I want to run a 12V (65W) fridge from a battery circuit on a boat. However, I want to protect the battery from being drained and potentially damaged by the fridge if the voltage drops too low so I need a switch that will open the circuit if the voltage drops below ~11.5V and close it again once it reaches 12V. What is this sort of switch called so I can search for one?
Electronic – Running a 12v fridge and protecting the battery from low voltage
12vcircuit-protectionlow-battery
Related Solutions
Although your circuit may produce 12 V for your relay in principle, I suspect that the impedance of your R-C network is too high at your load current. I have no calculations to back that up, though.
The issue that I see is danger, in the case of a component failure. In your circuit, a diode failing to short-circuit could put mains onto your low-voltage relay and risk a fire. Unless it is critical to your application to save space or parts cost, I would strongly recommend using a step-down mains transformer and bridge rectifier instead.
As a component, the reliability of a transformer will be extremely high and it will ensure that your load circuit cannot receive a dangerously high voltage. Otherwise you are avoiding a fire or similar damage only if every component works well for the lifetime of the circuit. The cost savings will look very unimportant in that light.
As I imagine you know, electronics and circuit design are only a part of engineering. Electrical safety and reliability are just as important, as are cost, environmental safety, suitability for test, suitability for servicing and a good few more.
I have designed equipment for approval to worldwide electrical standards, and before CE marking made it much easier. We used to export my designs of office equipment into 45 countries. To meet many of these approvals, our mains circuitry (PSUs, cabling) had to stay safe and protected under Single Point Of Failure (SPOF) conditions. This means that an engineer from an approving body would visit and open-circuit any single PSU/mains component or solder a short-circuit across it. The equipment would be switched on and operated and was required to not suffer a fire or present any safety hazard to the user.
In that environment, designing mains circuitry with SPOF protection becomes a lifelong discipline that you always follow. Unfortunately, your circuit is far from safe under such testing. I hope this information helps.
The most cost effective solution, since your processor has an ADC, is to monitor battery voltage and pull enable low when the battery voltage gets to a certain point (determined by code).
simulate this circuit – Schematic created using CircuitLab
When the regulator is enabled, M3 is ON, which turns on M2. When M2 is on, the processor can sense battery voltage through the ADC_IN. When the voltage gets too low, the processor drives GPIO1 high, which turns on M1, pulling enable low, and turning off the power to everything. In this state, the regulator is off, and the battery load is zero (other than the regulator quiescent current).
To turn the processor back on, user pushes button SW1, forcing enable high, and starting up the processor. If you want SW1 to work like a normal power button (on and off) then you need to add more circuitry. I thought it was complicated enough as is, so I didn't do that. See if you can trace everything through as-is.
Obviously this is untested, but I have done similar things in the past in products that shipped in high volume.
Best Answer
What you want to implement is called deep discharge protection for battery. You could find a simple cutoff relay over the internet or build such switch using a Microcontroller which can monitor 'battery's charge status with much higher accuracy. A good way to do this is by measuring both voltage of battery terminals and current. With this peice of information you could esitmate battery's internal resistance that is generally better correlated with depth of discharge.
The key challenge with estimating battery's charge status only by looking at voltage is the fact that terminal voltage is strongly varies greatly depending on the connected load. If a power hungry device draws a lot of current(e.g start of a compressor/motor), the battery's terminal voltage could instantly dip to a pretty low value and eventually recover. While the voltage dropped this does not mean the battery was fully discharged and would cause a false alarm/activation of your relay.
Additionally the device you would use to cutoff battery must implement some form of a voltage hysteresis. Think of it this way, Let's say your battery's voltage dropped to 10v and your circuit cutts off all load, as early as you cuttoff the load the voltage would jump back to 11v, the load would power on again, eventually your load would just keeps turning off/on repeatedly possibly causing more harm to the device you are powering.
Hope this helps.