That's very simple. It is the barrier potential. Under forward bias if the external voltage exceeds just the barrier potential the current shoots linearly. Thus even the potential changes a little the current is always the same maximum. Thus the base always stays at 0.7.
0.7 V is the barrier potential at the junction of a typical PN junction, once the supply voltage increases beyond this, the voltage drop won't increase much from 0.7 V. Only current will increase. (This means resistance adjusts itself to maintain only 0.7 V drop across the junction). In metals resistance won't change much , so drop across the conductor changes.
@@udaykiran2798 hey... I have a doubt,please clarify, firstly, don't understand, what is the work with 0.7v there???... In previous videos it is mentioned that V input =0.7 v. But now.. I don't understand... There are two voltage... Separately Vi and 0.7.two voltages are there..... Why... Instead of 0.7 v... Why didn't they connect Vi directly to circuit??
@@penumalasunitha14 hi sunitha, i request you to go and check your connections with barrier potentials of semiconductors, if that doesn't clear your doubt then i will surely help you.
@@udaykiran2798 I have a doubt that the vin is alternate voltage so when it is between 0 to 0.7V it acts as off switch so how would it amplify the input voltage
Just a quick question though Isn't RB usually supposed to be greater than RC that should indicate a reverse amplification, or does beta make the change ?
Probably literally the only vid online explaining input and output properly without asking me to mug up CE BE ICE IBE ABCD ETC. BECAUSE OH MY GOD WOULD IT PAIN THE PHYSICS BOOK AUTHORS TO USE LESS CONFUSING TERMS UGHH
Yes and it's just a way to tell the voltage at a specific point in the circuit to make things easier, it doesn't matter which point in the circuit is grounded or if it even is grounded, just simplification
They've explained more clearly the relation of Vi and Vo here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-cuOFG_ISHno.html They've also explained about those voltage drops
What if I want the output voltage (V_o) to be equal to the input voltage (V_i) multiplied by a constant (k), and not just their deltas? i.e. V_o = k * V_i
Could you elaborate your doubt please? I am guessing the doubt is how is the output voltage only 3V? The amplification requires electrical energy. That energy is given by the supply. So if the output supply voltage is 3V, then the maximum output voltage cannot exceed 3V
@@KhanAcademyIndiaEnglish I am only telling that if the amplification of voltage is done from i/p to o/p then how could there be a drop from +3v to amplified output
