Mosfet Linear Mode

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But MOSFETs designed for switch-mode power supplies can experience a wide temperature variation over different areas of their die. Optimized for on/off switching, they typically don’t work well in their linear region. A typical failure mode for a MOSFET is a short between source and drain.

Meanwhile, the MOSFET drivers also need to drive the MOSFETs between switching states as fast as possible to minimize the amount of time a MOSFET is in linear mode—the state between cut-off mode and saturation mode where the MOSFET is neither fully on nor fully off and conducts current with a significant resistance, creating significant heat.
MOSFETs working in linear mode can withstand very high power dissipation levels, due to simultaneous high current (I D) and the voltage across it (V DS). When a MOSFET works as a switch, it passes continuously from OFF state (high V DS but zero current) to ON state (ohmic or R DS(on) region). During these transients, the device stays in linear mode for a.

For someone trying to learn about FET’s, it seems there is no consistency in terminology (unless i am reading things wrong?)

Trying to understand a FET as a switch (for power applications). It has become clear that you want to operate a switch in the ohmic region. To get to the ohmic region you would need to go from cut off -> saturation -> ohmic region.

TI refers to it like this in an app note: https://www.ti.com/lit/ml/slua618a/slua618a.pdf?ts=1613979703351&ref_url=https%253A%252F%252Fwww.google.si%252F

From this I implied that the ‘linear’ region is in fact rather confusingly the saturation region of the FET as shown all across the web for a FET curve:

However, when i go to wikipedia i came across the following image:

This is saying the exact opposite of what i understood which is saying the ohmic region is the linear region?

Infineon has an app note around FET switching, titled linear mode operation (https://www.infineon.com/dgdl/Infineon-ApplicationNote_Linear_Mode_Operation_Safe_Operation_Diagram_MOSFETs-AN-v01_00-EN.pdf?fileId=db3a30433e30e4bf013e3646e9381200#:~:text=Many%20applications%20exist%20with%20the,to%20source%20(VGS)%20voltage.)

They refer to the linear mode as saturation region as shown from this :

So what is it then?

Also just to be clear is the reason we want to minimise time in the saturation region as a FET because here we have a VDS and an ID current flowing through the FET? Whilst as soon as the fet reaches linear region, it’s VDS would fall close to 0 and hence minimise losses?

When we say the FET is saturated, it means for a given vgs, vds makes no difference to drain current (because channel can’t let more electrons to flow?) Shouldn’t the ohmic region be called saturated? Since increasing VGS makes no difference to drain current as the ID is now limited by the circuit and not the FET?

Edit: Answers in electronic stack exchange say it is the saturation region:

Mosfet Linear Amplifier

link2mosfetfetsaturationlinear-regionShareCiteEditFollowFlagedited 2 hours agoasked 2 hours agoHasman40414111 silver badge99 bronze badges

#Hasman404, When I first read the second paragraph of your question, my impression was that you have mixed up (messed up :)) all the terms/concepts! The root cause of your confusion/inconsistency is that you have not differentiated between linear mode and linear region. So you won’t appreciate that for a MOSFET, you can operate in linear mode in the saturation region. This picture hopefully helps: i.imgur.com/CG5leI1.jpg. Happy thinking! Cheers. – tlfong016 mins ago Edit

2 Answers

ActiveOldestVotes3

So what is it then?

This is the correct graph: –

Taken from this wiki page.

Saturation refers to the channel being saturated and, as you said, no matter what VDSVDS you apply, current remains constant. It is sometimes also referred to as the active region (not to be confused with the MOSFET being activated or ON).

The linear (triode) or ohmic region is when the MOSFET is used as a switch (ON). The top graph is basically wrong because it doesn’t correctly show the different slopes in the ohmic region when you apply different gate voltages. This region is called linear because there are different slopes that are governed by the gate voltage and means the MOSFET can act like a variable resistor hence, it gets the name linear but, different texts use this term rather loosely.

Mosfet Linear Model

Yes, we want to minimize time in the saturation region because that is when the MOSFET is dissipating the most power and potentially operating below it’s ZTC (zero temperature coefficient) and may suffer rapid thermal runaway.

Shouldn’t the ohmic region be called saturated?

No, because in the ohmic region the channel isn’t saturated. However, for a BJT, that equivalent part of the characteristic is called the saturation region but, for different reasons; for a BJT, it is the base that becomes saturated. Same name, different mechanism, different part of the characteristic.ShareCiteEditFollowFlagedited 2 hours agoanswered 2 hours agoAndy aka330k1818 gold badges266266 silver badges575575 bronze badges

The two links i have provided refer to saturation as the linear region. Also is TI in the snippet and link to the app note not referring to the linear region as the saturation region when they say ‘it must go through it’s linear region’. – Hasman4042 hours ago
The so-called “linear” region does tend to be misused in some texts. And, I think that’s because in a BJT, the saturation is sometimes referred to as the linear region. People forget that BJTs and MOSFETs (despite having similar looking characteristics) have a different saturation mechanism. – Andy aka2 hours ago
That is really annoying.. Infineon has a FET application note called linear operating mode (infineon.com/dgdl/…) They also refer to saturation as the linear region! – Hasman4042 hours ago
Many share @Hasman404 ‘s confusion. I think it stems from the simple fact that (on your graph) the linear region is populated with curved lines (non-linear), whilst the saturation region is populated with straight lines (linear). It’s just exacerbating the confusion… – Paul Uszak1 hour ago
@PaulUszak those straight lines you mention are the most non-linear part of the MOSFET characteristic. The curved lines (despite them being curved) are more linear in terms of ohms law. – Andy aka1 hour ago

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Your answer is approximation.

The following is a graph of output characteristics for an arbitrary 2N7000 MOSFET:-

Assume a fixed gate drive VGSVGS of 5V and ignore all the other curves. So it’s fixed. Consider the MOSFET then as an opaque component with only two legs, Drain and Source.

As VDSVDS (voltage across component) changes, IDID (current through component) changes. And they change almost linearly if you squint sideways at my graph (purple line). Like a resistor. The first part increases linearly, and then goes flat at saturation when it no longer behaves as a resistor. These are approximations, but form the basis of the terminology in your question. The difference is simply real world characteristics being more complex than theoretical approximations. And confusing graphs.

Actually the initial slope is ∝VGS−VT∝VGS−VT where VTVT is the threshold voltage.ShareCiteEditFollowFlaganswered 50 mins agoPaul Uszak6,37633 gold badges2828 silver badges5656 bronze badgesAdd a comment

Ask QuestionAsked 3 days agoActive todayViewed 1k times43

I am trying to understand the curves of a MOSFET. Sorry if the question is very basic.

Where the red point is is the saturation zone of the MOSFET, therefore the source drain voltage must be 0V because at this point the MOSFET is saturated conduction at maximum current, because on the X axis of the graph called Vds marks 10V for the red point.

transistorsmosfetShareCiteEditFollowFlagedited Mar 10 at 9:49JRE46.3k88 gold badges7474 silver badges125125 bronze badgesasked Mar 10 at 7:28Mario9566 bronze badges

5therefore the source drain voltage must be 0v No, the red dot is at the point where VDSVDS = 10 V, see the X-axis of the graph. The source drain voltageis VDSVDS. Look up when a MOSFET is in saturation, there is an equation which tells you that VDSVDS needs to be larger than a certain value. – BimpelrekkieMar 10 at 7:57
5You may be confusing “saturation” in a bipolar transistor with “saturation” in a MOSFET. Unfortunately they mean practically the opposite phenomenon but have the same name. – Brian Drummond2 days ago
2@BrianDrummond Indeed. When I first learned this stuff (in the 1960s) the term I encountered was “pinch-off” rather than “saturation”. I think “pinch-off” is both closer to the physics and less confusing. – John Doty2 days ago
2@JRE What is the question now? – CGCampbell2 days ago
2@CGCampbell: There never was a question, just statements. There was a question mark, but it was at the end of a statement rather than a question so it was simply improper punctuation. – JRE2 days ago

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4 Answers

ActiveOldestVotes7

therefore the source drain voltage must be 0v because at this point the mosfet is saturated conduction at maximum current

No, you have this wrong. Maybe you were perhaps thinking of the BJT saturation region (when the collector-emitter voltage is close to 0 volts)? If so, then you’d be correct but, it’s the other way round for a MOSFET – the channel is saturated rather than the base/collector on a BJT.

From Wiki on MOSFETs: –

ShareCiteEditFollowFlaganswered Mar 10 at 9:12Andy aka331k1818 gold badges268268 silver badges577577 bronze badges

1That is …, if I measure the voltage with a multimeter between drain and source when the mosfet is saturated, does it not give close to zero? – Mario2 days ago
2No, the saturation region for a MOSFET is not the region where you can measure low on-resistances. The saturation region is the part of the characteristic where if you increase the drain-source voltage, the current barely changes at all. What you are talking about is the triode region @Mario – Andy aka2 days ago
1I am in a simulator with the following circuit: applying 5v to the gate of a mosfet whose threshold voltage is 1.5v, applying 10v to drain and source through a 300 ohm resistance, the voltage between drain and source measured with the multimeter are 580 mv and 9.40v resistance, I am now in the triode region?, because it looks like the saturation region of a bjt – Mario2 days ago
2You have to put the drive voltage between gate and source. Gate and source is the input port. 9.40v sounds more like a voltage and not a resistance @Mario – Andy aka2 days ago
1ok, thanks for the help – Mario2 days ago

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Question

What do the curves and the red dot represent in the following MOSFET Id vs Vds and Vgs characteristic graph?

Answer

Part A – Meaning of the curves and the operation point

The green (update: pale cyan) region is the “saturation” region.
The yellow region is the “linear”, or “ohmic”, or “triode” region.
In the saturation region, the thick horizontal (well, slightly tilting upwards) straight lines (well, OK, curves) represent the (connected) points in the region of a particular Vgs value.
So for example, the curve that the red dot sits represents the points of Vgs = 2.5V.
The vertical lines 0, 5, 10, 15, 20 mean the voltage across Drain and Source, Vds.
Now the red dot operating point says this: If (a) Vds = 10V, and (b) Vgs = 2.5V, then (c) Ids = approx 16A.

Part B – Meanings of “Linear region”, “Saturated region” and “Linear mode” and why the MOSFET can be “operated in linear mode at the saturated region”

(1) Meaning of “linear region”

When I first look at a curve in a two dimensional graph, I almost always look at the labels of the X and Y axis. For example, if (a) X axis is labelled “voltage across a resistor, Vr”, and (b) Y axis is labelled “current through the resistor, Ir”, and (c) The Ir vs Vr “curve”, is a straight line starting from origin and goes, say 30 degrees, upwards, then we can conclude that Ir is proportional to Vr, or in mathematical terms, Ir is a function Vr, ie, Ir = f(Vr), where function f, the proportional constant, is a linear function. This is the mathematical definition of a linear function.

Now let us go back to National Semi’s EE engineer Locher’s Ir vs Vds graph (Fig 8) and focus only at the straight line labelled “liner” in yellow, we should conclude that the straight line should represent a linear function, Ir = f(Vds), or Ir = (1/R) * Vr, where R is a constant, the resistance value in Ohms, of course obeying Ohm’s Law.

We might now ask ourselves: “OK, the straight line represents a linear or “Ohmic” function, but how come this linear “straight line” becomes a linear “region”?

Well, Prof Jaeger gives the answer with the following graph:

/ to continue, …

References

(1) Microelectronic Circuit Design, 4th Ed (free eBook) – Richard C Jaegar, Travis N Ballock, McGraw Hill 2011

(2) MOSFET Characteristic Curves (Ohmic, Triode & Saturation Region) (25 min YouTube) – Dr Sunanda Manke, Barkatullah University India 2020sep03

(3) Linear Mode Operation and Safe Operating Diagram of Power-MOSFETs – J Schoiswohl, Infenion, App Note V1.1 2017may

(4) AN-558 Introduction to Power MOSFETs and Their Applications, Doc No SNVA008 – Ralph Locher, National Semi, 1988dec (page 5 for the description of linear and saturation regions)

Mosfet Safe Operating Area Explained

Appendices

Appendix A – Recommended reading list of the Jaeger book

Part 1 Solid State Electronics and Devices

Chapters

Chapter 4 Field-Effect Transistors page 145,
Chapter 5 Bipolr Junction Transistors page 217

Sections

Saturation of the I-V characteristics, Section 4.2.4, Page 154, Fig 4.8
Mathematical Model in the Saturation (Pinch-off) Region, Section 4.2.5, Page 155, Fig 4.10
NMOS Transistor Mathematical ModelSummary (Cutoff region, Triode region, Saturation region, Threshold voltage) Chapter 4, page 160.

Appendix B – Clarifying concepts and terms in MOSFET characteristics graph

Appendix C – Comparing and Constrasting between MOSFET and BJT

Introduction

MOSFET and BJT, by their structure and operation mode, cannot be easily compared, though can be more easily constrasted. The following discussion is limited to NPN BJT and N-channel MOSFET, and are over simplified and therefore potentially misleading.

1.1 BJT is basically a “current device”. So we talk about (a) current amplification gain Ic/Ib and (b) current switching.

1.2 MOSFET is basically a “voltage device”. We change Vgs which causes a change in Rds and therefore Ids and Vload. So the amplification is more indirect.

Appendix D – Linear Region vs Saturation Region

/ to continue, …ShareCiteEditDeleteFlagedited 1 min agoanswered Mar 10 at 9:00tlfong011,57411 gold badge66 silver badges1111 bronze badges

7Please, please make your answers to the point without screen grabs, “appendices”, etc etc..! – awjlogan2 days ago
1You call the yellow area “linear” in point 2, but the first screenshot says “linear mode” refers to the saturation region, and not the ohmic region. Does “linear mode” and the “linear region” refer to opposite areas of the chart? – mbrig2 days ago
1@awjlogan This answer may look intimidating due to the extras, but it gives every possible answer in the form of bullet points right at the beginning. The screengrabs are for convenience, so people who are reading can understand the answer given further if they wish so. To be completely fair I did not even understand what question OP was making at first until this answer rewrote it out of courtesy. So, if anything, this type of answer is ideal. – lucasgcb2 days ago
1You call the right part of the graph “green”. I would argue it is more of a blue? – jusaca2 days ago
2@tlfong01 They don’t look “intimidating”, they’re just a mess of screenshots, colourings, irrelevant text etc etc. Point to a couple of references, fine, but SE is meant to be to the point Q+A, not this wall of noise. – awjlogan2 days ago

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I am trying to understand the curves of a MOSFET.

You also need to understand the (output) curves of other transistors – JFET, BJT, etc. Interestingly, however, they are very similar in that in the area of the red dot they are almost horizontal. This means that when the (drain-source or collector-emitter) voltage changes over a wide range, the (drain or collector) current hardly changes. Elements with such behavior are current-stabilizing nonlinear elements… and they are used to make the very useful constant-current sources. But how do they do this magic? The general idea behind them can be explained by the concept of “dynamic resistance”.

Think of the output part of the transistor as a variable “resistor” that, in contrast to the humble “static” resistor, changes its resistance in the same direction and rate when the voltage across it varies. For example, if Vinc increases, Rinc increases, and v.v., if Vdec decreases, Rdec decreases as well. So, in Ohm’s law, both the numerator and denominator increase simultaneously and the current does not change – I = Vinc/Rinc = Vdec/Rdec = const.

In this way, transistors behave as “dynamic resistors” that keep the current constant.ShareCiteEditFollowFlagedited 2 days agoanswered 2 days agoCircuit fantasist6,75011 gold badge99 silver badges3131 bronze badgesAdd a comment1

If you will ever find a magic MOSFET that has a drain-source voltage drop of zero at any measurable current through the channel at any operation mode then let me know immediately. That would be a straight way to a near 100% efficient DC-DC converter circuit and to an enormous success on the power supply market. Your graph only shows how wide the channel is open at different constant gate voltages. Obviously it is the more the gate voltage the lower the channel resistance within the “saturation” region.ShareCiteEditFollowFlaganswered 2 days agomrKirushko1911 bronze badge New contributor

Depletion Mode N Channel Mosfet

1I think that you too are getting confused with what the saturation region is in a MOSFET. – Andy aka2 days ago
It is true that the whole operation region from the graph where small gate voltage variation provides significant conductivity variation is generally referred to as the linear region (mode of operation). But I guess we have to be a bit flexible here and do not stick to the precise terminology too much. – mrKirushkoyesterday
No, we have to stick to the terminology or confusion will reign. – Andy akayesterday