Resistor/LED relationship in a 3PDT true bypass setting?

[skiraly017]

What part does the resistor play in this situation? I have a true bypass pedal with no resistor tied to the LED and everything seems to work fine. Wondering if someone could clue me in? Thanks.

[petemoore]

It's probably setup so the current throught the LED doesn't burn it up.
 Is it a 9v effect, and are you sure there's not a current limiting R on the board?

[vanhansen]

There's a good chance the resistor is on the board rather than inline with the LED like we see in wiring diagrams.  I'm slowly starting to incorporate putting the resistor on the board in my layouts.

[Mark Hammer]

If the battery is in the autumn of itslife, and the LED is not especially bright, sometimes you CAN just connect the LED directly to V+, though you may notice the LED becomes warm to the touch.  There are 2 reasons why you'll want to have a current-limiting resistor between V+ and the status LED.  One is that the life of the LED may be compromised if it is fed too much current (and overheats - internal temperature is no different than applying your soldering iron to it too long).  A second is that the life of the battery may be compromised if too much current is being "wasted" on unneeded brightness in the LED.

There are so many different brightnesses of LEDs around, plus so many higher and lower visibility circumstances (e.g., in the dark against a dark chassis vs inbroad daylight, against a white chassis) that it is difficult to make a firm recommendation for a resistor value.  That being said, I have rarely seen a circumstance where a current-limiting resistor between the LED and V+ of less than 1k was called for.  If the LED is a more contemporary high brightness unit (let's say >2000mcd), you can often get away with a 10k resistor or greater (I've gone up to 18k) and still have quite usable brightness.

Although many folks are tickled with the aesthetic properties of those nifty chrome bezels for LEDs (like you see in many boo-teek pedals, along with a blue LED), I think there is much to recommend about the "classic" black plastic LED clasp.  The black surround provides a nice sharp contrast , against which even small changes in illumination are easily seen.  Stick that black surround on a similarly dark chassis and you can easily conserve a couple of ma in current consumption and stretch the battery life out by 10% more, even *after* switching to higher efficiency LEDs.

[skiraly017]

I'm at work at the pedal guts are at home, sorry if this doesn't make sense.

One lead from the LED is attached to the input jack and the other lead is attached to the 3PDT. There is an "unknown purpose" 3.3k resistor with one lead attached to the 3PDT and the other to the circuit board. The LED and 3.3k leads are not connected to the same lug on the switch. I cannot find the 3.3k listed anywhere on the schematic. Is the 3.3k resistor for the LED? Thanks again for the help.

[petemoore]

sounds like the ground is to the LED, then the other lead of the lED is to switch, then that connect goes through switch [when on], through the 3k3 then to V+ at the board.
 3k3 is right in the pocket for a likely LED current limiting Resistor value.

[skiraly017]

I kind of answered my own question along the way, but thanks for the help.

One more...is 8.2k too much resistor to use for the LED?

[Mark Hammer]

One more...is 8.2k too much resistor to use for the LED?

Depends on the supply voltage and LED.  If the supply voltage goes up, a larger value resistor is needed to provide the "target" current for the LED (and conversely, as supply voltage goes down, so should the resistor value).  At the same supply voltage, as inherent brightness of the LED (the mcd rating) goes up, a larger value resistor becomes affordable (in the sense that you don't HAVE to use it, but you can "get away with it").

As I noted, I've gone as high as 18k for a superbright illuminated by 9v at a brightness *I* thought was acceptable.  That same pedal may be woefully underilluminated for someone else or in their typical playing context, in which case I would drop down to 12k or so for that LED.

The simplest thing to do is to score yourself a 50k trimpot (or maybe a 50k you can't use because a lug snapped off or something), solder a 1k resistor in series, and vary the current from the supply to the LED until you reach what YOU feel is acceptable illumination.  Simply because human visual judgment is not an absolute kind of thing, I'd recommend starting from max resistance, and gradually decreasing the resistance of the trim/pot until you hit the magic zone, noting what the combined pot/resistor resistance is.  Then, start from max brightness (pot at zero ohms) and work your way in the *other* direction, until you hit what you feel is a decent brightness.  Measure again.  If the two measures agree, decide on the nearest common value resistor and use that.  So, if you find, going from dimmest to brightest, that a value of 3.43k is good, and you find that, going from brightest to darkest, a value of 4.27k is good, then using a 3.9k resistor is probably what the doctor ordered....for that LED.

[Ge_Whiz]

For a typical red LED, an 8.2k resistor would allow a current of less than 1 mA to flow from a 9V source. Normally, this would not give sufficient brightness to be useful. However, there are some LEDs designed to emit brightly at 1 mA (usually over a narrow angle). Typically, an LED resistor will be 1k or less.

However, there are also some LEDs made that have a built-in current-limiter. These can usually be connected to any voltage between about 3V to 24V without any separate resistor needed. One advantage of these devices is that they maintain constant brightness as the voltage supply decreases (e.g. as a battery goes flat).

[skiraly017]

Much appreciated!