First let me start by saying that a 40 AH battery is a very large battery to be charging with a 250W PSU. Assuming your PSU supplies only 12V then you are getting about 20A of power out of it. At max charging potential you would be able to charge the battery in 2 hours, but since the load is going to demand some power as well it could take longer. You said that the load only draws 600mA from a 12V source so why do you need a 40 AH battery to power it? It's rediculously over powered and impractical unless you plan on running the load off of the battery for months at a time. Since you said that the PSU will be on for a while each day, you can use a much smaller battery that will charge more quickly and will still deliver the same amount of power needed for such a short time frame.
As for the LVD circuit, I will say
that in the second schematic you have posted there are a lot of things that need to be fixed. Starting at the battery charger end of the schematic, you have bypassed the battery charger with a diode so you are forcing the battery to charge whith what ever voltage you are supplying to the circuit. This is a good way to destroy your battery or to compromise the performace of your circuit. You need to have a diod from the positve terminal of the battery to the power bus.
I am also assuming that you are just using a lead acid battery to power your load, in which case you don't really need a complex circuit to charge your battery, you only need to make a trickle charger. If you are using NiCd or NiMH batteries, then your home made charger could destroy them and make them useless. If you are using any sort of lithium battery, you may want to have an emergency crew standing by for the fire that is likely to follow your first charge.
Secondly, your power MOSFET is impropperly biased, you need to swithc the source and drain pins other wise the internal power limiting diode is just bypassin your battery and the MOSFET is completely useless.
A low voltage detector is more practical on a battery powered aplication than one that usese a PSU for power. That being said, your circuit would be much more efficeint if you simply had a 12V battery that was charged with a 15V PSU and having the load connected across the battery. Then have the battery constantly powering the load, even when the PSU is connected. While the PSU is on, it will charge the battery and power the load at the same time like you want it to, but when it's removed the battery will begin powering the load without any hickup in voltage. If you try to use an LVD with a battery in that configureation, then the voltage across the load is going to drop a considerable amount untill it hits the trip point of the LVD. Instead, use the LVD to detect if the battery voltage has droped too low and have it disconnected from the circuit so that it can charge back up again once the power is applied.
Here is an example of one of my power supply designs that uses this principal:
https://www.electro-tech-online.com/attachments/radio-power-jpg.56332/
The only instance I can think of that would have a load drawing a specific wattage would be a light bulb of some kind, or a heater. In both cases the wattage demand of the load will decrease as the voltage decreases unless there is a circuit in place to modify the current going through the load. It would help if we could get some more info about the load attached to the power supply. The only thing I can think of that would run with both 12V and 4.2V is maybe digital logic if there was a voltage regulator in line. Other wise the lower voltage would cause the circuit to behave in very strange ways or not even work at all. Just because the load's specificatons at 12V are 600mA and 7.2W doesn't mean that they are the same at 4.2V. It is much more likely that the load will draw 2.52W at 4.2V if the current is fixed. Other wise the current draw will increase from 600mA to 1.7A if the power is fixed. Keeping this in mind your battery is a 40AH battery so it will deliver power for 40/1.7= 23.5 hours or 40/0.6= 66 hours. So your battery is more than large enough.
You are using Lithium batteries in your supply, this is a good choice because of the power they deliver for their size and weight, but be carefull not to charge or discharge them too quickly because the can catch fire or explode. If the do, you won't be able to put it out. Burning lithium has it's own oxidizer so it can burn just about anywhere. The good thing about your lithium batteries is that they have plenty of power to give to your load.
You only need the diode that is bypassing the battery charger if you have components in the PSU that are sensitve to voltages when they are turned off. It's probably a good idea to include the diode anyway.
When you are trying to switch the voltage from the PSU to the battery you have to consider the two possible devices you can use in your circuit, a power transistor or a power MOSFET. The choice will depend on what is more importatnt to your load, voltage or current. If current is more important then you will want to use the power tranisistor since it does not have a fixed current drop across it but it does have a fixed voltage drop. Depending on where you have the transistor located it can be either a 0.7V drop or a 1.4V drop. If voltage is more important to your load then MOSFET's are the way to go. The don't have a fixed voltage drop but they do have an internal resistance that will handicap your load at higher currents. This can be worked around by connecting several MOSFETs in parallel so that only a fraction of the total circuit current is going through them. It is exactly the same as putting resistors in parallel to reduce the voltage drop and increase current. If you are simply using a diode, then there is no real need to have any sort of a sensor control the source of power. As soon as the PSU voltage drops below 3.5V the battery will kick in and start driving the load (4.2-0.7=3.5V).
If your load needs a specific voltage or current you can make better use of the batteries by taking advantage of a buck/ boost system. If you connect all of the batteries in parallel then you can boost their voltage from 4.2V to 12V using a boost converter if you are willing to sacrafice some current (pretty much irrelivent right not). If you connect all of the batteries in series you can get an output of 12V or 4.2V and get more current from the conversion. Buck and boost converters have about a 95% efficiency in their conversion so they waste very little power. A voltage regulator on the other hand will waste anywhere from 1W to 30W depending on the conditions.
I would not put a switch between the battery and the load especialy if it needs to switch over very quickly. As I said just having a diod there will allow the battery to kick in and continue powering the load once the PSU has been turned off. If you do, you don't need a voltage comparator to activate the battery, it will activate itself.
The circuit that I attached is a power supply that I am working on for my radio. I have the load connected directly to the battery (not bypassing the sense resistor for the battery charger). The relay that is shown imidiately after the battery is a bypass relay. As long as the unit is pluged into the wall, the bypass relay is engaged and will allow current to flow around the switch that is in series with the load. The switch is there so that I can turn the unit off when it is running off of the battery to make it last longer. The MOSFET that is connected to the negative terminal of the load is there to ensure that the battery has adequet protection from excessive discharge. If the voltage drops below 9V it warns that the voltage is getting low, if it drops below 8V it disconnects the battery from the load so that it can be recharged once power is reapplied. If the voltage drops below 1V per cell then the battery charger will not recharge the batteries. As long as there is AC power applied, the sensor will allow the load to be powered all the time. The microcontroller is simply there to make sure that everything is opperating properly. As soon as something goes wrong, it sends a signal to the rest of the radio about what has just gone wrong and the a request on how to fix it. The radio will respond appropriatley.
Regarding your orrigonal post, a MOSFET that is capable of handeling 20A is the one that I used in my schematic. The IRL3103 can handel 63A and voltages up to 32V. Just make sure you have an appropriate heat sing connected to it.
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