# Power Efficiency of Embedded system

I am designing an embedded system and 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) 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.

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. Additionally I am worried the the brightness of the LED would suffer from 3.3v.

Any help and advice would be greatly appreciated =) Thanks in advance

• Have you done a test of using those batteries to drive just your LEDs with a simple linear regulator, to see how long they last? Even if you end up choosing a different regulator, the test result will not be different by much, from whatever you choose, especially since the test would not include the additional current for the MCU etc. The CR1620 batteries are not designed or rated for 3x20mA = 60mA continuous drain. Extrapolating (likely with great inaccuracy) from the published data, I guess the battery would reach their "end point" (2V as spec by the manufacturers) after max 10 mins [...] Apr 12, 2016 at 5:25
• [...] with that load, assuming 20 deg C - dropping more quickly at low temperatures. If you are expecting a much longer run-time than your test result from my suggested test above, then IMHO something has to change about the constraints (e.g. chosen LED behaviour and chosen battery type) to reach a longer run-time. Beware - you could get some unpleasant results (e.g. leakage, heating, or worse) when draining those batteries at 60mA - one CR1620 datasheet I just picked, listed the maximum continuous drain = 1mA. So if bad things happen, you have no recourse to the manufacturer :-( Apr 12, 2016 at 5:28

Using a lower voltage for the LED is mostly fine. You will waste less power in the resistor, but you trade this for less tolerance in matching the resistor and LED. The edge case is obviously where the resistor has to be 0 Ohms, and the forward voltage of the LED is 3.3V -- at this point, the U-I curve becomes fairly steep, so manufacturing tolerances will heavily affect light output.

Whether a linear regulator or a buck converter is more efficient for the processor and accelerometer depends mostly on the currents involved -- for small currents linear tends to be better.

You can usually mix 5V and 3.3V components with resistor dividers on the data lines (these are high-impedance, so the divider works), but you need a proper regulator for 3.3V power, and if you design that to be efficient, you are likely to be better off using a single voltage everywhere.

If all you need is full brightness, and cost isn't that much of an issue, the Allegro A4490 could work for you:

• using the voltage across the series resistor as the feedback voltage ($\frac{800\mathrm{mV}}{20\mathrm{mA}} = 40\mathrm{\Omega}$) gets you efficient current regulation
• turning the channels on with the high-impedance enable inputs reduces the current draw of the MCU, so you can use a linear regulator here (ideally, the MCU does not drive anything).

The downside here is BOM -- you need three resistors, inductors and flyback diodes as external components for that IC, and since your current demands are fairly small, it is most likely that there are better ICs out there.

• wonderful thank you =) this is exactly the answer I was hoping for! Apr 12, 2016 at 4:02