Question

(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i) 10 µA, 100 µA, 1 mA, and (ii) −5 × 10−12 A. (b) Repeat part (a) for IS = 10−13 A and part (a) (ii) for −10−14 A.

Answer 1

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

note: always use  $V_T=25.9mV$

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.