Boost converter with complimentary N channel

[step_s]

Hi guys.
I'm messing around with the SX1308 boost chip http://www.datasheet-pdf.com/PDF/SX1308-Datasheet-Suosemi-921054 and wanted to test something out, but can't get it to work.
As you can see, I have uploaded a rough design sketch of the setup. The point here is that the additional N channel mosfet in the bottom, will disconnect the internal mosfet body diode from ground, making it impossible for current to travel from the ground and upwards. This N channel will ofc be on when the IC is on, and off when the IC is off.
My problem here is, that the gate of this mosfet needs to driven very high, higher than output voltage in order to work, or how so?

Would be awesome of someone could just point it out to me, since I'm confusing myself right now

[Poneey]

Why are you making things more complicated, when the IC has EN pin, which is stands for ENable? Basically it switches the IC on and off by applying voltage to it (+0.4v and less to turn off, +1.5v and more to turn on the IC), so you don't need an external MosFET to do that. Anyways, if you want to use a MosFET, just connect EN pin to the IN pin (input voltage) and drive MosFET with an external voltage source, or use MosFET with lower gate voltage threshold.
Also

My problem here is, that the gate of this mosfet needs to driven very high, higher than output voltage in order to work, or how so?

you need to drive MosFET's gate by voltage that is specified in MosFET's datasheet, usually around 3-5v or higher.

[step_s]

Thank you for posting

The point here is not to turn the chip off, but to prevent current from flowing through the body diode of the internal MOSFET switch in the IC. That's why the other N channel MOSFET is in reverse direction.
This reminds me of the Lithium cell protection circuitry, to completely cut off current in both directions.

Since the source pin of the mosfet is now opposite, that was why I was wondering about the voltage.
I have tried this approach, but when connecting +4V directly from a Lithium cell to the gate of the N channel, the boost converter will work, but gives up under load, and simply turns off :/

[Buriedcode]

Not sure why you would want/need this, because the body diode's anode is at ground, so current would only flow to the output (when the chip is disabled) when the output is pulled below ~0.6V.  A much more common concern with boost converters is the ability to disconnect the load, because when the chip is disabled, current from the input can flow through the inductor, then the diode to the output.  This is why there are many synchronous boost converters than not only add slightly higher efficiency for low output voltages but also true 'load disconnect'.

[acolomitchi]

...This N channel will ofc be on when the IC is on, and off when the IC is off.
My problem here is, that the gate of this mosfet needs to driven very high, higher than output voltage in order to work, or how so?

Would be awesome of someone could just point it out to me, since I'm confusing myself right now

Yes, you can make a NMOS to conduct source to drain (in reverse) - this is how the schematics pictures it.
But if you don't get it into saturation, that body diode will conduct anyway - you did consider the presence of our body diode, didn't you? That pesky body diode... the "be off when the IC is off" may not hold.

If you are sure that the body diode doesn't screw your logic, to open up a nmos in reverse one needs a voltage about 0.8-1.2V above the V_gs(th) - if my memory serves. I'm desperately trying to find again that wonderful page with explanations and formulae by the heaps, page on which I stumbled about 6 months ago.

[step_s]

The reason is because I have some other components in the same circuit, and it draws current through the body diode of the IC, when the thing is off.
That's why I wanted to completely disconnect the IC, and have the two body diodes in reverse to eachother, in order to prevent ANY current from touching the chip.

@acolomitchi, that's what I'm currently wondering. . . The setup I have drawn, is what I have tested, but the NMOS doesn't seem to turn on completely, and make the Boost IC not work under even smaller loads.
So in theory the source pin shouldn't get very high in this scenario, but if the mosfet doesn't turn on all the way, then the source WILL be pulled higher than 0, so maybe it's an evil spiral?

[acolomitchi]

The reason is because I have some other components in the same circuit, and it draws current through the body diode of the IC, when the thing is off.

I don't know your IC, but are you sure that, if the IC is off, you do have a current towards ground? Is that current large enough?

@acolomitchi, that's what I'm currently wondering. . . The setup I have drawn, is what I have tested, but the NMOS doesn't seem to turn on completely, and make the Boost IC not work under even smaller loads.
So in theory the source pin shouldn't get very high in this scenario, but if the mosfet doesn't turn on all the way, then the source WILL be pulled higher than 0, so maybe it's an evil spiral?

As presented in the attached schematics, the nmos'es source should not get any higher than the voltage drop of the body diode.
If you have other components in series with the NMOS between IC's ground pin and the ground of the circuit, the above may not be true.

[Buriedcode]

Maybe I'm dumb but I still don't get it.   I don't see how current can flow from ground, to the output, unless its input power is removed and the output has a load with a ground lower than the ground of the converter.  Perhaps a schematic of your setup - not the specifics of the converter, just the modules/blocks you're connecting. If the input power isn't removed, and the chip is disabled, then current still won't flow through the body diode, because it will be reverse biased.

It really does sound like you want to disconnect the load, in which case the body diode isn't involved, because if there is power to the input, and the chip is disabled (EN - low) current will flow through the inductor, then diode, to the output.  A common way to do that is to use a PMOS driven by an NPN, which in turn is driven by the EN signal.  When EN is high, the NPN conducts, turning on the PMOS on the output.