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I have the following circuit hooked up on a breadboard.

enter image description here

I vary the gate voltage using a potentiometer. Here is what confuses me: according to wikipedia, the MOSFET is in saturation when V(GS) > V(TH) and V(DS) > V(GS) - V(TH).

If I slowly increase the gate voltage starting from 0, the MOSFET remains off. The LED starts conducting a small amount of current when the gate voltage is around 2.5V or so. The brightness stops increasing when the gate voltage reaches around 4V. There is no change in the brightness of the LED when the gate voltage is greater then 4V. Even if I increase the voltage rapidly from 4 to 12, the brightness of the LED remains unchanged.

I also monitor the Drain to Source voltage while I'm increasing the gate voltage. The drain to source voltage drops from 12V to close to 0V when the gate voltage is 4V or so. This is easy to understand: since R1 and R(DS) form a voltage divider and R1 is much larger than R(DS), most of the voltage is dropped on R1. In my measurements, around 10V is being dropped on R1 and the rest on the red LED (2V).

However, since V(DS) is now approximately 0, the condition V(DS) > V(GS) - V(TH) is not satisfied, is the MOSFET not in saturation? If this is the case, how would one design a circuit in which the MOSFET is in saturation?

Note that: R(DS) for IRF840 is 0.8 Ohms. V(TH) is between 2V and 4V. Vcc is 12V.



Here is the load line that I plotted of my circuit.

enter image description here

Now, from what I've gained from the answers here is that in order to operate the MOSFET as a switch, the operating point should be towards the left of the load line. Am I correct in my understanding?

And If one imposes the MOSFET characteristic curves, on the above graph, then the operating point would be in the so called "linear/triode" region. Infact, the switch should reach that region as quickly as possible in order to work efficiently. Do I get it or am I completely wrong?

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    \$\begingroup\$ Yes in order to operate as a switch the MOSFET should be in linear/triode region and yes you want to get to that region as fast as possible in order to minimize loses. \$\endgroup\$ – mazurnification Sep 1 '11 at 16:23
  • \$\begingroup\$ Thank you very much. And finally, if one makes an analogue Class A amplifier out of a MOSFET - he would be operating in the "saturation" region? The operating point ought to move WITHIN the saturation region on the load-line? \$\endgroup\$ – Saad Sep 1 '11 at 16:26
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    \$\begingroup\$ Yes, that is correct - for the Class A amplifier the MOSFET should operate within saturation region. \$\endgroup\$ – mazurnification Sep 1 '11 at 17:07
  • \$\begingroup\$ I think mazurnification's comment should actually be the accepted answer, as it's succinct and correct :-) \$\endgroup\$ – Jon Watte Apr 27 '13 at 19:35
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First of all, "saturation" in mosfets means that change in VDS will not produce significant change in the Id (drain current). You can think about MOSFET in saturation as a current source. That is regardless of the voltage across VDS (with limits of course) the current through the device will be (almost) constant.

Now going back to the question:

According to wikipedia, the MOSFET is in saturation when V(GS) > V(TH) and V(DS) > V(GS) - V(TH).

That is correct.

If I slowly increase the gate voltage starting from 0, the MOSFET remains off. The LED starts conducting a small amount of current when the gate voltage is around 2.5V or so.

You increased The Vgs above Vth of the NMOS so the channel was formed and device started to conduct.

The brightness stops increasing when the gate voltage reaches around 4V. There is no change in the brightness of the LED when the gate voltage is greater then 4V. Even if I increase the voltage rapidly from 4 to 12, the brightness of the LED remains unchanged.

You increased the Vgs making the device conducting more current. At Vgs = 4V the thing that is limiting amount of current is no longer transistor but resistor that you have in series with transistor.

I also monitor the Drain to Source voltage while I'm increasing the gate voltage. The drain to source voltage drops from 12V to close to 0V when the gate voltage is 4V or so. This is easy to understand: since R1 and R(DS) form a voltage divider and R1 is much larger than R(DS), most of the voltage is dropped on R1. In my measurements, around 10V is being dropped on R1 and the rest on the red LED (2V).

Everything looks in order here.

However, since V(DS) is now approximately 0, the condition V(DS) > V(GS) - V(TH) is not satisfied, is the MOSFET not in saturation?

No it is not. It is in linear or triode region. It behaves as resistor in that region. That is increasing Vds will increase Id.

If this is the case, how would one design a circuit in which the MOSFET is in saturation?

You already have. You just to need take care for operating point (make sure that conditions that you have mention are met).

A) In linear region you can observe following: -> when increasing the SUPPLY voltage, the LED will get brighter as the current across resistor and transistor will rise and thus more will be flowing through the LED.

B) In saturation region something different will happen -> when increasing SUPPLY voltage, the LED brightness will not change. The extra voltage that you apply on the SUPPLY will not translate to bigger current. Instead it will be across MOSFET, so the DRAIN volage will rise together with supply voltage (so increase supply by 2V will mean increasing drain volage by almost 2V)

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  • \$\begingroup\$ Thank you very much for this exhaustive answer. You state that "You already have. You just to need take care for operating point (make sure that conditions that you have mention are met)." - please see my edit on the original question. Am I correct in my understanding that in order for the MOSFET to work as a switch, the operating point needs to be located towards the left side? Since normally, one does not vary the supply voltage this means that the gate-voltage ought to be as high as possible? \$\endgroup\$ – Saad Sep 1 '11 at 16:24
  • \$\begingroup\$ Yes and Yes (biggest VGS feasible to decrease Rds_on and for the device to work as a switch mosfet should be in linear region) \$\endgroup\$ – mazurnification Sep 1 '11 at 17:09
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I interpret the meaning of 'saturation' in the context of the Wikipedia article as follows:

The datasheet for a MOSFET will show a graph with curves showing a particular \$I_D\$ for a particular \$V_{DS} \$ at a particular \$V_{GS}\$, usually for a number of different \$V_{GS}\$ values.

MOSFET Id vs Vds curves - from Wikipedia MOSFET article

In this example, the red parabolic line separates what's referred to as the 'linear' region from the 'saturation' region. In the saturation region, the \$I_D\$ lines are flat - the current does not increase any more as \$V_{DS}\$ increases. In the linear region, as the drain current increases, \$V_{DS}\$ increases - the MOSFET acts sort of like a resistor.

In your situation, assuming your part has similar curves to the example, technically 'no', the device is not in the saturation region. That being said, your \$I_D\$ is so low that the \$V_{DS}\$ drop is miniscule compared to the series resistor. No matter what \$V_{GS}\$ rises to, the 'linear' drop of the MOSFET is tiny compared to the \$390 \Omega\$ resistor, and "looks" saturated.

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Other answers here give a good explanation of the term "saturation" as applied to MOSFETs.

I'll just note here that this usage is very different from what is meant for bipolar transistors and some other classes of device.

The term is correctly used for MOSFETs where

  • V(DS) > V(GS) - V(TH)

BUT it never should have been.
But it is, so be aware of it.

A bipolar transistor (and NOT a MOSFET) is "in saturation" when it is turned hard on. The equivalent condition in an enhancement mode MOSFET (the most common kind) is when it is "fully enhanced" BUT the proper term for this has already been stolen.


Added:

A MOSFET is "turned on" by voltage applied to the gate relative to the source = Vgs.
The required Vgs where the FET starts to turn on and conducts a defined amount of current is known as the 'gate threshold voltage' or just 'threshold voltage' and is usually written as Vgsth or Vth or similar.
Vth gives an indication of how much voltage is going to be needed to operate the FET as a switch BUT actual fully-enhanced Vgs is typically several times Vgsth. Also, Vgs required for full enhancement varies with desired Ids.

This graph, copied from Madmanguruman's answer, shows that at Vgs = 7V the Ids/Vds realtionship is about linear up to about Ids = 20A so the FET is "fully enhanced" and looks like a resistor up to about this point. For this FET Vds is about 1.5V at about 20A so Rdson is about R = V/I = 1.5/20 = 75 milliOhms.
For this FET there is a curve at Vgs = 1V so VGSth = Vth is probably in the 0.5V-0.8V range at say 100 uA.

enter image description here

enter image description here

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  • \$\begingroup\$ Yes. That is what I remember learning as well. But here is the wikipedia article. You'll need to scroll down to "Saturation or Active Mode" heading. en.wikipedia.org/wiki/MOSFET Do you think this is wrong? \$\endgroup\$ – Saad Sep 1 '11 at 13:28
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    \$\begingroup\$ @saad - the confusion is that they are using the term "saturation" to mean something like "linear region". The meaning in English of saturation implies being at a maximum so their usage is poor at best and misleading. This MAY be std usage, or not, but it's not good. \$\endgroup\$ – Russell McMahon Sep 1 '11 at 14:04
  • \$\begingroup\$ Thanks. That article is now very confusion. Would you be king enough to point me to a book or article where I could learn more about MOSFETs? Would definitely prefer to avoid confusing terms! \$\endgroup\$ – Saad Sep 1 '11 at 14:11
  • \$\begingroup\$ The fact that the "saturation means something different is really confusing. So what is the correct term for "turned hard on" for a MOSFET, and how do you figure out what gate voltage is needed for a given MOSFET? \$\endgroup\$ – Duncan C Aug 17 '14 at 19:05
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    \$\begingroup\$ "Hard on", "Turned on completely", and "Fully enhanced". Why do I feel like I'm in a bad E.D. treatment ad? "Enhance yourself to your full potential! Feel the rush of current!" :) \$\endgroup\$ – Duncan C Aug 18 '14 at 2:13
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What you need to do to see saturation, is supply enough voltage till eventually the rise in voltage make no difference to the current.
To do this, set your Vgs to a static on (>Vth) value, then raise the voltage across Vds and measure current. Initially it will rise quite linearly, being in the ohmic or linear region, but it will eventually flatten out and despite raising further the current through the MOSFET will stay the same.

As regards the definition of saturation, I understand the saturation/linear in MOSFETs to mean roughly the opposite of what they do in a BJT. This document (under MOSFET characterisation a few pages in) suggests similar, though as long as you understand how they work and what you mean by the term, then you should be okay (at least until you are discussing transistors with someone :-) )

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    \$\begingroup\$ Does that mean that a MOSFET in saturation acts like a current limiter? \$\endgroup\$ – Duncan C Aug 17 '14 at 19:07
  • \$\begingroup\$ indeed it does, JFETs too, there are JFET based current limiters available. EG: 1N5298 en.wikipedia.org/wiki/Constant-current_diode \$\endgroup\$ – Jasen Dec 23 '18 at 4:55
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http://www.falstad.com/circuit/e-nmosfet.html

There is a good MOSFET simulator applet in this page. I hope it helps.
Also I asked a similar question a while ago; you may refer to it as well.

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B) In saturation region something different will happen -> when increasing SUPPLY voltage, the LED brightness will not change. The extra voltage that you apply on the SUPPLY will not translate to bigger current. Instead it will be across MOSFET, so the DRAIN volage will rise together with supply voltage (so increase supply by 2V will mean increasing drain volage by almost 2V)

How so? Increasing the supply should increase the V d-s only by Id X Rds(on). Considering the LED will have almost same forward voltage drop, then the increased voltage will have to be shared by the series resistor and device. Since the resistor has a much larger value (390 ohms compared to 0.8 ohms of the device), the major share of voltage drop has to be across the resistor. Moreover there defenitely will be an increase in drain current with increase in resistance. MOSFET losses are calculated at stady state as current squared multiplied by Rds(on). So the observation "DRAIN volage will rise together with supply voltage (so increase supply by 2V will mean increasing drain volage by almost 2V)" is incorrect

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