# Simple? LED problem!

#### Electronics & Photo Fan

##### Dedicated Cruncher
Hi, can anyone help me on this?

I've got two Blue LEDs, 3mm, these.

Forward voltage (Vf): 3.2V typical
Current: 30mA

I'm using a supply voltage of 12V (surprise, surprise)

I've connected them in parallel, and used the formula R = (Vs-Vf)/Current,
to calculate the resistor. I doubled the current, so the total supply to the two LEDs would be 3.2V, 60mA.

I calculated the resistor to be 176ohms, so I used a potentiometer, and connected it up to a mains adaptor to test. The current through the whole arrangement was almost exactly 60mA, but the voltage across both LEDs is 3.9V. They are getting warm!

Turning up the resistance simply reduces the current, the voltage across both stays the same???

I am fairly experienced in electronics, and I'm a second year A-Level student.

This has not happened to me before, When you increase the resistance the voltage across the potentiometer should increase, so the voltage across the LEDs should drop! (as well as the total current in the circuit dropping.)

Anyone got any ideas? (I figured more brains are better than one!)

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#### floppybootstomp

##### sugar 'n spikes
Moderator
Have you tried connecting them in series?

It's years since I done this stuff, I forgotten most of it

#### Electronics & Photo Fan

##### Dedicated Cruncher
No, I haven't, I will have to try that in the morning.

I'm pretty thrown as to why it's behaving like this in parallel, though.

#### floppybootstomp

##### sugar 'n spikes
Moderator
Hmm, ok, it's late, but do diodes have a resistance? I can't remember.

If so, two in paralell = low, 2 in series = high.

I probably completely off beam though

#### Electronics & Photo Fan

##### Dedicated Cruncher
That's what I was thinking... I'll have to scout around on the net a bit.

Without going into math too much will this help I wonder. Its too late to start using a calculator

http://led.linear1.org/1led.wiz

Solution 0: 2 x 1 array uses 2 LEDs exactly

+12V --LED--LED--R = 220 ohms
The wizard says: In solution 0:
• each 220 ohm resistor dissipates 198 mW
• the wizard thinks 1/2W resistors are needed for your application
• together, all resistors dissipate 198 mW
• together, the diodes dissipate 192 mW
• total power dissipated by the array is 390 mW
• the array draws current of 30 mA from the source.
Does that help?

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#### Electronics & Photo Fan

##### Dedicated Cruncher
Ah. I may have found the solution - it appears the specs for voltage, current etc. Maplin posted on their website and in the catalogue are totally wrong. I downloaded the manufacturers' technical spec PDF and it says a forward voltage of 3.6V, current 20mA are needed.

Just going to try it now.

#### floppybootstomp

##### sugar 'n spikes
Moderator
Well, lol, that doesn't surprise me about Maplins, they're crap

Once, circa 1982 - 84, they were good.

Now, all they're interested in is selling toys.

#### Electronics & Photo Fan

##### Dedicated Cruncher
Lol yes that's about right! I recalculated and it now works fine:

3.65V across LEDs
42mA total current in circuit.

Nice and bright, too!

Faint burning smell, though, after a few minutes....
0.3W through a 0.25W potentiometer. Didn't even remember to think of that.

Anyway thanks for everyone's help!