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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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An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
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Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

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