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mina [271]
3 years ago
13

Consider a CMOS inverter which has ideal transistors with the following characteristics: PMOS transistor: W/L = 2; Mobility (up)

= 72cm/V*s; threshold voltage = -0.4V nMOS transistor: W/L = 1; Mobility(un)= 180cmP/V*s; threshold voltage = 0.4V 180nm process; gate oxide capacitance/unit area = 8.6e-7 F/cm²; Vpp = 1.8V a) Calculate the B for each transistor, including the unites (3 points) b) What modes of operation is each transistor in when Vin=OV, 0.9V, and 1.8V? (3 points) c) Estimate the current through the inverter if Vin=0.9V. List any assumptions you make. (8 points) d) Would you expect the current to be higher or lower if the inverter was implemented in a 130nm process? Explain your answer. (6 points)
Engineering
1 answer:
yaroslaw [1]3 years ago
6 0

Answer:

The transistor will be in amplifier mode and we will expect current will be higher than expected.

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Which allows a user to run applications on a computing device? Group of answer choices Application software CSS Operating system
sveticcg [70]

Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user.  The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

Cheers.

4 0
3 years ago
Who is/are the founder/founders of transistor? ​
den301095 [7]

Answer:

William Shockley, Walter Houser Brattain and John Bardeen.

Explanation:

It was built in 1947 and they won the novel peace prize in 1956

7 0
3 years ago
Read 2 more answers
British standered institution
Scorpion4ik [409]

Answer:

The British Standards Institution, is the national standards body of the United Kingdom. BSI produces technical standards on a wide range of products and services and also supplies certification and standards-related services to businesses

6 0
3 years ago
Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)
AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

  • Since V_{ds} > V_{gs} - Vt because V_{gs} > Vt.
  • By the saturation region the MOSFET is operating.
  • A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
  • MOSFET is a transistor which is a device of semiconductor vastly used for the electronic amplifying signals and switching in the devices of electronics.
  • The core of this is integrated circuit.
  • It is fabricated and designed in an individual chips due to tiny sizes.
7 0
3 years ago
Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,
Lyrx [107]

Answer:

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0

Explanation:

v_0 = Initial velocity of ball A

v_A=v_0\cos45^{\circ}

v_B = Initial velocity of ball B = 0

(v_A)_n' = Final velocity of ball A

v_B' = Final velocity of ball B

e = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

Coefficient of restitution is given by

e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0

v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0

Adding the above two equations we get

2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0

\boldsymbol{\therefore v_B'=0.6364v_0}

(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0

From the conservation of momentum along the plane of contact we have

(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}

v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}

Velocity of ball B after impact is 0.6364v_0 and ball A is 0.711v_0.

5 0
3 years ago
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