Need help understanding S7V8F3

Hello,

Background:
I am attempting to battery power an atmega328p with a PIR and RFM69 radio and I am wondering if using the step up/down regulator S7V8F3 makes sense. I am going to put the controller and radio sleep and wake them periodically or when the PIR registers motion.

I am trying to understand if you gain more overall battery usage with say, 3 standard AA batteries and the S7V8F3 or powering directly from AA batteries with no regulator.

From how I read the specs and understand how this works, the max quiescent current of the S7V8F3 is less than .1mA even with the regulator always on and without the SHDN pin driven low, correct? If that’s the case its seems the extra 1700 or so mAH plus being able to go down in voltage would net a significant amount of extended battery life.

Thanks

It is not totally clear exactly what you are asking and I am not sure how you calculated your 1700mAH figure. Could you explain where it came from?

It sounds like you want to use the S7V8F3 regulator in your system to be able to turn off power to your microcontroller and/or radio while you wait for your PIR sensor to detect. If that is the case, using the regulator will likely extend the life of your battery, since those parts will not be drawing current most of the time.

If instead you are just asking if powering all three components from that regulator will let you get more energy out of your battery, that is a little trickier. The regulator might allow you to discharge the batteries to a lower voltage and get more energy out of them; however, that might not actually get you a significant increase in battery life because of the shape of the batteries’ discharge curve, and the regulator introducing its own additional losses.

-Claire

That helps. I think I see why my post is confusing. I think I was incorrect in thinking that by having three AA batteries in total as opposed to just two AA batteries I would gain more mAH (1700mAH being the assumption assuming a single AA battery is 1700mAH). After reading some more that does not seem like that is the case? Adding an additional AA battery will net a higher voltage but not provide me the additional capacity of another battery, correct? I don’t need the extra voltage from the third battery for my project, as it only requires 3.3v to power all devices.

Your last paragraph is what I was originally asking. I understand the regulator will add its own additional losses but under my original flawed thinking, any regulator inefficiencies would perhaps be outweighed significantly by the capacity from an additional AA battery (does not seem things work that way).

I was planning on just powering all three components from the regulator and using a sleep library to put the micro controller and radio to sleep while not in use. Given all this, it appears the regulator may not actually net any additional battery life. I suppose an advantage to using it would be a more consistent voltage.

I understand how to measure current draw of a component such as a sensor with a multi meter but how do I measure the additional current draw of the regulator itself, both under load and quiescent current, so I can see what the additional losses of using it will be? If I can measure that, I can try a few different types out and see what works best.

Thanks for the help and sorry for the confusion.

Batteries in series don’t increase amp hours, but you do get more watt hours.

Adding an extra AA cell will not increase the mAh of your battery pack, but it will increase Wh (the total energy available from the supply). By adding a switching regulator that lowers the voltage, you will also lower the current drawn from the battery pack, so you will get an increase in battery life with 3 cells compared to using 2 cells.

It is not really practical to measure the current drawn by just the regulator. Instead, we generally characterize the overall efficiency of our regulators at different operating points, and that information can be found in the graphs on the product page. Using the efficiency and an estimate of the current draw of your system you can calculate the current that would be drawn from the batteries if you use a regulator with the following efficiency equation:

eff = (Vreg × Isys) / (Vbat × Ibat)

Then you can calculate the time the batteries will last in both the regulator and batteries only setups by dividing the capacity (mAh) by the current drawn from the batteries.

-Claire

Ok, perfect. That equation helped so much, thanks!