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3 years ago

12

A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:

(a) vSG = 2 V, vSD = 3 V; (b) vSG = 5 V, vSD = 2 V.

1 answer:

Ksju [112]3 years ago

7 0

The answer is b ! Hope I helped

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Write an ALP to separate odd and even numbers from an array of N numbers; arrange odd

Marta_Voda [28]

Below is the program to separate odd and even numbers

<u>Explanation</u>:

<u>L1:</u>

mov ah,00

mov al,[BX]

mov dl,al

div dh

cmp ah,00

je EVEN1

mov [DI],dl

add OddAdd,dl

INC DI

INC BX

Loop L1

jmp CAL

<u>EVEN1:</u>

mov [SI],dl

add Even Add,dl

INC SI

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Loop L1

<u>CAL: </u>

mov ax,0000

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mov al,OddAdd

mov bl,EvenAdd

MOV ax,4C00h

int 21h

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The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.

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4 years ago

Describe how the Rotary Engine works.

LekaFEV [45]

Answer:

Rotary engine was early known by the name of internal combustion engine. It convert heat from a high pressure of combustion. The main advantage of rotary engine is that it can be operate with less number of vibration. It works on the principle of converting pressure into rotating motion. In rotary engine the expansion pressure is applied on the flank rotor.

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3 years ago

Tech A says that bleeding an electronic brake control system is just like bleeding a non-electronic brake control system. Tech B

PtichkaEL [24]

Answer:

Tech A is correct.

Explanation:

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So,

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3 years ago

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crys

vredina [299]

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

$\cos \phi = \frac{(1,0,0) \cdot (1,1,0)}{1 \times \sqrt2}$

$=\frac{1}{\sqrt2 }$

$\cos \lambda = \frac{(1,0,0) \cdot (1,-1,1)}{1 \times \sqrt3}$

$=\frac{1}{\sqrt3 }$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

ii. Slip plane --- (1 1 0)

Slip direction -- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, -1, 0)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iii. Slip plane --- (1 0 1)

Slip direction --- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, 1)}{1 \times \sqrt2} =\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, 1, -1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction ---- [1 1 1]

$\cos \phi = \frac{(1, 0, 0) \cdot (1, 0, -1)}{1 \times \sqrt2}=\frac{1}{\sqrt2}$

$\cos \lambda = \frac{(1, 0, 0) \cdot (1, -1, 1)}{1 \times \sqrt3} =\frac{1}{\sqrt3}$

τ = σ cos Φ cos λ

∴ $50= \sigma \times \frac{1}{\sqrt2} \times \frac{1}{\sqrt3}$

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

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Answer:

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Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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3 years ago