I've been rethinking the circuit and have made some improvements, before answering.
I attach the new circuit "AutomaticBackupSystem_v2.png".
I think the new circuit will do exactly what I want:
- WHEN POWER IS ON: Connect the PSU to the load, connect the charger to the battery. Not connect the battery to the load.
- WHEN POWER IS OFF: Connect the battery to the load with a boost converter for supplying 12 V at the load.
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.
The load is a kind of router that works with a fixed voltage of 12 V. This means that we must ensure 12 V at the output! I attached in this post the new version of the circuit with a "block" called Boost Converter. This block will transform the 4.2 V battery output to 12 V fixed. The load current varies depending on router consumption (its activity, people connected, etc..). Let's assume that the maximum current is 1 A.
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.
Thank you. I will be carefull.
Due to space constraints, I can only put a single 4.2 V battery, which necessarily need a boost converter (or similar voltage regulation) to provide 12 V to the 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.
As you can see in the attached file "AutomaticBackupSystem_v2.png", I put a diode before the bifurcation of the charger / battery, to prevent reverse power from the battery. I think this is no longer necessary. Is that correct?
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).
I will use a MOSFET because load is VOLTAGE FIXED. On the other hand, I think is more interesting to use a comparator and a MOSFET (instead of a simple diode) because I can improve the circuit by changing the minimum voltage threshold to activate the battery. Also interesting is the effect "hysteresis" of the comparator, which the diode does not provide.
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 will try to use a boost converter because is much more efficient. Do you recomend any one? I'm looking to supply 1 A and to convert 4.2 to 12 V. Better to integrate the MOSFET and the clock inside the chip. Better to be a Through Hole chip..
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 last circuit was bad designed. I need a switch to connect the battery to the load, not to desconnect the battery from the charger!
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.
I've been studying a little bit your circuit. I understand that ICL7665 do almost all the work regarding the LVD. Doesn't it? Why not to use a MOSFETs replacing the relay? I guess you do not have many space constraints... Relay is wasting energy when power is on, isn't it? Very advanced circuit...
Which program do you use to draw the schematics?
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.
Thank you very much! It is a good MOSFET.
Can I use correctly an NMOS in the new position (between the battery and the boost)? I am trying to get a good switch, but here is not necessary so much current so I think is a little oversized to put two NMOS (for losses and for saving the heatsinks). I want the circuit not too large. I have space constraints.
