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Reply Mon 14 Feb, 2005 09:12 pm
Hmmm, i have got a circuit made up of 20 LEDs... and i am wondering whether do all the LEDs need a seperate resistor, or can i just put 1 resistor in the middle of the circuit?
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Type: Discussion • Score: 1 • Views: 907 • Replies: 8
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timberlandko
 
  1  
Reply Mon 14 Feb, 2005 09:58 pm
Yer gonna hafta do yer own math, but Don Klipstein's LED FAQ will give the formulas ya need to do it. There's all sortsa neat stuff and links to a buncha other neat stuff of interest to the electronics hobyist on his Main Page . Another fascinatin' site, with lotsa info and projects, is Bowden's Hobby Circuits . One of Bowdens handy toys is this LED Series Resistance Calculator, which'll do some of the math for ya.
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farmerman
 
  1  
Reply Mon 14 Feb, 2005 10:19 pm
gonna wire up a sensor circuit and impedance is a factor so BM
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roger
 
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Reply Mon 14 Feb, 2005 10:30 pm
You can't design it as a d.c. circuit, and get rid of that darn impedance?
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neil
 
  1  
Reply Wed 16 Feb, 2005 09:47 pm
If you are running the LEDs at considerably less than their maximum ratings and they are closely matched, one resistor in the common lead might be ok. It depends somewhat on the circuit. You might try a 100 ohm variable resistor (rated 10 watts or more), and reduce the the resistance slowly until the first LED lights. As you further reduce the resistance, instal 33 ohm or there abouts only in series with the brightest LEDs.
Obviously if the circuit turns off 19 of the LEDs, the one remaining lit will fail due to excess current, unless the remaining LED has a separate resistor. Neil
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stuh505
 
  1  
Reply Wed 23 Feb, 2005 06:58 pm
you have all brought something to the table, but I don't believe any of you have answered electricelectric's question yet:

Quote:
and i am wondering whether do all the LEDs need a seperate resistor, or can i just put 1 resistor in the middle of the circuit?


The answer is no, you do not need a separate resistor for each LED assuming they are all wired in series.

Becasue current in a circuit is caused by the flow of electrons, which travel nearly the speed of light, the obersvable effects in your circuit will be steady-state ones as opposed to transient ones.

That is to say, you will not observe the LED's light in series just because they are wired in series...rather, you will see them all light at once.

By the same token, since resistors add in series, having one resistor will be the equivalent of any number of resistors placed in the loop whose sum is the same.

Remember that an LED is itself a resistor.

If your LED's are wired in parallel, you can still evenly control the electric potential across each LED by using 1 resistor in series before you have any parallel connections. Alternatively, you could use one equal resistor for each parallel connection...but that would be a waste of parts and circuit space.
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DrewDad
 
  1  
Reply Thu 24 Feb, 2005 09:34 am
Electricity propagates at nearly the speed of light... Even in electric power lines the electrons themselves travel at about walking speed.
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timberlandko
 
  1  
Reply Thu 24 Feb, 2005 10:53 am
stuh505 wrote:
you have all brought something to the table, but I don't believe any of you have answered electricelectric's question yet...


Mebbe not directly, but I'll venture to say that among them, the websites I linked to not only more than answered the question but offered design info specifically pertinent to circuitry involving large numbers of LEDs. Of particular note in regard to the application at study in this topic would be multiple series-parallel configuration. Doesn't seem to me anybody here has picked up on that, so, since to my mind that configuration has distinct advantages over the so far here-proposed lashups, I figure I oughtta mention it.
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stuh505
 
  1  
Reply Thu 24 Feb, 2005 12:19 pm
Quote:
Electricity propagates at nearly the speed of light... Even in electric power lines the electrons themselves travel at about walking speed.


Good point...

I think the electrons still have very high velocity but the actual speed in the direction of current is low due to bouncing around etc
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