Power mosfet can be used as bidirectional?

[tomy983]

Hello,

I have two power mosfets ( http://docs-europe.electrocomponents.com/webdocs/0dcb/0900766b80dcb1de.pdf )
They are connected in parallel to each other and they act as a "switch" for a lipo battery. They work as intended, and provide a soft flow of current to the motor controller large capacitors while charging a capacitor on the gate on power on.

I would like to know if I can charge the battery trough them, with current flowing from drain to source..


 

[Nerull]

A mosfet will conduct in reverse. It will conduct all the time through the body diode and it will conduct better if you turn it on. It will be incapable of blocking reverse current, if you need that ability. You cannot turn it completely "off" in reverse.

[tomy983]

Greate, thank you!
Do you think I can charge my battery at 15A without turning it on and, most important, without damanging it?

[suicidaleggroll]

Well you'd be burning off 15*Vsd, so 10+ watts.  Would it burn the FET out?  Depends on your thermal situation.  Why you'd want to skip turning on the FET and waste that much power is another matter.

[max666]

Most mosfets are largely symmetrical, so yes, a lot of them can be used bidirectional. However almost all mosfets will have a body diode, making them able to block current in only one direction.

But you said

I would like to know if I can charge the battery trough them, with current flowing from drain to source..

which is the normal current direction. Did you mean current flowing from source to drain?
If you meant current flowing from source to drain in order to charge your battery, then yes you can do that and charge current will flow through the body diode anyway, and you can turn on the mosfet too in order to reduce voltage drop across the body diode, which would be important for charge monitoring of the cell voltage if you aren't monitoring the voltage right at the cells.

From the datasheet the Continuous Source Current of the body diode is 350 A, so 15 A won't be any trouble, provided you have at least a little heatsinking. But you will be dissipating about 0.7 V * 15 A = 10 W of power in the body diode this way, if you turn on the mosfet too the power dissipation while charging will trop to about (15 A)^2 * 1.85 mOhm = 0.4 W and the voltage drop will drop from about 0.7 V to about 0.03 V.

[wraper]

If you need that switch working in both directions, then you need 2 mosfets in series like on the picture. With only one mosfet, it can block the current flow only in one direction. In other direction current will flow through the parasitic body diode (it cannot handle as much current as the mosfet current rating, usually specified in the datasheet).

[tomy983]

Sorry, I meant to write source to drain

My thermal situation is bad, the mosfets the are enclosed and without heatsink. The act as a switch for my ebike. And it is two of them sharing the load. I normally turn them on with 48v on the gate charging a capacitor (can't even understand if this voltage on the gate is ok or not from the datasheet, can you please advise me on this as well?).

I would like to charge trough them without turning them on because the battery pack is made of two packs in series, and while charging the series is open, so I will not have the votage between gate and source anymore because I take the 48v to the gate from the second pack.

10w looks too much for them, even shared between the two. I think it will be better for me to rewire everything...

[suicidaleggroll]

My thermal situation is bad, the mosfets the are enclosed and without heatsink. The act as a switch for my ebike. And it is two of them sharing the load. I normally turn them on with 48v on the gate charging a capacitor (can't even understand if this voltage on the gate is ok or not from the datasheet, can you please advise me on this as well?).

Those FETs are only rated to +/- 20 Vgs

[max666]

48 V is way too much Gate-to-Source Voltage!!! The datasheet says 20 V Absolute Maximum Ratings. But on the bright side, if your mosfets are holding 48 V Gate-to-Source Voltage, then you can enjoy some nice low R_DS(on) 

10 W power dissipation or even 5 W is way to much without any heatsinking, we are talking roughly 10 W * 40 °C/W = 420 °C (or half of that) junction temperature above ambient!
Edit: Actually I think the body diode has a negative temperature coefficient maybe, don't know for sure. Which would mean in parallel one of them will runaway for sure and carry almost all the current, especially if the aren't thermally connected.

[tomy983]

I made a mistake again.. That switch I made it 3 years ago, and I did put a 67k resistor between V+ and the gate. I now measured my vgs and it is 11v. I should be good then!

And I remenber back then I choosed that mosfet because of the low resistance!

Well, thank you again for your help, I will not reverse the current in the mosfet, even though now I know it is possible with due precaution.