The instability of Collector current(Ic) of a transistor depends on three factors viz. Icbo, Vbe & Beta. Can you explain why Vbe decreases(by 2.5 mV/°C) with increase in temperature?
"Can you explain why Vbe decreases(by 2.5 mV/°C) with increase in temperature?"
To me, this is not the correct wording and can lead to misunderstandings.
No - Vbe does not decrease. The other way round: When the temperature goes up, the collector current increases. However, if you want to keep the collector current constant you must (externally) decrease the Vbe voltage by 2.5mV/K.
This tempco is attributet to the temperature sensitivity of Icbo only.
This link leads you to a note written by one of the best-known electronic engineers (Robert A. Pease): https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=9&ved=2ahUKEwiyz5qugKfpAhVmShUIHdoDBM0QFjAIegQIAxAB&url=ftp%3A%2F%2Ftelescript.denayer.wenk.be%2Fpub%2FCD-Microcontrollers%2FIIW%2FABA%2FCD_analoog_e%2Fbijlagen%2Ftransistoren%2FVbe_stuff.pdf&usg=AOvVaw2Z5cjXa6fAcZPkYwGkO7F3
The diode current has the formula
Idiode = Is * e ^ [Q * Vdiode / (K * T)]
WIth the T being temperature, and a common voltage at 1milliAmp being about 0.6 volts, we can simply divide that 0.6 volts by room temperature of 290 degree K, and we have about 2.1 milliVolts change per degree. This however requires the WRONG POLARITY of voltage change.
Digging further, wikipedia says I am wrong, and that the change of Is will be the far greater (and opposite polarity) effect. Check out Wikipedia "diode equation".
examining "reverse saturation current" on Wikipedia, we see the useful statement "a rule of thumb is that the current doubles for every 10 ° C". We can use that, along with 18 milliVolts being adequate to cause Idiode to double, and see this predicts 18/10 == 1.8 milliVolts decrease.