Electronic Devices - Semiconductor Diodes - Discussion
Discussion Forum : Semiconductor Diodes - General Questions (Q.No. 3)
3.
It is not uncommon for a germanium diode with an Is in the order of 1–2 
A at 25°C to have leakage current of 0.1 mA at a temperature of 100°C.
A at 25°C to have leakage current of 0.1 mA at a temperature of 100°C.Discussion:
14 comments Page 1 of 2.
RAJALINGAM said:
1 decade ago
Can anyone please explain the above?
Kachi said:
1 decade ago
Can someone throw more light to this?
Anuktha said:
1 decade ago
I can't understand. Can anyone explain for the above question?
Neha said:
1 decade ago
Because leakage current doubles itself for every 10 degree rise in temperature.
Marichike said:
1 decade ago
Germanium diode can work fine upto 200C, so its not Uncommon for 100C.
Tejas said:
1 decade ago
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.
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.
(1)
Aditya Sharma said:
1 decade ago
I know:
i(T2) = i(T1) * 2^((T2-T1)/10).
i(T2) = i(T1) * 2^((T2-T1)/10).
Priyank saxena said:
1 decade ago
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.
Now we can calculate.
(1)
Anshul jain said:
1 decade ago
For every 10 degree rise in temperature reverse current becomes twice.
Srimanta bijali said:
1 decade ago
Is it common for all types of diode?
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