Switching Power Supply or Linear Voltage Regulator (battery)

Hello all, newbie questions here.

Project description:
Arduino Nano is connected to a PIR sensor (HC-SR501), a LED, and a piezo buzzer.
When PIR detects motion, LED lights up and piezo buzzes.

Goal is to make the project portable by:

  • Bypassing the Voltage regulator
  • Eliminating the power led of the Nano
  • Utilizing the POWER DOWN mode and related Nick Gammon's super tutorial.
    So while the Nano sleeps, the PIR is powered all the time, waiting for motion to wake the Nano to light the LED and sound the siren.

Question:

  • HC-SR501 quiescent current is supposedly around 50μA.
  • Nano power down current will also be in the μA league by implementing above described plan.
    Initially I was under the impression that using a 9V battery and a buck converter to take 5V for both the Nano and the PIR would be super efficient.

By reading around the web I realised (I think) that the current consumption of the almost idle buck converter would cancel out all my other power saving efforts since the efficiency of these drops for low currents (experts are welcome to chime in :).

Then I saw a couple of tutorials regarding LDO linear voltage regulators like this (Low Power Arduino! Power Supply Considerations p1 - YouTube) and thought: That's it! I got it now...

But would something like this: http://ww1.microchip.com/downloads/en/DeviceDoc/21373C.pdf
work with 4 1.5V AA batteries? In my mind it would..

What do you guys think? Which would be the most enery efficient thing to do? Besides quitting the project, ha, ha!

Since the Nano, et al will be active only in short bursts, the inefficiency of a Linear Regulator is minor as a result.

But, lets do some math:

The product datasheet for an EverReady 9V 522 [Energizer MAX] battery [http://data.energizer.com/pdfs/522.pdf] indicates that, at 10mA continuous draw [the lowest current case offered], the battery has a capacity nearing 600mAh

A tabulation of Current:

  • PIR 50uA
  • LDO -- the MCP1702 has a quiescent current of 5uA
  • Nano [sleeping] -- unknown -- but, an Atmega328P in Power-Down sleep mode with the Watchdog Timer running draws an impressive 6uA! But, because of other components on the board, a more realistic number might be: 60uA.
  • LED -- zero, because it is only lit when the Nano wakes up
  • Buzzer -- zero, for the same reason as the LED

So, that's a total of 50 + 5 + 60 = 115uA

So, the battery should [theoretically] supply from 9V, to 4.5V, for this long: 590mAh/115uA = 5130 hours, which is 5130/24 = 214 days. Or, a little more than a half a year. That time period will of course be reduced by the thing waking up and flashing that LED and running that buzzer. But, hard to do any math, there, because of missing information, like amount of current drawn by the buzzer, and the LED, duration of LED flash, and buzzer sounding, expected amount of foot traffic, etc.. Also, as the 9V battery runs down, it's internal resistance will increase, and assuming a 9V battery is capable of supplying enough current to run everything, that might change as the battery drains, so length of service might be shorter because of this.

Here's some math using an EverReady AA E91 [Energizer MAX] battery [http://data.energizer.com/pdfs/e91.pdf]:

The AA case is a little harder to determine, since the capacity is specified to a discharge voltage of 0.8V, which would place a 4 cell battery voltage at: 3.2V. So, targeting a 4.5V minimum battery voltage, and estimating the capacity by looking the other graphs, I come up with a capacity of: 2200mAh.

So: 2200mAh/115uA = 19130 hours or 797 days or roughly 2.2 years.

Again, pulsing the LED and Buzzer will reduce that, but there's a LOT more leeway! But, AA's have a lot more current delivering capability, so greater chance of being capable of running that buzzer.

So, I would go with the AA batteries.

BTW: If you went with 6 AA batteries, that would be 0.8V * 6 = 4.8V as the minimum battery voltage as specified by the datasheet, so you would get the full 3000mAh, for a maximum operating period of: 3 years [less energy needed to flash & buzz].

Also, keep in mind, there are a few assumptions made here, so these numbers are, probably, only ballpark.

Also, I'm realizing now, that I forgot to factor in the LDO dropout voltage, which for the MCP1702 is anywhere from 50mV [or less] for currents at or below 50mA TO as much a 650mV for currents on the order of 250mA. But, a strategically sized capacitor at the output of the Regulator would take care of any Dropout problem, since this regulator goes to 0 dropout, at 0 current, so it's capable of charging a capacitor to the full regulator output voltage, even if the input voltage is the same. Of course, if that Buzzer is a current hog, that capacitor may need to be rather large to achieve this, and there's the issue of recovery time...so...

I'm puzzled by the PIR's datasheet, which says power consumption is 65mA. Maybe that's what it could source through the 3.3V output? I understood it to be a low-power device.

If you bypass the Nano's regulator, you may also have to cut its output pin so current doesn't flow back from Vcc through the regulator to ground. Also remember that you have a USB adapter that draws some power even when not connected to USB.

So for the final version, you may want to switch to a 5V Pro Mini, which has no USB adapter. If you remove the regulator and power LED, shut down everything but the pin interrupt (see another one of Darrah's videos for those things), you can get the Power_Down current down to about 0.4µA. So it would be that plus the quiescent current of the regulator plus the idle current of the PIR for the total sleep current. LEDs and buzzers are relatively high current devices, but it almost doesn't matter if they aren't on much, assuming you don't have constant foot traffic setting things off frequently.

If you could find a PIR and a buzzer that work with power of 3.2-4.2V, you could power a 3.3V 8MHz Pro Mini and everything else directly from an 18650 battery, with no regulator at all.

Another vote for the MCP1702 regulator. Typical quiescent current is about 2µA.

First of all thank you guys. Sharing your knowledge is much appreciated!

ReverseEMF, so buck converter (I have the Mini360) is out of the question you think? Thanks for the MPC1702 tip.

ShermanP, MPC1702 vote counted. Pro Mini will be the final choice. A 18650 and 3.3V Pro Mini sounds great and makes all problems dissapear, but in reality this project is a dummy project for me to comperhend and put in practice the low power stuff and the way to power them.. So I would like to understand which and why would be the best (battery duration-wise) solution if e.g. I had to use 5V.

I think you both clearly pointed one way. I will digest the info and return mightier!