8
\$\begingroup\$

I'm trying to make a circuit that will run 1-3 LED in a parallel, powered by a 12V power source. I used a voltage divider calculator and it said to use a 100ohm and 33ohm resistor. when I wire them in correctly, the 100ohm resistor overheats and eventually fails. How would I change this circuit to keep this resistor from overheating?? Both resistors are 1/2 Watt. LEDs are 3.6v max.

\$\endgroup\$
  • 1
    \$\begingroup\$ Please edit your question, hit Ctrl-M and add a circuit diagram. Also add information about the current and voltage drop you want in for each LED's. Is a series circuit not an option? \$\endgroup\$ – jippie Apr 21 '13 at 6:42
  • \$\begingroup\$ Exactly why are you running them in parallel? And Why are you using a voltage divider? \$\endgroup\$ – Passerby Apr 21 '13 at 13:42
  • \$\begingroup\$ are you sure the resistor that is failing is 1000 ohms? or that it is actually failing? 12v across a 1k resistor results in .144 watts of power dissipation which is well under the rating of the resistor, there should be no way for that resistor to overheat/fail based on what you have told us \$\endgroup\$ – Gorloth Apr 21 '13 at 19:28
  • \$\begingroup\$ I concluded using a voltage divider would be the easiest and most compact way of getting 12 volts down to approximately 3 volts. As far as running the LEDs in parallel it uses less electricity, based off what I have seen. \$\endgroup\$ – Robert Joseph Rowley Apr 22 '13 at 16:15
16
\$\begingroup\$

By decoding your question, this is what I arrive at:

You are using the two resistors as a voltage divider, with the LEDs wired in parallel connected to the junction of these two resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

This is not a valid approach, the voltage divider will not provide the 1:4 (approximately) voltage division your calculations indicate.

Firstly, either apply the three LEDs in series, with one resistor in series, or if you must have them in parallel, each will need a series resistor.

schematic

simulate this circuit

The above indicative values of resistors are assuming 20 mA per LED as optimal. Do your own calculations for your specific LEDs.

For a ready to use online calculator, use an "LED resistance calculator" instead of a voltage divider one. This one even provides the correct schematic or wiring diagram as you prefer, to wire the LEDs up.

\$\endgroup\$
2
\$\begingroup\$

The issue is that 3.6V is (very likely) not the maximum voltage for each LED, and therefore you don't need a voltage divider.

The 3.6V max you are mentioning is most likely the voltage drop across the LEDs when the LED is conducting. It is not the maximum voltage, LEDs, don't really care about voltage (this is an approximation, but a useful one). What LEDs do care about is current: they must be protected from is excessive current, and they cannot do this for themselves, so to speak, because they have almost no resistance. Most regular LEDs "like" 20mA maximum, otherwise they melt, shrivel, and die. I don't know about your LEDs, but let us assume 20mA as a good value for current (incidentally, you can find this in the datasheet or on ebay as "Maximum (forward) current", usually given in milliamps; for many LEDs the maximum forward current is 25mA, so 20 is a good value to shoot for). With these assumptions here's what the situation looks like:

After the 3.6V voltage drop across the LED, you have 8.4 volt potential, and the current must be kept at a bit under 20mA. According to Ohm's law (\$V=IR\$), we work out that you need 8.4/0.02 = 420 Ohm current-limiting resistors, assuming the LEDs are in parallel, and a resistor for each LED. How much power is that? Well, power is just the product of current times the potential, i.e \$V*I\$, or volts*amps. We have a 8.4V drop across each resistor, and 20mA current so we get 8.4*0.02 = 0.17W for each resistor. Since your resistors are 1/2 Watt, 0.17W is well within the limits of the resistor.

On a side note, 420 Ohm resistors are less common that 470 Ohm resistors, so just use 470 Ohm resistors. Your LEDs will be a bit less bright, but a bit more safe.

Again I am assuming a parallel setup like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
0
\$\begingroup\$

In your application you haven't mentioned the current each LED is drawing. Since you are connecting LEDs in parallel then the effective resistance of all the three LEDs became small. It draws more current from the source this might be causing the problem in your application. Try to add some resistor in your circuit before parallel connection of LED as this will increase the effective resistance and limit the current.

\$\endgroup\$
0
\$\begingroup\$

it would appear some confusion is taking place. the circuit in question says R1 is 100 ohm not 1K ohm. As for the rest, series is the way to go and the current needs to be defined. you could be pushing 10 amps. @ 12V or 1 amp. The voltage divider would be of use if your'e running something else along with the LED's. ...In short the above advice is sound but the question is incomplete.

\$\endgroup\$
0
\$\begingroup\$

There are a few things to consider when dealing with automotive circuits that are powered directly from battery. Most car batteries operate at 12V nominally - but can sit anywhere between 9V and 16V during normal operation.

Consider this circuit (note Vbattery = 16V):

schematic

simulate this circuit – Schematic created using CircuitLab

If your LEDs drop ~3.5V and draw 100mA of current, you are burning .35Watts (Power = Current x Voltage) in the LED - no big deal. You still have 12.5V to drop somewhere else. In this case it across the resistor. Presistor = (16V - 3.5V) * 100mA = 1.25W. That's quite a bit.

Nominal Calcluations (i.e. Vbattery = 12V):

Pled = 3.5V * 100mA = .35W (same as before)

Presistor = 8.5V * 100mA = .85W (still could be problematic)

I recommend going with this circuit, as to avoid dealing with excessive voltage drop across one resistor. Be aware that if your battery drops very far below 12V, your LED string will probably shut off. 3.5V + 3.5V + 3.5V + Current*Rresistor = pretty close to 12V.

schematic

simulate this circuit

You can also spread the power dissipation across two resistors in each string instead of just one. Each of these resistors will need to be half the resistance. See the schematic below.

Extra Credit: If you do want a parallel LED application, you can try a more complex approach.

Another consideration is current matching, which isn't critical in most hobbyist applications but this approach will also help spread out the power.

Consider this circuit:

schematic

simulate this circuit

At the bottom of each string are BJTs. The left-most string has a BJT configured with the base shorted to the collector - and the emitter to ground. All bases are tied together. This is called a current mirror as it forces the current in each of the strings to be the same. (I'm doing a lot of hand-waving here. It's also responsible to note that it isn't a guaranteed, perfect current match between strings because of thermal differences, process variations between BJTs, etc. - but in this case it's good enough.)

What IS important here, is that you can put a BJT that is capable of handling a bit of power to help you drop your voltage "more safely." Instead of putting a huge 2W resistor, you can get a little fancier and drop some of the voltage across the BJT - decreasing the amount of power your resistors need to burn. It won't be a whole lot. You can choose a BJT with a Vbe = 1V and drop .1W across the BJT (or a Vce = 2V and drop .2W). You also get the added benefit of ensuring your LEDs are all the same brightness.

Hope this helps!

\$\endgroup\$
  • \$\begingroup\$ +1 for mentioning that unregulated auto voltage supplies can be interesting. \$\endgroup\$ – user65586 Nov 4 '15 at 15:32
-1
\$\begingroup\$

I recommend to you to add even 1 led to 3 leds =4leds and put 100 ohms.series connection available.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.