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I'm working with a bunch of mechanical engineers on an electrical engineering project and we're totally stumped.

The main problem is to drive a small robot chassis using the power generated from solar panels. The concept we came up with is to charge a large capacitor by sitting in front of the provided work lamp, then use that stored charge to drive the motors to drive up a small ramp.

As this project still has yet to work, the min viable product of charge time to drive time doesn't really matter, just has to be 15 minutes in total.

We have six Maxeon monocrystalline cells in series which generate 3.3-3.4V when tested. We have been testing with a very large 62F, 18V, 2.8Wh capacitor but also have some smaller ones. We are using two 12V DC motors, which we only need to drive at ~6V for adequate speed.

To get the solar voltage to a workable higher voltage, we also have MT3608 boost converter.

This is the basic circuit we've been working with:

Solar Cell - Capacitor Circuit Diagram

We are also using a motor driver which requires 5V to operate.

Our line of thought is using the booster, we can charge the capacitor to the set voltage, even if it takes 10 minutes with the very low current, which then can be used in motor drivers, etc. in a much smaller time period.

In experiments, even after setting the booster to 8V, the capacitor will not charge more than 3.3V that the solar cells are outputting.

We've also tried placing the booster after the capacitor, thinking we can boost the low voltage to be usable, but that also resulted in no effect.

We do not need to power a PWM input or the Arduino used to control with these cells, only the power input to the drivers.

The distance to drive is pretty short, the plan is to make multiple trips and charge and discharge the capacitor each time so time to charge when designing was important.

JRE
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Drew Fernandez
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    Why are you not using rechargeable batteries? Capacitors seem like the wrong thing for the job for such a high load as a motor, for such a long time as 15 minutes. – Hearth Apr 07 '23 at 21:00
  • We thought that rechargeable batteries would take too long to charge, as the distance to drive is very short. The initial plan is to make multiple trips, charging and discharging the capacitor each time. – Drew Fernandez Apr 07 '23 at 21:08
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    The other question is, are you using an MPPT converter for your solar panels? Because you should be. Give us more details on the motors, too. – Hearth Apr 07 '23 at 21:13
  • No we aren't using an MPPT... wasn't sure if it would help or be necessary. Will edit with motor info. – Drew Fernandez Apr 07 '23 at 21:17
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    Keep in mind that 2.8 wH figure is at 18 V. At 3 or even 8 V, it will be far far less. – evildemonic Apr 07 '23 at 21:18
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    32 F at 3.4 V is only 0.05 wH – evildemonic Apr 07 '23 at 21:24
  • Those motors (like all motors) require a lot more current to start up than they do in operation; see where it mentions stall current? That's the current you need to supply when they're not spinning, to get them to start (or the current they'll pull if prevented from turning). Your tiny DC-DC converter is nowhere near capable of outputting that much current--that "2 A" is the switch current, and will be divided by the boost factor (so a 2x boost circuit won't be able to output more than 1 A--it's actually going to be slightly less than that, too). – Hearth Apr 07 '23 at 21:32
  • I see, so boosting from capacitor to driver would fry the booster. Would this effect boosting from the solar cells to the capacitor? And would using an MPPT transform enough current to power the booster? I'm just not really sure how the MPPTs work after researching them – Drew Fernandez Apr 07 '23 at 21:52
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    What is the useful power output (in watts) of the solar panel? You've measured its *open circuit* voltage under the lighting conditions...but 3.4V @ 0A is 0 watts. Now make a graph of current vs voltage by sticking resistor "dummy loads" across it. With resistors, if you know ohms and volts that gives you amps. – Harper - Reinstate Monica Apr 08 '23 at 04:20
  • @evildemonic that would be Wh, not wH. – winny Apr 08 '23 at 07:01
  • What you need is an MPPT battery charger which can accept your open circuit voltage, and a battery. – winny Apr 08 '23 at 07:03
  • You should provide the datasheet for the booster: https://www.olimex.com/Products/Breadboarding/BB-PWR-3608/resources/MT3608.pdf – PStechPaul Apr 08 '23 at 07:10
  • Is the _work lamp_ just to provide a means to charge the robot via solar cells, or is the lamp also required to provide light for the robot and/or humans? Just wondering if you had considered inductive charging or not. From a quick search not sure of the relative efficiency of solar .vs. inductive charging. – Chester Gillon Apr 08 '23 at 18:49
  • You could use the capacitor more like a battery by putting a charge controller between the panels and the capacitor, and then drawing power through a voltage regulator. That way you could use the full storage of the capacitor at 18V, and easily provide stable output. https://www.analog.com/media/en/technical-documentation/data-sheets/3750fa.pdf – le3th4x0rbot Apr 09 '23 at 05:28
  • `the capacitor will not charge more than 3.3V that the solar cells are outputting` Did you try adding about 1 mF at the boost input? – greybeard Apr 09 '23 at 05:58

5 Answers5

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The chances are that your solar cells cannot deliver enough current when illuminated with a normal work lamp. The boost converter will require a minimum current to work properly. If that is greater than the current the cells can deliver, then it won't boost.

Simon B
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  • Do you have any pointers to where we could modify the circuit to get enough current or adequately charge the capacitor? – Drew Fernandez Apr 07 '23 at 21:25
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    As mentioned in a comment above, you need to determine the maximum power point of the solar cells, by measuring voltage on various resistive loads. A work lamp probably will not provide sufficient insolation for the cells. A single solar cell can produce only about 600 mV, so you will need a lot more than six in series. – PStechPaul Apr 08 '23 at 07:22
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    @DrewFernandez You need to find out how much current the solar cells are producing. Then check the specification of the boost converter. Your only realistic option for modifying a boost converter is to find teh data sheet for the chip that controls it and see if it allows an option to work at a lower current. – Simon B Apr 08 '23 at 16:28
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You should break this problem into 2 parts.

  1. Make the robot work under battery power. Use a 1S or 2S lipo.

  2. Build a circuit that can charge this battery from your solar panels. If your lipo is 2S you will need to include a balancing IC.

I believe that by using capacitors instead of batteries, you're needlessly complicating the problem.

Drew
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Your problem is one of impedance. The capacitor is a short circuit load so the charge current will be limited by the short circuit current of the panel. (You can measure that with a DMM.) This will be nowhere near the maximum power point of the panel and is probably extremely low as a consequence. It is likely you will require a SEPIC converter with some intelligent MPPT software to make this project work. But first you need to establish what the maximum power is that can be extracted from your illumination and determine if that is sufficient.

Transistor
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Fourtytwo
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You need a few steps. Use a conventional power supply to figure out how much current the device needs. Then figure out how much power your solar arrangement can provide. Convert to power. If the solar panel can't produce enough power, you need a bigger solar panel, or a different strategy.

Transformers and boost converters can only go so far. They can't produce power out of thin air.

Scott Seidman
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    It's not a question of power, because they can idle while the robot charges. Their real requirement is to fully charge the robot using the meager power from the solar panels. – Drew Apr 08 '23 at 06:08
  • @Drew, I can't even understand what's charging from your ccx question – Scott Seidman Apr 08 '23 at 17:26
  • I think you may have me confused with the asker. – Drew Apr 09 '23 at 17:44
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From your description, I can't understand what and how you have connected right now. Whatever way I look at it, it doesn't add up.

I did some napkin math and for 3.4V available from the panels, charging a 62F capacitor to 8V means pumping it with 3968J of energy, Watt-seconds, so about 66 Watt-minutes. Assuming 100% efficient voltage/current converter in the middle, powered by 3.4V, it means about 19.5 Amp-minutes. You mentioned 10minutes of charging, so that means ~1.95A being drawn from the panels. Over 10 minutes. To charge the cap to 8V. In perfect conditions, no losses, etc.

Have you checked if your solar panel array can provide this much current at 3.4V (so about 6.6W of electrical output power) from a mere worklamp?

I have my doubts since you said "even if it takes 10 minutes with the very low current", and 2.0A isn't a "very low current".

If on the other hand the panels are 3.4 per one, and all are in series, then you'll get max 20.4V output and the math will change slightly, 66 Watt-minutes are stil a thing, but now 3.24 Amp-minutes. 10 minutes, so 0.324 Amps. Ok, that's getting somewhere near 'low current', and maybe you could get 300mA from panels under a worklight.. but having a 20V at input and not being able to charge a cap to more than ~3.5V sounds very odd. Maybe the load is engaged? Are the motors and all other power hungry devices like processor properly cut off while charging? Maybe they are draining the power that should go to the cap?

But then, anyways - why a boost converter when you have 20V available and want just 8V at the cap? If you have a strong light to provide these watt-minutes, then you have a high voltage already, so you will need a buck converter here, to lower the voltage to the desired one. Or rather a buck-boost converter, to cover both cases, both of strong light and weak light..

quetzalcoatl
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  • A "Joule thief" type circuit might be useful to extract most of the energy from the fully charged capacitor. But if it is 18V, it should be mostly depleted at about 3 volts. Energy is proportional to square of voltage. – PStechPaul Apr 08 '23 at 23:48