Electronic Devices - Semiconductor Diodes - Discussion

We know that,

Id=Is e^kT/V, where k is the Boltzmann's Constant and k=1.38x10^-23, the T is the temperature in kelvin = T in degree Celsius+273 and V is the voltage across the diode. As it is the germanium diode, we take V=0.3V. Then calculate the value of Id at 298 kelvin i.e. at 25 degree Celsius. Then the value of Id doubles itself for every 10 degree rise in the temperature. Thus, the value of Id obtained will be equal to 0.1x10^-3A. Also, the diode can operate at this temperature. Therefore, it is not uncommon to have this reading.

Because for every 10 degree rise in temperature leakage current doubles. Now on reaching 25 to 95 we have to increase the 6X10+5.

Now we can calculate.

Use Shockley equation with vd of germanium is 0.3v and change your temperature to degree kelvin.

Because leakage current doubles itself for every 10 degree rise in temperature.

Germanium diode can work fine upto 200C, so its not Uncommon for 100C.

For every 10 degree rise in temperature reverse current becomes twice.

I can't understand. Can anyone explain for the above question?

I know:

i(T2) = i(T1) * 2^((T2-T1)/10).

Can someone throw more light to this?

Can anyone please explain the above?