# Perhaps a silly Question

Discussion in 'The Projects Forum' started by iONic, Oct 20, 2010.

1. ### iONic Thread Starter AAC Fanatic!

Nov 16, 2007
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If I were to run a constant current source of 20mA through 2 series LED's rated at If = 20mA, then am I correct to assume that the voltage across each LED would be aprox. 3.2V each as the datasheet suggests, even though I have a initial supply voltage of 9V?

In other words do I still need a resistor to drop the remainder of the voltage even though theoreticaally the LED's should not be overdriving?

??

Thanks

2. ### beenthere Retired Moderator

Apr 20, 2004
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Of course you need the resistor. By the figure you post, the total voltage drop is 6.4 volts across the LEDs in series. With a 9 volt battery, that leaves 2.6 volts. For 20 ma, divide the voltage by the current and get 130 ohms for the current limiting resistor.

3. ### steveb Senior Member

Jul 3, 2008
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Yes.

Although not necessary, it might still be a good precaution to put a current limiting resistor that prevents the current from going over the maximum rating, which would be higher than the operating value (see the datasheet). You don't want to drop all of the remaining voltage, but leaving the full 2.6V of headroom can result in blowing the devices if there is a circuit problem.

beenthere, I think you missed the fact that he is using a current source and not a direct connection to the voltage.

4. ### hgmjr Moderator

Jan 28, 2005
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beenthere,

Since ionic is proposing the use of a true constant current source wouldn't that preclude the need for a resistor. The constant current source should be able to insure that both LEDs get the same 20 mA current drive and the total voltage drop across the LEDs would be 6.4V based on the assumed Vf of the LEDs. The current source would dissipate 52 milliwatts.

hgmjr

5. ### SgtWookie Expert

Jul 17, 2007
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Not a silly question at all, and the answer really needs some qualification.

If it is a true constant current source, then a series resistor should not be needed; it would only dissipate power.

Caveat emptor; many switching supplies have capacitors on the output stage; values exceeding 1000uF are not uncommon. If you have a current-limited voltage supply, and attempt to connect the LEDs when the supply is already powered up, you will have a LARGE surge current through your LEDs until the output capacitors discharge the excess voltage. This will likely fry your LEDs.

Ideal current sources have infinite output impedance; they will keep the current constant no matter what the output voltage is. Switching supplies by their very nature don't have such high impedance outputs; generally they are very low-impedance.

Ideal voltage sources have zero output impedance; they will keep the voltage constant no matter how low the impedance of the load, as long as it is non-zero.

If you want to be safe about it, discharge the output of your supply (and keep it shorted out) prior to connecting the LEDs, then once they are connected, remove the short.

6. ### beenthere Retired Moderator

Apr 20, 2004
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Yes, I missed the current source part.

7. ### Kermit2 AAC Fanatic!

Feb 5, 2010
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R = 1.25V / I

Example: We will supply 3 lumileds 1W power rated in serie with a 12V battery. The nominal current for the leds is 0,35A We can find the proper resistor value with the formula above:

R = 1.25V / I

R = 1.25V /0,35A = 3.57 ohm

Thus we need a resistor of 3.57 ohm but will not find one with this value. To solve this problem take a value that's higher. Here in the example we will take one of 3,9 ohm.
The real current will be then I = U / R = 1,25V / 3.9ohm = 0,32A what not will be a problem (it extends the lifetime of the leds)

Last edited: Oct 21, 2010
8. ### DonQ Active Member

May 6, 2009
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The purpose of a resistor in series in a LED circuit driven by a voltage source is to allow only a certain current to flow. If you have a current source, you have already taken care of the 'certain current'. The current source takes the place of a resistor.

The circuit by Kermit2 is a simple but effective way to drive LEDs. As long as the power dissipated by the LM317 is within reason, this works across a very wide range of input voltages. This works even better than a resistor because as long as you have enough voltage over that required by the string of however many LEDs for the current source to work (about 2 1/2 volts total for the LM317 and R), the current stays the same even as your voltage changes, say as batteries run down. If you use a LDO regulator, you can reduce the voltage overhead even further.

9. ### iONic Thread Starter AAC Fanatic!

Nov 16, 2007
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Very nice info Sgt. The method I thought of employing was with the Tl341AC IC, T-92 pkg. I can not drawthe circuit out as I have no software on this new computer yet.
I was looking at figure 22 and figure 26. I'm having trouble understanding how toset the resistor values for operating a 20mA
LED. In particular the unnamed resistor. And from what I figured the Rcl would be 0.05 Ohms!!

http://pdf1.alldatasheet.com/datasheet-pdf/view/25402/STMICROELECTRONICS/TL431AC.html

Last edited: Oct 22, 2010
10. ### SgtWookie Expert

Jul 17, 2007
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Take a look at the LM317L; it's the "baby brother" of the LM317M and LM317T.

http://www.national.com/mpf/LM/LM317L.html

Basically, the LM317 is capable of up to 1.5A out; LM317M, 500mA; LM317L, 100mA.

The LM317L is available in a TO-92 package; and has a maximum power dissipation rating of 625mW.

One caveat of the LM317x is that they have a minimum 3v dropout when used as a current regulator. There is a 1.7v dropout from IN to OUT when used as a voltage regulator, and then Vref (voltage at OUT using ADJ as the reference point) is added to the 1.7v dropout. Since Vref can vary from 1.2v to 1.3v, you must assume the worst case (1.3v). 1.3v+1.7v = 3v total dropout.

Another caveat is that the LM317T requires a minimum 10mA current flow to provide guaranteed regulation; the LM317M and LM317L only require >= 5mA current flow.

There are LDO regulators available that won't require as much headroom. However, linear regulators are really quite inefficient; as much power is dissipated as heat instead of being delivered to the load.

11. ### iONic Thread Starter AAC Fanatic!

Nov 16, 2007
1,420
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Yes, I may even have a couple of them on hand. It may very well be the best option.

But take a look at my previous post as I added some info. Still would like to understand the comcept of this IC\Circuit and run a test or two.

i

Last edited: Oct 23, 2010
12. ### SgtWookie Expert

Jul 17, 2007
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OK, in your previous post, you mention Figure 22 and Figure 26, but the datasheet you linked to does not contain those figures!

The TL431 datasheet that I have from Texas Instruments DOES have figures with those numbers, but the circuits shown would not make sense for an LED current limiter.

Post a link to the exact datasheet that you are looking at, preferably on TI's site, so we can all be on the same page.

Apr 5, 2008
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Hello,

For a low current source you can also use a JFET or a LM234.

Bertus

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14. ### mike8675309 New Member

Sep 25, 2010
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Look at figure 2.1 in this blog post from Bill Marsden.

I've built the current source in that figure and it works great driving 4 superbright red LED's at 20ma with each led having a 1.7fv and a varied input voltage from 16vdc to 10vdc with no change in LED intensity.

Radio Shack has LM314T units in their stocking supply.

15. ### iONic Thread Starter AAC Fanatic!

Nov 16, 2007
1,420
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Please see my previous post for updated info...

i

16. ### SgtWookie Expert

Jul 17, 2007
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Ionic,
You linked too deeply into AllDataSheets.com, so when you go to the link, it does a search and comes up with ST Microelectronics' datasheet, which does not have figures 22 and 26.

It is better if you link to the current datasheet on the manufacturer's site. That way you know that you have the most current datasheet, and manufacturer's don't play silly games with links like AllDataSheet does.

Here is the Fig 22 from the NXP datasheet:

However, there is no figure 26, so I'm not looking at the datasheet you are looking at.

I'm not going to waste any more time trying to track down the datasheet that you are looking at. If you want an answer, then post either the datasheet you are looking at, or a direct link to the datasheet on a manufacturer's website, not AllDataSheet.

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18. ### SgtWookie Expert

Jul 17, 2007
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Gee, the formulas are right there, and Vref is specified in the tables.

I've attached figure 22. As you will see, the output current is not precisely constant, as it varies somewhat with input voltage. There is also a rather large minimum dropout of around 3.2v-3.3v, along with requiring more parts than an equivalent LM317L circuit.

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19. ### iONic Thread Starter AAC Fanatic!

Nov 16, 2007
1,420
68
Thanks for the info SGT. The drop-out for this TL431 is about the same for the LM317X when used as a current regulator. These Dropouts are way too large. I have various Linear Vregs that boast very small DO as Voltage regulators and will have to see what they might offer as Iregs.

I guess a good question might be: Is it better to use Voltage regulators as LED drivers or Current Regulators?

iONic

20. ### SgtWookie Expert

Jul 17, 2007
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Well, it's slightly higher for the TL431 than the LM317x.
For the LM317, you have Vin-Vout+Vref, which is at least 3v total.
For the TL431, you have Vref (2.495v) + the Vbe of the pass transistor, which is used as an emitter follower. Even under fairly light load, you'll have about a 0.7v drop.

Now, you can increase the linearity of regulation by increasing R1 somewhat, but that also increases the dropout.

Linear regulators are just plain not very efficient as current regulators (or voltage regulators either, for that matter). If you really want efficiency, you need to look at switching regulators or dedicated switching LED current controllers. Some are so simple that they only require an external cap, resistor, diode and a small inductor; and can be up to 97% efficient.