I have been working on a design based on solar power. While browsing through some circuits on the Internet I came across these designs belowenter image description hereenter image description here

In these designs, why was a diode added between the solar panel and the voltage regulator chip? I understand that a diode will be usually used along with a Solar panel to prevent the reverse flow of current which in turn might damage the panel under dark conditions.

But I really don't understand its use in the above designs. Is there a possibility of reverse current flow from these regulator chips to the panel? I can only imagine that happening when the panel is connected directly to a battery.

  • \$\begingroup\$ There are a number of scenarios that will result in reverse current into a panel. Since a panel costs around $200, a $2.00 investment in a diode, is certainly money well spent - even if you don't intend to hookup a battery! \$\endgroup\$
    – Guill
    Aug 25 '17 at 3:06

Almost all voltage regulators do have a path backwards from output back into its input because they are not designed to handle blocking reverse current. So your battery would drain back into the solar panel when there is insufficient light falling on the panel.

For example, here's the internal circuit of an LM338 in your first schematic:

enter image description here

(From here)

While there is a path from the output of the regulator back into the input through diodes D2 and D3, it is through high value resistors. Your biggest problem is that you'll be feeding battery voltage back through R22 160 Ohms into the base of Q26 into its collector and out to your solar panel without the blocking diode. And there are other paths like that through other parts of that circuit.

So besides reversed panel connections like @analogsystemsrf pointed out, it prevents your battery from discharging back through your regulator back into your solar panel.

However, in your second circuit, the diode is redundant because the battery already has a series diode. But the diode at the solar panel can serve to prevent damage from reverse connections.

  • \$\begingroup\$ Re. redundancy in 2nd circuit: I don't know how fragile the panels are, but could the capacitor C1 cause problems without the diode? \$\endgroup\$
    – TripeHound
    Aug 21 '17 at 12:37
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    \$\begingroup\$ I doubt it. C1 is so tiny and solar cells are basically a series stack of diodes with a HUGE amount of surface area so I'm sure they can handle the back current. The VREG does need at least a tiny bit of capacitance for stability so can't get rid of C1. It would really be the voltage regulator that would be the most vulnerable part if you end up back-flowing current into the solar panel. \$\endgroup\$ Aug 21 '17 at 16:35
  • \$\begingroup\$ Thanks for the answer it really helps. But here is an alternate scenario where I intend not to use a battery in my project as it might make things bulky ( I know that it is not the standard way of doing things). I intend to power a simple circuit using Solar panel directly just by stabilizing the voltage using a capacitor. But am thinking of using a big capacitor say around 10uf to provide moderate stability in the absence of battery. So should I still need diode after the panel @VincePatron . Also does IC chips get affected when underpowered? \$\endgroup\$ Aug 22 '17 at 4:26
  • \$\begingroup\$ I don't think 10 uF will be a problem at all. Worst-case is full sun and then suddenly turns dark (like within 0 microseconds), all 10 uF * Volts (energy in Coulombs) gets dumped into solar cell. No biggie. Cell has huge area to absorb the current. Yes, ICs get affected if undervoltage. Switching speed gets slower. Logic threshold shift. Circuits can have logic errors. MCUs are usually smart enough to hold themselves in reset but could reset, restart, reset loop. Read the datasheets and specs of your components and ICs. \$\endgroup\$ Aug 22 '17 at 18:26
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    \$\begingroup\$ Thanks a ton @VincePatron marked your answer as accepted. And thanks everyone for offering help \$\endgroup\$ Aug 24 '17 at 4:11

The Diode is used to prevent backfeed into the cell from the battery. While daytime, the solar cell generates a voltage higher than the battery voltage, therefore current is flowing into the battery. When the cell is covered or it's nighttime, the voltage across the cell is 0V, so the battery, with it's higher voltage, would try to feed current into the cell, therefore draining and possibly damaging the cell. The diode blocks current in this direction, so the battery cannot discharge into the cell.

  • \$\begingroup\$ How much is it possible to damaging a cell? Recently i detect my 22000uf capacitor was discharged back in reverse bias. Capacitors has a very small discharge time. \$\endgroup\$
    – payam_sbr
    Apr 26 '18 at 10:47

To protect against REVERSED panel connections.

And, as other answers state, to allow failed panels (or shaded panels) to automatically "disconnect" from the total current output.

  • \$\begingroup\$ Can you please elaborate more. \$\endgroup\$
    – Mayank
    Aug 21 '17 at 5:23
  • \$\begingroup\$ &Mayank investigate what will happen if the panel connections are reversed and there is no diode fitted. also, when there are multiple panels in series - bypass diodes are fitted in case a panel fails. \$\endgroup\$
    – Solar Mike
    Aug 21 '17 at 5:45
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    \$\begingroup\$ I doubt this is the case. Solar panels are not intended to be connected and disconnected twice a day by unqualified folk who cannot be bothered to observe the polarity. \$\endgroup\$ Aug 21 '17 at 14:36

The first diode is used to protect against reverse panel connection. The second diode to the right is used for stopping regulator back powering from the battery.


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