I read somewhere that in a BJT transistor, the Vbe vary with temp by -2mV/°C.
Is this a rule which can be demonstrated, then how to demonstrate ?
Or is it an experimental results?
Thanks for your hep.
Anyways, if you connect the transistor as a diode (base shorted to collector) and pass some current through it the voltage betveen base and emitter is Vth and if you change the temperature while keeping the current constant you can see it drops with an increase in temperture.
Value -2mV/C though a good approximation and exact value depends on the process and close to this value.
Nowadays IC temperatures (like pentium uP) are measured by this property of a diode inside an IC.
This value of -2...-2.2mV/0C is for both silicium and germanium junctions, and fairy linear. So, diode and transistor (connected as diode) can be used for temperature measuring within its operating temperature range. However, transistor gives the better linearity than the diode. To measure low temperature within the narrow range, this is the perfect sensor.
Remember the current relation for a BJT:
Ic = Is*exp(VBE/VT), where the thermal voltage VT=kT/q. k is Boltzmanns constant and q the electron charge.
Solving this for VBE and differentiating with respect to the temperature gives
dVBE/dT = k/q*ln(Ic/Is)
For example with Ic = 1 mA and using a typical value Is = 10^-14 gives -2.2 mV/C. This is only an approximate value but maybe good enough. -2.2 mV/C (or just -2 mV/C) is used as a rule of thumb. Actually Is is also temperature dependent in some more complicated way. More exact relations are found in books on semiconductor physics.