Power LED questions

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arpruss
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Power LED questions

Post by arpruss » Wed Oct 11, 2017 3:06 am

Does anyone know how much current the power LED is using on the cheap dev board (black pill in my case)? Can I just safely take flush-cutters and cut the power LED in half to save current?

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RogerClark
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Re: Power LED questions

Post by RogerClark » Wed Oct 11, 2017 3:12 am

Blue pill has 500 ohms in series with the LED

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

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ahull
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Re: Power LED questions

Post by ahull » Wed Oct 11, 2017 9:11 pm

Removing the power LED on the bluepill boards is relatively easy. You might also consider removing the current limiting resistor instead, if it is easier to access, or if you are feeling particularly brave, substituting it for a higher value, though this will require considerably more skill than simply removing it.

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.
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RogerClark
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Re: Power LED questions

Post by RogerClark » Wed Oct 11, 2017 9:39 pm

Andy

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.

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ahull
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Re: Power LED questions

Post by ahull » Thu Oct 12, 2017 7:48 am

RogerClark wrote:
Wed Oct 11, 2017 9:39 pm
Andy

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.
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.
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Pito
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Re: Power LED questions

Post by Pito » Thu Oct 12, 2017 11:33 am

The aprox. current taken by an LED is following (3.3V power voltage, R is the series resistor):

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.
Last edited by Pito on Thu Oct 12, 2017 12:43 pm, edited 3 times in total.
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ahull
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Re: Power LED questions

Post by ahull » Thu Oct 12, 2017 12:27 pm

To put that in perspective, if the LED lights at 0.1mA, it will eat a CR2032 (nominal capacity of around 200mAh) in around 80 days.

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 :D
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arpruss
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Re: Power LED questions

Post by arpruss » Thu Oct 12, 2017 5:26 pm

I want to continue to power over USB, but I'd like to be able to run it from a phone via OTG without draining the battery too much. My SMD soldering skills are very rudimentary (i.e., I once managed to solder a SOIC-8), so I don't want to add any new components.

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.)

stevestrong
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Re: Power LED questions

Post by stevestrong » Thu Oct 12, 2017 5:55 pm

If the edge of the cutter is wide enough, it should be safe.
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MarkB
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Re: Power LED questions

Post by MarkB » Fri Oct 13, 2017 7:20 pm

arpruss wrote:
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.

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