In saturated region its either on or off, so there's either lots of current and very little volt drop, or no current and all the volt drop - either way P=I*V dictates that very few watts are dissipated,

In linear mode the device can simultaneously be dropping significant voltage and carrying significant current I*V gives you the resulting watts.

Ian,

I think you're a bit confused here on terminology. You're mixing up linear and saturated.

In linear mode the device is a switch and the current through the device is found from Ohm's law: I = V/R or V*gm in this case.

In saturation the MOSFET is acting as an amplifier. The output impedance is high (ideally infinite) and the power is I*V. In this case the MOSFET most certainly does not follow Ohm's law, since ideally, as long as it is in saturation, the current is constant (it is set by Vgs alone).

 

First time hearing this, that very few watts are dissipated in the saturation region. Everything I've been told is to put them in the linear region, when you can, to keep them from heating/dissipating watts. Can you give a reference to this concept? I'm constantly in learning mode.

Good on ya for being in constant learning mode. There is often confusion on this site I've noticed but don't let that deter you.

Linear region is just a resistor, the power depends on the voltage drop and the current. In digital CMOS, one of the devices in an inverter is in linear and the other is in cutoff if the gate isn't switching. Digital CMOS burns no static power except leakage (a more advanced concept). So that tells you that a linear region MOSFET can in some circumstances dissipate almost no power (no DC path to ground).

Saturation the power depends just on the current (since power is iv of course). There are very little resistive losses in a MOSFET. If you have two saturated MOSFETs in series, as you might in a current source, the power is still i*v even though they are carrying the same current.

Edit: slide 7 of the attached describes the different regions pretty well.
http://www.engr.uky.edu/~elias/lectures/06_MOSFET_Operation.pdf

 

Analogdesign, that , or one very much like it, was one of the things I had read last month. But I'm really just a neophyte in this stuff, so could you put it in that reference?
Is the most heat made in a mosfet when the Vgs is under the data sheet value for the drain current, usually 10Vgs?

 

Analogdesign, that , or one very much like it, was one of the things I had read last month. But I'm really just a neophyte in this stuff, so could you put it in that reference?
Is the most heat made in a mosfet when the Vgs is under the data sheet value for the drain current, usually 10Vgs?

I'm sorry I'm not sure what you're asking by "put it in that reference".

As for heat, yes you're right the most heat is when the MOSFET is at its max drain current which will occur at its max VGS. The datasheet will have something in it called "thermal resistance" which tells you how whats are transferred to degrees by the device. You calculate the Watts the by multiplying the power supply with the VDS. (if you have multiple MOSFETs in series it's a bit more complicated).

Hope this helps.

 

On a somewhat related note, I have never once seen in a datashee the classic characteristic curves for a transistor -- the family of curves for Ic as a function of Vce for different base currents that is almost always how transistors are introduced. I've conclused that no matter how useful this might be from a teaching standpoint, it must not have enough practical utility to warrant inclusion.

I remember that. It was in the days when we had pencils, rulers & slide rules. We used to draw load lines and predict output voltage swings, gains and the like. Do they still teach like that?

 

I remember that. It was in the days when we had pencils, rulers & slide rules. We used to draw load lines and predict output voltage swings, gains and the like. Do they still teach like that?

What you are talking about is called "load line analysys". I'm not sure if it is still taught or not. I hope so. But even if it is, it is only taught as a means of introducing the concept of two devices finding a mutually consistent operating point and of solving a nonlinear problem. It is not taught with the notion that it will actually be used in practice. But then that was pretty much the case when I went through the pipe nearly thirty years ago.

 

I'm sorry I'm not sure what you're asking by "put it in that reference".

As for heat, yes you're right the most heat is when the MOSFET is at its max drain current which will occur at its max VGS. The datasheet will have something in it called "thermal resistance" which tells you how whats are transferred to degrees by the device. You calculate the Watts the by multiplying the power supply with the VDS. (if you have multiple MOSFETs in series it's a bit more complicated).

Hope this helps.

The "put it in that reference", was in my being a neophyte.

This whole thing is again blowing my mind. Every thing I've read both on AAC and every where else has said, "when ever possible use 10Vgs (for a standard nmos) to prevent excess heat in the mosfet". Now you guys are saying it's OK to use a lower Vgs, as long as it gives enough current for the circuit, because it gives less heat rise in the mosfet. The sound you hear is my brain exploding!

One of the first circuits I wanted was a DC motor controller. It was to run a motor a a certain speed. When I was on another forum, I asked about just giving a lower than 10Vgs to a mosfet to control the speed. You would have thought I was committing a mortal sin. Everyone said "only do that if you want to burn up the mosfet." They convinced me to use PWM because it would keep the mosfet from self destructing. This wasn't a variable speed thing, just needed a slightly slower speed.

All of the tutorials on mosfets recommend "turn on a mosfet as fast as you can, to get it through the saturated region fast. To prevent mosfet heating" Is that now a wrong thought? Not trying to be a troll, just trying to understand this.

 

First time hearing this, that very few watts are dissipated in the saturation region. Everything I've been told is to put them in the linear region, when you can, to keep them from heating/dissipating watts. Can you give a reference to this concept? I'm constantly in learning mode.

A google search as broad as "MOSFET switching circuits" should get you started.

 

Ian,

I think you're a bit confused here on terminology. You're mixing up linear and saturated.

In linear mode the device is a switch and the current through the device is found from Ohm's law: I = V/R or V*gm in this case.

In saturation the MOSFET is acting as an amplifier. The output impedance is high (ideally infinite) and the power is I*V. In this case the MOSFET most certainly does not follow Ohm's law, since ideally, as long as it is in saturation, the current is constant (it is set by Vgs alone).

Don't take my word for it - try reading a book or two.

 

...
On the other hand, the term "linear" would not be a suitable descriptor for the active region in a MOSFET because, here, the source-drain current goes as the square of the gate-source voltage. ...

Does this mean that the linearity of MOSFET is not good as BJT?

 

Does this mean that the linearity of MOSFET is not good as BJT?

When you talk about linearity, you need to specifiy what is linear with respect to what. In the active region, the collector current of a BJT is quite linear with respect to base current, which is often considered to be the controlling variable. But it is exponential with respect to the base-emitter voltage, which is arguably the true controlling variable (much debate can be had on this point, so let's not go there). For the MOSFET in the active region, the controlling variable is the gate-source voltage and the drain current goes as the square of Vgs.

But once you put either device into a circuit, you are seldom concerned about the linearity of these variables, but rather some other output variable with respect to some other input variable. Many other factors are more significant in determining that.

 

A google search as broad as "MOSFET switching circuits" should get you started.

I've spent more time trying to understand this than anything else in my life. Every time I think I'm getting a handle on it, someone says some thing new.

 

This isn't that hard to get a handle on.

The power dissipated in a MOSFET is, like any other device, the product of the voltage across the device and the current going through the device. So it is

P = I_ds * V_ds

For a given Vgs, you get to either dictate I_ds or dictate V_ds, the device will dictate the other.

As you initially increase V_ds from 0V, you are in the triode region and the I_ds is related to the V_ds by the following basic relationship:

\(
I_{ds} \, = \, K \, \left[ \left( V_{gs} \, - \, V_{th} \right) V_{ds} \, - \, \frac{V^2_{ds}}{2} \right]
\)

This continues until you transition to the active region, which occurs at about the point where V_ds = (V_gs - V_th). At this point the current is basically

\(
I_{ds} \, = \, \frac{K}{2} \, { \left( V_{gs} \, - \, V_{th} \right) }^2
\)

If we define the overdrive voltage as

\(
V_{ov} \, = \, \left( V__{gs} \, - \, V_{th} \right)
\)

Then the prior two equations become

\(
I_{ds} \, = \, K \, \left( V_{ov} V_{ds} \, - \, \frac{V^2_{ds}}{2} \right) \ ; \ triode: \ V_{ds}<V_{ov}
\)

\(
I_{ds} \, = \, \frac{K}{2} \, V_{ov}^2 \ ; \ active: \ V_{ds}>V_{ov}
\)

To get the power, we just need to multiply these by V_ds

\(
P_{triode} \, = \, K \, \left( V_{ov} \, - \, \frac{V_{ds}}{2} \right) V^2_{ds}\ ; \ triode: \ V_{ds}<V_{ov}
\)

\(
P_{active} \, = \, \frac{K}{2} \, V_{ov}^2 V_{ds} \ ; \ active: \ V_{ds}>V_{ov}
\)

We can now define a parameter α that is the ratio of V_ds to V_ov

\(
\alpha \, = \, \frac{V_{ds}}{V_{ov}}
\)

and replace V_ds using

\(
V_{ds} \, = \, \alpha V_{ov}
\)

This results in

\(
P_{triode} \, = \, \left[ \alpha \left( 2 - \alpha \right) \right] \cdot \frac{K}{2} \, V_{ov}^2 V_{ds} \ ; \ triode: \ \alpha < 1
\)

\(
P_{active} \, = \, \frac{K}{2} \, V_{ov}^2 V_{ds} \ ; \ active: \ \alpha > 1
\)

From this it can be seen that, at any give overdrive voltage, V_ov, the power dissipation will be less if the device is operated in the triode region since the multiplying factor, α(2-α), is always smaller than one when α<1, which it always is in the triode region.