What are the values of K and VTR for a MOSFET with given v-I characteristics?

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Discussion Overview

The discussion revolves around determining the values of K and VTR for a MOSFET based on its v-I characteristics. Participants explore the relationships between the gate-source voltage (VGS), drain-source voltage (VDS), and the drain current (iD) in different operational regions of the MOSFET.

Discussion Character

  • Homework-related, Technical explanation, Exploratory

Main Points Raised

  • One participant expresses confusion about how to derive K and VTR from the provided v-I characteristics and notes that the book gives specific values without showing the derivation process.
  • Another participant suggests that the absence of VDS in the equation indicates a specific operational region that needs to be identified.
  • A participant proposes using the point where the graph becomes horizontal to find VTR, indicating that for VGS=4V, VTR appears to be 2V.
  • Further discussion reveals that using different VGS curves can yield different VTR values, with one participant noting that using the VGS=8V curve suggests VTR is around 3V.
  • Participants discuss the importance of selecting the correct operational region (triode, saturation, constant current) and the corresponding equations for accurate calculations.
  • One participant confirms they figured out the problem after applying the method of using two different curves to derive K and VTR.

Areas of Agreement / Disagreement

Participants generally agree on the method of using different curves to find K and VTR, but there is no consensus on the exact values of VTR, as different curves suggest different results. The discussion remains unresolved regarding the definitive values of K and VTR.

Contextual Notes

Participants note that the equations used depend on the operational region of the MOSFET, and there are limitations in determining VTR based solely on visual inspection of the graph without considering the specific conditions of operation.

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Homework Statement


A MOSFET has the set of v-I characteristics shown in Fig. P5.5.

What are the values of K and VTR?

Homework Equations


iD=K(VGS-VTR)2


The Attempt at a Solution


I have looked in the book and all of the example problems give VTR, but it doesn't show how to actually get the K and VTR values based on the graph. I really have no idea what to do.

The answer in the book says K=0.25 mA/V2 and VTR=2V. Is VTR 2V because VGS increases by 2V?
 

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First job for you is to figure out where on the Vds axis (the x axis) you're operating at.
Hint: notice that Vds does not appear in your equation. What does that suggest?
 
The transition from triode to constant current uses the equation VDS=VGS-VTR. So, substituting into the equation I gave in my first post:

iD=KVDS2

I also noticed that I can find VTR using the point where the graph goes horizontal. For VGS=4V, the point would be at 2V. Then 4-2=2V is what I want (verified this by looking at an example in the book where I was given VTR). This seems to work fine, but not sure if it is the "correct" way.
 
OK, there are 2 ways to go:
1. Use VT = Vgs - Vds. So look where the curves bend flat. As you point out, using the Vgs = 4V curve it looks like VT ~ 2V.

But if you used the Vgs = 8V curve you'd get more like VT = 3V (the curve bends at about Vds = 5V so 8 - 5 = 3). The Vgs = 6V and 11V curves would also compute to about VT = 3.

A better way to go is to use your original equation of Ids = k(Vgs - VT)^2. Use it twice, on two different curves, to solve simultaneously for k and VT.

But as I said, first you have to decide where on the Vds axis you're sitting. That equation of yours holds only if Vds > (Vgs - VT).
 
Do I just pick a region (i.e. constant current, saturation, triode) and use the corresponding equation?
 
hogrampage said:
Do I just pick a region (i.e. constant current, saturation, triode) and use the corresponding equation?

Yes, but remember what I said about the region where that equation is valid.

There are separate equations fdepending on the region of the i-V characteristics you're looking at.

For example, in the triode region your equation would not be appropriate.
 
Last edited:
Ah, gotcha. I figured it out after using two different curves, as you said.

Thank you for the help.
 

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