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Circuit

Graph Looking at the problem mathematically, we see the equations

$$I_\text{ds}=k_n'(V_\text{gs}-V_t)^2$$

$$V_\text{ds}=V_\text{dd}-I_\text{ds} \times R_o$$

Following this, it would be predictable that \$V_\text{ds}\$ vs. \$V_\text{gs}\$ would have a parabolic relationship. However, that is not the case, proved by the graph attached. What would be the reason?

Null
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Janhvi
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    A hand-drawn graph doesn't "prove" anything. Just because you drew the curve roughly linear doesn't mean it is in fact linear. We only approximate as linear in the small signal regime, for which we do a first order Taylor series expansion of the quadratic term. – LetterSized Feb 23 '22 at 19:23
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    If you take a large parabola and plot a small segment of it on graph, it will look like a straight line. So you have to bring a precisely scaled and plotted graph before trying to defy the underlying mathematics. – Mitu Raj Feb 23 '22 at 19:40

1 Answers1

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You are looking at VTC (voltage transfer characteristics) for the inverter shown and not the device characteristics.

It is because the saturation region for MOSFETs has linear characteristics while the triode region is parabolic.

Read: Fundamentals of Microelectronics by Behzad Razavi (link) Page 280-290, from where I have cobbled together the following collage.

For triode region, there is a parabolic dependence on \$V_{DS}\$ in eq 6.9. As saturation is reached (eq 6.35) the dotted curve is not followed any longer and the dependence on \$V_{DS}\$ becomes linear modeled by the \$\lambda\$ parameter.

enter image description here

Syed
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