Swapping a 2.2V on/off LED with a low-current 1.7V one

[mc50]

Hello, and nice to be here (it's my first post here),

I've recently built an Si Fuzz Face kit from Musikding. As far as I can tell, it came with this "regular" red LED, that the website claims is rated at 2.2V. At least that's what I can tell from the parts list.

This being a Fuzz Face and all I'd like to stay traditional and use a battery, and am thinking about swapping that LED with something like this one. That one only uses 2mA - but it runs at 1.7V.

Is it safe to Just Swap Them (TM), or should I also change a resistor in the schematic?

Sorry, I know this is probably a beginner question. That's 'cause I'm a relative beginner. 

Thanks!

[idy]

Welcome!

I would replace the resistor.  Called a "current limiting resistor" or CLR, the schematic shows 1.5k  That's pretty small, especially for a superbright. Some of us use 10k resistors on superbrights. The old fashioned value you see on schematics for old fashioned LEDs is 4.7k. For superbright I use 8....something or other k.

The long answer is find the spec sheet for your new led, which will tell you the maximum current it can take. They just say it "needs" .002amp. figure that if your LED is dropping 1.7 v the resistor is dropping the rest, 7.3v. Ohm's law E=IxR: 7.3=.002xR. 3.6k would give you that value, 2ma. That will give you the brightness specified on the data sheet...

I don't know the max current it can take, that would give you a minimum safe R value. The minimum for old fashioned LEDs at 9v seemed to be less than 1k....

What we usually want is the biggest R that gives enough brightness, without draining the battery. And those ultra bright can be blinding in the dark.

[iainpunk]

well, JUST swapping the LED's will give the new LED a higher current, instead of a lower one.
the original current is:
(9v-V_LED)/R_LED=I_LED
(9-2.2)/1500=4.5mA
if you keep the resistance the same with the new led, the current gets higher
(9-1.7)/1500=4.8mA
if you want that 2mA you'd have to flip the formula to:
(9v-V_LED)/I_LED=R_LED
this gives
(9-1.7)/0.002=3650ohm
if you throw in a 3k9 resistor, it would be fine.
or use a higher value if its too bright (i usually use a 2k2 and a 15k trim-pot in series)

cheers, Iain

[davent]

I had super brights that needed 47k with a 9v source.

Before you solder in the new led test it with various values of resistors and a 9v battery. Some of us have built testers for this with a rotary switch to switch between resistor to find what will be best. For a super bright I'd start with a 10k resistor, adjust bigger or smaller as you see fit.

Doing the math i usually end up with less than 1mA current for the super brights.

[mc50]

Got it, crystal clear now. Thanks!