mosfet 2 source selector help

[johnny78]

hi Guys

i need to select between 2 battery souces to power some lamps

is thery any problem using this circuit or anything else to consider about ?
is it ok to connect the both drains of

the mosfets this way?

thanks for help
Johnny

 

[KlausST]

Hi,

there may be some issues.

1) your schematic includes no GND symbol, thus we don´t know where to reference voltages.

2) a MOSFET decides ON/OFF according it´s V_GS. V_GS means voltage between gate and source.
Since the source in your case is at 12V, you need to drive the gate HIGHER than 12V to make the MOSFET switch ON.
We don´t know your SW1 input voltage ... thus we don´t know if your circuit will operate correctly.

3) As shown in the symbol a MOSFET includes a diode. Thus if you SWITCH ON the higher battery voltage it may automatically cause to "charge" the lower voltage battery. Maybe you want it this way, maybe not.

We don´t know your load current, thus we can´t validate proper circuit operation, nor can we give detailed assistance / recommendations.

We don´t know how your switch is driven. If it is a real mechanical switch, then just use it to select between batteries. no MOSFETs, resistors required.
If it´s not a real switch, then tell us how it really is

***
Btw: the schematic is hard to read. The batteries are drawn upside down. Usually the most negative (supply) voltage is in the lower part of the schematic, the most high (supply) voltage is at the upper part of the schematic. Singal flow, power flow, informatin flow .. should be left-to-right.


Klaus

 

[johnny78]

Hi,

there may be some issues.

1) your schematic includes no GND symbol, thus we don´t know where to reference voltages.

no common Gnd is needed every mosfet source & battery has its seperate negative
in my case if the power source which is bat 1 or bat 2 is unavailable the Gnd for that mosfet source isnt important

2) a MOSFET decides ON/OFF according it´s V_GS. V_GS means voltage between gate and source.
Since the source in your case is at 12V, you need to drive the gate HIGHER than 12V to make the MOSFET switch ON.
We don´t know your SW1 input voltage ... thus we don´t know if your circuit will operate correctly.

the sw1 input will be the positive of the desired battery & i've faced this issue before using mosfet so i put small resistor 56ohm & 10kpull down one
i hope this is ok

3) As shown in the symbol a MOSFET includes a diode. Thus if you SWITCH ON the higher battery voltage it may automatically cause to "charge" the lower voltage battery. Maybe you want it this way, maybe not.

no i dont want it to charge the lower voltage battery & i dont want to use a diode wich will drop voltage as i know
but as i checked the mosfet the internal diode will prevent passing voltage to the other bat
would you please check it again?

We don´t know your load current, thus we can´t validate proper circuit operation, nor can we give detailed assistance / recommendations.

the load current isnt decided yet but this circuit only for the idea if it works i mean the common positive of batteries & load

We don´t know how your switch is driven. If it is a real mechanical switch, then just use it to select between batteries. no MOSFETs, resistors required.
If it´s not a real switch, then tell us how it really is

no i will drive it through MCU & optocouplers

***
Btw: the schematic is hard to read. The batteries are drawn upside down. Usually the most negative (supply) voltage is in the lower part of the schematic, the most high (supply) voltage is at the upper part of the schematic. Singal flow, power flow, informatin flow .. should be left-to-right.

im sorry for that i will enhance it next time

Klaus

 

[FvM]

Presume the battery symbols are drawn correctly, negative terminals at MOSFET source. Then MOSFET diode will charge the other battery if its voltage is at least 0.6 V lower than active battery. To prevent parasitic charging, you need bidirectional switches (back-to-back MOSFETs).

 

[johnny78]

Presume the battery symbols are drawn correctly, negative terminals at MOSFET source. Then MOSFET diode will charge the other battery if its voltage is at least 0.6 V lower than active battery. To prevent parasitic charging, you need bidirectional switches (back-to-back MOSFETs).

so i need 2 mosfets for each channel
is there another solution like mosfet without internal diode ?
as no common point between the inputs can i put mosfets in reverse ?
one drain as outut & the other as input
what do you think ?

thanks for reply

reading this now
https://electronics.stackexchange.c...to-back-mosfets-common-source-vs-common-drain