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I'm trying to make a high power UPS, I have a 13.6V PSU and an Inverter and a Lead acid battery.

Everything would be connected this way:

enter image description here

There's a problem with that circuit, if the battery is too discharged the initial charge current will be too high for it, so I need to limit the current while charging and not while discharging.

The simplest solution would be this:

enter image description here

But that solution isn't very efficient. So now the next thing that comes to my mind is using a mosfet as diode, like the reverse polarity protection circuits. I only have N channel fets, so the end result should be this:

enter image description here

However according to LTspice that circuit doesn't work. As long as it is the default Nmos it does work, but if I try to replace it with with any Nmos in ltspice's catalog (Mosfets with a VGS above 20V), the MOSFET still conducts when it shouldn't. Why?

By the way, the MOSFET I want to use is an IRF3205.

Nick Alexeev
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  • Power MOSFETs have a [body diode](https://electronics.stackexchange.com/a/160179/7036) between source and drain. – Nick Alexeev Jul 30 '18 at 00:31
  • Sorry but is that a problem? Putting the body diode in the schematic shows that the circuit should work: https://i.imgur.com/OuJzEl5.png The diode blocks current from the PSU into the battery. Basically I want the mosfet to be on when the PSU is off, and to be off when the PSU is on. Once the mosfet is off current will go thru and be limited by the 1 ohm resistor. – Phoenix1 Jul 30 '18 at 00:41
  • Your N-FET is in reverse. – Ale..chenski Jul 30 '18 at 03:38
  • Are you sure? https://i.imgur.com/lpTzd9t.png There are several images that explain how to use an Nfet for reverse polarity protection. I have it according to those. – Phoenix1 Jul 30 '18 at 04:01
  • Anyway, I placed the Nfet the other way around and according to ltspice it is still conducting current from the supply to the battery. – Phoenix1 Jul 30 '18 at 04:20
  • That's because you have a 1\$\Omega\$ shunt resistor in there which conducts no matter the MOS. Besides, your earlier schematics would not have worked, either, since you place two voltage sources in parallel (unless you specify some `Rser` parasitics into at least one of them). Also, you may wish to change the default NMOS with a readily available model, or at least with one of your own (or subcircuit, whatever works for you). – a concerned citizen Jul 30 '18 at 07:39
  • The point is that the resistor is to limit the charging current of the battery (reduces the peak current from several tens of amps to a few amps), however the mosfet still conducts in reverse, so no reduction in the charging current. I want the mosfet to be close when current is drawn from the battery to the Inverter, and to be open when the PSU is on. So that the charging current that goes to the battery is limited by the 1 ohm resistor. – Phoenix1 Jul 30 '18 at 08:14
  • And the second example does work, when the PSU is on, current flows from it to the inverter, and to the battery thru the 1 Ohm resistor. When the PSU is off (power failure) current goes from the battery, thru the diode (and some thru the 1 resistor as well) to the inverter. Now since the diode is forward biased, the current from the battery isn't limited, however there's still the schottky diode voltage drop that adds inefficiency to the circuit. The solution is to replace the diode with a low RDS on fet, however it allows current from the PSU to the battery when it shouldn't right? – Phoenix1 Jul 30 '18 at 08:25

1 Answers1

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The reverse polarity circuit you're basing it on will protect against reverse voltage. It doesn't care which way the current is flowing.

Your FET is always on because the gate is always positive with respect to the source. You need to sense current separately and switch the FET depending on the direction of the current.

  • I see, thanks. I was thinking about turning the mosfet based on the supply voltage, like this: https://i.imgur.com/tDjopZj.png If the supply voltage exceeds (needs to be ajusted) 13V (that means the PSU is on) then the bjt turns on and drops the voltage to the gate of the mosfet, limiting the current to the batteries thru the 1 ohm resistor. Now the problem I see with that method is that the PSU would need to bring the batteries to 13V briefly for it to work, it may still not work. Could you please explain more deeply about current sensing in this case? – Phoenix1 Jul 30 '18 at 15:55
  • That looks good. Check your BJT B-E voltage over the conditions to see how variable your threshold is. – Cristobol Polychronopolis Jul 30 '18 at 16:02