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I'm designing a circuit which occasionally requires small amounts of power (it's switched off most of the time). I want to use small solar panels to charge a supercapacitor, and the cap then serves as an energy reservoir in the absence of full sunlight.

I have already set up a basic circuit with a EDLC supercap (VINAtech, 100F, 3V), a small solar panel (3V, 270mA) and a 1N4001 diode. It seems to work fine, the supercap voltage appears to stabilise at around 2.85V with the panel pointed at the sun, full sunshine and the panels clean. Such ideal conditions will be rare though, the panel may be shaded most of the time. I know that an MPPT charger would be more efficient but I want this to stay as cheap and simple as possible.

schematic

simulate this circuit – Schematic created using CircuitLab

In this schematic, V1 is the solar panel, C1 is the SuperCap and U1 is a generic boost converter based on a CE8301A50T. I'm not sure about the final circuit yet, but there will certainly be a microcontroller (Atmega16A) and an HD44780-based LCD with LED backlight. SW1 will only be closed for a few minutes at a time (no more than 15 minutes), very few times a day (probably not at all for several days in a row).

Datasheets:

I just want to know if there might be any pitfalls I might have missed. The circuit will operate outdoors, i.e. it will have to survive temperatures of -25°C (-13F) with no problems. Unfortunately, manufacturers don't seem to test the behaviour of their supercaps at low temperatures (as far as leakage current or change in capacitance goes) so I'm not sure if there might be issues. I don't think temperatures will rise above 50°C (122F) in the summer.

Any thoughts?

Thanks!

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  • \$\begingroup\$ Welcome to EE.SE! Have you thought about a Schottky diode instead? Lower forward voltage drop at the cost of more reverse leakage. \$\endgroup\$
    – winny
    Commented May 20, 2020 at 9:29
  • \$\begingroup\$ Thanks for the suggestion. I have considered using a Schottky diode, but it would mean that the supercap would probably get charged beyond the 3V limit. \$\endgroup\$
    – merendo07
    Commented May 20, 2020 at 19:40
  • \$\begingroup\$ Oh! What’s the open circuit voltage of the panel? \$\endgroup\$
    – winny
    Commented May 20, 2020 at 19:46
  • \$\begingroup\$ I've seen 3.33 Volts at one time (blue sky, panel clean and pointed at the sun at 47°N where I live). \$\endgroup\$
    – merendo07
    Commented May 20, 2020 at 22:04
  • \$\begingroup\$ In open circuit configuration? It should be stated in the datasheet. \$\endgroup\$
    – winny
    Commented May 21, 2020 at 6:47

1 Answer 1

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Any thoughts?

The devil will be in the detail and you have provided little.

  • Start with defining your load and the current consumption profile. You also need to state how low the capacitor voltage can drop before your load fails on under-voltage.

  • define peak currents in terms of amps and duration - i.e. the full load profile.

  • Data sheets for everything is needed. Leakage currents for the supercap WILL be an important consideration.

  • Anything that doesn't support your temperature range is going to make decisions on the circuits ability to fulfill your needs extremely dubious.

  • Make no assumptions that something "should" work - prove it by due diligence.

and a 1N4001 diode

I'm unsure about this - I think you need to show your proposed circuit.

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  • \$\begingroup\$ The load on the 5V side will be about 100mW, so 20mA. I don't expect there to be any significant current peaks and if there are, I will take measures to flatten them. Let's assume the 20mA are constant. Using my bench power supply, I was able to go as low 0.9V with the boost converter maintaining the output voltage. I have modified my original post to include a schematic and links to datasheets. \$\endgroup\$
    – merendo07
    Commented May 22, 2020 at 8:17
  • \$\begingroup\$ Why do you think you need the 1N400x diode in series with the panel? \$\endgroup\$
    – Andy aka
    Commented May 22, 2020 at 8:39
  • \$\begingroup\$ @Andyaka If nothing else, that diode is keeping the voltage within the maximum for the capacitor. Take it out, and the voltage could go up to about 3.45V. \$\endgroup\$
    – Simon B
    Commented May 22, 2020 at 9:37
  • \$\begingroup\$ OK so if you can suffer a drop down to 0.9 volts from a fully charged situation of (say) 2.7 volts you have a "dv" of 1.8 volts. Current will be less at the start and will get worse as the voltage drops because you are feeding a boost converter. At 2.7 volts the current taken will be approximately 40 mA (give or take) and could be as high as around 120 mA when the capacitor voltage is near enough depleted to 0.9 volts. Take a worst case average of 100 mA. So 100 mA = C * dv/dt or, dt = C*dv/0.1 = 100*1.8/0.1 = 1800 seconds. Is this long enough? If you want a more accurate result use a sim tool \$\endgroup\$
    – Andy aka
    Commented May 22, 2020 at 10:05
  • \$\begingroup\$ 1800 seconds is more than plenty. Thanks! Doesn't the diode stop the cap from being discharged through the solar panel if it receives less than full sunlight? Either way, I do need it to keep the voltage with limits. I suppose I could get a panel with a lower voltage though. \$\endgroup\$
    – merendo07
    Commented May 22, 2020 at 12:47

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