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3 years ago

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I need help asapppp!!!!!

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11Alexandr11 [23.1K]3 years ago

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I don’t see the the picture?

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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

$i_d = I_se^\frac{V_d}{V_T}-1$

$i_d$ is the current going through the diode
$I_s$ is your saturation current
$V_D$ is the voltage across your diode
$V_T$ is the voltage of the diode at a certain room temperature. by default, you always use $V_T=25.9mV$ for room temperature.

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.

$i_d = I_se^\frac{V_d}{V_T}-1$

$i_d\approx I_se^\frac{V_d}{V_T}$

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

Rearrange your equation to solve for $V_D$ .

$V_D=V_Tln(\frac{i_D}{I_s})$

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

at $i_d=1mA$ , $V_D=V_Tln(\frac{i_D}{I_s})=(25.9\cdot10^{-3})ln(\frac{1\cdot10^{-3}}{10\cdot10^{-11}})=.417V$

<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

4 years ago