I understand there are two models \$Ic=\beta Ib\$ and \$Ic=I_s exp (\frac{V_{BE}}{Vt})\$ for a bipolar junction transistor.
However (correct me if I am wrong), I have not seen any textbook derivation which actually finds \$\beta\$ as a function of the circuit itself.
Is there some reason for this? I am asking this question because I wanted to know if there could be an expression for this behaviour:
Taken from BC547 datasheet.
In other words, given only \$I_C\$,\$V_{BE}\$,\$I_S\$ (is \$V_{CE}\$ required?) can you work out \$\beta\$ without going out and experimentally finding it?
To begin with (in active region), I think there is a forward biased pn junction between the base and the emitter. Thus it would be a case of finding the diode current (\$I_B\$) for the applied voltage (\$V_{BE}\$)? And then relating this to the \$I_C\$?
Small related question, is the \$I_S\$ in the \$Ic=I_s exp (\frac{V_{BE}}{Vt})\$ stated anywhere in a transistor datasheet?
Best Answer
The equation you are thinking of is from the Ebers-Moll BJT model, simplified for the forward-active operating region:
$$I_E = I_{ES}\left(\exp\left(\frac{V_{BE}}{V_T}\right) - 1 \right)$$
Note this gives the emitter current, not the collector current.
The other two equations in the Ebers-Moll model are
$$I_C = \alpha{}I_E$$
$$I_B = (1-\alpha)I_E$$
Unfortunately these equations don't help you find \$\beta\$ because \$\alpha\$ is just \$\beta\$ in disguise:
$$\beta=\frac{I_C}{I_B}=\frac{\alpha}{1-\alpha}$$
It's usually not in the datasheet, but if a SPICE model is available for your device, you can get a "typical" value from there. Look for the
ISparameter in the MODEL card.