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

The equation used to solve a diode is

`i_d = I_se^(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^(V_d)/(V_T)-1`, you'd notice that the
`e^(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^(V_d)/(V_T)-1`

`i_d\approx I_se^(V_d)/(V_T)`

Therefore, use
`i_d= I_se^(V_d)/(V_T)` to solve your equation.

Rearrange your equation to solve for
`V_D`.

`V_D=V_Tln((i_D)/(I_s))`

a.)

i.)

You're given
`I_s=10^(-11)A`

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

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

at
`i_d=1mA`,
`V_D=V_Tln((i_D)/(I_s))=(25.9\cdot10^(-3))ln((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.