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Perhaps a Simple Question, but maybe Im misunderstanding "Voltage" drop in terms of LEDS.

So I have a Circuit wired up with a 9V Battery, one 270ohm Resistor and 3 Blue LEDS (that draw around....20ma and 2.0v forward voltage. (this is for testing purposes, I would always put a resistor with each LED)

I put in 1 Led and Read the Ground and Positive Leg of the LED. Multimeter says: 3.52 V ok.....

2 Leds (same LED): 3.35v

3 Leds : 3.20v

enter image description here

It's hard to see on my super crude MS paint, but im labeling each LED with what the Voltage on the Multimeter reads when I measure that LED leg.

Whats going on here? the LEDS get slightly dimmer (and I mean VERY slight).....but shouldn't the voltage difference after each one be alot more?

Also the Entire System is drawing 30mA exactly (+/- .2-3 mA) regardless of How many LEDS I have on there(and im checking the mA after each additional LED leg to be sure). Clearly im missing something? Is this because the LEDS are in Series? I just kinda assume "Ok more LEDS = MORE mA drawn and voltage drops among each LED.

And I know theirs a Similar question about LEDS in Series, but im more curious about WHY there is no voltage drop or mA change per LED in Series than if it's a good or bad idea (Im sure it's not a great idea to keep LED in series like that of course).

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    \$\begingroup\$ 3.20v across what exactly? Across each LED, and all 3 the same? Or across the whole set of 3 LEDs ? \$\endgroup\$ – Rocketmagnet Apr 18 '12 at 21:42
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    \$\begingroup\$ Schematic please!!! I can't understand what's happening \$\endgroup\$ – clabacchio Apr 18 '12 at 21:47
  • \$\begingroup\$ I wish I could draw a schematic but I don't have any schematic drawing software. What I meant was On Measuring the Voltage of the First LED (with 3 of them on there) the first one shows 3.52v, when I measure the second it shows 3.35v and when I measure the third it shows 3.20v. They ALL show 30mA. EDIT: Im gonna take a picture RQ \$\endgroup\$ – user3073 Apr 18 '12 at 21:52
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    \$\begingroup\$ It's hard to see, but are those LEDs in parallel ? \$\endgroup\$ – Rocketmagnet Apr 18 '12 at 22:02
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    \$\begingroup\$ @Rocketmagnet, yes, I do think those look in parallel as well. \$\endgroup\$ – Kris Bahnsen Apr 18 '12 at 22:06
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You're getting the expected result. What you see is the normal behavior of diodes in series and it's completely normal to have one resistor and a string of LEDs connected after it.

What's basically happening is this: When they told you that the forward voltage is 2 V, they lied. It actually depends on the current going through the LED and you can consider the 2 V some sort of nominal value, but the exact drop should be read in the datasheet (if it's available).

In general case when you want to connect diodes in series, you use this formula for resistor:

$$ R= \frac {V_{supply}-NV_{f}}{I_{f}}$$ where the N is number of diodes you have.

This way it turns into simple Ohm's law. But in your case, you're approaching the border at which the above formula will not hold. You basically have a circuit with one branch only and the current going through that branch isn't going to much change with the number of LEDs if the voltage of the supply is high enough to be higher than LED forward voltage.

Take a look at this diagram from Wikipedia:

enter image description here

Notice the point marked \$ V_d\$. For this diode, once the voltage at the diode terminals reaches that point, the current will start quickly increasing with only a small change in voltage. That is why adding more LEDs doesn't immediately affect current. The voltage is high enough that all LEDs will conduct. Should you for example put 10 LEDs in series, the voltage will be too low and they will either show barely noticeable light levels or stay off.

Next, let's take a look at the different voltages you got at the LEDs. Again take a look at the curve for the diode from the Wikipedia. The \$V_d\$ point for each diode made is different and there are some tolerances here. So some diodes of same model number will at same current have a bit larger voltage drop and others will have a bit smaller voltage drop.

Next about LEDs in series. There is nothing wrong with that, but you're still not doing it right. Using the formula I provided, you should set the resistor so that the LEDs will be within their rated current. If you fulfill that condition, there's absolutely nothing wrong with having multiple LEDs connected in series, should you have voltage to spare.

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  • \$\begingroup\$ @AndreaKo - An excellent answer. \$\endgroup\$ – Rocketmagnet Apr 18 '12 at 21:59
  • \$\begingroup\$ Ok.....So im trying to understand this correctly. But basically what your saying is, until I reach some "Threshhold" point (IE like 10 LEDS or something) they are all going to conduct without much change in voltage, but once I go over that Threshold they will either be SUPER Dim or not conduct at all? \$\endgroup\$ – user3073 Apr 18 '12 at 22:05
  • \$\begingroup\$ @Sauron Yes, that's right. The threshold is related to the forward voltage drop of the diode and the voltage you have. Basically as long as the voltage of the source is larger than sum of the voltage drops at the LEDs, you're going to have some light. Adjust the resistor to control the amount of light. Once the source voltage reaches the sum of forward votlage drops of diodes, you're going too get to the extremely dim phase. So in your case, once you connect 5 LED such as the ones you have they should become very dim. \$\endgroup\$ – AndrejaKo Apr 18 '12 at 22:10
  • \$\begingroup\$ Ok apparently I had them in Parallel anyways, once I had them in series........they were dim and I measured each leg and it showed around a 2.3 voltage drop after each one. I guess I got parallel and series voltages confused. \$\endgroup\$ – user3073 Apr 18 '12 at 22:13
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    \$\begingroup\$ @Sauron I got it using the formula I mentioned. Drop at each LED is 2 V. Resistor is 270 \$\Omega\$ and source voltage is 9 V. So we get this: \$I=\frac{ 9-3*2}{270}\$ and that's around 11.1 mA. I didn't use a calculator when I obtained the 10 mA I posted. \$\endgroup\$ – AndrejaKo Apr 19 '12 at 17:18
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It's a little hard to tell from the photo, but it looks like you have the LEDs in parallel, rather than series. This is the difference.

Series or Parallel

I'm sorry if I miss read your photo, and this is patronising. You should connect the LEDs in series to work correctly.

Why did they work when connected in parallel?

Because the LEDs have some internal resistance, they will share current if connected in parallel. However, because you are working close to the steep part of their voltage/current curve, any slight variation in them may cause one to draw noticeably more current than the others. You can often get lucky and have 3 LEDs working in parallel. But you can just as easily be unlucky. And you may also find that, as the LEDs age, they do so differently, and that a year from now, one is brighter than the othes.

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  • \$\begingroup\$ Yeah you were right, once I made them in series they showed a 2.3 voltage drop after each one. How come Multiple LEDS in parallel are able to show the same voltage however? \$\endgroup\$ – user3073 Apr 18 '12 at 22:16
  • \$\begingroup\$ @Sauron - I added some explanation, but now that I think about it, I'm not sure that I have actually answered your question. You ask why they show the same voltage, but on your photo, you have labelled the LEDs with different voltages. \$\endgroup\$ – Rocketmagnet Apr 18 '12 at 22:26
  • \$\begingroup\$ What I meant was "Practically" the same voltage. Like there was only a .10 difference or so inbetween them, but now that i realize they are ACTUALLY in parallel it makes sense. \$\endgroup\$ – user3073 Apr 18 '12 at 22:39
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It seems you did not understand the equivalent circuit of a Light Emitting Diode, is a Diode. !! and have an effective series resistance.
The diode voltage for Blue and white is ~ 3.2V while red yellow ~ 1.6V. The ESR depends on current and size but is around 20~40Ω for 5mm. So use Ohm's Law to solve for current in each LED and assume they are matched from the same batch which are usually binned in 0.2V range. If you want to run 3 in series on 9V battery, it may be not be enough, but 2LED is plenty in series and use 20mA for one & 40mA for two LEDs so your current limiting resistor needs to change and increase in power dissipation.. ( Not reliable) consider Lithium coin cells 3.2V work well on 3.2V BLue/white LEDs direct drive. or for 3 LEDs in parallel use 3 R's in parallel for current limit. or 2 for 2, 4 for 4 etc

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  • \$\begingroup\$ I guess I just assumed more LEDS would be drawing more current? but I guess since im measuring each one individually.....they probably are all drawing near the same amount. is there any way to see how much current ALL of them are drawing together? \$\endgroup\$ – user3073 Apr 18 '12 at 22:08
  • \$\begingroup\$ yes.. ohms law on series R drop \$\endgroup\$ – Sunnyskyguy EE75 Apr 18 '12 at 22:10
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    \$\begingroup\$ A LED is a diode True, but not necessarily helpful. \$\endgroup\$ – m.Alin Apr 18 '12 at 23:17
  • \$\begingroup\$ If you think of diodes like battery voltage, it makes it simple to think of series parallel solutions and current limiting resistance. \$\endgroup\$ – Sunnyskyguy EE75 Apr 19 '12 at 0:37
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An LED is basically a diode (thus they have equivalent models in a circuit). Current through an LED will not be affected very much, and the voltage drop across an LED is some constant number (in your case, around 0.15V).

I think you might be confusing an LED with a resistor, which has a variable voltage drop due to the circuit / current provided.

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