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2 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Engineering

1 answer:

Murljashka [212]2 years ago

5 0

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$

also note that ; Pactive ∝ Nd2 (

tD = K . $\frac{Vdd}{(Vdd - Vt )^2}$ since K = constant

Hence : Nd ∝ rt

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3 years ago

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Answer:

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Explanation:

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i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?

solution

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Standard error = $\frac{sd}{\sqrt{n}}$

Standard error = $\frac{12}{\sqrt{16}}$

Standard error = 3

so here we get Z value for 3 units away are from mean are

mean = -1 and + 1

so here

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probability P = P( z < -1 or z > 1)

probability P = 0.1587 + 0.1587

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