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denis23 [38]
3 years ago
13

Design complementary static CMOS circuits with minimized number of transistors to realize the following Boolean functions (hint:

you may want to try logic minimization before implementing it in circuits): F = (ABC + D(A+B)) F = AC + BD F = ABCD + ABC + ABCD + ABC + ABCD+ABCD.

Engineering
1 answer:
Pie3 years ago
8 0

Answer:

as pull up network. the metteing point of pull down and pull up is the point where we take the output

note 1: if two n-mos are connected in series it gives logical AND and p-mos paralle gives logical-AND

note 2: if two n-mos are connected in parallel it gives logical OR and p-mos series gives logical-OR

note 3: output is always complement of what we implement

example Y= (AB)'

image attached

A) F = (ABC + D(A+B) )'

pulldown:

this can be realize by takeing three n-mos in series which gives ABC ,two n-mos are parallel which in series with another n-mos whic gives D(A+B), now connect ABC and D(A+B) in parallel

pull up

this can be realize by takeing three p-mos in parallel which gives ABC ,two p-mos are series which is in serires with

another p-mos whic gives D(A+B), now connect ABC and D(A+B) in series

the out put will be (ABC + D(A+B) )'

so we require total 6-mos and 6-pmos total 12mos transistors

B) F = AC + BD

pull down

this can be realize by takeing two n-mos in series which gives AB ,two n-mos are in series

which whic gives BD, now connect AC and BD in parallel

pull up

this can be realize by takeing two p-mos in parallel which gives Ac ,two p-mos are in parallel

which whic gives BD, now connect AC and BD in series

the output is (AC+BD)'

to avoid the complement we have to connect the output to c-mos inverter then we get AC+BD

so we require 5-nmos, 5-pmos total 10 mos transistors

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krek1111 [17]

Answer: E-cycle used computer equipment. Recycle Printer Cartridges and Drum Units. Properly Dispose of Old Technology.

Hope this helps!

8 0
3 years ago
Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi
Hitman42 [59]

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

  • 3-SAT ≤p TSP due to any  complete problem of NP to other problem by exits of reductions.
  • If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
  • If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
7 0
3 years ago
Gshyevdnmdcicm gngjc
Len [333]

This is not a valid question. Please try again.

4 0
3 years ago
Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by
irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

5 0
3 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit
8090 [49]

Answer:

a) The change in Kinetic energy, KE = -1.95 kJ

b) Power output, W = 10221.72 kW

c) Turbine inlet area, A_1 = 0.0044 m^2

Explanation:

a) Change in Kinetic Energy

For an adiabatic steady state flow of steam:

KE = \frac{V_2^2 - V_1^2}{2} \\.........(1)

Where Inlet velocity,  V₁ = 80 m/s

Outlet velocity, V₂ = 50 m/s

Substitute these values into equation (1)

KE = \frac{50^2 - 80^2}{2} \\

KE = -1950 m²/s²

To convert this to kJ/kg, divide by 1000

KE = -1950/1000

KE = -1.95 kJ/kg

b) The power output, w

The equation below is used to represent a  steady state flow.

q - w = h_2 - h_1 + KE + g(z_2 - z_1)

For an adiabatic process, the rate of heat transfer, q = 0

z₂ = z₁

The equation thus reduces to :

w = h₁ - h₂ - KE...........(2)

Where Power output, W = \dot{m}w..........(3)

Mass flow rate, \dot{m} = 12 kg/s

To get the specific enthalpy at the inlet, h₁

At P₁ = 10 MPa, T₁ = 450°C,

h₁ = 3242.4 kJ/kg,

Specific volume, v₁ = 0.029782 m³/kg

At P₂ = 10 kPa, h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg, x₂ = 0.92

specific enthalpy at the outlet, h₂ = h_1 + x_2 h_{fg}

h₂ = 3242.4 + 0.92(2392.1)

h₂ = 2392.54 kJ/kg

Substitute these values into equation (2)

w = 3242.4 - 2392.54 - (-1.95)

w = 851.81 kJ/kg

To get the power output, put the value of w into equation (3)

W = 12 * 851.81

W = 10221.72 kW

c) The turbine inlet area

A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2

3 0
3 years ago
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