Characteristics of MOSFET: How Does i_{DS} Vary with v_{GS} in the ON State?

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The discussion focuses on the characteristics of a specific MOSFET with a threshold voltage (V_T) of -1 V, highlighting the ON and OFF states based on gate-source voltage (v_{GS}). In the ON state, the current (i_{DS}) is modeled using a switch-resistor (SR) approach, where i_{DS} equals zero for v_{GS} below V_T. For v_{GS} above V_T, the variation of i_{DS} as a function of v_{GS} is unclear, leading to confusion about whether i_{DS} remains constant or varies. Participants clarify that the MOSFET operates in different regions, with the linear region being where i_{DS} decreases with increasing drain-source voltage (V_{DS}). The key takeaway is the need to plot i_{DS} versus V_{DS} for various v_{GS} values to accurately represent the MOSFET's behavior.
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Homework Statement



A specific type of MOSFET has V_T = -1\ \rm V. The MOSFET is in the ON state when v_{GS} \geq V_T. The MOSFET is in the OFF state when v_{GS} < V_T.
a) Graph the i_{DS} versus v_{GS} characteristics of this MOSFET.

Homework Equations



The Attempt at a Solution



My doubt concerns part a. I will use the switch-resistor (SR) model (the ON state of the MOSFET is modeled as a resistance R_{ON} between the drain and the source). So, the graph of i_{DS} versus v_{GS} would have i_{DS} = 0 for all v_{GS} < V_{T}.
But what about the case where v_{GS} \geq v_T? The problem is that I don't know how i_{DS} varies as a function of v_{GS}.
I know that i_{DS} = \dfrac{v_{DS}}{R_{ON}}; if this value doesn't vary with v_{GS}, then, for v_{GS} \geq v_T the graph would have just a horizontal line with y-value i_{DS} = \dfrac{v_{DS}}{R_{ON}}.
What am I missing?
 
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pc2-brazil said:

Homework Statement



A specific type of MOSFET has V_T = -1\ \rm V. The MOSFET is in the ON state when v_{GS} \geq V_T. The MOSFET is in the OFF state when v_{GS} < V_T.
a) Graph the i_{DS} versus v_{GS} characteristics of this MOSFET.

Homework Equations



The Attempt at a Solution



My doubt concerns part a. I will use the switch-resistor (SR) model (the ON state of the MOSFET is modeled as a resistance R_{ON} between the drain and the source). So, the graph of i_{DS} versus v_{GS} would have i_{DS} = 0 for all v_{GS} < V_{T}.
But what about the case where v_{GS} \geq v_T? The problem is that I don't know how i_{DS} varies as a function of v_{GS}.
I know that i_{DS} = \dfrac{v_{DS}}{R_{ON}}; if this value doesn't vary with v_{GS}, then, for v_{GS} \geq v_T the graph would have just a horizontal line with y-value i_{DS} = \dfrac{v_{DS}}{R_{ON}}.
What am I missing?

You need the equations for i as a function of Vds and Vgs. The most common way is to graph i vs. Vds so you'll have a family of curves, one for each Vgs where usually Vgs varies from 0 to some max. number like 10V in 2V increments (i.e. 6 curves). The Vgs = 0 curve is of course the i = 0 axis.

There are three regions of operation for the MOSFET. I am attaching a pdf file for you. Use the "square law" equations at the top of page 1 and assume μCoxW/L and VT are constants. VT = 1V in your case. A typical value for μCoxW/L might be 0.025 A/V2.
 

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rude man said:
You need the equations for i as a function of Vds and Vgs. The most common way is to graph i vs. Vds so you'll have a family of curves, one for each Vgs where usually Vgs varies from 0 to some max. number like 10V in 2V increments (i.e. 6 curves). The Vgs = 0 curve is of course the i = 0 axis.

There are three regions of operation for the MOSFET. I am attaching a pdf file for you. Use the "square law" equations at the top of page 1 and assume μCoxW/L and VT are constants. VT = 1V in your case. A typical value for μCoxW/L might be 0.025 A/V2.

Thank you for the help, but, actually, the book I'm using doesn't present the square law model in the chapter where it asks this problem. So, I think I should be able to graph the approximate behavior of the MOSFET only by using the SR model (the MOSFET acts like a resistor R_{ON} in its ON state, for sufficiently small values of V_{GS}).

By the way, the book is "Foundations of Analog and Digital Circuits" by Agarwal and Lang. This problem is from Chapter 6.
 
To get a single value of IDS for specific VGS, I think you have to assume that VDS is large enough to saturate the MOSFET.
 
mfb said:
To get a single value of IDS for specific VGS, I think you have to assume that VDS is large enough to saturate the MOSFET.

What exactly do you mean by saturating the MOSFET?

In this situation, would I have a single value of i_{DS} = \dfrac{v_{DS}}{R_{ON}} for v_{GS} \geq V_T, which would then be plotted as a horizontal line in the i_{DS} versus v_{GS} graph?
 
pc2-brazil said:
What exactly do you mean by saturating the MOSFET?

In this situation, would I have a single value of i_{DS} = \dfrac{v_{DS}}{R_{ON}} for v_{GS} \geq V_T, which would then be plotted as a horizontal line in the i_{DS} versus v_{GS} graph?

On the contrary, I think what's intended here is not the saturated region but the 'linear' region. (The saturated region is when Vsd > (Vgs + VT). In that region, increasing Vds does not materially affect i if Vgs is held constant). The linear region is where i decreases with Vds for a given Vgs.

It's confusing nomenclature since the linear region is where the device is used as an on/off switch, which is thought of as the device being "saturated". I.e. Vds is about as low as it can go which is what you want when the device is "on". It's 'saturated' in the sense of minimum Vds which also implies minimum Ron.

The plot you need is i vs. Vds for various values of Vgs. For each value of Vgs, the value of i/Vds is approximately constant & varies only with Vgs. These values of i/Vds = 1/Ron. You will find that i/Vds increases as Vgs increases.

PS - com jeito vai! Eu era Carioca entre 1956-58!
 
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