Is VBEn turn on voltage, if so why is it changing?
It says when \$V_I\$ is increased, voltage at the base of Qn increases and \$V_O = V_I + V_{BB}/2 -V_{BEn}\$
Can you please explain me how does increase in \$i_{Cn}\$ result in increase in \$V_{BEn}\$ voltage? Firstly, I didn't get why was it increased because \$V_{BEn}\$ must be constant, because it is turn on voltage. Secondly, If it is able to change, why was it increased but not decreased?
Thank you
Edit: new link of the circuit
Base and emitter together forms a pn-junction. And hence the current through it varies exponentially with the voltage across it. $$I=I_0\exp(v/v_T)-I_0$$
If you increase the voltage across this junction from 0V, the current will be negligible till some voltage. After that you can observe sudden increase in current. This voltage is what you are referring to as "turn-on voltage". This is not actually a constant.
The voltage across the junction will not vary much after turn-on voltage (only \$\approx 60\,mV\$ change can be seen for 1 decade change in current) because of the logarithmic variation of voltage with current. That's why people assumes it as a constant.
Read this answer also: Difference in real diode characteristics vs shockley equation
\$V_{BE}\$ is the voltage drop across the base-emitter junction. Note, for example, when an NPN is in active region the base-emitter acts as a diode and drops about 0.7V.
The voltage \$V_{BE}\$ has a logarithmic relationship with the collector current, whch is given roughly $$V_{BE}=60mV.log(\frac{I_C}{I_S})$$
According to above, we have, $$V_{BE_2}-V_{BE_1}=60mV.log(\frac{I_{C_2}}{I_{C_1}})$$ or $$V_{BE_2}=V_{BE_1}+60mV.log(\frac{I_{C_2}}{I_{C_1}})$$ So now if \$I_{C_2}\$ increases by ten times such that \$I_{C_2}=10I_{C_1}\$, the base-emitter voltage increases by $$\Delta V_{BE}=60mV$$.
As you see the voltage across the base-emitter junction doesn't change much per decade of collector current. Therefore the base-emitter voltage is a very weak function of the collector current.
