MOSFET as a two-way-switch?

[logicnibble]

Hi!

I'm building a circuit that is powered by mains and have a backup battery (12V sealed lead acid).
The battery powers the circuit when the mains is off (situation 1) and is charged when the mains is on (situation 2).

I would like to switch off the battery in case its voltage reaches its minimum to avoid battery deteoration.
For that I have thought using a MOSFET as a switch in series with the battery.

But, as you might figure it out, in situation 1 there is current flowing FROM the battery and in situation 2 there is current flowing INTO the battery.

Does a single MOSFET (turned on) allow current in both direction, or do I have to use two MOSFETs for both situations? In that case how should be the driving circuit?


Thanks!

 

[ericgibbs]

Hi!

I'm building a circuit that is powered by mains and have a backup battery (12V sealed lead acid).
The battery powers the circuit when the mains is off (situation 1) and is charged when the mains is on (situation 2).

I would like to switch off the battery in case its voltage reaches its minimum to avoid battery deteoration.
For that I have thought using a MOSFET as a switch in series with the battery.

But, as you might figure it out, in situation 1 there is current flowing FROM the battery and in situation 2 there is current flowing INTO the battery.

Does a single MOSFET (turned on) allow current in both direction, or do I have to use two MOSFETs for both situations? In that case how should be the driving circuit?
Thanks!

hi,
Connect the 12V battery via diode to the circuit voltage regulator[+5v.?]

Do you follow.?

 

[logicnibble]

For each situation? 1 or 2?
Doesn't the MOSFET conduct both ways?

 

[ericgibbs]

For each situation? 1 or 2?
Doesn't the MOSFET conduct both ways?

hi,
The MOSFETs are designed as N or P channel, that is a positive or negative Drain supply.
Why would you want to switch FET's when a diode will do the job, easier and a lot cheaper.

 

[Ubergeek63]

Hi!

I'm building a circuit that is powered by mains and have a backup battery (12V sealed lead acid).
The battery powers the circuit when the mains is off (situation 1) and is charged when the mains is on (situation 2).

I would like to switch off the battery in case its voltage reaches its minimum to avoid battery deteoration.
For that I have thought using a MOSFET as a switch in series with the battery.

But, as you might figure it out, in situation 1 there is current flowing FROM the battery and in situation 2 there is current flowing INTO the battery.

Does a single MOSFET (turned on) allow current in both direction, or do I have to use two MOSFETs for both situations? In that case how should be the driving circuit?


Thanks!

A single PFET is fine. If you set it up to turn off the load the parasitic substrate diode will still allow the battery to charge.

 

[logicnibble]

Thanks Eric,

But imagine if there is no mains and the battery has a voltage so low that if it continues to power the rest of the circuit it will be damaged. How would I switch it off?
As in your circuit, there is no way to do that.

Maybe I should rephrase my question:

Is it possible for a MOSFET to conduct both ways?
I've used a circuit simulator (electronics workbench with real MOSFET part number) and it says it's possible. I just can't find any datasheet or theory that suggest it's possible.
Does anyone know about it?

 

[on1aag]

Hi LogicNibble,

A MOSFET will conduct current in both directions but if you
switch it off remember that there is a structural diode inside
the transistor that doesn't switch off !

on1aag.

 

[ericgibbs]

Thanks Eric,

But imagine if there is no mains and the battery has a voltage so low that if it continues to power the rest of the circuit it will be damaged. How would I switch it off?
As in your circuit, there is no way to do that.

Maybe I should rephrase my question:

Is it possible for a MOSFET to conduct both ways?

I've used a circuit simulator (electronics workbench with real MOSFET part number) and it says it's possible. I just can't find any datasheet or theory that suggest it's possible.
Does anyone know about it?

hi,
Look here:
https://www.electro-tech-online.com/custompdfs/2008/06/72952.pdf

I have never used a single MOSFET in a bi-directional mode, I would imagine knowing the Gate/Source voltages it would difficult to achieve.

You could use a relay powered via a zener diode to do a low volts drop out switch.

 

[logicnibble]

Thanks Eric and on1aag!

on1aag:
Yes, you're right, the pn junction acts like a diode and allows current in one direction. That's ok if I make that direction the battery incoming current. My main concern is about battery supplying power when it's deeply discharged.

However, MOSFETs datasheets don't have characteristics for the reverse current operation, even the datasheet that Eric mentioned doesn't have any information about that.
That would be usefull at least knowing the PN juntion forward voltage @ Inom.

 

[ericgibbs]

Thanks Eric and on1aag!

on1aag:
Yes, you're right, the pn junction acts like a diode and allows current in one direction. That's ok if I make that direction the battery incoming current. My main concern is about battery supplying power when it's deeply discharged.

However, MOSFETs datasheets don't have characteristics for the reverse current operation, even the datasheet that Eric mentioned doesn't have any information about that.
That would be usefull at least knowing the PN juntion forward voltage @ Inom.

hi,
Do you mean the pn Vfwd of the diode.?
If yes, about 0.7V to 1.2V depending upon the diode and current thru the diode.
Look at diode junction graphs.

A simple solution is use a low power relay, with say a 6V coil, in series with the coil have a zener diode, about 6V2
Connected between the input [+12v] to the Vreg and 0V.

Adjust the current thru zener/relay so that the relay de-energises at Vbty of 10.8V.
A little trial and error will be required to find the correct current that cause the relay to de-energise.
As the relay de-energises, the relay contacts disconnect the supply to the Vreg and relay coil.
To RESET a manually operated momentary switch is required across the relay contact, which causes the relay to enegise.

Do you follow.?

 

[Ubergeek63]

Thanks Eric and on1aag!

on1aag:
Yes, you're right, the pn junction acts like a diode and allows current in one direction. That's ok if I make that direction the battery incoming current. My main concern is about battery supplying power when it's deeply discharged.

However, MOSFETs datasheets don't have characteristics for the reverse current operation, even the datasheet that Eric mentioned doesn't have any information about that.
That would be usefull at least knowing the PN juntion forward voltage @ Inom.

Actually most newer ones do since they are characterizing them more and more as switching diodes. You get the FET switching characteristics and the intrinsic diode switching and voltage drop.

Many times they are even avalanche rated. These are often shown as having zeners for intrinsic diodes in the symbol.

 

[logicnibble]

Eric,

Your suggestion makes sense, but that would be more expensive than just a MOSFET. Furthermore the MOSFET would be controlled by a uC which reads the voltage by an ADC.

You must be talking about the freewheeling diode that most MOSFET have built-in. That Vfwd must refer to it.
I meant the PN junction between Drain and Body (which is connected to Source).
I thought on1aag was talking about it. Wasn't he?

 

[logicnibble]

I've been doing some breadboard testing and check these results:

P-channel MOSFET:

MOSFET on:
Forward current = 800mA : Vds = -0.156V
Reverse current = 200mA : Vds = 0.050V

MOSFET off:
Forward current: not possible
Reverse current = 200mA : Vds = 0.700V

Comparing Reverse current for ON and OFF conditions, I must conclude that in OFF the freewheeling diode is conducting, having the Vds ~= Vfwd (from the datasheet) while in ON the Vfwd is much lower. But this low Vfwd (0.05V) seems to be using the same path as the forward current, thus the same Rds(on). Is it?

 

[ericgibbs]

hi logic,
I would like to see a circuit diagram of how you plan to connect these FET's within your project... any chance of that..?

 

[logicnibble]

Here it is a schematic of what I was thinking.

If you have mains and battery, the MOSFET will be ON and the curent will flow INTO the battery.
If you turn off the mains, the MOSFET keeps ON but now the current will flow FROM the battery into the Load.

Now check the results I've posted before.

Do you follow?

 

[ericgibbs]

Here it is a schematic of what I was thinking.

If you have mains and battery, the MOSFET will be ON and the curent will flow INTO the battery.
If you turn off the mains, the MOSFET keeps ON but now the current will flow FROM the battery into the Load.

Now check the results I've posted before.

Do you follow?

hi,
Thanks for the drawing, I'll look thru it.
Have you built and tested it yet.?
Which P MOS FET type are you using.?

Lets know how it performs.

 

[logicnibble]

Hi!

I've tested it already with the results posted before.
I've used the P-MOSFET Infineon SPP08P06P.

It actually works, but I'd like to know if it's ok to the MOSFET to conduct reverse current (500mA maximum) or not.

 

[ericgibbs]

Hi!

I've tested it already with the results posted before.
I've used the P-MOSFET Infineon SPP08P06P.

It actually works, but I'd like to know if it's ok to the MOSFET to conduct reverse current (500mA maximum) or not.

Hi,
I have downloaded and looked at the datasheet and found this extract.

It suggests that 500mA would be OK.

 

[logicnibble]

Hi!

The extract you've posted has the free-wheeling diode characteristics.
I believe the free-wheeling diode is put inside the chip along with the MOSFET to protect it when switching (dI/dt).

But from my testing results, it looks like its not this free-wheeling diode that conducts the reverse current but the MOSFET itself.
If the diode is conducting, then why the dropout voltage is different whether the MOSFET is ON or OFF?

 

[ericgibbs]

hi,
As there appears to be no obvious data about this, perhaps an empirical approach may be the best way, providing you dont mind sacrificing a FET.

If I was trying to determine whether it would tolerate 500mA continous I would start at a low current and increase the current in steps upto say 750mA.

Note any changes in the FET and any heating effect.

I suppose if the 'current' did exceed the operating envelope of the FET the Reverse diode would kick into conduction.???

The only other way I would suggest is to contact the manufacturers of the FET by email.

I would be interested in what you discover..