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# Understanding the curves of a MOSFET

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.

• 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 voltage is 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. – Bimpelrekkie Mar 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 Drummond 2 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 Doty 2 days ago
• 2@JRE What is the question now? – CGCampbell 2 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. – JRE 2 days ago

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: –

• 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? – Mario 2 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 aka 2 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 – Mario 2 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 aka 2 days ago
• 1ok, thanks for the help – Mario 2 days ago

Question

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

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”

(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

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

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

• 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? – mbrig 2 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. – lucasgcb 2 days ago
• 1You call the right part of the graph “green”. I would argue it is more of a blue? – jusaca 2 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. – awjlogan 2 days ago

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

• 1I think that you too are getting confused with what the saturation region is in a MOSFET. – Andy aka 2 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. – mrKirushko yesterday
• No, we have to stick to the terminology or confusion will reign. – Andy aka yesterday