NMOS Diode and current consumption

Hi guys,

I am working on a design with two (continuous) power supply : 12V from external supply and 12V from a battery.
I was thinking about protecting each power supply using a NMOS wired as a diode, in order to reduce current consumption. I have heard of this solution, but never tested it.
I already have BSC028N06NS in my design so I would like to use them.

In case there is a failure in power supply, the battery will then be the only supply. Then the design will be :



I have two questions :
- Is it possible to reproduce the NMOS-Diode behaviour in LTSpice? In the design above, this does not seem to work. I will like to be sure this design is working, as I have never tested it before.
- In this case, how can I calculate/simulate the current consumption of the NMOS transaistor. Shall it be evalute from the value of Rds(on) ?

Regards,

That's no useful diode configuration because it involves several volts voltage drop. An ideal diode a gate bias voltage.

Hi

using a NMOS wired as a diode, in order to reduce current consumption.

Click to expand...

I doubt this reduces current consumption...maybe it reduces voltage drop.

Klaus

Thank you for the answers. I am quite lost on this point to be honest.

I wanted to use a Schottky but the power loss might be too important. I heard about this "ideal" diode model and thought it might help.

I am out of idea, do you have a simple solution ? Linear provided a nice solution with the LTC4357, I have to take a closer look at that.

Regards,

Hi,

You only talk about 12V....but what current? What expected power loss?

Klaus

I expect the load current to be 8A when both powers supply are ON.
As soon as the battery is alone, the current will not exceed 4A at first. My aim is to lower the load current (on the battery) as maximum(>1mA), by shutting down most of the load components.

The LTC4357 provided the perfect solution when power is ON, but it supply current is about 0.5mA to 1.25mA which too much when I run on battery only.

A NMOS ideal diode circuit needs a charge pump, thus the quiescent current. You can use e.g. LTC4416 PMOS ideal diode controller with lower quiescent current. Or design a circuit on your own.

If you expect a 4A load why are you quibbling about the
1.25mA for a charge pump?

What 12V battery are you using, what capacity (Ah)?
Do a calculation of how much you'd lose, of that
capacity, by bleeding a couple of mA out of it over
the required battery hold-up time. Like, if it's a 12V
7Ah battery and you wanted one hour at 4A, you'd
be paying a "tax" of 0.00125Ah besides. Perspective.
What you choose to challenge yourself to do, is of
course your own business. But set some priorities.

These are just expectation as I was thinking both supplies will share the load in normal mode, so 4A each.

My main objective is to protect the power supplies from each other.
But as soon as I run on battery, all functionnalities will be shut off in order to reduce the load as much as possible. I have to last as long as possible in some sort of "saving mode"

Currently, saving mode current should be around 0.8mA, and so having a 1.25mA quiescent current is not acceptable for me.

OK, you want what you want, but I see no evidence
of quantification driving what's "acceptable".

You might consider "rolling your own" peak/hold style
driver for PMOSFETs (although you'd need also to pay
attention to reverse conduction when AC supply is
down) or you could try to make it so rather than
switching from one source to the other, the battery
is always connected and the AC source simply acts
like a "charge controller" and falls back in a benign
way when "unlit". That could be well simpler, choice
of the right AC-DC supply. As an example I have
sitting here, a RV charge/source unit that takes
wallplug power and maintains a 12V lead-acid battery
while also powering vehicle accessory 12V (to the
tune of 100 amps) when plugged in, direct connect,
no switching. Cost less than a good deep cycle battery,
on eBay.