32 diodes
I was thinking about the series blocking diode addition to my recent solar panel purchase:
https://wb9kzy.blogspot.com/2026/04/make-or-buy.html
Certainly there is a drop across the diode when the panel is charging, less than 250 mV with low current. But is it really needed ?
Looking at the solar panel itself again there are 6 wafers, each wafer has 6 groupings of photodiodes with 6-5-5-5-6 diodes each or 32 total diodes in series.
Per Wikipedia the output voltage of a silicon solar cell (diode) is .5 to .6 volts. Multiply 32 diodes times .56 volts gets to roughly 18 volts which is about the rated voltage:
21.6 volts divided by 32 is .675 volts so this sounds reasonable for a no-load situation.
Now since the panel is charging a lead acid AGM / gel-cell battery the highest voltage that the battery could apply to those 32 diodes is roughly 13 volts:
source: https://voltagebasics.com/lead-acid-battery-state-of-charge-voltage-chart/
So unless more voltage is applied, not that much current should flow until the "knee" of the diode IV curve is reached, right ? If we assume an even voltage drop across the diodes: 13 volts / 32 diodes = .41 volts across each diode. Unfortunately we don't have the IV curve for forward conduction on the solar panel.
Enough talk, this requires measurements ! I waited until dark, then covered the panel just to make sure that no light was present. Then I shorted the diode and measured the leakage current through the solar panel. The battery voltage was 12.9 volts (measured via Morse code with the DC Beeper kit, see:
https://wb9kzy.blogspot.com/2026/04/make-or-buy.html
Certainly there is a drop across the diode when the panel is charging, less than 250 mV with low current. But is it really needed ?
Looking at the solar panel itself again there are 6 wafers, each wafer has 6 groupings of photodiodes with 6-5-5-5-6 diodes each or 32 total diodes in series.
 |
| count the wider horizontal dark lines to determine the number of solar cells |
Per Wikipedia the output voltage of a silicon solar cell (diode) is .5 to .6 volts. Multiply 32 diodes times .56 volts gets to roughly 18 volts which is about the rated voltage:
21.6 volts divided by 32 is .675 volts so this sounds reasonable for a no-load situation.
Now since the panel is charging a lead acid AGM / gel-cell battery the highest voltage that the battery could apply to those 32 diodes is roughly 13 volts:
source: https://voltagebasics.com/lead-acid-battery-state-of-charge-voltage-chart/
So unless more voltage is applied, not that much current should flow until the "knee" of the diode IV curve is reached, right ? If we assume an even voltage drop across the diodes: 13 volts / 32 diodes = .41 volts across each diode. Unfortunately we don't have the IV curve for forward conduction on the solar panel.
Enough talk, this requires measurements ! I waited until dark, then covered the panel just to make sure that no light was present. Then I shorted the diode and measured the leakage current through the solar panel. The battery voltage was 12.9 volts (measured via Morse code with the DC Beeper kit, see:
the leakage current was 5.2 mA. 
With the diode back in series the leakage current was .083 mA:
5.2 divided by .083 => about 63 times as much leakage with a direct connection - bottom line: keep the diode in there at least for now.
With the diode back in series the leakage current was .083 mA:
5.2 divided by .083 => about 63 times as much leakage with a direct connection - bottom line: keep the diode in there at least for now.
I should also check the other solar panel for leakage, it may or may not have a series diode.
Best Regards,
Chuck, WB9KZY
http://wb9kzy.com/ham.htm
Best Regards,
Chuck, WB9KZY
http://wb9kzy.com/ham.htm
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