ledraspberry pi
I am try to make an LED circuit that can be controlled with pything using a Pi Zero.
The problem is when I try to turn the LED off, it retains a faint glow (image).
Equipment:
I believe the circuit is wired correctly because to turn the LED "off" the output has to be set to 5V.
What's Happening:
Internal to the module, there are some diodes to protect the inputs. Typically these are ESD diodes, but they will conduct DC current if you reverse bias them. It's not a strong power source, and you have resistor limiting of it, so it's not very bright. The current flows from D1, into the driver VDD (weakly powering the module and LED), to the GND net, through the internal ESD diode, out to R1, and then to ground either through the DAC or R2. Note that this will make the DIM pin voltage less than the effective VSS of the module, which would command a full intensity. It may also be possible that the base-emitter junction is going into avalanche and your current is going that way. Typical V_eb max values are about 5V, and if the circuit worked you could be seeing up to 24V there.
When Q1 is shorted, that increases the current flowing through R1 by placing R_L in parallel with the module. This effectively reduces the voltage available to the module by a bit, and pushes the available supply below the minimum required for operation.
What should you do:
This module has an enable pin! You can get the desired functionality with minimal changes by putting a pull-up to 3.3V or 5.0V (either one works), and then using your microcontroller-enabled MOSFET. If you want to have the pull-up with the 24V supply, then you'll need to make a voltage divider to reduce the voltage (the PWM pin can't tolerate 24V). You could also get rid of R5 and replace it with a 3.0V Zener diode to be safe. Since most microcontrollers start with inputs disabled, you might be able to do away with everything except the voltage divider resistors. In that case, you could omit M1 and R2 in the schematic below, and directly connect your PIN5 to the enable.
simulate this circuit – Schematic created using CircuitLab
Other notes:
(source)
When looking at the connections for common Cathode / Anode LEDs you can see that once you connect the common node appropriately (i.e. Vdd/Gnd) then you just need to drive the RGB lines High for Common Cathode and Low for Common Anode to illuminate the LEDs.
UPDATE: If (as OP states) you're prepared to wire-link the common node to Vdd or Gnd then you just need to make the RGB drive signals selectable positive/negative logic (i.e. 1= ON or 0=ON).
You can do this easily with an XOR gate or a MUX.
(source)
Here the MUX selects line selects A or !A to so you can drive common Cathode / Anode respectively.
Looking at the XOR version you can see B is inverted if A=1 .
(source)
Best Answer
tl; dr: the LED has a leak path between 3.3V and 5V. Either don't use 5V for the LED, or add a transistor, MOSFET or true open-collector gate to isolate the GPIO and drive the LED.
What’s going on?
If you use a 3.3V GPIO pin to drive a 5V-tied LED, when the GPIO is high it will not be fully 'off': there will still be leakage current between the pin's IO protection and the LED.
Here's what that looks like, and how to fix it:
simulate this circuit – Schematic created using CircuitLab
You’d think that because 5V-3.3V = 1.7V is lower than the LED forward voltage, or Vf, that the LED would not conduct and thus not light up. This is not the case: there will still be some small leakage current that will result in a faint glow. This leakage has a name: subthreshold conduction, and is present in practically all semiconductor devices, including LEDs. Just tens of microamps can make it light.
A note:
How do I fix It?
The solution options depend on the LED's forward voltage:
All these solutions eliminate the leakage path.
Why use 5V at all?
Using 3.3V for a 3V-ish Vf LED (e.g., white, blue) isn't a good idea as it's very hard to get the current-limiting resistor right: with only 300mV or less IR drop in the resistor, a very small R value must be used, and so the LED current is very sensitive to forward voltage variation.