# Computing stuff tied to the physical world

## Eneloop with solar top-up

In Hardware on Jun 8, 2012 at 00:01

Here’s another idea in the continuing search for long autonomous JeeNode run times:

The basic circuit is an Eneloop AA(A) cell, driving the AA Power Board to boost its voltage to 3.3V. There’s a 1 kΩ in series with the battery, as well as a Schottky diode to limit the voltage drop to about 0.3V during times “high” current consumption. I’ll explain why later on.

On the input side is a really simple circuit: a solar cell with a series diode, simply feeding the Eneloop battery when there is solar energy available.

The solar cell I’m using is that same 4.5V @ 1 mA cell I’ve been using all the time in my recent experiments. It is surprisingly good at generating some electricity indoor, even behind the coated double-glazing we have here.

The 1 kΩ resistor in series will let me measure the actual current flowing across it – 1 µA will read out as a 1 mV drop (that Ohm’s law again, of course!). So with a charging current of up to say 200 µA, this conveniently matches the 200 mV lowest range of most multimeters. And 0.2V is not a dramatic voltage drop, so the circuit should continue to work – almost the same as without those measurement resistors included.

A similar 1 kΩ resistor has also been inserted between the battery and the AA Power Board, but in this case we have to be more careful: a JeeNode will briefly pull 25 mA while in transmit mode, and the 1 kΩ resistor would effectively shut off input power with such currents. So I added a diode with minimal forward drop in parallel – it’ll interfere with my readings, but I’m really only interested in the ultra-low power consumption phases.

Here’s my “flying circus” concoction:

I’ve added some wires to easily allow clipping various meters on.

Now, clearly, 4V is way over the 1.3V nominal of an Eneloop battery. But here’s why this setup should still work:

• this solar cell is so feeble that its voltage will collapse when drawing more than a fraction of a milliamp
• solar cells may be shorted out – doing so switches them from constant-voltage to constant current mode

As for the Eneloop, my assumption is that it doesn’t really care much about being overcharged at these very low power levels. In the worst case of continuous sunshine for days on end, it’ll be fed at most 1 mA, even when full. That will probably just lead to a tiny amount of internal heating.

So let’s try and predict how this will work out, in terms of battery lifetimes…

I’ll take a JeeNode + Room Board as reference point, which draws about 60 µA continuous, on average (50 µA for the PIR, which needs to remain always-on). That’s on the 3.3V side of the AA Power board. So with a (somewhat depleted) AA battery @ 1.1V, than means the battery would have to supply 180 µA with a perfect boost regulator.

Unfortunately, perfect boost regulators are a bit hard to get. The chip on the AA Power Board does reasonably well, with about 20 µA idle and about 60..70% conversion efficiency. Let’s just batch those together as 50% efficiency, then the continuous power draw for a Room Node would be about 360 µA. Let’s round that up to 400 µA.

An Eneloop AA battery has about 1900 mAh capacity, but it loses some energy due to self-discharge. The claim is that it retains 85% over 2 years, so this battery can effectively give us about 1600 mAh of power.

The outcome of this little exercise, is that we ought to get some 4000 hours run-time out of one fully-charged AA cell, i.e. 166 days, almost six months. Not bad, but a little lower than I would have liked to see.

If the solar cell were to generate 4 hours per day @ 0.5 mA, when averaged over an entire year (that might be optimistic), then that’s 4 x 365 x 0.5 = 730 mAh. That comes down to an average current of 83 µA.

IOW, roughly one fifth of the total power needs could be supplied by the solar cell. Not enough for total autonomy, but still, it’s a start. Note that most of these last figures were pulled out of thin air at this stage: I don’t know yet!

Yet another idea would be to add an extra diode from the solar cell straight to the JeeNode +3V pin. IOW, when there is sufficient sunlight, we off-load the boost circuit altogether and charge up a capacitor of say 100..1000 µF on the JeeNode itself. No more losses, other than the AA Power Board’s quiescent current consumption.

1. Hehe.. you are crazy :) in a good way

2. JCW, how would you decouple the AA cell if it is fully charged and the solar cell would directly power the JeeNode / charge a cap through a diode?