Given (as others have said), I think you will need some hysteresis and in all probability at least a comparator for the "mains switch", I would use a solid-state relay to switch the mains side, as its "coil" will only require 1-2mA
The DC switching for your up-converter, using two relays seems unnecessarily wasteful too. You could do away with one relay for a start, by using one pole (as indicated by others) only, leave the (say) ground side connected on input and output and switch + through the relay contacts.
Check your up-converter - some (but not all) draw negligable current when they have no output load. If yours is one of them, you can probably leave the input connected permanently.
Again, not universal, but some supplies have internal output isolation so applying power to the OUTPUT doesn't actually draw any power or cause any harm. IF your supplies (both the PC supply, and the 12->20V converter) come into this category, you can dump the output switching relay entirely and just hard-wire the outputs in parallel
The other option might be to use FETs to switch your DC side. At the fairly modest currents you are indicating, the FETs will have very low resistance and waste very little power. Downside is that their "very little waste" will be for the whole time the system is running on DC.... so do your maths carefully - if the system will be running on mains for say 0.1% of the time, even 0.1 watts wasted in the fets may be more than the coil would take for the 0.1% of the time it is running...
The mains relay I've selected is 140mW (ie about 12mA) to drive the coil, and the main advantage here is that when there is NO power on the DC side it switches the mains on.
A agree that if the inverter and the mains adaptor both are protected against back EMF then I could hardwire them in parallel, but they are both black boxes that I can't open and don't wish to break!
Your FETs idea is a good one; I'll certainly think about it again before actually placing the order for the relays.
Do you have any particular parts in mind or an example circuit? I'm not sure that I have ever actually used a stand-alone power FET though I've drooled over the data books enough, sad git that I am! B^>
Rgds
Damon[ Parent ]
OK, I'm getting my act together slowly, and I can run the laptop via the DC/DC converter and it draws ~21W from the 12V battery, nice!
Problem is, the start-up current of the converter is so high that it immediately triggers the load short-circuit protection on the solar controller. So I can't just treat it as a normal load from the controller.
Therefore I'm going to have to modify my circuit be its own low-voltage drop-out with power drawn direct from the battery, and I think I'll have to use FETs as you suggest.
Can I simply put a P-channel FET in the positive line from the battery, with the gate pulled up to the battery positive voltage with a resistor, and 'let go' to cut the current to the load (and pull down with an open-collector output to allow current to the load)? Or will that move the gate voltage too slowly and cause it to overheat?
My DC/DC converter has a peak load of 150W (though I shall never go over about 60W) and an input fuse of 15A, so allowing healthy margins all round I've picked out an Infineon SPU30P06P 30A (60V max), and would pull up its gate with something like a 10K resistor. I'd pull it down via an open-collector 'voltage OK' output in series with an LED for a small 'current on' indicator.
I propose to set the drop-out voltage at 12V and the reconnect voltage at 12.8V to 13V for some decent hysteresis.
Does that sound sane?
Never used a bare p-channel power FET before... Your guidance would be much appreciated.
