might have to think about DC non distribution devices.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 31, 2014, 12:47:22 pm
content: Understanding of power supply source resistance is how you work out the fault current.If these guys are designing AC power distribution, even distribution within a building then they understand source resistance.End of story.

author: jlmoon
author_rank: Unknown
timestamp: July 31, 2014, 02:12:05 pm
content: Quote from: HackedFridgeMagnet on July 31, 2014, 12:47:22 pmwouldn't that be 'source impedance'.. albeit involving AC power distribution..

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: July 31, 2014, 03:26:47 pm
content: Yes that's the term.