An NMOS amplifier is to be designed to provide a 0.20-V peak output signal across a 20-kΩ load that can be used as a drain resis
tor.
(a) If a gain of at least 10V/V is needed, what gm is required?
(b) Using a dc supply of 1.8V, what values of ID and VOV would you choose?
(c) What W/L ratio is required if μnCox=200 μA/V^2? If Vt =0.4V, find VGS.
(a) If a gain of at least 10V/V is needed, what gm is required?
(b) Using a dc supply of 1.8V, what values of ID and VOV would you choose?
(c) What W/L ratio is required if μnCox=200 μA/V^2? If Vt =0.4V, find VGS.
1 answer:
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Answer:
The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons
Explanation:
We know that for two point charges of magnitude the magnitude of force between them is given by
where
is constant
is the separation between the charges
Initially when the charges are separated by 2.4 meters the force can be calculated as
Now when the separation is reduced to 0.7 meters the force is similarly calculated as
Applying value of the constant we get
Thus
Answer:
a) 740 W b) 6.2 A c) 8.1%
Explanation:
We need first to get the total energy spent during the 30 days, that can be calculated as follows:
1 month = 30 days = 720 hr
If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:
80 $/mo = 0.15 $/Kwh*x Kwh/mo
Solving for the total energy spent in the month:
Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:
b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:
c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:
P = 740 W - 60 W = 680 W
The new value of the energy spent during the entire month will be as follows:
E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo
The reduction in percentage regarding the total energy spent can be calculated as follows:
ΔE =
⇒ ΔE(%) = 8.1%