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Will the mosfet turn on? I say it will

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pete16

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Hello

Looking at the image will the P-Channel Mosfet conduct through the channel , through the body diode or both after the switch is turned on?
Assume Vgs threshold is -1.2V. and the switch is initially off.
Question.jpg
 
Solution
Hi,

To turn ON the MOSFET V_GS needs to be less than -1,2V. More negative. -3V for example.
V_GS is V_gate - V_source.

Now gate is connected to GND = 0V
Thus - to turn ON the MOSFET: V_source needs to be higher than +2V. (More positive)

Without the diode .. it's difficult.
But due to the diode, the expected source voltage will be about 12V - V_diode = about 11.4V.

V_gs = V_gate - V_source = 0V - 11.4V = -11.4V.
This is more negative than -1.2V thus the MOSEFT will be ON ... driving V_source even higher in direction of +12V.

Klaus

KlausST

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Hi,

To turn ON the MOSFET V_GS needs to be less than -1,2V. More negative. -3V for example.
V_GS is V_gate - V_source.

Now gate is connected to GND = 0V
Thus - to turn ON the MOSFET: V_source needs to be higher than +2V. (More positive)

Without the diode .. it's difficult.
But due to the diode, the expected source voltage will be about 12V - V_diode = about 11.4V.

V_gs = V_gate - V_source = 0V - 11.4V = -11.4V.
This is more negative than -1.2V thus the MOSEFT will be ON ... driving V_source even higher in direction of +12V.

Klaus
 

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FvM

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The Vgs = 0V so the Mosfet is turned off.
Did you notice that it's a PMOS? Respectively it turns on with positive battery voltage because Vgs is at least equal to -(Vbat - Vdiode). A commonly known power supply reversal protection circuit. With negative battery voltage Vgs = 0.
 

crutschow

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The current will initially flow through the substrate diode until the voltage reaches the Vgs(th) of the MOSFET,
Then the MOSFET will start to turn on and reduce the on-voltage to the MOSFET on-resistance times the load current.
This works because a MOSFET conducts equally well in both directions when ON.

LTspice simulation below:
Note the kink in the output voltage (green trace) due to the transition from diode conduction to MOSDFET conduction at the MOSFET threshold voltage after the voltage goes positive.


1643266632721.png
 

james.qiu

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it is a standard reverse polarity protection circuit in automotive electrionic, when battery on, mosfet will will be turned on and if reverse the battery mosfet will be turn off.
 

dick_freebird

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DMOS devices are asymmetric but at low current / voltage
the VT should be symmetric. The "neck" and drift region
account for voltage standoff. Inverted operation will have
dismal low breakdown as there's neither on the source
contact side.

Planar LDMOS similarly.

Even single-diffused planar MOSFETs may be made with
asymmetric attributes (drain extension for breakdown, a
"poor boy LDMOS" in effect. But standard CMOS devices
do tend to be built fully symmetric outside of special
cases. I've seen some processes where you can have
no extension, drain extended, double extended (and
you could extend the source alone, by the rules, but
the only time I've seen that useful is in trim networks
where normal operation is "right side up" but during
zener zapping or fuse blow flyback, the operation is
"upside down" and the source needs to stand off the
"event voltage".
 

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