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
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
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 ?
They've explained more clearly the relation of Vi and Vo here: ruclips.net/video/cuOFG_ISHno/видео.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
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
The best video about this subject so far I can find on RUclips
same
I like the way you explain and calculate. I need more of these.
Amazing job as always, but there really aren't enough topics for you, there's a lot more topics in this I need 🥺
Please, Could you tell me what the name of teaching
tool or software? I really like it. It's very nice and great video.
Its Sketchbook
Thanks. Best explanation
you are an amusig teacher. Thanks
I have a doubt sir.
How does the voltage 0.7 remain constant ?
I mean..isn't it changing with Vi ?
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
what is meant by grounded,connected to the ground or a referrence point.
when a circuit end is grounded/earthed it means it has an electric potential of 0V
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
Nice explaination.... Basically... Purely based on mathematical analysis
khan academy is the best
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 ?
can you make the video on logic gates?
I ve a question, In the circuit what is that + 3 V , Where does it come from?
The +3V is the DC power supply. In simple terms it can be thought of as a 3V battery
Great Explanation! Thanks Sir Mahesh! :)
Sir, I can't understand this part : 11:24
(Higher potential drop, lower potential drop)
They've explained more clearly the relation of Vi and Vo here: ruclips.net/video/cuOFG_ISHno/видео.html
They've also explained about those voltage drops
Pls make video of logic gate
Sir,.7 v is already an input votage. So,why we added extra input voltage? I dont understand this❤️❤️
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
You should not do this. But if you really want to relate them without deltas then your equation will be like V_o = k*V_i + c
Remember there is no fixed 0 potential , we define it on our comfort.
how can you take voltage drop from +3v to Vo since Vo is the amplified voltage and will be much higher than +3v supply
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
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
i dont understand the final graph .we needed to amplifie the input voltage but we get. reversed amplification. why
Hi if we can amplify voltage , why don't we use it as a big generator of electricity?? 🤔🤔🤔
If V0 goes below 0.7V... Doesn't it become forward biased?
They've explained more clearly the relation of Vi and Vo here: ruclips.net/video/cuOFG_ISHno/видео.html
Your Doubt : 11:15
My brain is now yougurt