LED question

[daz]

It's not battery powered tho. It's powered via the amp.

[Mark Black]

Can you measure the open circuit voltage at the stock LED location with both LED's removed, and then with just the stock one in place? And is one of the stock LED legs tied to ground or is it floating in the middle of the circuit?

Not knowing anything about the drive circuit makes things a little tougher. With an abundance of SMD components present they may have done something sort of fancy.

Edit: Of course the open circuit voltage measurement would have to be when the LED is supposed to be 'on'. :-)

[Enzo]

The LED will have a natural voltage across it. It is the current through it that determines its brightness.

A typical LED circuit is the LED and a series resistor, the resistor sets the current. If you remove the LED and measure voltage at its pads, the missing LED makes an open circuit, wwhich means zero current, and so there will be zero voltage drop across the resistor. SO under those conditions, you will just measure the power supply voltage to the circuit.

[dstrat]

Hello , first post here. been lurking for a while.
Everyone is making good points but ... current follows path of least resistance so yes the first led will be brighter.

[daz]

Hello , first post here. been lurking for a while.
Everyone is making good points but ... current follows path of least resistance so yes the first led will be brighter.

there is no 1st LED, they're in parallel.

[dstrat]

well the one closest to the source.

[The Dude]

As daz says, they're in parallel. Neither is "closest to the source."

[RODNEY]

Sounds like the original LED is a low current type so the "current limiting resistor" has been calculated accordingly which is to large for the newly added LED. Use ohms law....measure the voltage across the "current limiting resistor" (in volts) and divide it by the resistors' value (in ohms) , this will give you the actual current going through the LED so you can choose which type of LED you need (i.e standard or low current). Low current LED can show full luminosity down to about 2mA.

[daz]

Sounds like the original LED is a low current type so the "current limiting resistor" has been calculated accordingly which is to large for the newly added LED. Use ohms law....measure the voltage across the "current limiting resistor" (in volts) and divide it by the resistors' value (in ohms) , this will give you the actual current going through the LED so you can choose which type of LED you need (i.e standard or low current). Low current LED can show full luminosity down to about 2mA.

Thanks, i'll try that if needed. But for now i have found a LED that lights fully but i need to try a resistor with it because it is now allowing the switching to be reliable. And also the original pedal LED goes a bit dimmer when they're both on. So i'm going to put a trimmer in series with the one i just tried and turn it down to the point it's just bright enough and see if everything is good then. If not i'll try what you said. Thanks.

[daz]

Solved. Looks like the intermittent on/off switching i was experiencing wasn't the LEDs after all, it was the switch and i've taken care of that. So this LED is fine. Both light up plenty bright. Thanks for all the replies.

[Enzo]

Hello dstrat, welcome to the forum.

[Mark Black]

Glad to hear you found a solution!