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I'm simply trying to find Vt and W/L for a given practice exam problem shown below:

Practice Exam Problem

The solution is given as:
enter image description here

Initially, I was trying to use the equation as shown in line 1 of the solution to develop 2 equations with 2 unknowns and solve for each, but there appears to be a much faster way to arrive at the solution which I'm having trouble understanding. Specifically, I don't understand line 2 of the solution.

What is the logic behind taking the ratio of Id (0.225mA/0.1mA = 2.25)?

The book defines VGT as Vgs-Vt, I don't understand which numbers/logic is used to determine Vgt1/Vgt2 = 1.5. I suspect VGT is not equal to Vgt (lower case), but can't find a formal definition to support my suspicion.

And Finally, how is Vt = 0.6V? The most I can rationalize is that 1.5/2.25 = 0.66667, but doesn't seem right.

I appreciate any help. Thank you.
*This is a practice exam, not hw

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Why don't you simply divide the two the equation?

$$\frac{I_{D1}}{I_{D2}} = \frac{\textrm{k}(V_{GS1} - V_T)^2}{\textrm{k}(V_{GS2} - V_T)^2}$$

And the result is

$$\frac{I_{D1}}{I_{D2}} = \frac{(V_{GS1} - V_T)^2}{(V_{GS2} - V_T)^2}$$

$$\sqrt{\frac{I_{D1}}{I_{D2}}} = \frac{(V_{GS1} - V_T)}{(V_{GS2} - V_T)}$$

The End.

PS. Of course, this is true only if \$V_{DS1} = V_{DS2} = V_{DS}\$ otherwise : https://www.sccs.swarthmore.edu/users/06/adem/engin/e77vlsi/lab4/

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The ratio in currents is useful here because current is proportional to the square of overdrive voltage (called \$V_{gt}\$ here and \$V_{ov}\$ in some other courses)1. By taking the ratio in currents, you easily get the ratio of overdrive voltages, which is useful in some problems such as this one.

The notation appears to be a result of computing \$V_{gt}\$ at the two provided operating points which were implicitly numbered 1 and 2. \$V_{gt1} = 0.9\,\text{V} - V_t\$ and \$V_{gt1} = 0.8\,\text{V} - V_t\$.

Given \$V_{gt1}/V_{gt2} = 1.5\$, you know that \$\frac{0.9\,\text{V}-V_t}{0.8\,\text{V}-V_t} = 1.5\$. This can be easily seen to be satisfied by \$V_t = 0.6\,\text{V}\$ since 0.3/0.2 = 1.5.

1 As a simplified model that holds well for strongly-inverted FETs.

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  • \$\begingroup\$ Hi, thank you for responding to my questions. I'm going to see what I can learn about overdrive voltage right now, but I wanted to say that Vgt1/Vgt2=1.5 isn't given in the question explicitly, but only seems to be shown in the solution. Is there another way you were able to deduce the ratio is equal to 1.5? \$\endgroup\$ – InterestingGuy Aug 26 '19 at 20:44
  • \$\begingroup\$ @InterestingGuy The ratio of the two provided currents was 2.25. Since the ratio of the overdrive voltages is the square root of the ratio of currents assuming everything else is unchanged, we obtain Vgt1/Vgt2=1.5. \$\endgroup\$ – Reinstate Monica - ζ-- Aug 26 '19 at 21:10

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