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I have a solar charger on a vehicle which I need to protect:

Schematic is down there... The charger connects to the main battery, and the solar panel, but the manual specifically states that you MUST connect the battery 1st, THEN the panel, or the controller can be damaged.

The issue is that there is an external safety battery cutoff that allows the battery to be disconnected for maintenance - so someone could knock the cutoff and remove the battery without having disconnected the solar panel.

Now, it's easy enough to imagine putting relay in the solar panel line, but the question is how to control it - once the panel is connected & the sun's shining, the controller is feeding charge out to the battery so both sides of the cutoff will be getting volts, and for safety the battery side of the cutoff should really be isolated & dead when the switch is open as it likely means the cables are removed & dangling.

To clarify the situations we can experience are:

On the controller side of the cutoff we may be getting battery voltage coming out of the controller, or we may not if it's dark.

On the battery side of the cutoff, we may get full battery voltage (battery connected but cutoff open), or we may have a complete open circuit (battery removed).

We need to detect the cutoff being OPEN, NOT the absence of a battery

I'm looking for suggestions that are simple, low-power, and robust (automotive systems can have serious currents & transients, cutoff is rated to handle 100A). If it helps, there is a 2nd supply available (vehicle has two batteries) so the circuit can use some 12v power from elsewhere.

Thanks in advance!

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Can you control the solar charge controller in any way or is it fixed? \$\endgroup\$ – winny Jun 1 '16 at 13:40
  • \$\begingroup\$ It's fixed - it is dual-voltage hence needs to sense battery voltage (12v or 24v) before the panel is connected, otherwise (I assume) the output could float up to 24v+ and damage a 12v battery which was then connected. \$\endgroup\$ – John U Jun 1 '16 at 14:19
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    \$\begingroup\$ Why not connect the solar panel and charger directly to the battery? \$\endgroup\$ – Dwayne Reid Jun 1 '16 at 16:19
  • \$\begingroup\$ Mount an auxiliary switch to your cutoff, or use a two pole cutoff, one delayed close. This can be done either by a simple disconnect switch, or an interlock plug with different length terminals. \$\endgroup\$ – R Drast Jun 1 '16 at 18:28
  • \$\begingroup\$ Yes. What Dwayne said. Just put the solar panel on the battery side of the cutoff switch. Problem solved. \$\endgroup\$ – mkeith Jun 1 '16 at 20:43
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Why not control a relay via a comparator, which changes state if the voltage on the battery itself drops, which is what would happen if the Controller was disconnected, ie normally closed relay, then if the comparator is activated, it opens the relay.

Edit:

Ok - to clarify:-

1) Place a normally closed relay between the panels and charger, then use a comparator/opamp to drive the relay.

2) The 'sense' inputs to the opamp come from either side of the Battery Cutoff switch, -ve to the battery side, +ve to the Charge Controller (CC) side.

3) If the Cuttoff switch is opened, the voltage difference activates the relay and protects the CC.

End edit.

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  • \$\begingroup\$ Can you clarify what you mean - at night, when the solar is not generating voltage & the battery is being used to power lights, the battery voltage will also drop. \$\endgroup\$ – John U Jun 1 '16 at 11:46
  • \$\begingroup\$ Please see my edit. If you cannot understand, I´ll add a schematic, but it is really very simple. The battery drop at night doesn´t matter, as both sides of the comparator will show the same voltage, so no change. I thought you were just trying to protect the CC from accidental disconnection of the battery? \$\endgroup\$ – F. Bloggs Jun 2 '16 at 17:05
  • \$\begingroup\$ That makes more sense, the way your original post was written sounded like you were relying on detecting VBatt < Vcc rather than VBatt != Vcc. I can't see how one comparator can work both ways though (output high if VBatt > Vcc OR Vbatt < Vcc). I can see how to do that with two comparators and just OR the outputs to drive the relay. \$\endgroup\$ – John U Jun 3 '16 at 9:32
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I believe there is no generic, universal solution. You have to look at your charger -- why does it require a battery to be always present? One possible reason is that without load, the solar panel will produce voltage which is too high, and thus exceed max input voltage rating. Another possible reason is that the charger will have to dissipate all the solar energy internally and thus will overheat. A third reason is that the controller needs a low-impedance load to be stable -- this is the easiest case, just add a big cap to the output. I cannot really tell without knowing what kind of charger you have.

In any case, your circuit should have two parts:

  • first part will detect if the controller is about to be damaged because the battery is disconnected while the sun is shining (depending on your controller, add a voltage sensor to the input, voltage sensor to the output, and/or temperature sensor)
  • the second part will keep the relay off when there is no voltage on battery input (either because battery was disconnected while it was dark, or because the first circuit has triggered).
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  • \$\begingroup\$ See my comments above, the controller needs to see the battery 1st so it can decide if it's 12v or 24v. \$\endgroup\$ – John U Jun 1 '16 at 15:58
  • \$\begingroup\$ If the statement is your comment ("output could float up to 24v+ and damage a 12v battery which was then connected.") is true, then the circuit is trivial -- you just need a relay which only closes when voltage on BATT terminal is between 5 and 20 volt (and has been this way for few seconds). This way, if the battery is disconnected while the sun is shining, the output voltage quickly increases above 20V and shuts of the panel. If the battery is disconnected while its dark, the output voltage quickly decreases below 5V and solar panel is also shut off. Either way you are safe! \$\endgroup\$ – theamk Jun 1 '16 at 20:49
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enter image description herebattery_presentSense the battery voltage via diode and have a bleeder on the inside of the diode?

The reason being that your normal battery input already have a large cap on it, so if you remove the battery with no load, there will be a voltage present for a long time. Solution - add a new path from the battery with no cap or just a very small filter and a bleeder to drop that voltage down to zero of the voltage has been removed.

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  • \$\begingroup\$ Can you elaborate on that? I can see how that works if the battery is removed, but when the battery is still connected but the cutoff is open you have full battery voltage on the terminal. \$\endgroup\$ – John U Jun 1 '16 at 11:04
  • \$\begingroup\$ If you put a diode in the battery line, how would you ever be able to charge the battery? \$\endgroup\$ – brhans Jun 1 '16 at 11:32
  • \$\begingroup\$ @brhans It's forward biased. If you can guarantee that it's never connected in reverse, you are fine in that sense but you still have a large cap on the battery input of your solar regulator which will be 12 V for a while even the battery has been disconnected. Hence the diode. \$\endgroup\$ – winny Jun 1 '16 at 12:15
  • \$\begingroup\$ I can't work out how your circuit would connect to my system, and how it would sense the case where there's volts present on both sides of the cutoff but the cutoff is open? \$\endgroup\$ – John U Jun 1 '16 at 12:17
  • \$\begingroup\$ I can't see it either. Where would the solar charger go and what would its current path be when charging the battery? \$\endgroup\$ – brhans Jun 1 '16 at 12:19

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