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I use the following circuit to store solar energy harvested by photodiodes into a supercapacitor. The cap is rated at maximum 2.8V. The zener diode is rated at 3V, but depending on the harvested solar current it goes up to 3.15V. The schottky diode shall prevent the supercapacitor from discharging into the photodiodes.

Problem is: the voltage drop of the schottky diode is simply too low, which means in direct sunlight the capacitor gets overcharged up to 2.9V -> bad! I do not want to take a lower rated zener diode, as the current one drives the photodiodes nicely at the MPPT in sunny conditions.

Now the question: Are there any disadvantages of using a second Schottky in series, just to get another small voltage drop of around 200mV? That would be sufficient to not overcharge the supercap. But I'm unsure about other side effects. The whole circuit is extremely low power, so I want so save every possible uA.

I have seen schottkys in series, but rather for backup reasons (in case one fails).

Circuit

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  • \$\begingroup\$ " I do not want to take a lower rated zener diode, as the current one drives the photodiodes nicely at the MPPT in sunny conditions." - Where is the MPPT connected in your circuit? Which Schottky diode are you using? \$\endgroup\$ Mar 12, 2020 at 2:40
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    \$\begingroup\$ Hi Bruce. MPPT was the wrong word. There is no dedicated circuit, but the zener diode's zener voltage is roughly around 0.8 of the open circuit voltage of the solar cells. That is the MPP (without T, that was a mistake, it is not tracking). Below the open circuit voltage the solar cells can be modeled as a current source, so the zener diode will regulate the voltage depending on the current coming from the solar cells. Best, Timm \$\endgroup\$
    – timmbuktu
    Mar 13, 2020 at 9:16
  • \$\begingroup\$ Ah sorry, the schottky diode is a NSR0140P2T5G. \$\endgroup\$
    – timmbuktu
    Mar 13, 2020 at 9:18
  • \$\begingroup\$ Which supercap are you using, and which Zener diode? What are the specs of the solar panel? What is the minimum voltage and current you need to harvest? " I do not want to take a lower rated zener diode, as the current one drives the photodiodes nicely at the MPPT in sunny conditions." - but that overcharges the capacitor, right? Working at MPP is pointless if it produces too much voltage, since you will have to dump the excess. \$\endgroup\$ Mar 13, 2020 at 20:53
  • \$\begingroup\$ Hi Bruce, yes, this overcharges the supercap! But I conducted a couple of tests during sunlight and decided to simply let the MCU waste energy once the voltage gets too high. The charge current @2.7V in bright sunlight is 0.4mA maximum, so I have plenty of time to react! Thanks for you input though! The solar panel produces up to 3.4V open circuit. \$\endgroup\$
    – timmbuktu
    Mar 16, 2020 at 8:54

3 Answers 3

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Voltage rating of the super capacitor is as important as the operating temperature specification. The higher voltage leads to significantly higher self discharge.

Adding a second diode will definitely help to keep the voltage applied to the super capacitor within its operating condition. I don't see any disadvantage because of the secondary diode addition for the presenter circuit.

One suggestion is to employ a low leakage diode instead of two Schottky if feasible. It will less discharge the capacitor when the solar voltage is not present.

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  • \$\begingroup\$ Thanks for your answer! I use a NSR0140P2T5G schottky diode, with 3V reverse voltage it leaks less than 10nA. One disadvantage which came to my mind is: I will lower the voltage even in low light conditions, so my charging efficiency is getting worse. My zener diode is rated at 3V, so I thought about using a 2.7V zener diode instead -> that would be more efficient, right? \$\endgroup\$
    – timmbuktu
    Mar 13, 2020 at 9:23
  • \$\begingroup\$ that would do. check the tolerance of zener diode and the capacitor again. 2.8 V is mentioned in the question. it means there is less power wasted across the diodes. \$\endgroup\$
    – User323693
    Mar 13, 2020 at 9:35
  • \$\begingroup\$ Okay, thanks a lot! \$\endgroup\$
    – timmbuktu
    Mar 16, 2020 at 8:51
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After some testing today:

The voltage drop of shottkey diodes goes down the less current is running through them. That means it's entirely possible for the photocells to charge the capacitor up to the maximum open circuit voltage of the photocells even with the diodes in place, since their voltage drop will go to near zero at very low currents.

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I know it's been a while and you've probably already figured something out.

  1. the comment above about forgetting about the MPP voltage for a simple solution is correct. The capacitor is going to drag the voltage down anyway. For a highly efficient MPP charge, a stepdown DC-DC converter would be required and off the shelf ones will not work for this application.

  2. Running diodes in series will waste power, the voltage drop times the current flowing through them will be dissipated as heat. I'd suggest using the lowest voltage drop diode available for this application, they're available down to around 0.1v.

  3. OK, we cannot over volt the capacitor. Two solutions: One, reduce the number of photocells such that their maximum open circuit voltage is at or below the maximum rating of the capacitor. Two, use a zener diode rated at the maximum voltage of the capacitor to bleed off excess voltage.

I'm making a similar simple system, but with larger components (120F 16v capacitor array) and have decided to use both of the methods... making sure the panel maximum open circuit voltage is lower than the capacitor max rating AND using a zener diode in case some unpredictable environmental conditions occur that raise the PV voltage.

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