I'm just wanting to check if I'm thinking this problem through correctly:
Let's say we have 3x AA batteries (let's each one is Alkaline, 1.5V/1500mAH) in series to yield a total of 4.5V/1500mAH, or 1.5V/4500mAH
These batteries are regulated down to 3.3V via an LDO to some small load. Let's say this load draws a constant 5mA at 3.3V for convenience (obviously more complicated than that in real life)
My understanding is the following:
The efficiency of the buck conversion in this case would be V_out / V_in = 3.3V / 4.5V = 73.3%
At 3.3V we would have a pro-rated 4500mAH x (1.5V / 3.3V) = 2045mAH which is the same as 1500mAH x (4.5V / 3.3V) if the 3x AA's are in series
This means the batteries should last (2045mAH * 0.733) / 5mA = 300 hrs
Again, please note that these calculations don't take into consideration changing voltages, etc. I'm just using a "nominal" value that should yield worst-case.
Is that logic correct or will the LDO also have it's own efficiency that I have to take into consideration besides the efficiency of the buck conversion? And should I consider using a switching regulator instead of linear regulator?
My understanding is the following:
- The efficiency of the buck conversion in this case would be V_out / V_in = 3.3V / 4.5V = 73.3%
Sorry, no, efficiency is based on Energy or Power In and Energy or Power Out.
Measuring voltage in and out will not give you converter efficiency.
It's going to help if you get the terminology right.
An LDO is not a buck converter. It's a linear regulator which throws power away as waste heat.
A buck converter IS a switching regulator. A step-down one. It's a lot more efficient than a linear regulator.
As a first approximation assume that with a good switching reg you will be able to use around 90% of the energy your battery contains.
Ahh OK, sorry, I was a bit loose with that terminology, thanks!
So it sounds like I need to consider both the "power efficiency" (73%) as well as whatever the LDO's efficiency is, which will further reduce the battery life.
Only problem with switching regulators is that they are noisier and have higher quiescent current. My impression is that they're mainly good for higher-current applications and not for low-power devices that draw only a little bit, especially if they're "sleeping" most of the time, because then the quiescent current will outweigh the high efficiency benefits.
Also, how would I find the efficiency of an LDO that doesn't have efficiency charts in the spec sheet? Just curious.
And if I measure the actual current being consumed by the device, that should account for all inefficiencies right? Like the regulator inefficiency, power inefficiency of stepping down, etc?
For example, if my device is measured with precision tools (like a uCurrent Gold or something) to draw exactly a constant 1mA directly from the 3x AA batteries, does that mean I can just use the 2045mAH capacity of the 3x AA's at 3.3V but not worry about the regulator inefficiency?
Use formula #1 in reply #1 to calculate the efficiency of a linear regulator.
OK, that equation does make sense. However, if I already measure the current of the system like this:
[device drawing power @3.3V] ---- [ammeter shows 1mA] ---- [3x AA's @4.5V total]
can I simply neglect the regulator efficiency because the actual measurement already takes that into consideration? Then since this power draw is directly from the batteries, should I only be using the 1500mAH since the AA's are in series? So basically,
battery life = (1500mAH @ 4.5V) / (1mA @ 4.5V) = 1500 hrs
Yes, that will get you the value you want.
(also - always take battery manufacturers' capacity claims with a grain of salt; they range from "accurate under specific and favorable conditions" (for reputable brands with detailed datasheets) to "total lies" (for chinese 18650's, where the number of the label has nothing in common with the actual capacity other than being measured in mAh))
Typically an LDO datasheet will specify a quiescent current (current drawn when no or virtually no load is present), and a second, higher current dependent on the current drawn from the regulator (sometimes called the "ground current" - the current through the ground terminal of the regulator, which represents power used by the regulator in the course of regulating); this may be shown with a graph.
For example, in the AP2114, the quiescent current is ~60uA, and increases almost linearly with current to ~300uA at 1A output. Quiescent current is shown on the electrical properties table, and total ground current (including quiescent current) is shown on the graphs on page 14.
Thanks! I think I have what I need.