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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.

enter image description here

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    \$\begingroup\$ therefore the source drain voltage must be 0v No, the red dot is at the point where \$V_{DS}\$ = 10 V, see the X-axis of the graph. The source drain voltage is \$V_{DS}\$. Look up when a MOSFET is in saturation, there is an equation which tells you that \$V_{DS}\$ needs to be larger than a certain value. \$\endgroup\$ Mar 10 at 7:57
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    \$\begingroup\$ You 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. \$\endgroup\$ Mar 10 at 13:41
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    \$\begingroup\$ @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. \$\endgroup\$
    – John Doty
    Mar 10 at 17:09
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    \$\begingroup\$ @JRE What is the question now? \$\endgroup\$
    – CGCampbell
    Mar 10 at 17:58
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    \$\begingroup\$ @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. \$\endgroup\$
    – JRE
    Mar 10 at 18:04
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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: -

enter image description here

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    \$\begingroup\$ That 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? \$\endgroup\$
    – Mario
    Mar 10 at 18:49
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    \$\begingroup\$ No, 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 \$\endgroup\$
    – Andy aka
    Mar 10 at 18:51
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    \$\begingroup\$ I 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 \$\endgroup\$
    – Mario
    Mar 10 at 21:23
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    \$\begingroup\$ You 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 \$\endgroup\$
    – Andy aka
    Mar 10 at 21:28
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    \$\begingroup\$ ok, thanks for the help \$\endgroup\$
    – Mario
    Mar 10 at 21:31
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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?

mosfet graph


Answer

Part A - Meaning of the curves and the operation point

  1. The green (update: pale cyan) region is the "saturation" region.

  2. The yellow region is the "linear", or "ohmic", or "triode" region.

  3. 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.

  4. So for example, the curve that the red dot sits represents the points of Vgs = 2.5V.

  5. The vertical lines 0, 5, 10, 15, 20 mean the voltage across Drain and Source, Vds.

  6. 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"

confusing terms

(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:

linear region

One thing to clarify is that the linear region shown above is a zoom in view of the left bottom corner of the big I-V characteristics. It is only at the zoom in that we can see the linear curves separated. In the big picture, the curves are merged and becomes just one straight line. Here, the y-axis shows current in order of 0.1mA, and voltage in 0.1V, so the resistance is approx 0.1mA / 0.1V or of the order of 1 milliOhm.

/ 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)


Appendices

Appendix A - Recommended reading list of the Jaeger book

Part 1 Solid State Electronics and Devices

Chapters

  1. Chapter 4 Field-Effect Transistors page 145,

  2. Chapter 5 Bipolr Junction Transistors page 217


Sections

  1. Saturation of the I-V characteristics, Section 4.2.4, Page 154, Fig 4.8

  2. Mathematical Model in the Saturation (Pinch-off) Region, Section 4.2.5, Page 155, Fig 4.10

  3. 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

Linear Mode Operation and Safe Operating Digram of Power-MOSFETs - J Schoiswohl, Infenion, App Note V1.1 2017may

mosfet concepts and terms


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.

bjt characteristics


Appendix D - Linear Region vs Saturation Region

linear vs saturation


/ to continue, ...

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    \$\begingroup\$ Please, please make your answers to the point without screen grabs, "appendices", etc etc..! \$\endgroup\$
    – awjlogan
    Mar 10 at 13:03
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    \$\begingroup\$ You 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? \$\endgroup\$
    – mbrig
    Mar 11 at 4:03
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    \$\begingroup\$ @mbrig, you might find the following picture helpful to clarify things. I might try to write up a more detailed explanation of why you can operate in linear mode in the saturation (ie non linear) region. MOSFET Linear Transfer Characteristic in Saturation Region - National Semi: i.imgur.com/QaP9LUt.jpg Note: the sentence in pink is the crux of the matter. Cheers. \$\endgroup\$
    – tlfong01
    Mar 11 at 6:53
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    \$\begingroup\$ @mbrig, so you see, (1) In the Ohmic or linear region, the function Isd = f(Vds), where f is linear (i/v = constant = ohm). (2) In the saturated region, Isd = g(Vgs) where g is also linear (read the picture's sentence in pink), (3) Confusion arises because the two linear functions have two different independent variables: Vds and Vgs. Cheers. \$\endgroup\$
    – tlfong01
    Mar 11 at 7:14
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    \$\begingroup\$ @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. \$\endgroup\$
    – awjlogan
    Mar 11 at 9:08
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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.

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    \$\begingroup\$ I think that you too are getting confused with what the saturation region is in a MOSFET. \$\endgroup\$
    – Andy aka
    Mar 10 at 18:53
  • \$\begingroup\$ 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. \$\endgroup\$
    – mrKirushko
    Mar 11 at 17:22
  • \$\begingroup\$ No, we have to stick to the terminology or confusion will reign. \$\endgroup\$
    – Andy aka
    Mar 11 at 17:33
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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.

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