# LED lamps getting hot

Discussion in 'Homework Help' started by MingJae, Feb 20, 2010.

1. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
hi, i'm a newbie just asking a simple question and maybe someone will help me to solve this problem

case:
1.orange/red/green 3mm LED getting hot while connected to 3v DC battery
2.orange/red/green (3-20 pieces) + white LED 3mm = white led get dim...
3.I need a scheme of orange+red+green+white LED with only 2x1.5v batteries

questions:
1.how many resistor (Ω)should I use to optimize the lights of orange,red,and green LED?(I already use the transistor calculator but have no idea..)

2.can I make all the LED turn on with one switch?(one button to turn ON all LED)with scheme pic pls?

sorry if my english is not correct and thx b4

2. ### Audioguru New Member

Dec 20, 2007
9,411
896
Ohm's Law is used to calculate the value of a current-limiting resistor for an LED. It is based on the voltage dropped across the resistor that determines its current and the current of the LED in series with it.

You need to know the actual voltage of your battery (it is 3.2V when brand new and might be 2.4V when used a little plus you need to know the actual voltage of your LED (not its maximum voltage). The red LED might be 1.8V, the green LED might be 2.2V and the white LED might be 3.3V. You also need to know the typical current of your LEDs which might be 20mA.

For example:
1) The battery is 3.0V, the red LED is 1.8V and its typical current is 20mA.
2) 3.0V - 1.8V= 1.2V. The resistor value is 1.2V/20mA= 60 ohms which is not a standard resistance value. Use 56 ohms or 62 ohms.
3) But the battery might be new at 3.2V and the LED might actually be 1.7V so its current with a 56 ohm resistor is 26.8mA which will be OK.

If the white LED is 3.3V then it will be extremely dim with a 3.0V battery.

You must select a battery that is big enough to last as long as you want with all your LEDs lighted.

3. ### retched AAC Fanatic!

Dec 5, 2009
5,201
313
You may want to use two 3v lithium batteries, like the CR123a. They don't take up as much room as the AAs. That will give you greater life and more reliable voltages over a longer period.

[ed]
Here are some non-rechargeable for 92 cents (us) each.
http://www.cr123batteries.com/product.php?p=ryder_cr123a&product=171675
These type batteries should be available everywhere,
[/ed]

Last edited: Feb 20, 2010
4. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
at first, thank for all the explaination...

I try to connect 6 of white LED with 2x1.5volt(in series batts)in parallel and they are lit well, but if I connect 5 orange LED+1 white=white LED get dim than 6 white..(the all 5 orange still get hot),
is it better if I use 4x1.5v batteries with orange+red+green+white+resistors to light them on together? if so, I don't know to place the resistors...pls help

ok thanks, but in my region a single 3v battery is uncommon in store,
how about 4x1.5v batteries in series? is it still good to my LED lifes?

5. ### retched AAC Fanatic!

Dec 5, 2009
5,201
313
The white led is pulling to much current through your orange ones. You want to separate them on another line and e sure to use current limiting resisrots on the orange and white led lines.

And yes. If you can get the 4 x 1.5 lithium AA's that will help. However the alkaline will work, however you will have shorter battery life.

Different LED's have different ratings. You do not want to mix ratings in series. Each have different requirements that can kill the others in line.

6. ### Audioguru New Member

Dec 20, 2007
9,411
896
A current-limiting resistor must be calculated then connected in series with each LED.
Your LEDs get hot because their current is much too high without proper current-limiting resistors and your LEDs are burning out.

7. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
i try to mix 1xwhite + 1xblue + 2xorange + 1x47Ω resistor in 6volt parallel=white+blue led won't light...should I use more than one scheme and more switch to divide orange,blue and white LED?
is it possible if I only use 6v batts and a switch to lit them all?
whis is what i've got:

ok thank you,now i'll use resistor to protect my LED

8. ### PentodePuppy Member

Aug 19, 2009
12
1
It's important to understand the "nuts and bolts" of biasing an LED. The crucial parameter is current, not voltage. When a current is passed through an LED, a voltage does develop across it, but that is secondary.

The cheapest way of biasing an LED is with a series resistor, but that is not the best way to do it. The best way is with a constant current supply. Now, you can mimic a constant current supply with just a resistor and a voltage, but to get a truly constant current you need a voltage that is much higher than the voltage that is dropped by the LED -- the higher the better, with infinite voltage producing an exactly constant current.

Since it is impossible to achieve an infinite voltage (as well as impractical), something lower must suffice.

But, before I delve into current supply techniques, let me point out that LED voltage is a somewhat unreliable parameter -- especially for the more energetic LEDs such as the Blues/Whites (white is really blue with yellow phosphor), Violets and Ultra-violets.

For instance, I have sampled White LEDs and found them to range from 3.2 Volts to 3.6 Volts. The typical spec sheet for a white LED has 3.5 as typical and 4.0 as the maximum voltage. I have also seen the voltage of an LED change with time and with temperature. On the other hand, it has been my experience that the voltage on a Red LED (for instance) stays pretty constant, across batches and across time -- typically 1.7 to 1.8 for the original (dimmer) LEDs and typically 2.0 Volts for the Super-Bright versions. Also, the voltage on an LED tends to change with change in current (with yellow being the most stable -- e.g. yellow LEDs make fairly good voltage references for non-critical applications). This is why regulating the current is important when running an LED.

Here is an example to illustrate the effect of supply voltage on a series resistor current limiter:

Consider a circuit with a voltage source (such as a battery), a resistor, and an LED, all in series. Lets say the LED is a white which out of the box could have a voltage drop of anywhere between 3.2 and 4.0 volts. Lets try three cases:

1. Battery voltage = 5 volts.
2. Battery voltage = 6 volts.
3. Battery voltage = 12 volts.

Lets design each for the typical LED voltage of 3.5 volts and the nominal current of 20 ma, which would result in the following resistor values for each case:

1. 5.0V - 3.5V = 1.5V 1.5V/20ma = 75Ω
2. 6.0V - 3.5V = 2.5V 2.5V/20ma = 125Ω
3. 12.0V - 3.5V = 8.5V 8.5V/20ma = 425Ω

The following is a table of the LED currents for each of the 3 cases at each of three possible actual LED voltages (remember, we designed for the typical LED voltage of 3.5 volts BUT, in the real world, that voltage could be anywhere from 3.2 volts to 4.0 volts.) The formula used to compute the currents is: (Vs - Vled)/R where Vs = the Source Voltage, Vled = the LED Voltage, and R = the series resistance:

 Case # R 3.2V 3.5V 4.0V 1 75Ω 24.0ma 20.0ma 13.3ma 2 125Ω 22.4ma 20.0ma 16.0ma 3 425Ω 20.7ma 20.0ma 18.8ma

Notice how the current varies less as the source voltage is increased and also as the resistance is increased. An ideal constant current supply has an effective resistance of infinity! A practical constant(ish) current supply need only have a dynamic resistance that is sufficiently higher than the effective resistance presented by the range of possible LED voltages.

Now, in regards to wiring LEDs directly in parallel--not a good idea. Each LED is going to have a slightly different voltage, which may sound like "not a problem", but if you consider that there is an exponential change in current for every change in voltage in an LED (just like a diode, because, guess what, an LED is a diode), it becomes a serious problem. Even if you took the trouble to match the LED so the currents divide evenly between them, in time or because of uneven changes in temperature (such as if the temperature of one of the LEDs increases or decreases relative to the other), this balance can be compromised--greatly compromised.

The result of uneven voltages in parallel LED arrangements is an uneven distribution of current between the two or more parallel LEDs. It doesn't take much of a variance in voltage difference for there to be a significant difference in current through the devices. The way around this is to current limit each parallel leg--ideally with a separate constant current regulator, but in a low budget solution, with a sufficiently high source voltage and a series resistor on each leg -- i.e. one voltage source, sufficiently high for the leg with the highest LED voltage drop, and one resistor for each leg, calculated to supply the correct current for that leg based on the total typical LED voltage drop(s)). Each leg can have more than one LED in series, so when calculating the value of the current limiting resistor, you must add up the typical LED voltages for the LEDs in that leg.

An even better way to drive one or more LEDs is with a Joule Thief circuit. You can run a whole string of LED in series with one 1.5 volt battery, and it will suck nearly ALL of the energy out of the battery (i.e. until it discharges to around .4 volts)--thus the name "Joule Thief". For more on this technique, check out the following sites:

http://www.emanator.demon.co.uk/bigclive/joule.htm
http://www.joulethief.com/kit.php
http://www.prc68.com/I/JouleThief.shtml

Then, you can use an actual current regulator (either a constant current source or constant current sink). A simple technique is to use an LM317L. Apply the positive side of the source voltage to the LM317L Input and tie the Adjust and Output legs together and connect them to the anode (or positive side) of the LED. Connect the cathode to the negative side of the voltage supply. The drawback is the need for a fairly high voltage as it has to over come both the drop-out voltage (around 1.7 volts) and the reference voltage (1.25 nominal -- 1.3 max). So, for a 3.5 V White LED the supply vooltage would need to be at least 1.7 + 1.3 + 3.5 or 6.5 volts. A lot of power gets wasted to drop the voltage to 3.5 but the LED is driven correctly, with a constant current. Another way is with an OP Amp and a transistor. You an find that circuit in the Application Notes for a typical OP Amp (such as the LM741).

Another trick is to match the internal resistance of a battery to that required to drive an LED. This is best done with a Lithium coin cell. When properly matched, you can connect the LED directly across the cell and it won't get hot.

Last edited by a moderator: Feb 23, 2010
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9. ### PentodePuppy Member

Aug 19, 2009
12
1
This is totally wrong. The White LED is not "pulling" current through the orange ones.

The correct evaluation is that one of the white LEDs (the one dropping the most voltage) is setting the total voltage of the parallel circuit. For the orange LEDs to run at this voltage they must draw a much higher current than the white LED(s). There is an exponential relationship between the voltage and the current in an LED. Raise the voltage across an LED a little bit, and the current that it draws will increase considerably. This is what is happening. When the battery voltage is applied through the 47Ω resistor, the current that flows through the resistor divides across the LEDs. Because the White LEDs run at a higher voltage for the same current than the Orange LEDs, the Orange LEDs are forced to run at the voltage that White LEDs usually run at. Consider that an Orange LED, at 20 ma, typically runs at around 2.2 volts and a white LED at around 3.5 volts. And that there is an exponential relationship between voltage and current in an LED, what does that do to an Orange LED that is forced to run at more than 3 volts? That's an increase of 0.8 or more volts over the orange LED normal 20ma operating voltage. A 0.8 volt increase on an LED is HUGE. It causes a large increase in current.

You can prove this to yourself by doing the following experiment:

1. Connect a 10K variable resistor, a 330Ω 1/2 watt fixed resistor, one of your orange LEDs (I would use a new one, not one that has gotten hot before) and a current meter (one that can register from 1 ma to 30 ma) in series across a 12 Volt regulated power supply.
2. Connect a voltage meter across the LED
3. Adjust the variable resistor until 1 ma is flowing in the circuit and write down the current and the LED voltage.
4. Increase the current to 2ma and take a measurement then to 3ma and so on until you reach 30ma.
5. Notice how little the voltage changed compared to the current.
6. If you're brave, change the fixed resistor to around 100Ω or less and see if you can get the voltage of the Orange LED to 3 volts or more. Better use a 2 Watt or more resistor, and the variable resistor should be around 5 Watts or more. Also, the current meter will need to be able to read a lot higher -- perhaps as much a 100 ma. It will probably ruin the LED, but hey, science ain't always pretty
What is needed is a limiting resistor for each of the LEDs.

10. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
ok i got it, that;s why the orange LED light on but the white don't,
but I still confused with your experiment coz I'm a newbie...

btw,finally I found the way out to turn all the LED on...

this is what i do, trial and error...but there's still a problem...
the white LEDs are dim...but if i put one more 47ohm resistor on blue LED, white LEDs light increase...
Can u helm me to calculate the number of correct resistors?(ohm)

11. ### Audioguru New Member

Dec 20, 2007
9,411
896
I think your 6V battery is almost dead. It does not have enough power to light 5 LEDs.
If it is brand new then its voltage will always be 6V then the orange LEDs will need a 150 ohm resistor in series with each one and the white and blue LEDs will need a 100 ohm resistor in series with each one.

Every LED needs a resistor in series to limit the current or they will burn out.
If the 6V battery is new then two orange LEDs can be connected in series and in series with a 68 ohm resistor.

12. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
ups sorry...in my previous post, the white LED dim in 3v brand new 2x1.5v batteries.
but,if i use 4x1.5v brand new batteries, the orange LED light turn into red light...i don't know why maybe it gonna be to explode...

btw, in series(6 orange) + paralel(white,blue+resistor) in 6 v batteries, is it need 1.7x6 v power supply to optimized the series orange?
ok, i've tried this:

2x1.5v AA batteries= orange led light turn into red light(very-very hot)
1x3v micro lithium coin battery= orange LED light is very bright(hot but still orange)

I dunno, maybe a brand new AA batteries(just Audioguru said before) 3.2v, maybe 1.6v+1.6v,
but this single micro lithium cell is 3,1 v?
then 0,1v is value that much?

Last edited: Feb 23, 2010
13. ### PentodePuppy Member

Aug 19, 2009
12
1
In a parallel circuit each parallel path has the same voltage across it. So, if we add a resistor in series with each LED, then each resistor/LED pair has 6 volts across it. That makes calculating each resistor a simple algebra problem. From Kirchhoff's Voltage Law we get:
Vs + -I*R + -Vled = 0 [where Vs is the battery voltage]
or
Vs = I*R + Vled
Solving for R we get:
R = (Vs - Vled)/I
Assuming the battery is still 6 volts, and assuming you want to run each LED at 20ma, and assuming the LEDs drop the following voltages when run at 20ma:
Orange: 2.0 volts
White: 3.5 volts
Blue: 3.5 volts
Then the resistors are:
Rorange = (6.0V - 2.0V)/20ma = 200Ω
Rwhite = (6.0V - 3.5V)/20ma = 125Ω
Rblue = (6.0V - 3.5V)/20ma = 125Ω
You should probably measure the voltage across each LED with it running at 20 ma and use that voltage in the equation.

Also, you could put the two orange LEDs together in series and run them with one resistor, because their voltage drops would add up to a voltage less than 6 volts (e.g. 2 * 2.0 volts = 4.0 volts). But, that would sacrifice current regulation (i.e. it would be more vulnerable to LED voltage fluctuation/variation).

You really shouldn't be running your LEDs at a continuous current any higher than 20 ma (some LED spec sheets state a maximum of 30 ma, but it has been my experience that they burn out a lot faster when run over 20ma -- you can, however, run LEDs at a much higher current if you pulse them at the proper frequency and duty cycle).

Unless the battery voltage is being reduced by the load, your 47Ω resister is causing a current through the orange LEDs in the neighborhood of 70 to 80 ma -- way too high (unless you're not interested in longevity). And connecting the LED directly across the batterys with no limiting resistor is probably causing even higher currents (unless the internal resistance of the battery is limiting the current.)

BTW: I wouldn't be surprised if upon measuring the voltages across your LEDs that they are much higher than typical, because I suspect you have stressed them a great deal, and LED nominal voltages tend to increase when they are subjected to high currents (probably due to excess heating). This will cause them to run more dimly in a normal series resistor limiting circuit, like what you are using, because the current will be reduced by the LEDs high voltage drop.

14. ### PentodePuppy Member

Aug 19, 2009
12
1
It helps to know what the internal resistance of a battery is before throwing LEDs across it. Also, the reason the micro lithium coin battery caused the orange LED to get hot, is the Orange LED normally operates at around 2 volts and the Lithium cell is 3 volts. It didn't have enough internal resistance to limit the current when running a 2 volt LED. You'll have better success applying Blue, White, UV and even some Green LEDs across a lithium cell [i.e. LEDs that run at a safe current at around 3 volts]. And it occurs to me that it's more about voltage than internal resistance, in this case. There is a direct relationship between voltage level and current draw in an LED. At the Lithum batteries voltage, there is a better chance of a White or Blue (etc.) LED running at a safe current level. BUT, this is tricky because small changes in voltage cause large changes in LED current. But, the internal battery resistance of a Lithium coin cell will limit that somewhat, making it a practical solution.

Fresh AA Alkaline batteries have EXTREMELY low internal resistance, so you're not going to get that effect with them. Probably only with Lithium coil cells as lithium batteries in general have higher internal resistance than, say Alkaline. Also, NiCad and NiMH batteries have even lower internal resistance, with NiCads having the lowest. Putting a white LED across a pair of AA Alkaline batteries will probably work (i.e. run the LED at a safe current level), but there isn't much room for error, and the LED will probably go dim long before the batteries fully run down.

15. ### PentodePuppy Member

Aug 19, 2009
12
1
Yup, for an LED 0,1v is huge! Exponential, in fact

16. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
ok thx for the explaination...
btw,how long will a 3v micro coin battery will survive with full load(eq.12 orange LEDs and 4 blue)? more than 3 hours?

woow,I am using the led calculator and the result for 2xorange(1.7v ea)
is:
3v=68Ω resistor
3.1v=82Ω resistor...Δ 14Ω
but,in fact my brand new AA batteries(2x1.5v) only using 47Ω resistor is good to my LED...it is not getting hot...should i still use 68 Ω resistor?

17. ### PentodePuppy Member

Aug 19, 2009
12
1
An LED will be damaged by currents that are far lower than what makes it feel hot. Dude, use an ammeter.

18. ### PentodePuppy Member

Aug 19, 2009
12
1
Try this:

Just vary the pot and watch how the voltage changes in relation to the current. Try it for several different LED colors (and try it with LEDs that you haven't fried -- i.e. LEDs you haven't touched yet )

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Last edited: Feb 26, 2010
19. ### Audioguru New Member

Dec 20, 2007
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896
You need to measure the voltage of the batteries while they are powering your LEDs. You also need to measure the current so you do not exceed the max allowed current for the LEDs.

AA cells can provide 3.2V at 200mA for a few seconds when they are brand new and have never been used.
A lithium coin cell is small and provides 3.1V at 20mA to 50mA for a few seconds when it is brand new and has never been used.

After a few seconds of use the voltage from the battery begins dropping.

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20. ### MingJae Thread Starter New Member

Feb 19, 2010
13
0
ow, 200mA vs 50mA...very usefull information thx!
allaboutcircuits.com = two thumbs up !