Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

MOSFET linear and saturation region operation

Status
Not open for further replies.

BoopathiS

Member level 2
Joined
May 18, 2018
Messages
44
Helped
1
Reputation
2
Reaction score
0
Trophy points
6
Activity points
483
Hi,

N channel MOSFET Saturation operation condition is Vds > (Vgs - Vth).
But in saturation region, MOSFET will act like closed switch. hence Vds is approximately 0V(Rds*I). Examble VGS = 10V and VTH is 2V, and if switch is closed means VDS= 0V, but above formula is not satisfied.

Anyone kindly clarify this ?

And How to drive the N channel MOSFET to act as variable resistor ?
I saw the condition Vds < (Vgs -Vth), but how to calculate Vgs for required resistor ?
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
691
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,011
i searched on "international rectifier fet gate charge" (google)

look at


International rectifier power mosfet basics
figure 13
Power MOSFET Basics - Infineon Technologies
https://www.google.com/url?sa=t&rct...57444e913f4f&usg=AOvVaw30EmxL7qVIhaoOEA610FJ-

International rectifier application note AN-944
figure 5
Use Gate Charge to Design the Gate Drive Circuit for Power MOSFETs ...
https://www.google.com/url?sa=t&rct...3ba630970081&usg=AOvVaw0CQyDMLtfZe6deoAi2juEK

those figures show the behavior of fet as a function of gate charge
there are clearly indicated regions - off, turning on, linear, and on (saturated)

to make the fet act like a variable resistor, apply a feedback circuit to control the gate based on the desired ???
sometimes it helps to add a resistor across the drain source to be the maximum resistance
then, as the fet moves through the linear regions, the resistance drops
the saturated fet gives the lower limit of the resistance
 

Chinmaye

Advanced Member level 4
Joined
Jan 18, 2016
Messages
119
Helped
0
Reputation
0
Reaction score
1
Trophy points
18
Activity points
944
MOSFET behaves like a resistor in linear region with resistance = 1/(un*cox*(W/L)*(Vgs - Vth)) for small values of Vds.
 

sherline123

Member level 2
Joined
Mar 9, 2016
Messages
46
Helped
1
Reputation
2
Reaction score
1
Trophy points
8
Activity points
328
Hi,

N channel MOSFET Saturation operation condition is Vds > (Vgs - Vth).
But in saturation region, MOSFET will act like closed switch. hence Vds is approximately 0V(Rds*I). Examble VGS = 10V and VTH is 2V, and if switch is closed means VDS= 0V, but above formula is not satisfied.

Anyone kindly clarify this ?

And How to drive the N channel MOSFET to act as variable resistor ?
I saw the condition Vds < (Vgs -Vth), but how to calculate Vgs for required resistor ?
MOSFET acts like a switch and it is opened when vgs < vth but closed when vgs > vth.
From simple MOSFET equation, ignoring body effect and etc.
When Vds < vgs - vth it acts like a variable resistor. Current allowed to pass through will be directly proportional to Vds.
But when Vds > vgs - vth, it will be saturated and the current allowed to pass through will be saturated, no matter how high is the Vds.
 

Akanimo

Advanced Member level 2
Joined
Aug 30, 2016
Messages
624
Helped
111
Reputation
222
Reaction score
109
Trophy points
43
Activity points
4,369
Hi,

In saturation region, with the constraint you provided, namely: Vg > Vth and Vds > Vds - Vth, the MOSFET is an amplifier.

This is a major difference between the MOSFET and the BJT.
 
Last edited:

crutschow

Advanced Member level 5
Joined
Feb 22, 2012
Messages
3,737
Helped
891
Reputation
1,780
Reaction score
869
Trophy points
1,393
Location
L.A. USA Zulu -8
Activity points
20,988
N channel MOSFET Saturation operation condition is Vds > (Vgs - Vth).
But in saturation region, MOSFET will act like closed switch. hence Vds is approximately 0V(Rds*I). Examble VGS = 10V and VTH is 2V, and if switch is closed means VDS= 0V, but above formula is not satisfied.

Anyone kindly clarify this ?
Perhaps this diagram from Wikipedia will help clarify:

Capture.PNG

And How to drive the N channel MOSFET to act as variable resistor ?
I saw the condition Vds < (Vgs -Vth), but how to calculate Vgs for required resistor ?
There's no easy way to calculate the Vds required for a required resistor value since there's a wide variation in the Vgs threshold voltage between MOSFETs, even those of the same part number.
One way is to measure the transistor you have to determine the resistance versus Vgs curve.
 

asdf44

Advanced Member level 4
Joined
Feb 15, 2014
Messages
1,006
Helped
353
Reputation
706
Reaction score
346
Trophy points
83
Activity points
8,410
One question I have when someone understands this stuff is why?

A mosftet goes from 'as off as it gets' -> partly on -> 'more on' -> 'mostly on' -> 'as on as it gets' as Vgs increases. But the Vgs threshold is so widely specified you can't count on anything in the linear region. So the exact relationship and the equations governing it are typically irrelevant. I've designed a lot of linear circuits and I've certainly never referenced any.
 

decaf14

Newbie level 5
Joined
Jul 15, 2019
Messages
10
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
173
The MOSFET acts as a zero-voltage drop device (ideally) IF it is connected to a load. If the MOSFET is by itself, it will of course have a voltage at the drain.
 

crutschow

Advanced Member level 5
Joined
Feb 22, 2012
Messages
3,737
Helped
891
Reputation
1,780
Reaction score
869
Trophy points
1,393
Location
L.A. USA Zulu -8
Activity points
20,988
One question I have when someone understands this stuff is why?

A mosftet goes from 'as off as it gets' -> partly on -> 'more on' -> 'mostly on' -> 'as on as it gets' as Vgs increases. But the Vgs threshold is so widely specified you can't count on anything in the linear region. So the exact relationship and the equations governing it are typically irrelevant. I've designed a lot of linear circuits and I've certainly never referenced any.
Because that describes how the MOSFET responds to the gate-source voltage in the linear region.
Just because you've never used that info doesn't mean no-one does.

Besides, the info in the "linear region" is typically used when the MOSFET is used as a switch or variable resistance.
For linear (i.e. AC) amplifiers the MOSFET typically operates in the saturated region.
(It can be confusing because that's more or less opposite of the BJT definitions for the two operating regions).
 

asdf44

Advanced Member level 4
Joined
Feb 15, 2014
Messages
1,006
Helped
353
Reputation
706
Reaction score
346
Trophy points
83
Activity points
8,410
Because that describes how the MOSFET responds to the gate-source voltage in the linear region.
Just because you've never used that info doesn't mean no-one does.

Besides, the info in the "linear region" is typically used when the MOSFET is used as a switch or variable resistance.
For linear (i.e. AC) amplifiers the MOSFET typically operates in the saturated region.
(It can be confusing because that's more or less opposite of the BJT definitions for the two operating regions).
Eh my point is that once you wrap it with feedback the exact equations become irrelevant.

And if you don't wrap it in feedback part variations mean you better not care about the exact relationship either.

Both cases are well served by understanding the shapes of the curves but don't necessitate plugging numbers into the equations.
 

crutschow

Advanced Member level 5
Joined
Feb 22, 2012
Messages
3,737
Helped
891
Reputation
1,780
Reaction score
869
Trophy points
1,393
Location
L.A. USA Zulu -8
Activity points
20,988
Eh my point is that once you wrap it with feedback the exact equations become irrelevant.

And if you don't wrap it in feedback part variations mean you better not care about the exact relationship either.

Both cases are well served by understanding the shapes of the curves but don't necessitate plugging numbers into the equations.
Okay.
But that's generally true for any parts in a feedback loop, not just MOSFETs.
It you don't need the equations, don't use them.
But they are necessary for a fundamental understanding of the devices.
 

std_match

Advanced Member level 4
Joined
Jul 9, 2010
Messages
1,083
Helped
409
Reputation
818
Reaction score
404
Trophy points
1,363
Location
Sweden
Activity points
8,276
I have designed power outputs with MOSFET's where the short circuit protection was implemented using the Ids/Vgs diagram in the datasheet.
By not switching on the MOSFET to reach the minimum Rds, it will act as a current limiter.
To allow for temperature and individual variations, we adjusted Vgs so the maximum current in the MOSFET would nominally be 3 times the maximum load current.
When a short circuit occurs, the voltage will increase a lot over the MOSFET and that is very easy to detect, and Vgs can be switched off early enough to keep the MOSFET within the SOA (Safe Operating Area).
 

Status
Not open for further replies.
Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top