Low Voltage Protector XH-M609 - prevent it from draining the battery in case of low voltage

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GIS

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Hello guys.

I frequently use the (probably Chinese made) XH-M609 boards to protect all kinds of batteries (especially 18V Power Toll batteries who serve as a self made bower bank).

These boards work fine for me. I can set cut-off voltage and hysteresis. All great.

The issue I have:
If the board disconnects the power due to low voltage in the battery, the board itself stays powered and will deep discharge the battery if I forget to disconnect the battery in time (days).

My thoughts so far:
I did get a momentary switch and a SS-Relay (picture attached). With the switch I wanted to enable power to the board for 2-3 seconds, until it checks the Voltage of the battery and could turn on. In case it turns on, I wanted to use the output of the board to feed the input, until battery is low and everything shuts off. To make this happen, I bought a DC SS-Relay (which can handle my source voltage (16-20V).
I however can not manage to wrap my head around the wiring. The issue I have is that the circuit diagram for the relay. It seems to expect the following wiring:
Battery+ -> Load -> SwitchedRelaySide Input [marked +] -> SwitchedRelaySideOutput [marked -] -> Ground. I currently am lost, can't manage to wrap my head around the wiring.

Question1:
What suggestions do you guys have to realize the board totally shuts off in case of low voltage battery?

Question2:
In case you find my approach with the momentary switch and the SS-Relay useful, can you give me some hints in regards to the wiring, please?


Thanks for reading this far


 

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Here is one way.
It requires a couple added diodes to make an OR gate and isolate the SSR control signal.

 
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crutschow said:
Here is one way.
It requires a couple added diodes to make an OR gate and isolate the SSR control signal.
Click to expand...
Ah, OK. Thank you.

The datasheet from the SSR seems to require that the load is before the relay (on the + side). Since I am lacking the understanding of the exact workings of SSR-internals: Do you think it matters if the load is before, or after the relay? Will attach the datasheet I fond for clarification.
 

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GIS said:
Do you think it matters if the load is before, or after the relay?
Click to expand...
No, the output is completely floating so makes no difference, as long the the proper polarity is observed for the SSR (current flow direction from + to - terminals).
 
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Another question, related to the above circuit, but moch more complexity:
(I might open a new thread, but maybe I could get a general guideline, since the circuit is already known here)

In case I would have 2 powers sources (2 power tool batteries):
Could I implement a circuit which would switch between two batteries in the following way:
Battery-Slot 1 and Battery-Slot 2 available. If at least one slow has a battery with high enough voltage: Switch on. If battery voltage drops too low, check voltage in other slot, and if high enough switch over. Stay with this slot, until voltage low and then check if other slot has high enough voltage. If so, switch over, otherwise switch off (without drawing any more power to prevent deep discharge).
 
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GIS said:
Could I implement a circuit which would switch between two batteries in the following way:
Battery-Slot 1 and Battery-Slot 2 available.
Click to expand...
That requires sequential logic decisions, which are likely best done with a microprocessor.
Are you familiar with programming one?
 
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GIS said:
In case I would have 2 powers sources (2 power tool batteries):
Click to expand...
Two appropriately large schottky diodes from the batteries to the switch circuit?
It would then feed from either or both, until the voltage drops to the cutoff threshold.
 
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rjenkinsgb said:
Two appropriately large schottky diodes from the batteries to the switch circuit?
It would then feed from either or both, until the voltage drops to the cutoff threshold.
Click to expand...
If possible, I would like that only one battery is discharged at a time, so the empty one can be exchanged for a full one, while the other one is discharging. With the schotty diodes, I would prevent that one (fuller) battery feeds the one with the lesser charge, but would always discharge both at the same time, correct?
 
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