Hi! I am pretty new to circuitry (working on my first few projects). One of my projects was an LED flashlight hat. I had wired it up and everything, and it worked great. I used two 9v batteries in series with 5 LEDs in series, meaning each LED was recieving approx. 3.6v each. I have used the LEDs with 6v ea. using two 3v batteries wired in series, and it just made the LEDs really bright so I figured if I added another 9v battery meaning a total of 27v, that would be 5.4v per LED and it would just make it really bright. So I added it in and... It blew the LEDs. All five of them. Can anyone explain why, I do not understand why 5.4 blew the LEDs when 6v didn't.

 

LEDs are not voltage devices, they are current devices.

LEDs, 555s, Flashers, and Light Chasers

If you show us your schematic we could be more help.

 

9v -- 9v -- 9v -- Switch -- LED -- LED -- LED -- LED -- LED --|
|____________________________________________________|

Pretty simple.

My original circuit was exactly this, but only two 9v batteries instead of 3. Three is when it blew.

9v Duracell Batteries
3.2-3.8v 30mA LEDs (yes, I know, but it worked with 6 volts just fine) (Specs)

 

You need to read that article I wrote. You can not use LEDs without resistors, you've seen the results.

Chapter 1 explains the basics, which is where you're at.

 

hahaha, i'm pretty sure that 95% of the guys on this forum started out like this...at least, i know i did haha. Well, for this circuit, as most simple circuits, you need to use ohm's law: I=V/R (Current = Voltage/Resistance). In this case, we need to find a resistor to limit the current in the circuit so the LED's don't blow out. So we arrange Ohm's Law to R=V/I (Resistance = Voltage/Current)

You're using 6 LED's that I'll assume require a forward voltage of about 3V at about 20mA each. You currently have the LED's placed in series, but I'd strongly recommend placing the LED's in parallel (this is when all the anodes (longer leads) are connected to the positive terminal of the batter and all the cathodes (shorter leads) are connected to the negative terminal of the battery). In this case, the voltage demand will be the same (3V) for all of the LED's, but their current requirements will be greater [.02A x 6 leds = .12A]. Now since we know the demands, we plug these values into Ohm's Law. Remember though (and I say it like this because I used to make the same mistake), when plugging the voltage into the formula, we don't put in the value of our voltage demand, but instead, the voltage that we don't need. (example, you're using a 9V battery and we only need 3V to drive the LED's so we want our resistor to block 6 of those volts).

So our equation becomes R=V/I: [R= 6/.12] R=50 ohms. So you want to place the LED's in parallel with a 50-Ohm resistor. This value will ensure maximum brightness without blowing out the load (in this case, the LED's.

This is about as simple as it gets, so if none of this makes sense, I definitely recommend reading everything in that link that Bill_Marsden posted. Hope this helps. -Joe

 

Parallel is OK, he just needs a 470Ω resistor in the circuit with them. 5 LEDs (if they are blue or white) work out to around 18V, with 3 9V batteries he has 9V headroom. Ohms law; 9V / 0.02A = 450Ω.

The color of the LED determines how much voltage it drops. Blue or white is around 3.6V.

 

The LEDs are white. The only reason I used 27v instead of 18v is because I thought it would make the LEDs brighter, as it did when I used 1 3v battery vs 2 3vs = 6v

 

I understand why you increased the voltage, but you need to understand that LEDs are not voltage controlled. You need only satisfy the forward voltage, then control the brightness via CURRENT.

LEDs will incur something called 'thermal runaway'. When they start to draw current, they heat. As they heat, they draw more current that causes them to heat further.

More heat..more current...POP!

Resistors will allow you to stack up batteries to lengthen the amount of time the flashlight will stay lit, while LIMITING the CURRENT to the LEDs so they do not go POP.

 

So I want a high current but only a 3.6 or so voltage?

 

I usually use 330 ohm resistors with a 9V battery. That's 27.27 mA, which is fine for a typical LED, though is approaching the limit.

 

For many LEDs it is over the limit, but not too much. I use 20ma as a target unless I have a data sheet (which is rare). If you drive an LED a little over you will shorten the lifespan a little, but LEDs have such a fantastic life span shortening that lifespan is often unnoticeable.

 

I think I am beginning to understand, but I just want to be completely clear. The voltage difference in the LEDs is not what is causing them to be brighter, instead it is the increase in amperage. Correct?

So that would mean that if I wanted them to be really really bright, I would just need to increase the amperage (not too high obviously). Correct? If so, how do I increase the amperage and what should the absolute maximum be if they are 30mA LEDs? I know this sounds like I am giving the answer, but the LEDs are rated 3.2-3.8 and they will take 6v, so I want to know if the amps will be similar.

P.S. Thank you to all that have helped so far!

 

It's not amperage, it's current.

If they are 30mA LEDs, the maximum is probably 100mA for a couple seconds, but the continuous limit is 30mA. I have some 30mA blue LEDs which I can run off 9 volts for a couple seconds... but they get really hot and start changing green. Generally, if the LED is hot to the touch, too much current is being used. Exceptions apply of course for power LEDs.

No problem, glad to help...

 

So just to be 100% clear, it does not matter what voltage I give it, as long as it is within the LEDs operating range? Correct? And what I want to do to increase it's voltage is increase the amperage? Correct? So how do I increase the amps to max out the LED's usage?

 

To increase the current, you need to reduce the resistance, or increase the source voltage.

An LED is a current controlled device, but current depends on the voltage supplied to the LED. All you need to understand is that a higher current results in more heat dissipation and more light output. Importantly, the relationship to current and voltage is a very much non-linear one, that is if at 3.2 volts current is 20mA, and if at 3.3 volts it is 30mA, it does not mean at 3.4 volts it is 40mA.

 

Just as a side note, I'm guessing the 3V batteries were coin cells, something like CR2032s. If so, they have high internal resistance so the voltage to the LEDs was probably less than 6V which is why they didn't burn. Still not the best idea for the LEDs.

 

Good point Mark, I tried connecting two LEDs to a single coin cell and they both dimmed. I think the maximum current to maintain a reasonable voltage, say 2.5 volts, would be around 50mA. Which makes sense; they are designed for computer clocks and small portable devices.

 

There is a maximum current for LEDs, as you may have guessed. The article I pointed you to explains how to calculate this current. LEDs usually don't just die, they dim out. So even if you are planning on slightly over driving LEDs it could still be contra indicated. Like I said earlier, I aim for 20ma, but only the LEDs datasheet has the real answer.

You don't need to put all these LEDs in series (it works, but it isn't the only way). Three 9V batteries make 27V, which in a rain storm can give you a bit of a shock (as in electrical shock). It is possible to have 5 LEDs (assume white), each with it's own resistor, powered by either 4 AA cells (6V) or one 9V battery.

Like I said earlier, you can not get by without a resistor for each LED chain, and you can not share resistors with LEDs in parallel (explained in the article).

 

So I want a high current but only a 3.6 or so voltage?

No.
The absolute maximum allowed current for your LEDs is only 30mA which is not a high current. New 9V batteries can supply 1000mA for a few seconds to burn out your LEDs if their current is not limited to 30mA or less with a resistor.

 

Is there an off-the-shelf chip that drives LEDs by current-regulation using a DC-to-DC switched-mode converter, using only a low-resistance sense resistor (internal or external) to sense current?