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3 years ago

12

) Assuming different AM regulations; the receiver is using mixer with subtracting format. The frequency selectivity ratio is app

roximately 10 and the AM range is from 750 kHz to 2600 kHz. The intermediate frequeney is most nearly:

1 answer:

Zarrin [17]3 years ago

8 0

Answer:

$F=710KHZ$

Explanation:

From the question we are told that:

Frequency selectivity ratio $R=10$

AM range 750 kHz to 2600 kHz

Therefore Bandwidth is

$B=2100-680$

$B=1420KHZ$

Generally the equation for The intermediate frequency is mathematically given by

Intermediate frequency=\frac{Bandwidth}{2}

$F=\frac{B}{2}$

$F=\frac{1420}{2}$

$F=710KHZ$

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3 years ago

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<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

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1 year ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

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3 years ago

(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i

kirill115 [55]

Answer:

The equation used to solve a diode is

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If you look at the equation, $i_d = I_se^\frac{V_d}{V_T}-1$ , you'd notice that the $e^\frac{V_d}{V_T}$ grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.

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Therefore, use $i_d= I_se^\frac{V_d}{V_T}$ to solve your equation.

Rearrange your equation to solve for $V_D$ .

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a.)

i.)

You're given $I_s=10^{-11}A$

at $i_d=10\mu A$ , $V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{10\cdot10^{-6}}{10\cdot10^{-11}})=.298V$

at $i_d=100\mu A$ , $V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{100\cdot10^{-6}}{10\cdot10^{-11}})=.358V$

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<em>note: always use</em> $V_T=25.9mV$

ii.)

Just repeat part (i) but change to $I_s=-5\cdot10^{-12}A$

b.)

same process as part A. You do the rest of the problem by yourself.

4 0

3 years ago