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I have two wall wart 240 - 12 V power supplies rated at 1.3 amps and 1.2 amps. My project uses a couple of solenoid valves, relays, a 12 V pump and some fans. All these will bring peak consumption to around 1.5 amps (and the pump may pull this a little higher if working hard).

My plan was to take these two power supplies and hook them up to an LM2596 buck regulator circuit in order to increase the available current. I say LM2596 because it's the only regulator IC I have at the moment, and it seems they can be run at 'unity' 12-12 V according to the datasheet here. Considering my circuit will be well within the total final current output, are there pitfalls to this approach? I actually will have some buck/boost regulator ICs soon, so would this be preferable? Thanks!

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    \$\begingroup\$ Do the wall warts output AC or DC? \$\endgroup\$ Jul 27 '16 at 20:00
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    \$\begingroup\$ What function do you intend for the LM2596 to achieve? \$\endgroup\$
    – brhans
    Jul 27 '16 at 20:09
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    \$\begingroup\$ You didn't link to the datasheet - you linked to a datasheet site instead. Put the direct link to the Linear site in your post. \$\endgroup\$
    – Transistor
    Jul 27 '16 at 20:30
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Putting power supply voltage outputs in parallel is not ideal. One device will always win i.e. it outputs 0.1 volts more than the other and the lower output supply will just shut down its output stage thinking that it has to find a way of lowering the output voltage.

My advice would be to connect them in series to produce ~24V then use a buck regulator to give you 12V. The maximum current into the buck will be limited to the lowest current but, even so, with an efficiency of maybe 90% you should easily get 2A from the 12V buck regulator.

Another consideration is what happens when one power supply is disconnected from AC. If in parallel you may stand a chance of damaging one or both. In series you can use bypass diodes to prevent this happening plus you have the voltage overhead to suffer a diode drop or two without breaking into a sweat.

My plan was to take these two power supplies and hook them up to an LM2596 buck regulator circuit in order to increase the available current.

This is madness and makes no sense. If the wall-wart(s) produce 12V then why bother with a buck regulator - the LM2596 needs a couple of volts higher on the input than the output anyway so your idea about 12Vin and out is fallacy. Read the DS.

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  • \$\begingroup\$ Please correct me if I am wrong but wont connecting the supplies in series will double the voltage but the current would remain at the lowest specd power supplies maximum current? EDIT: Nevermind.... P = IV so the voltage will double meaning the required current will halve. Is this correct? \$\endgroup\$
    – bitshift
    Jul 27 '16 at 20:32
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    \$\begingroup\$ Available current at 24V is 1.2 amps therefore power = 28.8 watts. Then lose 3 watts for inefficieny leaving ~26 watts so output current should easily exceed 2A at 12V. \$\endgroup\$
    – Andy aka
    Jul 27 '16 at 20:37
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My plan was to take these two power supplies and hook them up to an LM2596 buck regulator circuit in order to increase the available current. ... and it seems they can be run at 'unity' 12-12 V ...

If your converter input and output voltages are both 12 V you will need more current into the converter than what you draw out of it. Efficiency is always less than 100%. Again, you can't get more current out than you put in (at the same voltage).

I have two wall wart 240 - 12 V power supplies rated at 1.3 amps and 1.2 amps. My project uses a couple of solenoid valves, relays, a 12 V pump and some fans. All these will bring peak consumption to around 1.5 amps (and the pump may pull this a little higher if working hard).

Split the loads between the power supplies. The 1.3 A supply will run the pump. If the voltage sags a little on start the pump acceleration will be a little sluggish but as the current falls off it will pick up again. Power everything else from the other supply.

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