deskthority

I have to desolder the leds in my poker to swap the stems to clear, so while I am there I might as well put some new ones in. What "specs" do I need?

From what I can see all LEDs seem to be the same. Is this right?

I like talking to myself .
Anyways, the leds turned up this morning and are way too big so I have had to order more.
If I soldered an led into the right ctrl key would it work the same as the one in the caps lock?

LEDs are not quite all the same - I'd hazard as guess that the Poker uses 3mm, maybe smaller - but you can soon measure the width. Other factors aside from colour are whether they're just regular LEDs or have a resistor built in for use with a fixed voltage - and whether they're low current or not.

Are there any resistors on the Poker PCB near the LEDs?

Bear in mind you must connect LEDs the right way round.

It does use 3mm. So I have ordered these
http://www.ebay.co.uk/itm/170701248245? ... 1497.l2649
I don't know about the resisters but I don't remember seeing any when I took it apart to put a heavier spring and grey stem in my spacebar.

I found this on hardforum so I guess it is using resistors.

Another thing thats interesting is that the resistor's on the PCB seem to be running the blue LED's at half of their rated brightness.

Are the leds I bought fine?

The resistor would be there to provide a voltage drop as the LED only wants just over 2V across it. If you have a multimeter you might want to check the PD at the LED solder points before inserting a new one and possibly causing damage.

If the Poker provides enough current the LED will glow brightly. If an LED gets less current it will glow less brightly. Those you bought are listed as 20-30mA, which means that if they get, say, 15mA they will be dimmer. The current flow is related to the voltage and the resistor value in place.

V=IR where V=Volts, I=current in Amps, R=resistance in Ohms.

If you have an LED requiring 20mA with a forward drop of 2.2V and your power rail is 5V, you'd connect a resistor in series with the LED according to the formula, to drop 2.8V across the resistor:

2.8=0.02*R

2.8/0.02=R

R=140 Ohms (if my coffee has kicked in properly!)

To give a little more brightness, you might use a 120 Ohm resistor, for example. A 100 Ohm resistor would run the LED with almost 30mA of current, which would be its brightest.

You could get away with a very small resistor in there, if you need to replace one - the above example works out at only 1/20th Watt, so a 1/8th Watt type would be both small and solderable with normal (ie, non-SMD) kit.

Anyway, this is conjecture on my part, and only for illustration.

If you have a multimeter to test the Poker with, that would be handy.

I do have a multimeter that came with a soldering kit ages ago. That would be damage to the leds right not the pcb?
I don't know how to use a multimeter other than for testing leds work. How would I check the board?

I think we've cross-posted! I'll get back to this in a short while, unless someone else want to carry on.

I have taken it apart. How can i check the the pd?

when i put
http://www.ebay.co.uk/itm/360220886983? ... 1497.l2649
Into my spacebar the arrow keys lights don't work.

It seems the Poker runs off 5V, 100mA. You should be OK to use a regular 3mm LED as there appear to be resistors located on the reverse of the PCB for the purpose. Perhaps you've installed the LED the wrong way round.

If you have diode tester on your meter, check the LED with it. If you put it one way round, it will light; the other, not. The black lead on your meter is where you put the flat edge of the LED, or the short leg, and that goes to ground/0V, however it's labelled on the PCB, if it is.

Yeah I got it to work on both the poker and the meter. But when I used it in the poker the other leds turned off ( If you press fn+space it will light up the space and right windows context and ctrl) When the spacebar was the red 5mm led the others didn't work. I have ordered the 3mm and they should me here on wednesday.

It may well not be as simple as 5V .. resistor .. LED .. ground, and in any case if you are changing the colour of the LEDs, you would probably want to use a different resistor value. Also, using very high brightness LEDs, you probably want a much larger resistor value (unless you want the LEDs very bright!).

Deck has some good info about different LED forward voltages and resistor values. And for their backlighting, there's a bit of the circuit that feeds the LEDs with a variable voltage, which you can see in the top-right corner of page 1 of the schematic, labelled BL-SUPPLY.

I want them bright because the plan is to put them under non windowed doubleshots from round 4.

Interesting, Soarer, thanks for the Deck schematic.

I should clarify I don't know the Poker's circuit, I was just trying to explain something generic about LEDs in circuit. The Poker may well be more complex, but as long as a replacement LED has suitable specs, it should work. It's just a question of knowing what those specs are.

Yeah, I suspect the Poker doesn't do anything complex, but it's worth checking (by measuring the voltage across the resistor and LED together, and seeing if it's a solid 5V or not).

Full backlighting uses so many LEDs that, with regular LEDs, it's worth doing something to reduce the power consumed. In the case of red LEDs with a 5V supply, over half the power just goes into heating up the resistors! An easy solution is to regulate a supply that's a little more than the forward voltage, and then use much smaller resistors.

That is a neat idea. Also, low current LEDs. Those ultra-bright ones are thirsty. Though I doubt low-current ones are going to be visible through doubleshots...

I am not planning on having full back lighting, Just going to replace the leds that are already there with red ones. Is there a way of not having to play around with the resistors and just swap out the blues for reds because my electrical and soldering skills are quite basic.

That depends... let's say the blues were being run at 5mA, and have a forward voltage (Vf) of 3.5V. And that the reds have a Vf of, say, 2V. The resistor fitted for the blues would be about R = V / I = (5 - 3.5) / 0.005 = 300 Ohms. Swapping in a red would make the current I = V / R = (5 - 2) / 300 = 0.01 Amps = 10mA. Twice the current.

In fact, twice the current whatever current the blues are being run at.

So, it depends on what value of resistors it already has, which determines the current, and whether doubling that current will be too much for your new LEDs.

(OK, so those Vf values are approximate, but they're pretty close).

This is my multimeter how would I check what I need to know so I'm/we're not working on guessed numbers. Then I can go find suitable leds.

i exchange some led on my deck ice 82 and use:

3mm LEDs - Farbe: Rot
Leuchtstärke: 1560 mcd
Spannung: 1,8 - 2,4V
Gehäuse: Wasserklar

from this german seller (more details)
http://www.ebay.de/itm/350358887238

Resistor - with the board unplugged, set the meter to Ω, 2000 range: put a probe on each end of the resistor.

Voltages - with the board on, LED glowing, meter set to V (on the left), 20 range:
1. put a probe on each end of the resistor.
2. put a probe on each end of the LED.

It doesn't matter which way round you probe for any of that, it will just read a negative voltage if they're the 'wrong' way round (which doesn't bother us for this).

The voltage is 2.6 but when I set the multimeter to the ohms setting it just says 1

It probably uses a 1 (on the left of the display) to indicate infinite resistance - if you put the two probes together it should show 0 or close to 0 (on the right).

BTW, we want two voltages

OK got a reading on the multimeter is 1511

That sounds high-ish, but good news for your swap. That is with the board unplugged, isn't it?

Oh, and what is the other voltage?

Dammit I didn't check the voltage on the resistor and it's back together now. It will have to wait until tomorrow night.
And the resistance was with the board unplugged (it was done on the esc)

OK. If we assume that the voltage across the resistor is 2.4V, to make 5V in total, that would be I = V / R = 2.4 / 1511 = 0.0016 A = 1.6 mA. Sounds feasible, for a high-efficiency LED.

The 2.6V across the LED sounds low, but that's good for swapping in red LEDs - so far it's looking good for a straight swap without changing the resistors

Thank you. I will check the voltage across the resistor tomorrow afternoon.