NMOS Transistor

Thread Starter

RdAdr

Joined May 19, 2013
214
Is there a characteristic somewhere? I can't find it on google. I only see it for positive values of Vds.
 

Alec_t

Joined Sep 17, 2013
11,140
Many practical NMOS devices have a built-in diode across the drain and source. Which NMOS device are you studying?
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Not a particular NMOS, but NMOS in a general sense.

I have equations for vDS>0. Are there equations for vDS<0?

I can also say that if vgs>=Vt then the NMOS is ON. But "ON" is only a word. It does not have any meaning to me. The same with "pinched off". I would have wanted equations if there are any.
 

crutschow

Joined Mar 14, 2008
24,713
If by vDS<0 you mean a negative voltage, the yes the MOSFET will conduct if the Vgs is above the threshold voltage (ON or conducting).
MOSFET's conduct equally well in both directions when ON.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
So we have the same equations.
I mean, yes, it seems logical. The MOSFET does not care how you name its terminals. The electrons could go both ways through the channel in the same way.

Thanks.

And I suppose it does matter how you name the terminals if there is a built-in diode between the drain and the source. So, in this case, the MOSFET is constraining the electrons to flow only from source to drain (conventional current from drain to source).
 

Alec_t

Joined Sep 17, 2013
11,140
So, in this case, the MOSFET is constraining the electrons to flow only from source to drain (conventional current from drain to source).
The diode (reverse-biased in normal use) allows current to flow in one direction when the MOSFET's drain-source channel is non-conducting.
NFET.PNG
 

crutschow

Joined Mar 14, 2008
24,713
MOSFET conduction in the reverse direction is purposely used in some applications such as synchronous rectification or synchronous free-wheeling diode circuits.
The MOSFET is turned off to block conduction in the normal drain-source direction and is turned on to conduct in the reverse-direction (the forward direction of the drain-source diode) to give a much lower conduction voltage than a diode would.
 

shteii01

Joined Feb 19, 2010
4,644
Peeps you are talking to a man who is incapable of opening and reading a textbook. I have brought this point up in his last thread, the one dealing with linearity of Linear Time Invariant systems.
 
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Thread Starter

RdAdr

Joined May 19, 2013
214
Peeps you are talking to a man who is incapable of opening and reading a textbook. I have brought this point up in his last thread, the one dealing with linearity of Linear Time Invariable systems.
Look. You do not know me and I do not know you. In that topic you used some words that I decided to ignore. Now you again show a bad attitude and continue to offend me even though I did not offended you in any way. This shows that you do not have those 7 years from home when you should have learned to respect other people.

Papabravo gave me an answer which it did not resonate with my mind. Other answers that he gave in other topics did resonate. In this topic, crutschow's answer resonated more. You mean to tell me that in all your life all the explanations that you got from all the people were spot on? I do not think so. Some people like images, other people like equations, others like words, others like the material to be spoken to them, others like the material in written form. Everyone is different.

I'm sorry that I am not as smart as you. But at least I don't have your attitude.
Good day.
 

WBahn

Joined Mar 31, 2012
25,556
So we have the same equations.
I mean, yes, it seems logical. The MOSFET does not care how you name its terminals. The electrons could go both ways through the channel in the same way.

Thanks.

And I suppose it does matter how you name the terminals if there is a built-in diode between the drain and the source. So, in this case, the MOSFET is constraining the electrons to flow only from source to drain (conventional current from drain to source).
In general, a MOSFET can conduct in either direction. Whichever drain/source terminal is at the lower potential is the source terminal and the gate-source voltage is therefore the voltage between the gate and the lower-potential terminal.

But this is assuming a symmetric transistor layout. There is a parasitic diode between the body (bulk) region and the channel, which is indicated in the full MOSTFET transistor symbol.

NFET.png
In the case of an NFET, the bulk is the anode and the channel is the cathode. The bulk is usually either connected to the source terminal (in which case the layout is no longer symmetric) or to a sufficiently low-potential supply (in which case the layout might remain symmetric). In discrete transistors the bulk is almost always tied to the designated source terminal. Even so, the transistor will still function with negative Vds provided the reverse potential is kept small enough not to substantially turn on that parasitic diode.
 

WBahn

Joined Mar 31, 2012
25,556
Peeps you are talking to a man who is incapable of opening and reading a textbook. I have brought this point up in his last thread, the one dealing with linearity of Linear Time Invariable systems.
As a rule, what happens in one thread should be allowed to stay in that thread.
 

Thread Starter

RdAdr

Joined May 19, 2013
214
Thanks for answer.
In general, a MOSFET can conduct in either direction. Whichever drain/source terminal is at the lower potential is the source terminal and the gate-source voltage is therefore the voltage between the gate and the lower-potential terminal.

But this is assuming a symmetric transistor layout.
I understood this. And you say that if the bulk is connected to a sufficiently low-potential supply then we might have symmetric layout. I saw somewhere that the bulk is usually connected to 0V. So in this case the layout should be symmetric and the ids vs vds characteristic in the fourth quadrant looks the same as the one in the first quadrant. Right?

There is a parasitic diode between the body (bulk) region and the channel, which is indicated in the full MOSTFET transistor symbol.

View attachment 95508
In the case of an NFET, the bulk is the anode and the channel is the cathode. The bulk is usually either connected to the source terminal (in which case the layout is no longer symmetric) or to a sufficiently low-potential supply (in which case the layout might remain symmetric). In discrete transistors the bulk is almost always tied to the designated source terminal. Even so, the transistor will still function with negative Vds provided the reverse potential is kept small enough not to substantially turn on that parasitic diode.
So in discrete tranzistors the bulk is tied to source. Thus there is a parasitic diode (because of the asymmetry) from bulk/source to channel, i.e from source to drain!? by extension like the above posts say.
Thus, for negative vds, the channel conducts in the fourth quadrant like in the first quadrant until the diode kicks in. When the diode kicks in there is leakage current through the parasitic diode. Right?
 

WBahn

Joined Mar 31, 2012
25,556
You need to keep in mind the very important difference between saying that Vds is negative and saying that Vd or Vs is negative. If Vd = 1 V and Vs = 4 V, the Vds = - 3 V, but both Vd and Vs are positive and, if the bulk-channel diode is connected to 0V, then it is not in play. But if Vd = -2 V and Vs = 1 V, then that bulk-channel diode is forward biased. The same can be true even when Vds is positive, for instance if Vd = 1 V and Vs = -2 V, then Vds = 3V but the bulk-channel diode is forward biased.
 
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