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MrRa [10]
3 years ago
11

Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)

. Which of the following best describes the behavior of this transistor?a. The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the triode region. A value of VDS=12 V would not result in a higher IDS current. b. The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region. A value of VDS=12 V would result in higher IDS current. c. The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region. A value of VDS=12 V would result in a higher IDS current due to channel-length modulation. d. The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region. A value of VDS=12 V would not result in a higher IDS current.
Engineering
1 answer:
AveGali [126]3 years ago
7 0

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.
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What are the functions of each computer program
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3 years ago
Water is flowing at a rate of 0.15 ft3/s in a 6 inch diameter pipe. The water then goes through a sudden contraction to a 2 inch
Georgia [21]

Answer:

Head loss=0.00366 ft

Explanation:

Given :Water flow rate Q=0.15 \frac{ft^{3}}{sec}

         D_{1}= 6 inch=0.5 ft

        D_{2}=2 inch=0.1667 ft

As we know that Q=AV

A_{1}\times V_{1}=A_{2}\times V_{2}

So V_{2}=\frac{Q}{A_2}

     V_{2}=\dfrac{.015}{\frac{3.14}{4}\times 0.1667^{2}}

     V_{2=0.687 ft/sec

We know that Head loss due to sudden contraction

           h_{l}=K\frac{V_{2}^2}{2g}

If nothing is given then take K=0.5

So head lossh_{l}=(0.5)\frac{{0.687}^2}{2\times 32.18}

                                    =0.00366 ft

So head loss=0.00366 ft

4 0
3 years ago
An asphalt concrete mixture includes 94% aggregate by weight. The specific gravities of aggregate and asphalt are 2.65 and 1.0,
garik1379 [7]

Answer:

2.0%

Explanation:

Percentage of aggregate = 94%

Specific gravity = 2.65

Specific gravity of asphalt = 1.9

Density of mix = 147pcf = 147lb/ft³

Total weight of mix: (volume = 1ft³)

= (147lb/ft³)(1ft³)

= 147lb

Percentage weight of asphalt in<u> mix:</u>

100% - 94%

= 6%

Weight of asphalt binders

= 6% x 147lb

= 8.82lb

Weight of aggregate in mix:

= 94% x 147

= 138.18lb

Specific weight of asphalt binder:

(Gab)(Yw)

Yw = specific Weight of water

= 62.4lb

Gab = specific gravity of asphalt binder

= 1.0

(62.4lb)(1.0)

= 62.4 lb/ft³

Volume of asphalt in binder:

8.82/62.4

= 0.14ft³

Specific weight of binder in mix:

2.65 x 62.4lb/ft³

= 165.36 lb/ft³

Volume of aggregate:

= 138.18/165.36

= 0.84ft³

Volume of void in the mix:

1ft³ - 0.84ft³ - 0.14ft³

= 0.02ft³

<u>The percentage of void in total mix:</u>

VTM = (0.02ft³/1ft³)100

= 2.0%

8 0
3 years ago
IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The follow
Irina-Kira [14]

Answer:

13.9357 horse power

Explanation:

Annealed copper

Given :

Width, b = 9 inches

Thickness, $h_0=2.2$ inches

K= 90,000 Psi

μ = 0.2, R = 14 inches, N = 150 rpm

For the maximum possible draft in one pass,

$\Delta h = H_0-h_f=\mu^2R$

     $=0.2^2 \times 14 = 0.56$ inches

$h_f = 2.2 - 0.56$

     = 1.64 inches

Roll strip contact length (L) = $\sqrt{R(h_0-h_f)}$

                                             $=\sqrt{14 \times 0.56}$

                                             = 2.8 inches

Absolute value of true strain, $\epsilon_T$

$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$

Average true stress, $\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$ Psi

Roll force, $L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$

                                 = 788,900 lb

For SI units,

Power = $\frac{2 \pi FLN}{60}$  

           $=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$

           = 10399.81168 W

Horse power = 13.9357

6 0
3 years ago
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