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LED blacklight voltage Q

Downunderwonder

Downunderwonder

Aug 26, 2009
23,712
20,456
New Zealand
#1
I bought some 3.1V LED bulbs. Seller says I need a resistor to use them on 9V. I would have thought I put 3 of them in series and put 9V to them for 3V each:

Thanks for buying my items.

Just want to let you know that you definitly need resistors for your application. Each LED only drops the voltage by 0.7V, so the 'lightest' LED in the chain of 3 will still be getting 6.9V, not good :(

Anyway I will include 3 130 ohm resistors on the house, ideally if you can drop the voltage or obtain some resistors close to the value of 200 ohm (and still do the R-LED-LED-LED) that'd help out the circuit's reliability big time. Oh and these go open circuit from memory once blown.
Click to expand...

He's talking about a series of Resistor-LED-LED-LED getting 9V. Did he flunk highschool physics or did I not remember it right? Maybe they are really 0.7V LED?
 
#4
Downunderwonder said:
Yeah, that guy knows a lot more than when he started.

I think it comes down to what voltage the LED's I've bought actually need. One minute they are 3.1 next they are 0.7
Click to expand...

Well, there are no 0.7V LED:s. Red LED:s are usually around 1.6V, green and yellow around 2V, blue and white a bit over 3V. Variations are large, though. You will need at minimum one resistor to limit current. With three 3.1V LED:s in series 9V might not be enough, so you might need to go parallel.
 
#6
Downunderwonder said:
9V is pretty darn close to 9.3V isn't it?

Surely they don't "brown out" over missing 0.1V from 3.1V?
Click to expand...

Voltage is only a part of the equation. The resistors are there to limit current.

The voltate rating if the LEDs is the on-voltage, which is the voltage required to make the device emit light. There may be some imprecision there, so 3.0v may or may not be enough to light a 3.1v LED. Without a resistor, even if the devices are getting their proper voltage, you will still be vastly overpowering them in terms of amperage and wattage.
 
#8
It seems like either I'm missing something or you guys are simpletons.

If I need to chuck a resistor in series it's because the LED would burn up trying to dissipate the applied voltage. If I only apply the rated voltage I don't need a resistor. Right??

Am I missing something or misinterpreting V=IR ?
 
#9
Simpletons?

Go ahead and burn up your LEDs.

~

Yes, E = IR

But also, I = E/R

Without enough R (resistance), you are getting too much I (current), and you will burn up your LEDs. So you add a resistor of the appropriate value in series with the LEDs to limit the current.

Yay math!
 
#11
Downunderwonder said:
Am I missing something or misinterpreting V=IR ?
Click to expand...

For a diode, R (and therefore I)depends on V so it's a little more complex than that.

iv_charac_3.jpg

Current versus voltage for a typical silicon diode.


(Note: for your LEDs, the 0.7V point is actually 3.1V)

If diodes had constant resistance then this curve would be a straight line.

This means that LEDs aren't like regular bulbs. They have a minimum voltage for them to "turn on" and pass current. For yours, this *might* be 3V, but it might be a little less or a little more.

If it's a little more, then you won't be able to run 3 in series.

If it's a little less then you may have some voltage "left over", depending on whether your battery is really 9V exactly. If you have voltage "left over" and no series resistance, the only thing limiting the current is the internal resistance of the battery, which isn't going to be that high.
 
#12
If a regular bulb gets 10% more voltage it only passes 10% more current because it's a resistive load. (Actually even less than that).

If an LED gets 10% more voltage it can pass 3x as much current, depending on where the initial voltage was set.
 
#13
Downunderwonder said:
I conceded I may not understand it. Please explain why I'd burn them up if I'm only applying the rated voltage per LED.
Click to expand...

Because without a resistor to limit current, the low-impedence LEDs will not limit current from the battery to a level that is safe for them to continue to operate.

They may get the right voltage, but way too much current, which will cause them to burn and fail.
 
#14
Thanks, that's a pretty big gap somewhat filled.

I'm not quite sure where the compressed and expanded scales on the graph leave it but it seems you're saying the resistance of the LED bulb is neglible of itself and the "rated voltage" is just one that turns it on.

How do you figure what resistance to throw in? It all seems a bit bogus. Put too much resistance in and the applied voltage to the LED drops to less than the rated voltage. Put too little and the LED dies if the voltage goes over the rated voltage.
 
#15
Downunderwonder said:
Thanks, that's a pretty big gap somewhat filled.

I'm not quite sure where the compressed and expanded scales on the graph leave it but it seems you're saying the resistance of the LED bulb is neglible of itself and the "rated voltage" is just one that turns it on.
Click to expand...

The compressed/expanded stuff is for the "reverse biased" case, which isn't important here. Ignore it for now...

Downunderwonder said:
How do you figure what resistance to throw in? It all seems a bit bogus. Put too much resistance in and the applied voltage to the LED drops to less than the rated voltage. Put too little and the LED dies if the voltage goes over the rated voltage.
Click to expand...

Normally you work backwards from the current you want. As a simple example, say you're working on just one of the 3.1V LEDs, the desired forward current is 10mA, and the supply voltage is 9V.

Voltage across series resistor = 5.9V (9V - 3.1V)

Current through series resistor = 10mA (same as through LED, series circuit)

Series resistor = 5.9 / 0.01 = 590R

But what if we get it wrong and the LED forward voltage is actually 2.9V?

Well, the current goes to 6.1/590 = 10.3mA because the resistor is there and it's limiting the current. So we're still pretty safe.
 
#16
Well, if your LEDs are rated for 3.1v, you probably won't be able to run 3 of them in series from one 9v battery. You may have to run them in parallel. If you run them in parallel, each LED will need a 330 ohm resistor in parallel, assuming a forward current rating of 20mA (which is common) for the LEDs.
for
This site:

Link Removed

is good for this sort of thing.
 
#19
From the first answer on a google search for "UV LED forward current":

Emitted Colour : UV (ULTRA VIOLET)
Size (mm) : 5mm T1 3/4
Lens Colour : Water Clear
Peak Wave Length (nm) : 380 ~ 400
Forward Voltage (V) : 3.2 ~ 3.6
Typical Voltage (V) : 3.4
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : 2000 ~ 3000(Max)
Life Rating : 80,000 ~ 100,000 Hours
Viewing Angle : ±10°
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
Lead Soldering Temperature : 240°C (<5Sec)
Operating Temperature Range : -25°C ~ +85°C Preservative Temperature Range : -30°C ~ +100°C

For these (which won't be the same as yours) the key spec is "Max Continuous Forward Current". These ones take 30mA. Set your design to run at 80% of this for a bit of a safety margin and you end up at about 24mA.
 

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