I'm stucked---power Efficiency of Embedded system

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Amyniuo
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I'm stucked---power Efficiency of Embedded system

Post by Amyniuo » Mon Oct 24, 2016 9:01 am

Hello!
I am designing an embedded system and met some problems.
I am hoping to achieve maximum power efficiency as the system will be running off of 2x CR1620 batteries. In addition to power efficiency, I would like to keep the BOM as low as possible.
The system consists of 3 primary components
1x 5mm RGB LED
1x PIC16F18325 (5v Processor) or PIC16LF18325 (3.3v Processor). PIC16F18325 datasheet
1x MMA8452q Accelerometer (I2C and 3.3v and as advertised is NOT 5v tolerant)
The LED will consume the vast majority of the power as it will be on whenever the system is in use.
A primary design consideration, the LED's must be AS BRIGHT AS POSSIBLE regardless of consumption, it is the rest of the system that I need to optimize.
Image
My questions are based on these scenarios:
Scenario 1 PIC16F18325 (5v version is able to be powered directly by the 2x CR1620 batteries) 3x BSS138 (N-Channel FET) 1- to power the accelerometer (3.3v) 2- SCL 3- SDA
Scenario 2 PIC16F18325 (5v version is able to be powered directly by the 2x CR1620 batteries) 3x Resistor voltage dividers to power the accelerometer (3.3v), SCL and SDA ****Note,
I'm not even sure if this works******
Scenario 3 Combination of Scenario 1 and 2, maybe use 1x BSS138 for power and 2x voltage dividers for SCL and SDA or 1x voltage divider for power and 2x BSS138 for SCL and SDA
Scenario 4 PIC16LF18325 (3.3v MCU) using a LM3940 to regulate the voltage from 6v to 3.3v (at approx 55%-65% efficiency since its a linear regulator) and connecting everything directly.
So my questions are these: Which of these scenarios will work? (I'm pretty sure 1 and 4 works, but not sure about 2 & 3) Which of these scenarios is the most power efficient? In scenario 4
will the LED have any difference in brightness being driven by 3.3v vs 5v? assuming I adjust the resistor value to drive the RGB LED at 20mA on each channel
Ideally scenario 4 would be the easiest and most reliable way to design the system, but I'm worried about all the loss of power from the linear regulator.
Image
Additionally I am worried the the brightness of the LED would suffer from 3.3v.
Measuring power is hardware dependent: some embedded processors provide internal measurement capabilities; processor manufacturers may also provide "power calculators" which give some power information;
there are a number of power supply controller ICs which provide different forms of power measurement capabilities; some power supply controllers called VRMs (voltage regulator modules) have
these capabilities internal to them to be read over peripheral interfaces; and finally, there is the old-fashioned method of connecting an ammeter in series to the core power supply.

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Kilrah
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Re: I'm stucked---power Efficiency of Embedded system

Post by Kilrah » Mon Oct 24, 2016 9:18 am

What you don't say is what your "thing" is supposed to do and how often.

The biggest non-LED comsumption factor is the processor, the slowest this can run and the more it can be put to sleep the less it will draw. But a quick look shows that your processor and accelerometer will draw less than 1/10th of the LED in the worst case, so going to lengths saving on that is already pretty much useless.

Also note that CR batteries are not meant to supply more than a couple of mA, so drawing 20mA from them will perform very poorly. Don't expect more than an hour or 2 of operation.

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jhsa
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Re: I'm stucked---power Efficiency of Embedded system

Post by jhsa » Mon Oct 24, 2016 9:58 am

Maybe the microcontroller could be put to sleep when doing nothing. That is what it happens on my magnetic switch project. It then draws 0.5uA. But it uses an ATiny85. Don't know about the Pic.

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MikeB
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Re: I'm stucked---power Efficiency of Embedded system

Post by MikeB » Mon Oct 24, 2016 11:10 am

Both the processor and the accelerometer will work down to less than 2V so these may be operated directly from a single 3V cell.
A CR1620 has a capacity of around 80mAh, so 20mA into a LED would flatten the battery in 4 hours, and probably much faster as that capacity is at a rated current of around 100uA.

Search for "High Efficiency" LEDs. You can get bright ones that take only 2mA.

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