MOSFET operating zones

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hello.
(Note: this is not homework, I'm not at school anymore)

I know I've been around this in the past but the lack of practicing makes things go away.
I've been reading different sources about the operating zones of MOSFETs (n-channel, enhancement mode).
There are 3 operating zones (known to me)
Cut-off, Linear and Saturation zones.

First, about this, I see some sources saying that Saturation and Linear zones are the same:
https://www.electronics-tutorials.ws/transistor/tran_7.html
1640862813748.png

Why is this said? Can I assume it is wrong or at least not accurate info?
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Then, there is another detail I'm confused about. Enhancement and Depletion modes. Despite the fact that I think I don't need this knowledge, at least for now, when it's said that the mode depends on \(\displaystyle{V_{GG}}\) applied (positive or negative), it raises a question about the requisites I'm talking about below. I'll elaborate my questions about the modes below.
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Next thing is that different sources takes different starting points to classify the requisites to say a MOSFET is in one mode of operation or another.
Example 1:
https://www.tutorialspoint.com/basic_electronics/basic_electronics_mosfet.htm
This first example uses only words to describe the possible \(\displaystyle{V_{GS}}\) values applied to the MOSFET.
1640863853597.png

Example 2:
https://electricalvoice.com/mosfet-types-operation-and-applications/

It is similar to Example 1, but Example 1 says that even with \(\displaystyle{V_{GS} = 0}\) there will still be some current flowing from drain to source. However Example 2 says that there will be no current flowing from drain to source because there is no channel connecting both terminals. This is contradictory, at least for me, at a first glance!
1640864432547.png

Example 3:
Art of Electronics, 3rd Edition, Chapter 3, Section 3.1, A - FET curves, page 132

AoE also uses mostly word, no equations or values so that we get a more realistic and practical idea of what is going on. I don't fully disagree with what is said but I think there is huge inaccuracy (or not, that's what I'm trying to clear)
1640864711237.png


So, taking a look to what is said in the 3 examples, and knowing a bit more about the requisites for the operation modes, I am confronted with some questions I can't answer to myself.
For instance, I learned that, to turn a MOSFET ON, to the Saturation mode/region, conditions are:
\(\displaystyle{V_{GS} > V_{TH}}\) and \(\displaystyle{V_{DS} > V_{GS} - V_{TH}}\)

However, the examples only say that \(\displaystyle{V_{GS}}\) or the gate voltage must be positive with respect to the source. But if \(\displaystyle{V_{TH}}\) is, for instance, 2.4V and \(\displaystyle{V_{GS}}\) = 1V, are the above statements still true? Like, what I want to say/ask is that it is not enough that the gate voltage is positive with respect to the source. It also must be greater than the threshold voltage.
Am I wrong?

With this said, I'm not even sure how to write down the requisites for \(\displaystyle{V_{GS}}\) and \(\displaystyle{V_{DS}}\) values to get the possible operation modes.
 

DickCappels

Joined Aug 21, 2008
10,245
Your post is long and winds through numerous paths. I agree some of that which you copied sounds plain wrong or at least poorly written and edited.

May I suggest that you put down that book and look at the resource on this website?
https://www.allaboutcircuits.com/te...sulated-gate-field-effect-transistors-mosfet/

Or better yet, take out some datasheets.

Now a couple of comments I can't resist making.

• MOSFETS are bidirectional but the diode interferes with using them in a way we think of as "Bakcwards" that's ok because there are limited applications for this capability.

• Similar to what Wolfmore said about linear operation, saturation definitions vary, but saturation is only reached at threshold when driving an extremely light drain load. To get into saturation with any significant drain current you will need significantly more drive -you can see typical drain current vs VGS curves on the datasheets.
 

crutschow

Joined Mar 14, 2008
34,815
Below is a typical characteristic curve for an N-MOSFET.

Note that it is plotted for Vgs-Vth, as no significant current will flow when Vgs is less than Vth.

The red line shows the transition point between the linear and saturation region.

Unlike the definition for a BJT, the saturation region is not when the transistor is fully on.
Typical MOSFETs are fully on (Vds determined by the ON channel resistance) when Vgs is 5 or 10V.

1640876788329.png
 

BobTPH

Joined Jun 5, 2013
9,251
Notice from Crutshow's example that the saturation region is where a change in Vds does not change the current any more. In other words, it is saturated, it cannot pass any more current at that gate voltage.

Bob
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
1. Linear Region has different meaning from different sources. See App note. compare to MOSFET
Seems to me that that App note also only mentions 2 modes of operation.
The MOSFET doc also mentions only 2 operation modes. Linear and Saturation and the Linear mode matches the Ohmic region of the other doc. This just brings more confusion to the discussion.

Shouldn't they be 3 modes, as I mentioned? Cut-Off, Linear and Saturation?

Your post is long and winds through numerous paths. I agree some of that which you copied sounds plain wrong or at least poorly written and edited.

May I suggest that you put down that book and look at the resource on this website?
https://www.allaboutcircuits.com/te...sulated-gate-field-effect-transistors-mosfet/

Or better yet, take out some datasheets.

Now a couple of comments I can't resist making.

• MOSFETS are bidirectional but the diode interferes with using them in a way we think of as "Bakcwards" that's ok because there are limited applications for this capability.

• Similar to what Wolfmore said about linear operation, saturation definitions vary, but saturation is only reached at threshold when driving an extremely light drain load. To get into saturation with any significant drain current you will need significantly more drive -you can see typical drain current vs VGS curves on the datasheets.
The post was intended to go multiple paths. I had (and still have) different questions.
The link you mention from AAC, doesn't show typical equations for the 3 modes, which was what I wanted in the end!

Below is a typical characteristic curve for an N-MOSFET.

Note that it is plotted for Vgs-Vth, as no significant current will flow when Vgs is less than Vth.

The red line shows the transition point between the linear and saturation region.

Unlike the definition for a BJT, the saturation region is not when the transistor is fully on.
Typical MOSFETs are fully on (Vds determined by the ON channel resistance) when Vgs is 5 or 10V.

View attachment 256437
Yeah, so that means that the Linear Region is defined just by that red line?

Notice from Crutshow's example that the saturation region is where a change in Vds does not change the current any more. In other words, it is saturated, it cannot pass any more current at that gate voltage.

Bob
Thanks


My goal is to be able to identify the typical equations for the 3 modes. I don't even care about their names. I'll use the names I was taught. Cut-Off, Linear and Saturation.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
The 2 screenshots I shared above doesn't share the same equations. So, I'm still not sure which equations to use.
1st says:
Cut-Off
\(\displaystyle{V_{GS} < V_{t} }\)

Triode
\(\displaystyle{V_{DS} < V_{GS} - V_{t}}\)

Saturation
\(\displaystyle{V_{DS} >= V_{GS} - V_{t}}\)


while the second says:
Linear
\(\displaystyle{V_{DS} << V_{GS} - V_{t}}\)

Triode
\(\displaystyle{V_{DS} < V_{GS} - V_{t}}\)

Saturation
\(\displaystyle{V_{DS} >= V_{GS} - V_{t}}\)


So, only 2 names match, and the other name of operation mode doesn't match and their equations doesn't match either.
For the 2 matching operation mode names, one of the equations is slightly different. One says "<" and the other says "<<".

So, what should I consider at the end?
 

BobTPH

Joined Jun 5, 2013
9,251
There are 4 modes listed in the two descriptions: cutoff, linear, triode, and saturation. Each of them has left out one of the four modes. They both agree on the other two modes.

Bob
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
In most of the cases, we stick with two equations only. The triode region and the saturation
http://www.ittc.ku.edu/~jstiles/312/handouts/A Mathematical Description of MOSFET Behavior.pdf

But what do you want to achieve? Because we almost never use this equation in real-world design (using discrete transistors).
I want to eventually build a small UV Led driver for something like 20 Leds or so. So, I'm starting with the simplest I could. Going back to mosfet basics. But even this can be tricky, unless I go to my attic and try to find my graduation notes on Electronics and Mosfets! Which would be probably harder than trying to figure this out from the internet. :0

Going to take a look at the link.

There are 4 modes listed in the two descriptions: cutoff, linear, triode, and saturation. Each of them has left out one of the four modes. They both agree on the other two modes.

Bob
Indeed. More accurate. :)
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
@Jony130 I just read the link, and there are some things different in that document too, but I'll assume that one as the most accurate and I'm not going to dig more on this, because I think this is more than enough for my purpose.

Now, let's say I want to drive a green led (3mm), just to test this out and practice calculations.
I have an IRF640 (n-channel mosfet) here with me that I can use for testing. I know this is a power mosfet, and I hope it fits to drive a single led.

So, to do this, what are the steps I need to take to find my resistor values? I think I need to start with a (more or less) random resistor and try to work from there, no?
What I know:
1641120605944.png

Green Led Voltage drop in my case is around 2.2V. It doesn't match what I see "out there", no idea why. Green Leds should drop something like 3.5V but I used a 470 ohm resistor and 9V power supply and I got 2.17V drop on this led. So, I'll use 2.2V.
 

DickCappels

Joined Aug 21, 2008
10,245
It is not necessary to calculate anything more. Looks like you are done. Just connect the resistor & LED in series from battery + to the drain, connect the source to battery -. When you want the LED on connect the gate to battery + and when you don't want it on, connect the gate to battery -.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
Hi @ericgibbs ... Well, I looked to a few tables and any of the ones I looked, showed those complete values. Anyway, thanks.

@DickCappels I want to put there some resistors for limiting currents, etc, and also to make the switching a more smooth and faster "experience". Also, I want to use like 20mA to drive this green led. Then, as I said above, I want to drive a set of 20 UV leds or so!
 

Jony130

Joined Feb 17, 2009
5,539
So you wanna use a MOSFET as an ON/OFF switch. So you are interested in Vgs(th) and rds(on) and Vds_max >> Vspuly.
In the case of an IRF640, we can see Vgs(th) = 4V therefore to achieve a "switching action" (MOSFET as a switch) we need to apply Vgs >> Vgs(th) to the MOSFET gate.
9 volts is fine. Notice that the rds(on) = 0.15Ω is given when 10V is applied to the gate. But you have 9V so it won't be a problem, we can safely apply 9V across Vgs.
And we still can safely assume that rds(on) will be around 0.15Ω.
Now we pick the resistor for Green LED R_led = (9V - 2.2V)/10mA = 680Ω so, we can use a resistor between 470Ω...1kΩ in a real-life situation.

Are those UV LEDs are the power LEDs? Or just ordinary ones?
For the ordinary LED's we have 20*20mA = 0.4A. This current won't be a problem for your MOSFET.
As we can see here

untitled.PNG


0.4A is not a problem if Vgs >> 4.5V. And the power dissipation P = 0.4A^2*0.15Ω = 24mW (When Vgs arond 10V).
You could also add a 10kΩ pull-down resistor, between the MOSFET gate and GND.
 
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crutschow

Joined Mar 14, 2008
34,815
If all you want is to switch the LEDs on and off, they you don't any of the MOSFET equations you posted for that design.

Standard MOSFETs need a Vgs of 10V to turn them on, and logic-level MOSFETs typically need 5V.
0V of course, turns them off.

The other main parameter of interest is the on-resistance of the MOSFET, as this determines the MOSFET voltage drop and internal dissipation for the load current.

The graph posted by Jony are typical values and the MOSFET you have may not be typical.
 

Audioguru again

Joined Oct 21, 2019
6,777
Very old, fairly dim green LEDs were 2.2V. Modern very bright green LEDs use the same chemistry as very bright blue and white LEDs (about 3.2V) that were not even invented when the old green LEDs were used.
 

Thread Starter

PsySc0rpi0n

Joined Mar 4, 2014
1,776
So you wanna use a MOSFET as an ON/OFF switch. So you are interested in Vgs(th) and rds(on) and Vds_max >> Vspuly.
In the case of an IRF640, we can see Vgs(th) = 4V therefore to achieve a "switching action" (MOSFET as a switch) we need to apply Vgs >> Vgs(th) to the MOSFET gate.
9 volts is fine. Notice that the rds(on) = 0.15Ω is given when 10V is applied to the gate. But you have 9V so it won't be a problem, we can safely apply 9V across Vgs.
And we still can safely assume that rds(on) will be around 0.15Ω.
Now we pick the resistor for Green LED R_led = (9V - 2.2V)/10mA = 680Ω so, we can use a resistor between 470Ω...1kΩ in a real-life situation.

Are those UV LEDs are the power LEDs? Or just ordinary ones?
For the ordinary LED's we have 20*20mA = 0.4A. This current won't be a problem for your MOSFET.
As we can see here

View attachment 256644


0.4A is not a problem if Vgs >> 4.5V. And the power dissipation P = 0.4A^2*0.15Ω = 24mW (When Vgs arond 10V).
You could also add a 10kΩ pull-down resistor, between the MOSFET gate and GND.
That seems the way to go. The only thing I didn't understand was the 10mA you used to get the 680Ω resistor. I assumed half of that because I thought of 20mA to drive this green led.
About the 10kΩ resistor, it is already there. So, now, I have something like this:

V_DD -- LED -- 350Ω -- Drain
V_DD -- 100Ω -- Gate
GND -- Source

1641148914631.png

Now, should I also add some resitor between Source and GND? And why?

If all you want is to switch the LEDs on and off, they you don't any of the MOSFET equations you posted for that design.

Standard MOSFETs need a Vgs of 10V to turn them on, and logic-level MOSFETs typically need 5V.
0V of course, turns them off.

The other main parameter of interest is the on-resistance of the MOSFET, as this determines the MOSFET voltage drop and internal dissipation for the load current.

The graph posted by Jony are typical values and the MOSFET you have may not be typical.
I'm not sure about that "standard" value of 10V of Gate voltage to turn a mosfet ON. Isn't that specific to each Mosfet? Like, each mosfet may have it's own V_GS, so, I assume you people all use 10V just as a thumbs up value. Not that it is the absolute value used for any mosfet.

Another thing I didn't understand is the R_DS On parameter setting the mosfet voltage drop. What voltage drop? V_DS? Why is it important here if the resistance is so low (when in cut-off mode 0.15Ω) which would produce a neglectable voltage drop?

Very old, fairly dim green LEDs were 2.2V. Modern very bright green LEDs use the same chemistry as very bright blue and white LEDs (about 3.2V) that were not even invented when the old green LEDs were used.
Ok, however, this green led I have, drops around 2.2V.
 

Attachments

Jony130

Joined Feb 17, 2009
5,539
Now, should I also add some resitor between Source and GND? And why?
Why? This resistor is not needed.
The only thing I didn't understand was the 10mA
Because the modern LEDs are more sensitive than the old LEDs. New LEDs need less current but they are shining very bright.
So we don't need to waste a current, and force 20mA through the LED.
I'm not sure about that "standard" value of 10V of Gate voltage to turn a mosfet ON. Isn't that specific to each Mosfet?
For a power MOSFET 10V is a standard Vgs voltage to measure the rds(on) (when MOSFET is fully Turned-ON).
For a login level MOSFE, it will be around 5V.
https://www.infineon.com/dgdl/irlz44npbf.pdf?fileId=5546d462533600a40153567217c32725
Another thing I didn't understand is the R_DS On parameter setting the MOSFET voltage drop. What voltage drop?
When the MOSFET is fully Turned-ON (Triode region) the channel resistance is very low. And it's specified in the datasheet as a Rds(on) (MOSFET behaves like a resistance that is small compared with the load).
Therefore in switching application drain to source voltage is defined as Vds = ID * Rds(on).
And we can calculate the power dissipated in MOSFET.
Pd = ID^2 * Rds(on)
 
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