Question

A silicon diode has saturation current of 7.5μa at room temperature of 300k calculate saturation current at 400k

Answer 1

Answer:

I01

= 7.5

µ

A = 7.5 * 10

-6

T1 = 300 K T2 = 400 K

∆

T = T2 – T2 = 460 – 300

∆

T = 160 K I

02

= 2

(

∆

T/10)

* I

01

= 2

(160/10)

* 7.5 * 10

-6

I

02

= 0.49152 A

Explanation:

Answer 2

7.68mA

• In a reverse-biased diode, the majority of carriers will move from the junction. Appreciable current flow will be absent.  
• Minority carriers will move towards the junction.  
• There will be a flow of charges due to the minority carriers across the junction.  
• This minor current is said to be saturation current.

Given,

$I_{ot}$ = 7.5μa = 7.5* $10^{-6}$

$T_{1}$ = 300k = 27°

$T_{2}$ = 400k = 127°

ΔT = $T_{2}$ - $T_{1}$

     = 127-27

ΔT = 100

$I_{o2}$ = $2^{\frac{ΔT}{10} }$ * $I_{ot}$

    = $2^{10}$ * $7.5* 10^{-6}$

    = 7680 $*10^{-6}$

    = 7.68mA