To continue yesterday’s post, let’s go into that last puzzle:
Why does the 1x AA Power Board run out of juice 3 times as fast as a 3x AA battery pack?
The superficial answer would be: it’s one battery instead of three, so obviously it’ll last 1/3rd as long.
But that’s not quite the whole story…
The AA Power Board contains a switching regulator called a boost converter. Switching regulators are a lot more efficient than ordinary “linear” voltage regulators. They play games with energy conversion into electrical and magnetic fields. And by doing so, they mess with the rule that current is always the same everywhere.
But let me first explain what a linear regulator does:
Don’t laugh – that’s the essence of a linear voltage regulator: a variable resistor!
Well, it’s far more complex than that in reality. But the machinery inside a linear voltage regulator is all about wasting energy. The goal is to waste just the right amount to get the desired 3.3V on the output pin, regardless of changes in input voltage and current draw. Functionally, all the regulator does is continuously adjust its internal resistance to get the right output.
If you think about it, there’s in fact little else you can do with resistors. They exist to drop voltage, and by doing so, they generate heat, even if the amount is minimal and usually irrelevant.
The other type of regulator is the switching regulator. Huge topic, way beyond the scope of this post (and way over my head, in fact). I’m bringing it up because the AA Power Board uses a switching regulator to “boost” the voltage from say 1.5V to 3.3V.
So how does one boost voltage?
The hydraulic analogy would be to pump water from one level to a higher level using only water power (height and flow). There’s an ingenious pump called a hydraulic ram which can do that. The one I’ve seen in action works by letting water flow and then quickly interrupting that flow. All of a sudden, the water has nowhere to go and pressure builds up. All you need is an outlet pointing up, and the water will go there as only option.
The flow will quickly stop, so the trick is to repeat this cycle, and then – in a pulsating fashion – you actually can get water to climb up. Voilá, a higher voltage!
That’s also how the AA Power Board works. It contains an efficient switch, which pulses the current flow, and (in most cases) an inductor which transforms current changes into a magnetic field, and vice versa. The inductive “kick” is what makes it possible to play games with current vs. voltage without turning it all into heat.
But you don’t get anything for free. Apart from circuit losses, you lose the conversion factor w.r.t. power – drawing 10 mA @ 3.3V will require 30 mA @ 1.1V – with circuit losses increasing that slightly further. You can’t ever get more watts out of this than you put in!
So, roughly speaking, to get 3.3V from a 1.2V AA NiMH battery, you need to draw about 3 times as much current from the battery as what will go into the target circuit.
Which is why a 2000 mA single AA battery behaves roughly like a 650 mAh battery delivering 3.3V via the boost converter. With our circuit drawing 10 mA, that will last 650 mAh / 10 mA = 65 hours.
That’s roughly a third as long as the 3x AA battery pack. QED.
There are some losses and inefficiences. Even with a switching regulator, you should expect no more than 90..95% efficiency under good conditions. But this is nowhere near the inefficiency of a linear regulator with a high input voltage. On a 9V battery, the on-board regulator of a JeeNode will be less than 40% efficient.
Note that there are also down-converting switching regulators (called buck converters), and these do have much higher efficency levels. Even the AA Power Board is able to handle over 5V on its input, and still deliver a 3.3V output, by going into a buck conversion mode. In which case the input current will be less than 10 mA to deliver 10 mA @ 3.3V on its output – something a linear regulator simply cannot do.
Conclusion: if you want very power-efficient solutions, look carefully at what voltages to use for supplying your circuits, and use switching regulators when the voltage differential is substantial. Linear regulators can only drop voltage, and can only do so by wasting energy.
There is another benefit to using a boost converter: it lets you suck the last breath out of batteries. The AA Power Board can be used with batteries with only 0.85V or so left in them, and if kept connected and running, it’ll work all the way down to 0.6V or so. You can be assured that by the time an AA Power Board gives up, its battery will have been completely drained!
One last note about the AA Power Board: maximum efficiency is achieved with an input voltage between 2.4V and 3.0V, so if you want to optimize, consider using either 2 AA (or AAA) cells, or a 3V battery such as the CR123A, which is half the size of an AA and a great source of energy with about 900 mAh of oomph…
But just to put all this into perspective: if all you want is a good solid source of power for a JeeNode and everything attached to it, use a 3x (or 4x if you have to) AA battery pack. Or use power from USB.
Me, I’ll stick to single rebranded Eneloops @ 1.2 .. 1.3V.
I’m curious if you are achieving the efficiency specified in the data sheet. One of my company’s products has a 3V -> 5V boost converter and we are getting slightly less efficiency than the minimum the data sheet promised. (However, we used a different inductor than the specific recommended one, so that may be the reason.)
Been wondering about that too. I intend to build some good low-power measurement circuits one day, to be able to measure voltage & current on both sides of a power supply circuit. A few low-power op amps plus the 4-channel Analog Plug ought to do it. Trouble is, it needs to be accurate over a large dynamic range: 1 µA to 50 mA, if possible.
I did measure the no-load current on the battery side, it seems to be around 20 µA. That’s probably the main factor for a JeeNode which is in power-down mode 99% of the time.
The data sheets often report efficiency only around the best efficiency point, which is usually near full load current. The converters I’ve used have low % efficiency at light loads, due to the constant power used by the boost chip itself being a larger fraction of the total.