Potato batteries in series vs. Potatoes in parallel

Other Hardware General Electronics
1

Hello,
I'd like to see if I can power BLE modules to send out beacons every once in a while, using potato or lemon batteries. I was thinking having the potatoes charge up a capacitor, and then power a LDO w/ low quiescent current from the capacitor. The output of the LDO provides (hopefully) constant voltage and sufficient current to the BLE module. I have two questions:

I'll need potatoes in series to get up to at least 4 volts. But I might want potatoes in parallel also to increase the current available. But the series combination of potato cells won't have the same voltage, so it's a little bit like putting batteries with different voltages in parallel. I'm not sure what the practical impact is of having different voltage sources in parallel.

When I look at current consumption when a beacon goes off, I see this.

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Transmit: 12mA for about 1ms per beacon. Probably transmit 3 beacons per second for one second per event.

Wake: 6mA for 2ms. One wake per event.

Events are very far spaced apart.

I've thought about this before, and looking at some equations for capacitor and current, I found this:

C = I x T / V
I = excess current to be provided.
T = time to provide this extra current.
V = acceptable drop in voltage during this period.
C = capacitance in Farad to meet this requirement.
C = .012A * .005S/0.5V = .00012F = 120uF

I'm not sure if this equation can be applied to my situation. If it is, I'd need a 120uF capacitor.

2

Your "batteries" will have fairly high internal resistance so I wouldn't expect any problems putting them in parallel.

I have no idea how much energy you can get out of a potato or lemon battery...

3

An LDO regulator, is likely to be wasteful, unless you can match potato voltage with the voltage requirement, closely enough to take full advantage of the LDO feature.

Realize that Low Dropout only occurs in cases where the difference between the unregulated voltage and the output voltage is small enough to make using an LDO feasible. An LDO can have the same dropout voltage as any other non-LDO regulator. The thing that makes an LDO special is that it's MINIMUM dropout voltage is lower than a non-LDO.

For instance: an LP2950CZ-5.0 has a typical minimum dropout voltage of 350mV [at 100mA], whereas an LM78L05 has a typical minimum dropout voltage of 1.7V, so let's say I have a battery pack composed of 4 AA batteries in series. That's a nominal voltage of 6V. For a 5V regulator, that's a dropout of 1V [6V - 5V].

The LP2950 will work just fine, and in fact, the battery voltage can drop down to 5V + 0.35V = 5.35V, and the LP2905 will still produce a regulated 5V on it's output.

The 78L05 will not work, because it's minimum dropout voltage of 1.7V is more than 1.0V, so it won't be able to produce a regulated 5V on it's output. You would have to add another AA cell to that battery pack, for a nominal voltage of 7.5V. 7.5V -5V = 2.5V -- thus that will be the dropout voltage. And, once that 7.5V battery pack's voltage drops below: 5V + 1.7V = 6.7V, the 78L05 will no longer regulate. And that's what LDO's are all about--not having to add that extra battery!

Notice, though, if the LP2950 were used with the 7.5V battery pack, the dropout voltage will still be 2.5V! BUT, the battery pack's output voltage can drop all the way to: 5V +0.35V = 5.35V, before that LP2950 would stop producing a regulated 5V on it's output! But, it would be wasteful to add that extra battery in this scenario, because that extra power would be wasted as power dissipation in the regulator, because of the high dropout voltage [yes! Even with an LDO!!].

So, because of that, I suggest a Switch Mode Buck Converter AND, rather than putting potatoes in parallel, put them ALL in series and let the Buck converter sort it out! That way, you won't have to deal with parallel inequities or power loss in adjacent potatoes!

So, all potatoes in series with a capacitor across the whole potato battery, then feed that voltage to the input of a Buck Converter!

4

May work better if the potato is cooked; lower internal resistance. Experiment with different cooking times. I prefer microwaved, then baked (for eating).

5

But the series combination of potato cells won't have the same voltage, so it's a little bit like putting batteries with different voltages in parallel. I'm not sure what the practical impact is of having different voltage sources in parallel.

The impact is that the higher voltage stack will drive current into the lower voltage stack of potatoes. Nullifying most, but not all, the advantage of having parallel stacks. The conventional answer is to use a diode to combine them, but then you loose some voltage across the diode.
Using a Schottky Diode will result in a lower voltage drop than a normal silicon one. A normal diode will have a voltage drop between 0.6 to 1.7 volts, while a Schottky diode voltage drop is usually between 0.15 and 0.45 volts.

6

Perhaps the use of closely matched potatoes is required ? :slight_smile:

7

Use the potato power to charge a super capacitor, then take the output of the cap via a LDO or zener diode regulator. You may not need an external regulator if you supply the raw pin.

8

And actually, being less flippant, an energy harvester chip might be useful for this

I found this BQ25504 Solar Cell LiPo Charger - Electronics-Lab.com - the BQ25504 seems to start from a very low voltage and
may be a good fit for this.

9

Potatoes in series, feeding a Buck Converter -- most efficient way to do it!

PotatoBattery02.png
PotatoBattery02.png1176×912 4.28 KB

With Series Potatoes, voltages will add up, and when Buck converted, current will multiply according to the following formula:

I[sub]OUT[/sub] = I[sub]IN[/sub]*(eff/100)*V[sub]IN[/sub]/V[sub]OUT[/sub]

Where:
VIN is the total series voltage of the potatoes
eff is the efficiency of the Buck Converter in %
VOUT is the Output Voltage of the Buck Converter
IIN of course, is the current supplied by the series arrangement of potatoes -- which will be lower than the output current (if the output voltage is lower than the total potato stack voltage, less PS efficiency).

So, the more potatoes in series, the more current you can get out of this thing!

10

If you think about it a pair of potatoes in parallel is like a single potato with bigger electrode area, and , if the tatty is the limiting factor a bigger one.

11

ReverseEMF:
Potatoes in series, feeding a Buck Converter -- most efficient way to do it!

PotatoBattery02.png
PotatoBattery02.png1176×912 4.28 KB

With Series Potatoes, voltages will add up, and when Buck converted, current will multiply according to the following formula:

I[sub]OUT[/sub] = I[sub]IN[/sub]*(eff/100)*V[sub]IN[/sub]/V[sub]OUT[/sub]

Where:
VIN is the total series voltage of the potatoes
eff is the efficiency of the Buck Converter in %
VOUT is the Output Voltage of the Buck Converter
IIN of course, is the current supplied by the series arrangement of potatoes -- which will be lower than the output current (if the output voltage is lower than the total potato stack voltage, less PS efficiency).

So, the more potatoes in series, the more current you can get out of this thing!

I really like that! Is the potato battery a standard schematic symbol?

12

ChrisTenone:
I really like that! Is the potato battery a standard schematic symbol?

Thanks! Not that I know of. I made it myself ;D

But, it would be really funny if it was!

BTW: I modeled it after a Yukon Gold -- but, I suppose it could also be a Red.