2
\$\begingroup\$

My electric fence unit stopped working, but it is a nice case with integrated solar panel and battery holder. I figured I could repurpose it as a rechargeable general purpose 6 V power supply. The solar panel still works fine and I think the battery does (although I let it run down flat so maybe not). I pulled the printed circuit and tried to figure out what bit actually charged the battery as opposed to doing the electric fence stuff. I came up with this circuit: circuit I found

JP1 is the battery, JP2 is the solar panel. There are three Zener diodes as shown, all 1N5908 through hole parts and one SMD part which looks like a diode labelled M7 which, I gather, is a 1N4007 diode. I've put all this together on a PCB but I haven't grasped how it should work, and I'm not sure it does. When I tried it I did get some charge on the battery (got up to 2 V) but it discharged. I read up on Zener diodes 'cos I had only used ordinary ones before, and these seem to allow current through if the reverse voltage gets past 6 V. My solar panel goes up to 9 V max, that's what I've measured in full sun. I gather the battery (6 V SLA) likes about 6-7 V to charge. So I think that if the solar panel gets above 6v the D3 diode will send it to ground. But what I don't see is a way to cut off the charge when the battery is full. So three questions:

  • Is this circuit likely to work or have I misinterpreted what I found?
  • Is there something there that is cutting off the charge to the battery when it is full?
  • Why would it discharge? I thought D1 would prevent that. Nothing was connected on JP4.

Obviously I'm no expert, so I do appreciate any help. Thanks.

At @winny's suggestion I have redrawn the diagram enter image description here

The battery is 6v as shown. There are three Zener diodes (1N5908) and one ordinary diode (1N4007). The solar panel, marked on the right as a current source is a 9v panel. I have shown the load as a lamp, but so far I have not connected a load to this circuit, and it won't be a lamp.

\$\endgroup\$
2
  • \$\begingroup\$ Please draw the battery as a battery and the solar panel as perhaps a current source to simplify understanding of the schematic for your readers. Is the open circuit voltage of your solar panel 9 V? \$\endgroup\$
    – winny
    Commented May 25, 2023 at 8:06
  • \$\begingroup\$ @winny I have updated the diagram \$\endgroup\$ Commented May 27, 2023 at 21:44

1 Answer 1

1
\$\begingroup\$

The 1N5908 https://www.st.com/resource/en/datasheet/1n5908.pdf is a transient voltage suppressor (TVS) with a stand-off voltage of 5V, Breakdown voltage of 6V at 1 mA, and a clamp voltage of 7.6V at 30A. So D3 and D4 will limit the voltage of the solar panel to about 6.8 VDC. D2 will limit the battery charging voltage to about 6V, and D1 will limit the load voltage to V(batt) - 0.8V. There will be some reverse bias leakage from the battery to the solar array, but no more than about 1 mA.

The circuit is likely misdrawn, or components misidentified. Otherwise, it appears that it will not work very well.

schematic

simulate this circuit – Schematic created using CircuitLab

Voltages

Currents

\$\endgroup\$
1
  • \$\begingroup\$ Thanks for this @PStechPaul. You've improved my understanding of that diode. The circuit does seem to work so I think I have the components correct, but whether it works well is another thing. I had not looked at CircuitLab before, looks like something I should spend some time with, \$\endgroup\$ Commented May 28, 2023 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.