PLEASE HELP! What am I doing wrong with my parallel LED circuit?

[Starlit Rose]

Hi everyone. I know this is long, but I am trying to give as much info as I can. I am a complete newbie to electronics and am working on wiring up a simple parallel circuit with two LEDs. I thought I had everything fine and dandy, but the more I observe my circuit, the more I am finding that something is wrong.

Info:
I am using 2 blue LEDs (specs: 3.7 forward voltage; 20mA)
a switch
2 AA batteries (1.5v each, so 3v total)
I believe the wire is 18g (not sure; it's my friend's wire, and I don't have the packaging for it). Stranded wire.
Circuit is parallel
I am NOT using any resistors. Reason being, I am copying exactly what my friend did, who didn't have any problems with her circuit. Same type of wiring; same type of LEDs, same voltage source, etc. (I will probably get yelled at for this, but before you do so, please read all of my situation! Because I had this same problem when using resistors before.)

Problems:
1) The first red flag is, even though the batteries are new, when I turn the switch "on", the LEDs do not instantly light up. Rather, they gradually fade on. It takes about 30 seconds or so before they are fully lit. I thought this was because I was using 3v to power them, but I really don't know enough about this stuff to know if that's the cause. x_x I thought using 3v would just mean my LEDs won't be as bright as they could be.

2) Once the LEDs are fully lit, I've noticed that the outermost bulb shines noticeably brighter than the LED closer to the battery. (See photos.) I thought when wiring in parallel, that the brightness would not vary? I know that is the case with wiring in series, but I did them in parallel...right?? ;__;





3) This part has been inconsistent, but I tried leaving my circuit on for 5 minutes to see if anything would change, and I touched the batteries to see if they were hot. They were not hot. They were still cool. Yet, earlier today, I did the exact same thing (left the circuit on for 5 minutes), but after I turned the switch off, I left the batteries in the battery holder. When I went to touch the batteries maybe 30 minutes later, the batteries were hot! Even though the circuit was "off", and the batteries hadn't felt hot during the 5 minutes the circuit was "on" prior to that. Without removing the batteries, I tried flicking the switch "on" again, and the LEDs wouldn't light up at all. But the batteries still have juice (tested them later) and the LEDs still work too.

This is my second attempt at wiring a parallel circuit. The first time I attempted to do this, I ran into the exact same problem. With my first attempt, I was attempting to wire 16 LEDs, and I WAS using a resistor with each LED that time. My circuit lit up, but after I had them all hooked up and had the switch and battery source wired up, when I turned the circuit on, I saw that I had one very bright LED, and every single bulb after that gradually got dimmer and dimmer. When I flipped the switch to turn that circuit on a second or third time, the brightest bulb blew out. But the rest of the bulbs in the circuit remained lit. If it had been a series circuit, all of the bulbs would have stopped staying lit, right? But with parallel, if one goes out, the rest are unaffected. Right?? And yet I'm getting variations of light intensity, as if I had wired them in series, even though I didn't wire them in series. The only difference with that circuit experience was that the battery I used would get hot within seconds - like, too-hot-to-hold hot.

I hope someone can look at this and figure out what I'm doing wrong here. This is getting so incredibly frustrating. ;___; Since my first circuit had the brightest bulb blow out, I am afraid that the bulb that is the brightest with this 2nd circuit will do the same thing after turning it on enough times.

Thank you in advance!!!

 

[wonko]

The way you have it set up is less than ideal. First of all, you're not providing enough voltage to efficiently drive the diodes. If I were doing this I would run them with a 9 volt instead. According to the specs you gave, you would need one 270 ohm resistor for each diode in order to drive it and not have problems. Is there any reason why it has to be run in parallel? You really need more power than 3 volts if you want good, consistent light output, and there needs to be a resistor in there, even if it's only a 1 ohm.

Brian

 

[Starlit Rose]

Hi Brian,

Thanks so much for taking the time to read and respond!

I went with parallel because 1) I thought that was the way to go for equal brightness, and also because it requires less voltage, and 2) I was trying to copy my friend's circuit as closely as I could. She wired hers in parallel, 2 blue LEDs, source is 2 AA batteries, no resistors used. Yet she hasn't had any problems, lol. I thought copying a successful example would yield success for me too, but of course it can't be that easy!

So you're saying that I would need a 270 ohm resistor for each LED if I were to power them with a 9v? That brings me another newbie question for you. When I was shopping for LEDs with my first circuit, I saw resistors that had the same ohm number, but different wattage (like 1/8 watt, 1/4, etc). I do not know what watt requirement I would need, and I hadn't seen it mentioned in any of the tutorials I had been looking at. Could you possibly help me understand how I figure out what kind to get? x_x

After asking my friend about her circuit setup again, I noticed one thing was different - she wired her switch differently. This is what hers looks like:


She has her switch going through only one side (the negative I think?), whereas I wired mine so that both the positive and negative wires connected to it before connecting it to the battery. Would this be part of my problem?? Hopefully you can see how I wired my switch for comparison in my first post.

Thank you again!!

 

[Mcameron]

can you post a clearer picture of how you have your on/off switch wired up?

have you tried simply removing your switch and seeing if your symptoms persist?

 

[Scotophor]

This part has been inconsistent, but I tried leaving my circuit on for 5 minutes to see if anything would change, and I touched the batteries to see if they were hot. They were not hot. They were still cool. Yet, earlier today, I did the exact same thing (left the circuit on for 5 minutes), but after I turned the switch off, I left the batteries in the battery holder. When I went to touch the batteries maybe 30 minutes later, the batteries were hot! Even though the circuit was "off", and the batteries hadn't felt hot during the 5 minutes the circuit was "on" prior to that. Without removing the batteries, I tried flicking the switch "on" again, and the LEDs wouldn't light up at all. But the batteries still have juice (tested them later) and the LEDs still work too.

After asking my friend about her circuit setup again, I noticed one thing was different - she wired her switch differently. This is what hers looks like:
View attachment 362488

She has her switch going through only one side (the negative I think?), whereas I wired mine so that both the positive and negative wires connected to it before connecting it to the battery. Would this be part of my problem?

It's really difficult to diagnose the exact nature of the problems you're having without a clear picture of how your circuit is wired. Can you post a line-drawing (schematic) of how you have everything connected, especially the battery and switch terminals? Also, with regard to batteries getting hot-- it's not usual to switch both sides of a circuit like that, though it is sometimes done by people who know what they're doing, and have good reason to do so. I suspect that the way the switch works is different from the way you think it does, and so when you switch your circuit "off", you're really short-circuiting the batteries. This would also explain why the LEDs come up to brightness slowly; the batteries are having to recover from having their voltage dragged down to near zero.

Also, LEDs are current-dependent components, and often do not exactly match one another in various electrical characteristics. Often if you wire two in parallel without a resistor (or trying to "share" one resistor between 2 or more LEDs), one LED will "hog" more current and light more brightly than the other(s). The only way to be reasonably sure of getting near-equal brightness from two of the same part number when wired in parallel, is to give each its own resistor. For your 3.7 volt blue LEDs, you need a minimum supply of 4.5 volts (3 x 1.5 volt cells in series); 6 volts or more would be better. Calculate the resistance needed like this: R = (Vsupply - Vled) / Iled. Iled is the LED current (I is the standard symbol for current), which you usually have to convert from milliamps to amps (divide by 1000). So we have R = (4.5 V - 3.7 V) / 0.02 amps for a 3-cell supply. This comes to 40 ohms; the next higher common standard value is 47 ohms. For a 4-cell supply, the result is 115 ohms and the next higher standard value is 120 ohms.

I hope this helps.

 

[Scotophor]

Regarding the resistor power rating, calculate the power that the resistor needs to handle like so: P = I^2 x R. In this instance, you're expecting the LED current, and thus the current through the resistor, to be 20 mA or less. If you use 3 cells, then, remembering to convert milliamps to amps, you get (0.02^2) x 47, or 0.0004 x 47 which is 0.0188 watts. The current will actually be a bit less because remember from the previous calculation, the exact resistance to get 20 mA was 40 ohms, but since it really isn't worth the bother of combining multiple resistances to get exactly 40 ohms you use the next standard value of 47 ohms, which will drop the current a bit lower. Anyway, the smallest through-hole (with wire leads) resistor size available is 1/8 watt or 0.125 watts. This is more than 6 times the power that the resistor is actually expected to be subjected to, so it will handle it with no problem. Repeating this exercise for the 4-cell circuit version, the power the resistor needs to be able to handle is still only 0.048 watts, so again a 1/8 watt resistor is plenty big enough. But if you happen to have 1/4 watt or larger resistors available of the correct values, there's no need to go out and buy the tiny ones, unless space is at a premium and you want the smallest that will do the job.

 

[Starlit Rose]

can you post a clearer picture of how you have your on/off switch wired up?

have you tried simply removing your switch and seeing if your symptoms persist?

Yeah, here are a couple of close up photos of my switch wiring:


To answer your second question - no, I have not removed the switch yet. I was hoping to see if anyone would respond in regards to it before I took anything apart. But since the next person mentioned my wiring was indeed unusual, I think that is going to be my next step.

It's really difficult to diagnose the exact nature of the problems you're having without a clear picture of how your circuit is wired. Can you post a line-drawing (schematic) of how you have everything connected, especially the battery and switch terminals?

Ehhhh I hope this works LOL drawing is not my strong point:

Also, with regard to batteries getting hot-- it's not usual to switch both sides of a circuit like that, though it is sometimes done by people who know what they're doing, and have good reason to do so. I suspect that the way the switch works is different from the way you think it does, and so when you switch your circuit "off", you're really short-circuiting the batteries. This would also explain why the LEDs come up to brightness slowly; the batteries are having to recover from having their voltage dragged down to near zero.

So my switch wiring IS what's causing all of this? Yeah I definitely have no idea what I'm doing, so I had no clue that's how it worked when wired the way I did it.

So this is the way my friend wired her switch. Is this the proper way to do it?

Also, LEDs are current-dependent components, and often do not exactly match one another in various electrical characteristics. Often if you wire two in parallel without a resistor (or trying to "share" one resistor between 2 or more LEDs), one LED will "hog" more current and light more brightly than the other(s). The only way to be reasonably sure of getting near-equal brightness from two of the same part number when wired in parallel, is to give each its own resistor. For your 3.7 volt blue LEDs, you need a minimum supply of 4.5 volts (3 x 1.5 volt cells in series) and 6 volts or more would be better. Calculate the resistance needed like this: R = (Vsupply - Vled) / Iled. Iled is the LED current (I is the standard symbol for current), which you usually have to convert from milliamps to amps. So we have R = (4.5 V - 3.7 V) / 0.02 amps for a 3-cell supply. This comes to 40 ohms; the next higher common standard value is 47 ohms. For a 4-cell supply, the result is 115 ohms and the next higher standard value is 120 ohms.

Thank you so much!! I appreciate you breaking that down clearly for me. It's easy to get overwhelmed by this stuff. I don't necessarily mind if one LED is brighter than the other, so much as my biggest concern was blowing one of them out because of improper wiring setup, like I did with my first circuit. Speaking of which, since I wired my switch the same way with my first circuit, would that be the reason that the LED that shined the brightest burned out? I turned that circuit on and off several times without giving the battery a break since I was not sure what was going on. If simply changing the way I wired my switch fixed that circuit, I would cry tears of joy.

Back to this current circuit - I happen to have a 4 AA battery pack, so I may pick up some 120 ohm resistors tomorrow and go that route instead. I have a question though, about the wattage specs listed on packs of resistors. The last time I was shopping for them, I saw two different packages of resistors that had the same ohm number. One had like "1/4 watt" and another had "1/8 watt" or something like that, and otherwise had identical specs (as far as I could tell), and frankly I do not know what that means for a resistor, and thus I don't know which type of watt requirement I should be going with for my applications. So if I see two packages of 120 ohm resistors that each have a different watt amount listed, I do not know which one is right for me. (If only the guys that work at the store knew anything about this stuff, lol.)

Thank you again! Sorry that I am asking so many questions, but I am determined to get this right. I greatly appreciate your guys' help and patience, seriously.

 

[Scotophor]

Yeah, here are a couple of close up photos of my switch wiring:
View attachment 362546View attachment 362547

To answer your second question - no, I have not removed the switch yet. I was hoping to see if anyone would respond in regards to it before I took anything apart. But since the next person mentioned my wiring was indeed unusual, I think that is going to be my next step.



Ehhhh I hope this works LOL drawing is not my strong point:
View attachment 362559



So my switch wiring IS what's causing all of this? Yeah I definitely have no idea what I'm doing, so I had no clue that's how it worked when wired the way I did it.

So this is the way my friend wired her switch. Is this this proper way to do it?
View attachment 362564

OK, you only have a 2-terminal switch. It MUST be wired as in your second drawing. It does not really matter if the switch goes on the negative or the positive side of the circuit, but these days it's standard practice to put it in the positive side. The way you have it wired in your first drawing, instead of opening the circuit, it's creating a short circuit. A short circuit is when positive and negative are connected together with nothing in the path to limit the amount of current flowing, other than the internal resistance of the supply and the wiring. If you had wired it that way and used a power source with greater current capability, you might have made big sparks, started a fire, or even caused the batteries to explode (certain rechargeable battery types, such as NiCd and Lithium Polymer, can deliver enough current into a short circuit that the battery itself may explode and catch fire!)

If you want to switch off both the positive and negative sides of a circuit, you need a different style of switch with at least 4 terminals. This is effectively two switches in one. Most such switches have 6 terminals and are called DPDT (Double Pole, Double Throw) switches, but sometimes you can find switches with only 4 terminals. But the switch you have is sufficient for this project.

Thank you so much!! I appreciate you breaking that down clearly for me. It's easy to get overwhelmed by this stuff. I don't necessarily mind if one LED is brighter than the other, so much as my biggest concern was blowing one of them out because of improper wiring setup, like I did with my first circuit. Speaking of which, since I wired my switch the same way with my first circuit, would that be the reason that the LED that shined the brightest burned out? I turned that circuit on and off several times without giving the battery a break since I was not sure what was going on. If simply changing the way I wired my switch fixed that circuit, I would cry tears of joy.

Back to this current circuit - I happen to have a 4 AA battery pack, so I may pick up some 120 ohm resistors tomorrow and go that route instead. I have a question though, about the wattage specs listed on packs of resistors. The last time I was shopping for them, I saw two different packages of resistors that had the same ohm number. One had like "1/4 watt" and another had "1/8 watt" or something like that, and otherwise had identical specs (as far as I could tell), and frankly I do not know what that means for a resistor, and thus I don't know which type of watt requirement I should be going with for my applications. So if I see two packages of 120 ohm resistors that each have a different watt amount listed, I do not know which one is right for me. (If only the guys that work at the store knew anything about this stuff, lol.)

Thank you again! Sorry that I am asking so many questions, but I am determined to get this right. I greatly appreciate your guys' help and patience, seriously.

Regarding the circuit with multiple parallel LEDs where one LED burned out, I suspect that you somehow forgot a resistor for the one LED that died, or you inadvertently bridged across the resistor with wire or solder, thus effectively removed it from the circuit.

 

[Starlit Rose]

OK, you only have a 2-terminal switch. It MUST be wired as in your second drawing. It does not really matter if the switch goes on the negative or the positive side of the circuit, but these days it's standard practice to put it in the positive side. The way you have it wired in your first drawing, instead of opening the circuit, it's creating a short circuit. A short circuit is when positive and negative are connected together with nothing in the path to limit the amount of current flowing, other than the internal resistance of the supply and the wiring. If you had wired it that way and used a power source with greater current capability, you might have made big sparks, started a fire, or even caused the batteries to explode (certain rechargeable battery types, such as NiCd and Lithium Polymer, can deliver enough current into a short circuit that the battery itself may explode and catch fire!)

If you want to switch off both the positive and negative sides of a circuit, you need a different style of switch with at least 4 terminals. This is effectively two switches in one. Most such switches have 6 terminals and are called DPDT (Double Pole, Double Throw) switches, but sometimes you can find switches with only 4 terminals. But the switch you have is sufficient for this project.

Awesome!!!!!!!!!!!!!!!!! I feel like a fool for not knowing how to wire a switch correctly, but I'm glad it was a simple solution. It definitely makes sense to me now. I just re-wired my switch to try it out, and yup, it's working the way it's supposed to.

Regarding the circuit with multiple parallel LEDs where one LED burned out, I suspect that you somehow forgot a resistor for the one LED that died, or you inadvertently bridged across the resistor with wire or solder, thus effectively removed it from the circuit.

Oooooh that second one might be what happened. That project is a hot mess of wires right now, but when I get the time to sit down with it again, I will check to see if that's what happened.

Thank you soooooooooo much!! I am very grateful that you not only figured out my main issue, but also took the time to explain it for me. Will definitely not be making this mistake again.

 

[vidman]

here are a few calculators and circuit designing online programs to help you in the future...

Use them , they have helped me in the past immensely ,,,,;

http://www.hobby-hour.com/electronics/ledcalc.php

http://led.linear1.org/1led.wiz

http://led.linear1.org/led.wiz

the last 2 are different if you klik these links even though it looks like the same link.

 

[13doctorwho]

If you want consistant brightness series works better. The reason being they will have the exact same current, and current controls the brightness. In past I have used a 12V "lighter" battery. If you do that you can put them in series and set the brightness with the series resistor. There will always be some difference in the brightness of LED because of tolerancing in there design. Generally the difference is not noticeable.

It is a bad practice to hook LEDs up with out a current limiting resistor. As already stated, even a 1 ohm resistor is fine.

 

[OldSkoolEffects]

It's bad practice to use LEDs in parallel unless you've matched their forward voltages.

 

[Robiwon]

Just saw this. Glad you got it figured out and didn't start a small fire the way the switch was wired!