Using a single resistor with several LEDs
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Posted by wylbur37 on July 31, 2004, 4:39 am
Please log in for more thread options rating of the LED, a resistor is typically used. If more than one LED is to be used (let's say three), each one would have its own respective resistor, and the schematic would look something like this ... +----------+-----------+-----------+-------------- | | | | | | | | | LED-1 LED-2 LED-3 power | | | source | | | | resistor-1 resistor-2 resistor-3 | | | | | | | | +----------+-----------+-----------+-------------- But if all the LEDs were of the same voltage and amperage (e.g., if they were all of the typical 3.6V 20mA), isn't there a simpler way to do this by using only ONE resistor? In other words, can't it be done like the following ? +--------------+-----------+-----------+--- | | | | | | | | power LED-1 LED-2 LED-3 source | | | | | | | | | | | +---resistor---+-----------+-----------+--- And what would be the proper value of the resistor? (I'm guessing it would be one third of what it was in the first diagram because the required voltage reduction would be the same but the effective current draw of the 3-LED assembly would be three times what it was before. So if the power source were 6V and the LEDs were 3.6V and 20mA each, then the required resistance would be 120 ohms in the first case and 40 ohms in the second case. Correct?) | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Jerry G. on July 31, 2004, 9:11 am
Please log in for more thread options likely have a situation where one would light normally, and the rest would be dim, or not lit at all. You need a separate resistor for each one, or you should look for LED's with built in resistors. If they have built in ones, then you will be restricted to the voltage range indicated for the LED's. -- Jerry G. ========================== When using an LED from a power source whose voltage is higher than the rating of the LED, a resistor is typically used. If more than one LED is to be used (let's say three), each one would have its own respective resistor, and the schematic would look something like this ... +----------+-----------+-----------+-------------- | | | | | | | | | LED-1 LED-2 LED-3 power | | | source | | | | resistor-1 resistor-2 resistor-3 | | | | | | | | +----------+-----------+-----------+-------------- But if all the LEDs were of the same voltage and amperage (e.g., if they were all of the typical 3.6V 20mA), isn't there a simpler way to do this by using only ONE resistor? In other words, can't it be done like the following ? +--------------+-----------+-----------+--- | | | | | | | | power LED-1 LED-2 LED-3 source | | | | | | | | | | | +---resistor---+-----------+-----------+--- And what would be the proper value of the resistor? (I'm guessing it would be one third of what it was in the first diagram because the required voltage reduction would be the same but the effective current draw of the 3-LED assembly would be three times what it was before. So if the power source were 6V and the LEDs were 3.6V and 20mA each, then the required resistance would be 120 ohms in the first case and 40 ohms in the second case. Correct?) | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Sam Goldwasser on July 31, 2004, 1:03 pm
Please log in for more thread options Or if the supply voltage is higher than the sum of the LED voltage drops AND
they have similar current and light output ratings, put the LEDs in series with a single resistor. With the example you gave though, there is no choice but to have individual resistors for each LED. --- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/ Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/ +Lasers | Sam's Laser FAQ: http://www.repairfaq.org/sam/lasersam.htm | Mirror Site Info: http://www.repairfaq.org/REPAIR/F_mirror.html Important: Anything sent to the email address in the message header is ignored. To contact me, please use the feedback form on the S.E.R FAQ Web sites. > When using an LED from a power source whose voltage is higher than the
> rating of the LED, a resistor is typically used. > If more than one LED is to be used (let's say three), > each one would have its own respective resistor, > and the schematic would look something like this ... > > > +----------+-----------+-----------+-------------- > | | | | > | | | | > | LED-1 LED-2 LED-3 > power | | | > source | | | > | resistor-1 resistor-2 resistor-3 > | | | | > | | | | > +----------+-----------+-----------+-------------- > > > But if all the LEDs were of the same voltage and amperage > (e.g., if they were all of the typical 3.6V 20mA), > isn't there a simpler way to do this by using only ONE resistor? > In other words, can't it be done like the following ? > > > +--------------+-----------+-----------+--- > | | | | > | | | | > power LED-1 LED-2 LED-3 > source | | | > | | | | > | | | | > +---resistor---+-----------+-----------+--- > > > And what would be the proper value of the resistor? > > (I'm guessing it would be one third of what it was in the first diagram > because the required voltage reduction would be the same but the > effective current draw of the 3-LED assembly would be three times what it > was before. > So if the power source were 6V and the LEDs were 3.6V and 20mA each, > then the required resistance would be 120 ohms in the first case > and 40 ohms in the second case. Correct?) | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Michael A. Covington on July 31, 2004, 9:28 am
Please log in for more thread options Put the LEDs in series, not parallel. That is:
+ -----////-----LED-----LED----LED----- - They will all carry the same current but need not have the same voltage (e.g., you can have a 3.2-volt blue LED, a 2.1-volt green, and a 1.8-volt red, all in series). The power supply must be higher than the sum of the LED voltages, with the resistor calculated to make up the difference. Example: With the 3 LEDs I just mentioned, 3.2 + 2.1 + 1.8 = 7.1 volts. Suppose the supply is 12 volts and you want 20 mA. Then the resistor is (12 - 7.1)/0.020 = 245 ohms (use 270, which is a standard value). As someone pointed out, if you put the LEDs in parallel, they will not share current equally unless their voltages are *exactly* equal, which depends on temperature, manufacturing lot, etc., even if they're all the same color. | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Byron A Jeff on July 31, 2004, 12:07 pm
Please log in for more thread options >When using an LED from a power source whose voltage is higher than the
>rating of the LED, a resistor is typically used. Actually you still need the resistor even when the voltage is at the rating of the LED albeit a small one. It serves the purpose of current limitation in addition to dropping voltage. >If more than one LED is to be used (let's say three),
>each one would have its own respective resistor, >and the schematic would look something like this ... > > > +----------+-----------+-----------+-------------- > | | | | > | | | | > | LED-1 LED-2 LED-3 >power | | | >source | | | > | resistor-1 resistor-2 resistor-3 > | | | | > | | | | > +----------+-----------+-----------+-------------- Right. It's the correct way to get uniform brightness among the set. >
>But if all the LEDs were of the same voltage and amperage >(e.g., if they were all of the typical 3.6V 20mA), >isn't there a simpler way to do this by using only ONE resistor? >In other words, can't it be done like the following ? > > > +--------------+-----------+-----------+--- > | | | | > | | | | >power LED-1 LED-2 LED-3 >source | | | > | | | | > | | | | > +---resistor---+-----------+-----------+--- It serves the safety purpose, but then you get variability on the brightness because each LED will not draw exactly the same current. So some will be dimmer than others. The second thing is that the total current through the resistor will be a factor of the current through each LED. So for your example the resistor above would need to allow 60ma of current to flow through to feed the LEDs downline. The problem is that if one LED blows out then that current will have to be pushed through the other working LEDs, which can cause the others to fail also. >And what would be the proper value of the resistor?
Single resistor value divided by the number of LEDs across it. >
>(I'm guessing it would be one third of what it was in the first diagram >because the required voltage reduction would be the same but the >effective current draw of the 3-LED assembly would be three times what it >was before. >So if the power source were 6V and the LEDs were 3.6V and 20mA each, >then the required resistance would be 120 ohms in the first case >and 40 ohms in the second case. Correct?) Right. But see the caveat above. A better way to do it is to bump the voltage up and put the LEDs in parallel like this: +V -> resistor -> LED1 -> LED2 -> LED3 -> GND
You treat the equation the same as a single LED except that you add the Vf voltages of the LEDs. So for your example the total voltage of the LEDs would be 10.8V. So the resistor you'd use with a 12V +V would be: (12V - 10.8V) / .02A = 60 ohms. But there are significant differences: 1) Each LED will draw the same current, so you'll get uniform brightness. 2) If a single LED (or the resistor) fails, then the whole string goes out with only the single failed component, instead of going into cascade failure due to increasing overload. When doing a single resistor, that's the better way to handle it. BTW, thanks for crossposting this message. There are many who don't understand the concept. But I think I'll limit my replies to basics and misc, which are appropriate. BAJ | |||||||||||||||||||||||||||||||||||||||||||
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> likely have a situation where one would light normally, and the rest would
> be dim, or not lit at all.
>
> You need a separate resistor for each one, or you should look for LED's with
> built in resistors. If they have built in ones, then you will be restricted
> to the voltage range indicated for the LED's.
>
> Jerry G.