@@santosh-mb5wc In the active regions one of the junctions is reverse biased and the other is forward biased.. We apply input across the forward biased one which offers a low resistance path.. Now the current must pass through both the diodes and so as the other diode is reverse biased it offers a large resistance and hence when we take our output across it, the signal has been amplified.. NESO academy has explained the concepts in earlier videos in their playlist :)
Aapko kese thnx kahoon samajh nahni aa raha..... Literally aapka formula meri jindgi bacha liya jaha bhi dhoondta thaa sabke sab circuit draw karne mein lag jaate the lekin aap ne alga se samjhaya direct forumula bol diya love u bro... Many many thnx This is a love from Odisha God bless u And again thnx a lot....
You are awesome, dude. I cant believe i understand better an english explained video between an spanish explained video. Thank u and regards from chile =^)
Q1. it is known that parallel combination of resistor divides current not voltage, than why this is known as voltage divider? Q2. While finding V_th, total resistance was taken as R1+R2, but that's the case for series not parallel, why?
I really like the the way you instruct. I learned all of this at UMass, Lowell for my EE degree. At the time it seemed so hard. Now, it brings pleasure having my memory refreshed with all this great stuff. Do you do Signals and Systems, Complex Variables?
Your explanation is perfect 👍👍 But the only problem is with the sound. Your voice is clear but not audible properly. Just increase your voice and it will be very good. 👍
At 8:30 you miss a trick. Since Ic = β.(Vth - Vbe) / (Rth + (β+1).Re), once you have the condition Rth > 1, you can immediately derive Ic = (Vth - Vbe) / Re and thence calculate your operating points. By the way, Vce is *not* the output voltage as you claim at 9:35 - the output voltage is the voltage at the transistor collector and that is Vce + IeRe relative to ground. Students get confused when the build the circuit and find the output voltage they measure is different from Vce. All of the discourse from 10:00 onwards is unnecessary. In any real biasing scheme, R2 will be less than R1. If you ensure that R2
In this problem at first we very abrupt fill of the thevenin voltage and the circuit and the circuit simplification so I have very helpful for me your video I pay you very very grateful to you thank you
sir what did you mean by disadvantages in using a small Rb... I mean does it affect the circuit it anyway or is it just too much of a hassle to find a single part with that small of a resistance?
plz make a video series on artificial intelligence ... not only this i desperately seeking all the videos on cse course and programming languages from neso academy ... it will be really nice of you if you take consideration on my request...thanks a lot neso academy
Nope, they are in parallel. Assume the supply voltage is Vcc. The voltage at the junction of R1 and R2 is then Vcc.R2 / (R1 +R2) - this is what we call Vth. Now connect a resistance R3 from the junction of R1 and R2 to ground so that it draws current from the node (i.e. in parallel with R2). Let R be the parallel combination of R2 || R3 = R2.R3 / (R2 + R3). The voltage at the node is now Vcc.R / (R1 + R) , so the current drawn through R3 is I3 = (Vcc.R / (R1 + R)) / R3. Substitute for R = R2.R3/(R2 + R3): I3 = (Vcc.R2.R3 / (R2 + R3)(R1 + (R2.R3/(R2 + R3))) / R3. Cancel the R3 on the top with the R3 which is dividing: I3 = Vcc.R2 / (R2 + R3)(R1 + (R2.R3/(R2 + R3)). Multiply out the denominator: I3 = Vcc.R2 / (R1.R2 + R1.R3 + R2.R3) However, if you consider a voltage of Vth fed through a resistance of R1 in parallel with R2 to the resistor R3, the current through R3 will then be: I3 = Vth / (R3 + (R1.R2/(R1 + R2). Substitute Vcc.R2 / (R1 + R2) for Vth: I3 = Vcc.R2 / (R1 + R2)(R3 + (R1.R2/(R1 + R2). Multiply out the denominator: I3 = Vcc.R2 / (R1.R3 + R2.R3 + R2.R3). That is the same current as calculated above, which shows that the Thevenin resistance (the resistance in series with the Thevenin voltage) is exactly the same as the parallel combination of R1 and R2. Summary: Vth = Vcc.R2 / (R1 + R2) Rth = R1.R2 / (R1 + R2)
Because nobody in their right mind would design a common emitter stage that needed two different separate supply voltages when they could use just one. Having two resistors instead of one is hardly a "complicated circuit".
*_So, why can we separate the Vcc in 2 points? If they're the same thing, shouldn't they always be together or something?_* I didn't know at all we could separate them and make calculus separated. But I'd like to know why, if anyone can explain. Thanks.
Because it's a parallel circuit with series components. In a parallel circuit each leg has the source voltage. If Vcc is 10v, then each leg sees 10v. So if you split the circuit and give each leg it's own 10v source, it's the exact same thing.
Then the input impedance of the amplifier is low. That means that we have to use sources for the input signal to be amplified that are even lower impedance, otherwise we lose some of that signal before it even arrives at the base of the transistor. Think of the source impedance (Rs) and the amplifier input impedance (Rin) as a divider network - the signal arriving at the base will be reduced by a factor of Rin / (Rin + Rs). We need Rin >> Rs to avoid any significant attenuation.
Here you are taking both resustors are parallel for Rth(r thevinins but you are taking voltage divider across that both resistor how it possible .i mean that in parallel resistors voltage is same know.
you can apply simple kvl and then multiply the resultant current with the load resistor, actually voltage divider is basically the simplest form of kvl, so that is why when he did kvl it seemed like voltage divider.
if you still wanna know ,, this is because of thevenin theory which creates a short circuit temporarily in Vcc so resistors becomes connected in parallel but after the short is removed the resistors return to be in series again
sridhar reddy, in all bias circuits Except in voltage divider bias, the collector current depends on the current gain(beta) which is temperature dependent. So it is desirable to have a bias independant of (beta). So voltage divider bias is used
I'm afraid that @Kathiravan Sekar is completely wrong. The use of negative feedback from collector to base makes the operating points _more_ independent of transistor parameters and temperature, not less. Moving the upper point of the top base bias resistor (R1) from the supply voltage to the collector of the transistor (and approximately halving its value) produces a far more stable circuit at the cost of losing some gain (which can be compensated by lowering the base bias voltage a little and decreasing the emitter resistor). Generally, this circuit is rather harder to analyse, but actually is superior to the fixed voltage bias in most ways.
An ac signal can be superimposed on an a dc bias at the base of the transistor by coupling in the ac signal through an input capacitor. Similarly, an amplified ac signal can be taken from the collector of a common emitter circuit by coupling out the ac signal through an output capacitor. Clearly, the size of the ac signal must remain rather less than the size of the voltages present in the dc biasing. For a properly designed circuit, the voltage gain will be -Rc/Re.
In any practical transistor, β is a lot bigger than 1. For small signal transistors like BC547, it will be well over 100 at normal currents. Many types of transistor allow you to specify a high current gain part, such as the BC547C, which has a gain of 400 or more.
That may increase the base current, so that the operating point which is to be designed in the active region will now shift into saturation region... As a result transistor will not work as an amplifier
Quite simply, it makes the input impedance of the amplifier very small as well, so it attenuates any input signal whose source impedance is not even smaller before it can be amplified. That's not a good way to design an amplifier.
our university is shutdown during coronavirus and all classes are online, youre saving my degree! thank you
This is the beauty of neso academy. Explained the complex circuit in a simple way. Luv u sir. From NIT jalandhar
How does a transistor amplify an AC signal (when it works on DC and cannot change current direction)?
@@santosh-mb5wc In the active regions one of the junctions is reverse biased and the other is forward biased.. We apply input across the forward biased one which offers a low resistance path.. Now the current must pass through both the diodes and so as the other diode is reverse biased it offers a large resistance and hence when we take our output across it, the signal has been amplified.. NESO academy has explained the concepts in earlier videos in their playlist :)
@@santosh-mb5wc that sinosoidal wave without having any AC input signal is due to noise at the input side.
These videos will be kept in my hall of highly recognised to be extremely helpful videos.
Aapko kese thnx kahoon samajh nahni aa raha.....
Literally aapka formula meri jindgi bacha liya jaha bhi dhoondta thaa sabke sab circuit draw karne mein lag jaate the lekin aap ne alga se samjhaya direct forumula bol diya love u bro... Many many thnx
This is a love from Odisha
God bless u
And again thnx a lot....
calm voice with powerful lectures
i wish i have seen this video series before. having never encountered such helpful tutorial series so far .thank you
Thanks a lot! You have helped me and many others to pass their exams. You're a lifesaver.
Very nice explanation.Thank you sir!!.. Tomorrow is my exam.These videos really helped a lot
did u pass?>
@@neminemii6366 lol
@@Puneet-Sharma yeah very funny
Your explanation is awesome you r d best ......sir
I am giving much respect to you after seeing all d videos....!!!very clear cut explanation
No words for u sir,u are really amazing lecturer.
Time saver lectures..must follow these...
this lecture was a life saver! thanks a lot.
plz, Do u understand this video ? if i sent a problem, can u solving it for me?
@@mohamedadel9763 then why don't u start a channel and teach us..?
How does a transistor amplify an AC signal (when it works on DC and cannot change current direction)?
You are doing a great thing Bro. I m watching this 1 day before exam and you are the best.
I'm very happy that I have got detailed explanations on transistors. Thank you.
You are awesome, dude. I cant believe i understand better an english explained video between an spanish explained video. Thank u and regards from chile =^)
These videos are really great. Without these I wouldn't have understood a thing happening in my lab classes.
Its reallly d most specific nd amazing way to teach electronic......thnxxx a lot sir
He said ' if you don't know thevinenen's theorem rafer my video' but he said everything about thevinenen's theorem in this video only
great work,thanks for making our engineering simpler
Best explaination ever. I love you man, you're a life saver.
plz, Do u understand this video ? if i sent a problem, can u solving it for me?
Q1. it is known that parallel combination of resistor divides current not voltage, than why this is known as voltage divider?
Q2. While finding V_th, total resistance was taken as R1+R2, but that's the case for series not parallel, why?
Thank you so much from Chennai, Tamil Nadu
Superb explanation sir 😊😊😊😊😊
Really helpful videos! I can't follow my teachers and you save me!
thanks this video helped me the night before EXAM...THANKS..
Wonder full lecture
I really like the the way you instruct. I learned all of this at UMass, Lowell for my EE degree. At the time it seemed so hard. Now, it brings pleasure having my memory refreshed with all this great stuff.
Do you do Signals and Systems, Complex Variables?
i like the presentation quality it helps in quick prep of tech competitions
Thanks to your lectures, I was able to understand the concepts of my bee(basic electric and electronics) subject .Thank you very much !!!!!
Best explanation ever
Your explanation is perfect 👍👍
But the only problem is with the sound.
Your voice is clear but not audible properly.
Just increase your voice and it will be very good. 👍
New videos have better sound quality, thank you for the feedback.
At 8:30 you miss a trick. Since Ic = β.(Vth - Vbe) / (Rth + (β+1).Re), once you have the condition Rth > 1, you can immediately derive Ic = (Vth - Vbe) / Re and thence calculate your operating points. By the way, Vce is *not* the output voltage as you claim at 9:35 - the output voltage is the voltage at the transistor collector and that is Vce + IeRe relative to ground. Students get confused when the build the circuit and find the output voltage they measure is different from Vce.
All of the discourse from 10:00 onwards is unnecessary. In any real biasing scheme, R2 will be less than R1. If you ensure that R2
thanks alot for the much needed help 😇
your lecture as well as your voice so impressive
Very nice and insightful! Thank you very much
Thanks for very good information 🙏🙏🙏
In this problem at first we very abrupt fill of the thevenin voltage and the circuit and the circuit simplification so I have very helpful for me your video I pay you very very grateful to you thank you
Thank you sir ❤️
Solid video great graphics. Cheers for the help
Nice video ❤❤❤❤❤
I don’t understand Why people are disliking this video.This is a good lecture.Then y?
Pls use a good quality mic while recording lectures......U r the best
helped me for last minute revision for internals.
Best explanation sir
sir u are a great teacher
sir what did you mean by disadvantages in using a small Rb... I mean does it affect the circuit it anyway or is it just too much of a hassle to find a single part with that small of a resistance?
please make videos on various types of theorem
Hi,
Neso Academy,
Please make a video about Miscellaneous biasing Method.
Thanks.🙂
Best explanation
kindly explain CB CC CE configuration transistor in terms of movement of charge carrier in this way students will clear their concept deeply
Best video I've seen all week
These circuit simplification is more easier than other video lecture OG RUclips
Sir In stability factor... in the numerator it is not simply Rb it is (1+beta) Rb ..Is that also less than (1+beta) Rc ?
good explanation
Hello! I just want to know if there is any difference if there is a Vee of -5V? will the calculations be the same?
Thank you for explaining sir!
Very very thank u.
Best solution🙏
Clearly explained
thanks sir
Thanks a lot sir.. Now i can happily skip my online classes😁
U are best sir... thank u soo much...
Thank You Sir. 💯 Great Explanation
Thank you very much sir..
I have just one question. Which is the program you use to write the equations, to do the graphics and everything in the video?
plz make a video series on artificial intelligence ...
not only this i desperately seeking all the videos on cse course and programming languages from neso academy ...
it will be really nice of you if you take consideration on my request...thanks a lot neso academy
Do u want any PDFs about AI?
@@_mr.keys_2057 blud tried to get her contact details, but got ignored hahaha
Can we able to drive a 7812 ic from a 200 volt DC supply using voltage divider rule?
Thanks. Great tutorials
Thank u so much sir
for Rth r1 and r2 are in series
Nope, they are in parallel. Assume the supply voltage is Vcc. The voltage at the junction of R1 and R2 is then Vcc.R2 / (R1 +R2) - this is what we call Vth.
Now connect a resistance R3 from the junction of R1 and R2 to ground so that it draws current from the node (i.e. in parallel with R2).
Let R be the parallel combination of R2 || R3 = R2.R3 / (R2 + R3).
The voltage at the node is now Vcc.R / (R1 + R) , so the current drawn through R3 is I3 = (Vcc.R / (R1 + R)) / R3. Substitute for R = R2.R3/(R2 + R3):
I3 = (Vcc.R2.R3 / (R2 + R3)(R1 + (R2.R3/(R2 + R3))) / R3. Cancel the R3 on the top with the R3 which is dividing:
I3 = Vcc.R2 / (R2 + R3)(R1 + (R2.R3/(R2 + R3)). Multiply out the denominator:
I3 = Vcc.R2 / (R1.R2 + R1.R3 + R2.R3)
However, if you consider a voltage of Vth fed through a resistance of R1 in parallel with R2 to the resistor R3, the current through R3 will then be:
I3 = Vth / (R3 + (R1.R2/(R1 + R2). Substitute Vcc.R2 / (R1 + R2) for Vth:
I3 = Vcc.R2 / (R1 + R2)(R3 + (R1.R2/(R1 + R2). Multiply out the denominator:
I3 = Vcc.R2 / (R1.R3 + R2.R3 + R2.R3).
That is the same current as calculated above, which shows that the Thevenin resistance (the resistance in series with the Thevenin voltage) is exactly the same as the parallel combination of R1 and R2.
Summary:
Vth = Vcc.R2 / (R1 + R2)
Rth = R1.R2 / (R1 + R2)
Thank you!!
sir can u explain the advantages of collector feedback configuration over voltage bias configuration or vice versa ????
Sir if we connect a simple voltage source to base then it would be same then why we use that complicated circuit consist of R1 R2
11:01
Because nobody in their right mind would design a common emitter stage that needed two different separate supply voltages when they could use just one. Having two resistors instead of one is hardly a "complicated circuit".
thank you so much
good video sir.
*_So, why can we separate the Vcc in 2 points? If they're the same thing, shouldn't they always be together or something?_* I didn't know at all we could separate them and make calculus separated. But I'd like to know why, if anyone can explain. Thanks.
Because it's a parallel circuit with series components. In a parallel circuit each leg has the source voltage. If Vcc is 10v, then each leg sees 10v. So if you split the circuit and give each leg it's own 10v source, it's the exact same thing.
very well done
sir I have a doubt.
why we are calculating voltage across only R2 only
Perfect Voice!⚡
to see this video I feel more than one like to your video
Good one
What are the disadvantages of using low base resistance ?
big fan sir
Nice!! Thanks
thanks
Thank you so much for the great video!!!
What exactly are the disadvantages if we use low resistance Rb(Resistor-B)?
Then the input impedance of the amplifier is low. That means that we have to use sources for the input signal to be amplified that are even lower impedance, otherwise we lose some of that signal before it even arrives at the base of the transistor. Think of the source impedance (Rs) and the amplifier input impedance (Rin) as a divider network - the signal arriving at the base will be reduced by a factor of Rin / (Rin + Rs). We need Rin >> Rs to avoid any significant attenuation.
why we are not using VDR rather than thevinin's theorem?
sir please please make more videos on analog 2nd year course "analog CMOS
integrated circuit "
Thk u vry much
Here you are taking both resustors are parallel for Rth(r thevinins but you are taking voltage divider across that both resistor how it possible .i mean that in parallel resistors voltage is same know.
please reply for this query
you can apply simple kvl and then multiply the resultant current with the load resistor, actually voltage divider is basically the simplest form of kvl, so that is why when he did kvl it seemed like voltage divider.
if you still wanna know ,, this is because of thevenin theory which creates a short circuit temporarily in Vcc so resistors becomes connected in parallel but after the short is removed the resistors return to be in series again
Sir wo mai puchh raha tha ki agar Beta=90 diya hi to kya hum collector current is almost equal to base current man kar ques bana sakte hi???
Plse reply
Super sir
Thank you so much!!
what will happen when Rc is open in voltage divider bias? Is output voltage
can be obtained
we always prefer voltage bias configuration why not collector feedback??
sridhar reddy, in all bias circuits Except in voltage divider bias, the collector current depends on the current gain(beta) which is temperature dependent. So it is desirable to have a bias independant of (beta). So voltage divider bias is used
I'm afraid that @Kathiravan Sekar is completely wrong. The use of negative feedback from collector to base makes the operating points _more_ independent of transistor parameters and temperature, not less. Moving the upper point of the top base bias resistor (R1) from the supply voltage to the collector of the transistor (and approximately halving its value) produces a far more stable circuit at the cost of losing some gain (which can be compensated by lowering the base bias voltage a little and decreasing the emitter resistor). Generally, this circuit is rather harder to analyse, but actually is superior to the fixed voltage bias in most ways.
Thnx sir but one doubt
How does a transistor amplify an AC signal (when it works on DC and cannot change current direction)?
An ac signal can be superimposed on an a dc bias at the base of the transistor by coupling in the ac signal through an input capacitor. Similarly, an amplified ac signal can be taken from the collector of a common emitter circuit by coupling out the ac signal through an output capacitor. Clearly, the size of the ac signal must remain rather less than the size of the voltages present in the dc biasing. For a properly designed circuit, the voltage gain will be -Rc/Re.
Please tell me why the reverse bias Potential of C-B junction decreases as we decrease Rb, can somebody explain?
keep up the great work !
thanks
which is better among voltage divider bias and emitter bias?
Voltage divider
at 8:23 doesn't Ic depend on beta? it looks like it only becomes independent of beta if beta is significantly larger than 1?
In any practical transistor, β is a lot bigger than 1. For small signal transistors like BC547, it will be well over 100 at normal currents. Many types of transistor allow you to specify a high current gain part, such as the BC547C, which has a gain of 400 or more.
I have a Question ,What are the disadvantages of making Rb very small ??
That may increase the base current, so that the operating point which is to be designed in the active region will now shift into saturation region... As a result transistor will not work as an amplifier
Quite simply, it makes the input impedance of the amplifier very small as well, so it attenuates any input signal whose source impedance is not even smaller before it can be amplified. That's not a good way to design an amplifier.