http://wiki.stm32duino.com/images/c/c1/ … ematic.pdf
Not sure about the Black Pill, though its probably the same.
So the LED will only take 5mA.
If you want to remove it, I’m sure it would be easy to do with a soldering iron, as normally the heat transfers quickly though most SMD parts and they end up sticking to the iron and can be removed without too much trouble
The LED will probably light even at 1mA or less (some LEDs will produce light at well below 0.5mA, but it all depends on the type of LED in question). In theory, *any* current will produce some light, however the relationship between current and brightness is non linear.
The bluepills I used for battery operated projects all had the LED removed.
Another thing worth considering is replacing the voltage regulator with a more efficient one, or removing it all together and simply powering the bluepill directly from batteries. The later option is the only one that will allow you to run the STM32F103 in its most power efficient manner, since the regulator is not 100% efficient, and will still draw significant power from the battery, even if the STM32F103 is halted, and sipping a few uA.
I agree about removing the resistor, it would be my first option, as they remove very easily.
I wasnt aware that the regulator drew current though its output, but I’m not surprised that it does, since its not its normal mode of operation.
[RogerClark – Wed Oct 11, 2017 9:39 pm] –
AndyI agree about removing the resistor, it would be my first option, as they remove very easily.
I wasnt aware that the regulator drew current though its output, but I’m not surprised that it does, since its not its normal mode of operation.
It very much depends on the regulator, but generally they represent a small but significant resistive load. If you want to hit the low levels of current that the STM32F103 can snooze at, you need to remove all other significant drains on the battery.
Powering the board with the regulator from say 5V is obviously even less efficient. All of this matters if you want very long battery life (from say a data logger), or you want to power from something like a CR2032 coin cell.
I think there are a couple of other threads on here that explore this topic.
1. Red LED
I = (3.3V – 1.7V) / R
2. Yellow, Green LED
I = (3.3V – 2.0V) / R
3. Blue, White LED
I = (3.3V – 3.0V) / R
Example: Green LED and 500ohm resistor
I = (3.3V – 2.0V) / 500 = 2.6mA
From my measurements the modern LEDs lit nice with ~100 uAmps (or 0.1mA).
PS: Most devboard vendors do like ” a firework show”, therefore their LEDs lit such you get blind and thus you cannot read the silkscreen texts placed around.. Simply replace the resistor with 4k7 or something similar.
For every doubling of that LED current, you half the duration the battery will last.
If you remove the LED, obviously your battery will last a lot longer. That same CR2032 has a self discharge rate of around 2% per year, low enough to allow your device to sit sipping a few mA for a seconds per hour say, for a typical data logging, or IR remote control transmitter application, for several years.
As you can see the LED makes a significant difference in low capacity battery powered devices. Now you know why your TV remote doesn’t have a power LED ![]()
How about my thought about just taking flush cutters and cutting the LED in half? That shouldn’t lead to any problems? (I’ve done it in the past with a Digispark LED, but some time later the Digispark blew. Presumably unrelated.)
Learning by doing.
[arpruss – Thu Oct 12, 2017 5:26 pm] –
How about my thought about just taking flush cutters and cutting the LED in half? That shouldn’t lead to any problems? (I’ve done it in the past with a Digispark LED, but some time later the Digispark blew. Presumably unrelated.)
This would put a large mechanical load on the printed circuit traces, potentially causing them to separate from the board. If the LED is at the end of the traces, that might not matter.
The previous suggestion of removing the resistor with a soldering iron is preferable. One should be able to heat up both ends of the part until the solder melts and flip the part off the pads.
[MarkB – Fri Oct 13, 2017 7:20 pm] –[arpruss – Thu Oct 12, 2017 5:26 pm] –
How about my thought about just taking flush cutters and cutting the LED in half? That shouldn’t lead to any problems? (I’ve done it in the past with a Digispark LED, but some time later the Digispark blew. Presumably unrelated.)This would put a large mechanical load on the printed circuit traces, potentially causing them to separate from the board. If the LED is at the end of the traces, that might not matter.
The previous suggestion of removing the resistor with a soldering iron is preferable. One should be able to heat up both ends of the part until the solder melts and flip the part off the pads.
+1
Agreed. I wouldn’t take the risk. The traces on these boards are pretty fine, and the quality of the PCB material is unknown. If it were my call, I’d dab the resistor or LED off with a fine pointed soldering iron rather than risk damaging the board.
Removing the LED or resistor is relatively easy, compare with repairing the board, in fact if you want to make life even easier, dab a little flux on top of the component before hitting it with the iron.
If you pull off the traces, the board is probably for the bin, since a new board is only a couple of bucks, and that equates to about five minutes of my time. Repairing the board, assuming it were possible is going to cause more grief than its worth.


