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I'm interested in charging a 12V Sealed Lead-Acid battery from a single monocrystalline solar cell. The solar cell's voltage is only 0.5V at maximum power point, but it supplies 8A. I will need a 32X voltage multiplier to emulate a 12V panel to drive my SLA charger.

Could someone suggest an appropriate voltage multiplier design? I have read about Villard- and Dickson-type cascades, but they all will incur too much loss in the early-stage diodes.

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  • \$\begingroup\$ Possibly useful search term : "Joule Thief" \$\endgroup\$ May 6 '14 at 19:09
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Extracting power from such a low voltage is tricky because even otherwise small voltage drops are significant fractions of the input.

Fortunately, you have the 12 V battery available to power the control electronics of a boost converter. At its core, a basic boost converter will be a inductor, switch, and diode:

The switch closes, which makes the inductor current rise linearly with time. Once the inductor current gets to a decent level and a useful amount of energy is stored in it, the switch opens. At that point the inductor current can only go thru the diode, dumping current onto the output. The reverse voltage now on the inductor causes the current to decrease linearly over time, transferring the energy stored in the inductor to the output.

Since your input is 500-750 mV, the most critical part of this circuit is a switch that will only drop a small fraction of that. A bipolar transistor with maybe 200 mV saturation is not appropriate here. You want a FET with low on resistance (Rdson). Let's say you want to keep the drop accross the switch to 50 mV or less at 8 A. By Ohm's law, that means you need 6.3 mΩ Rdson or less. There are FETs that can do that, but for something like this paralleling a couple of them may be quite appropriate.

Fortunately, the drop accross the diode is relative to the output voltage, not the input voltage. A regular Schottky diode rated for the current should be good enough here. Let's say it drops 500 mV when dumping current onto the 12 V output. That's a 4% power loss, which in this case, especially considering you're asking here, I'd just swallow. To get around that you'd need to implement synchronous rectification, which is not a beginner concept unless you can find it included in a chip.

Since the battery is there, you can run the control circuit that turns the FET on and off from it. Many microcontrollers can do this. The micro would measure the input and output voltages, and produce the switch signal via built-in PWM hardware. The micro can power itself down when there is insufficient voltage from the solar cell to make it worth it to run the boost converter. With the right circuit, the micro can take less current than the internal leakage of the battery when powered down. It can then wake up every few seconds to check whether sufficient input voltage is available to run the converter, and go back to sleep when not.

You have to pay attention to low current design with the micro, and make sure things like the battery voltage measurement circuit don't take any current when off. Fortunately this is something a beginner can do, although it requires paying attention.

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  • \$\begingroup\$ old question: I know you know: Note 0.50 mV is 10% of PV voltage!. Lower Rdson FETs a really good idea here. 1 milliOhm FETs are affordable - no doubt more so than in May 2014. \$\endgroup\$
    – Russell McMahon
    Sep 10 '16 at 9:32
  • \$\begingroup\$ My 2016 comment should read " ... 50 mV ..." not 0.50 mV. \$\endgroup\$
    – Russell McMahon
    Oct 14 '19 at 6:43
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This is a useful enough question to be worth an update.

5+ years on Olin's answer is still relevant.
The following notes ways of achieving operation at such low voltages and lists some examples of modern MOSFETs that significantly improve on what Olin suggested "way back then".

As others note, very great care to manage low voltage drops everywhere is needed.
A single cell CAN be used, and people do it, but if you can divide a cell into say 4 parts (or even two) the task is much easier and, importantly, conversion efficiency will probably be usefully higher.
I have ssen commercial products which tout their single PV cell plus boost converter designs BUT the efficiencies of conversion were poor.

You could get the bootstrap voltage for a power conversion IC with no battery per se by adding a tiny multicell PV panel next to the main one.

The truly enthusiastic can get double the PV voltage by a "start" switch which transfers a cap across the panel to in series with the panel.
Using a FET as ~= zero drop rectifier helps charge a 2nd cap BUT various lawn-light ICs run from about 0.8 to 1.0 V so doubling VPV of about 0.6V with a manual switch will get the boost converter working "and it's away".

The overly enthused could drive a multistage voltage multiplier with multiple button presses - but the methods above seem more sensible.

_____________________________________________________

Suitable MOSFET switches.

A very low Rdson MOSFETs with adequate voltage and current rating is required.
The Ton:Toff ratio is very high - ~= Vout / Vin x 1/efficiency.
So the converter can operate in CCM (continuous conduction mode) with Ipeak close to Iaverage, or in discontinuous mode with Ipeak ~= 2 x Iaverage.

Here is a Digikey listing of only some of the more suitable FETS with Rdson under 3 milliOhms and adequate voltage and current ratings.

Rdson is often given under pulse conditions at low temperature.
As a 'rule of thunb', real world worst case Rdson is seldom twice the listed value and often less than twice.

So at 16A Ipeak and 2 x listed Rdson, a 3mΩ FET will drop
V = IR = 16A x 6 mΩ = 96 mV or approaching 20% of the PV single cell voltage.

The 0.45mΩ NVMTS0D4N04CTXG will drop V = IR = 0.45mΩ x 16A ~= 8 mV or only about 1.5% of the PV voltage.

Examples only:

The Vishay SIRA36DP-T1-GE3 prices & description
30V, 40A, 2.9 mOhm at 20A. Vgs = 2.2V. Vgs at lowest Rdson = 10V.
$US0.58/1. SO8 pkg.

Nexperia PSMN2R4-30MLD price & description
Datasheet
30V, 70A, N-channel 30 V, 2.4 mΩ logic level MOSFET in LFPAK33
$0.76/1

Serious but dear.
OnSemi NVMTS0D4N04CTXG price & description
Datasheet
40V, 80A, 40 V, 0.45 mΩ

Nexperia PSMNR90-30BL,118 Prices ...
datasheet
N-channel 30 V 120A 1.0 mΩ logic level MOSFET in D2PAK
$2.82/1

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  • \$\begingroup\$ Though this does not directly answer the question, you suggest the clearly reasonable solution to the problem. Anyone facing this problem should strongly consider a PV array to get the open-circuit voltage up to something reasonable. \$\endgroup\$ Oct 14 '19 at 15:58
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I would consider using a low power energy harvester chip such as the LTC3105 to generate a "start-up" 5V auxiliary low-power rail. This chip can produce 5V from a supply as low as 225 mV but can only realistically provide a few tens of mA.

I'd use this 5V as a "start-up" power source for MOSFET drivers in a synchronous boost regulator and, as the boosted power voltage headed above 5V towards the 12V area, I'd re-power the MOSFET drivers with the rising 12V (in order to drive the MOSFETs more efficiently).

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  • \$\begingroup\$ Thank you for addressing the bootstrapping issue left open by Olin's answer. This would allow the recovery of a fully dead battery. \$\endgroup\$ May 6 '14 at 14:03
  • \$\begingroup\$ @ChrisMerck I hadn't seen Olin's answer and presumed you wished it to work from below 0.5V unaided!! \$\endgroup\$
    – Andy aka
    May 6 '14 at 14:07
  • \$\begingroup\$ I suspect 12V Pb battery which does not able to deliver bootstrap voltage (e. g. 3V) is dead completely and is not usable. However method is applicable for supercapacitor applications. \$\endgroup\$
    – Vovanium
    May 6 '14 at 14:39
  • \$\begingroup\$ @Vovanium Yes, you are probably correct, thank you. \$\endgroup\$
    – Andy aka
    May 6 '14 at 15:01

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