Author Topic: I'm running 9 LED's off 21 volts with only a 100 ohm resistor in series.How can  (Read 2415 times)

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Offline raspberrypiTopic starter

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The circuit is 100Ohm then into 9 various color size LED's in series. It runs off of a variable LED driver PSU. The voltage off the neg terminal and the first positive lead of the  LED is 21 volts then it jumps down a few volts after each LED. Why doesn't the 21 volts fry the LEDs or at least the first one? What would happen if I did this in parallel? and parallel with out the resister paralleled? I though to power a bunch of leds you paralleled them with 3 volts and gave them as much current as they desire as they wont take more then then need. But with the volts they dont have the option of using less volts.  :wtf: I think I'm also understanding how the electrical 110 in your house can get out even when you use one side/leg of the 220V breaker more then the other.
I'm legally blind so sometimes I ask obvious questions, but its because I can't see well.
 

Online IanB

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Remember that voltage is a difference between two points. It does not exist by itself. So if you put your voltmeter across the legs of each LED, what voltage do you read for each one?

And if you measure the voltage across the 100 ohm resistor what do you see there?

Now by Ohm's law, what current is flowing through the resistor?

If everything is in series, what current is therefore flowing through each LED?

What current should flow through each LED to avoid blowing them up? (It probably should be less than 20 mA.)

Are the LEDs safe with the current they are actually passing?
 

Offline kg4arn

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https://en.m.wikipedia.org/wiki/Light-emitting_diode

On average each of your LEDs has a 2.3 Volt drop while on.  Look at the graph of the current vs voltage for an LED in the above reference. What you are observing is just the way they work.
 

Offline Wilksey

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If an LED has a voltage drop across it of say 3V, each LED will drop that input voltage by 3V if they are in series, so by the time you get to the last LED the voltage to plug into Ohms law has significantly dropped, so your 100R resistor will sufficiently limit the current to allow <=20mA to flow through the LEDs.

If you put the LEDs in parallel, with the 100R resistor, you will suddenly get around 200mA through the LEDs, in turn blowing the arse off them.

Which is why, when you see some circuits for Christmas trees or Valentines hearts they use a certain number of LED's, a 9V battery and no resistor, this is because the voltage drop (also called Forward Voltage, Vf) is significant enough to not require one, things like batteries have internal resistance, so bear this in mind also.
 

Offline Seekonk

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If the numbers don't come out right, that is because different color LED have different forward voltage.
 

Offline CraigHB

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I've got some Cree lighting LEDs in my stock of parts for projects.  The white ones drop around 3V @ 20mA where the red ones drop around 2V @ 20mA.
 

Offline SaintGimp

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It can feel non-intuitive to reason about this stuff at first, because it feels like voltage and current ought to be completely independent things but of course Ohm's Law tells us that they're not independent things at all.

You said "they [LEDs] wont take more [current] than they need".  The thing that makes that true is resistance.  Resistance limits current flow to a certain level, given a certain voltage, per Ohm's Law.  But an LED all by itself is not an ohmic device; the relationship between voltage and current in an LED is not linear so it doesn't predictably limit current flow all by itself.  It needs a resistor to do that job.

Water analogies are pretty useful here (as long as you remember that they eventually break down).  A circuit with a bunch of LEDs in series is like a bunch of small waterfalls on a river.  The energy you get from each individual waterfall is related to the height of that particular waterfall (not all of them together) times the amount of water flowing over the waterfall.  All of the individual waterfalls have the same amount of water flowing through them; it's not possible to have more water flowing over one than another because all the water must pass over all of them.  But if one waterfall is higher than the others, it will dissipate more energy than the others.  You can control the total flow of the river by using a valve (resistor) anywhere along it.

It can also be confusing because people writing beginner articles about LEDs might discuss "current-limiting resistors" or "voltage-dropping resistors".  Of course, Ohm's Law tells us that we can think of it either way and it's ultimately the same thing, but I've found that it's generally more intuitive in this case to think about it from the current-limiting perspective.  That is, you calculate all of the voltage drops that your LEDs give you, then you figure out how much voltage is going to be left over to get to ground, then you use Ohm's Law to calculate what size of resistor you need to limit the current to a safe level given that (leftover) voltage.

 

Online IanB

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I proposed some experiments for raspberrypi to perform to better understand what is going on. But there has been no response from the OP in this thread. Is this a case of a "drive by" posting?
 

Offline AlxDroidDev

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I don't think someone wants to build DUGA3 doesn't know Ohm's law.

Yeap! I've reading his posts and lost track of how many times he contradicted himself. That DUGA3 post (has it been deleted?) was some major trolling.

AFAIK, he's just a troll wasting our times.
"The nice thing about standards is that you have so many to choose from." (Andrew S. Tanenbaum)
 


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