Timestamps for the different topics covered in the video: 0:00 Introduction 0:51 Advantages of Differential Amplifier 3:00 BJT- Differential Amplifier circuit 4:18 DC analysis of Differential Amplifier 6:21 Response of Differential Amplifier to Common Mode input signal (Qualitative Analysis) 9:00 Response of Differential Amplifier to Differential Input signal (Qualitative Analysis)
why government open Universitys..just search for men like him...nd make videos like him...clear concept; 1 lakh time better than my university teacher....
Key to understanding: Remark : Transistors are connected according to the common emitter circuit, in the OE circuit, the voltage at the emitter, roughly speaking, is always equal to the base voltage. Since the potential at the point of connection of the emitters MUST be the same for both transistors, someone must give in .... This small voltage difference falls directly on the PN junction of another transistor, closing it, that is, roughly speaking, if we assume that the transistor is open when the voltage across the BE is 0.7V, then with a voltage difference of 0.1V, we have a voltage drop on the other transistor on the BE 0.6V is not a completely open transition !!! This is equivalent to the fact that we stupidly applied 0.6V to the base. What will be the base current at this voltage?? Look at the CVC of the diode, the current will be almost absent, so the author talks about Eberton Moll's formula. You can see the CVC of the diode, what current the LED will give out if its PN junction is not fully open, that is, if it is open at 2.7V, apply 2.6V to its legs and you will see that the current almost does not flow. So here, increasing the difference Like this comment if it was useful for you so that others can see it.
My friend, you literally have no idea how useful this video was to me, not just that, If I were a scientist learning this for the first time I'd say this connected alot of dots for me, I believe they'd say *"THIS IS REVOLUTIONARY"*
Vc = Vcc - Ic RC. But we are seeing only change in Vc1. Therefore, ΔVc1 = -ΔIc * Rc. (just think intuitively, if input voltage increases then base current will increase, and hence collector current increase, therefore, there will be more voltage drop across the resistor and hence less collector voltage. That's why three is a negative sign) And ΔVc2 = - ΔIc2*Rc. Now, if you put the value of ΔIc1 and ΔIc2 in these two expressions then you will get ΔVc1 = - gm ΔV Rc and ΔVc2 = gm ΔV Rc. I hope, it will clear your doubt.
Great videos, thanks for making them available! Just one thing, though, I will say after watching many of them now: Sound. It is pretty bad. Your voice recording on almost all your videos are clipping, and hence distorted. It is really the easiest thing to fix, just don't gain the audio as hard as you do, and if you want more level beyond that, use dynamic compression instead. Your level is already gained way louder than the average youtube video speak, so gaining it less would not be a problem, everyone can adjust their speakers etc. Otherwise great content!
A great video overall, thank you! But Where did alpha come from? @5:28 you state i_c=i_e which makes perfect sense. But then 10 seconds later @5:38 essential states i_c = alpha*i_e. Would appreciate help here, thanks!
Not sure if I missed it but it would be nice if it was explained that the input itself is assumed to be differential. This amp doesn't create a differential signal although I realize is referred to that in industry. It only maintains the differential. Otherwise it just looks like to npns in parallel. 6:26 is especially misleading?
It is a bit confusing on equations between 5:00 and 8:00. If you substitute Vcm - Vbe to Vp in 8:00, the polarity of Vbe is different from equation 5:00.
I didn't get what you mean to say. But at 5:00 I was referring to the DC analysis, while at 8:00, it is how differential amplifier will behave to the common-mode signal, where the common-mode signal is present at both inputs of the differential amplifier. At 5:00, VCM is 0.
@@ALLABOUTELECTRONICS yes you are right. But if you substitue Vcm - Vbe to Vp, the formula in 8:00 will become IEE = ( VEE - VCM + VBE ) / RE. So if Vcm = 0, what's left is IEE = ( VEE + VBE ) / RE . the sign of vbe is wrong.
Thanks for video. Only problem is that u are using way too much equations instead of just simply explaining how it works. This video with explaining could have last 2 minutes instead of 17. People come here to fastly and easy understand how something works, not to look at formulas. For formulas and equations we have our professors. You should just in most simple way describe how something works, if you can ofc. Anyways, this is just my opinion about it. Your work is great. Keep it up. :)
The man is volunteering, in his own time to show how these circuits work by equations and is brilliant at it. And you have contributed? Precisely nothing. You'll probably understand the requirement for formulas when you pass into adulthood.
Nice presentation. One doubt is that while analyzing DC, you say that the emitter current from each emitter terminal is IEE/2; why not IE and the total current to the RE is 2IE?
Hello Sir, I'm student of Diploma and I am studying in Nashik. I have request regarding to the videos . Can you make the special playlist for Audio Video engineering Basics and advance. I have got very less videos of this course. So can you make it ?. And I have also one request that can you make it in hindi?
I have an important question though. What happens if we want to delete a BG music in my signal, if I put it on the other side it will change as time passes, so if we apply the same locic, to cancel it from the output the voltage at Vp will be equal to that BG music, right?
This is why i hate electrical engineering. You EEs NEVER WRITE WHAT YOUR VARIABLES ARE REFERENCING! WRITE DOWN WHAT CURRENT, VOLTAGE, RESISTANCE YOU ARE REFERRING TO. It makes so damn difficult to follow what you are talking about when I have to guess which voltage or current you mean.
Some of the words you slaughter are very simple "OP" "AMP". 2 words and very simple to say. I know English isn't your first language. But, these very simple mistakes undermine the listeners' belief that you know what you're talking about. The second thing that is somewhat helpful is that you should not write on the bottom of the screen -- because we are watching the Closed Captioned English down there and it covers anything your write on the bottom 15% of the screen. Keep your writing above that point please.
Timestamps for the different topics covered in the video:
0:00 Introduction
0:51 Advantages of Differential Amplifier
3:00 BJT- Differential Amplifier circuit
4:18 DC analysis of Differential Amplifier
6:21 Response of Differential Amplifier to Common Mode input signal (Qualitative Analysis)
9:00 Response of Differential Amplifier to Differential Input signal (Qualitative Analysis)
One help sir please
sir where can we get all your work in the form of pdf or some notes. please provide us some source
Yes sir please provide some source to download all your notes
Sir please reply
@@raghavadevarakonda7592 If possible will provide the notes for the same. will let you know.
why government open Universitys..just search for men like him...nd make videos like him...clear concept; 1 lakh time better than my university teacher....
Because youtube can't provide a degree 😅
@@RajChauhan-sf9yqwell the one giving degrees are nothing in front of people like him
Учителя сами часто не понимают материал, который преподают.
My gratitude to you for making this video can't be explained in words 🙏
Key to understanding:
Remark :
Transistors are connected according to the common emitter circuit, in the OE circuit, the voltage at the emitter, roughly speaking, is always equal to the base voltage.
Since the potential at the point of connection of the emitters MUST be the same for both transistors, someone must give in ....
This small voltage difference falls directly on the PN junction of another transistor, closing it, that is, roughly speaking, if we assume that the transistor is open when the voltage across the BE is 0.7V, then with a voltage difference of 0.1V, we have a voltage drop on the other transistor on the BE 0.6V is not a completely open transition !!! This is equivalent to the fact that we stupidly applied 0.6V to the base. What will be the base current at this voltage?? Look at the CVC of the diode, the current will be almost absent, so the author talks about Eberton Moll's formula.
You can see the CVC of the diode, what current the LED will give out if its PN junction is not fully open, that is, if it is open at 2.7V, apply 2.6V to its legs and you will see that the current almost does not flow. So here, increasing the difference
Like this comment if it was useful for you so that others can see it.
Which country you are from?
❤
My friend, you literally have no idea how useful this video was to me, not just that, If I were a scientist learning this for the first time I'd say this connected alot of dots for me, I believe they'd say *"THIS IS REVOLUTIONARY"*
A simple like and subscribe is not enough to show my gratitude
@@Z-eng0 You can pay them lol
15:53 What is this "-" and "+" written below Vo1 and Vo2 respectively please?
It's just to indicate the polarity of Vo, just to show that Vo = Vo2 - Vo1
@@elielsilvaurra7689 Thank you🌹
Excellent yar....first time I m finding such beautiful content on differential amplifier
15:12 Why/How did you put a "-" in the expression for Vc1?and how a "+" in Vc2's expression here?
Vc = Vcc - Ic RC. But we are seeing only change in Vc1. Therefore, ΔVc1 = -ΔIc * Rc. (just think intuitively, if input voltage increases then base current will increase, and hence collector current increase, therefore, there will be more voltage drop across the resistor and hence less collector voltage. That's why three is a negative sign) And ΔVc2 = - ΔIc2*Rc. Now, if you put the value of ΔIc1 and ΔIc2 in these two expressions then you will get ΔVc1 = - gm ΔV Rc and ΔVc2 = gm ΔV Rc.
I hope, it will clear your doubt.
@@ALLABOUTELECTRONICS Awesome.....thank you for replying.....You're really a committed teacher....I can feel it in your reply......nice explanation
Great videos, thanks for making them available! Just one thing, though, I will say after watching many of them now: Sound. It is pretty bad. Your voice recording on almost all your videos are clipping, and hence distorted. It is really the easiest thing to fix, just don't gain the audio as hard as you do, and if you want more level beyond that, use dynamic compression instead. Your level is already gained way louder than the average youtube video speak, so gaining it less would not be a problem, everyone can adjust their speakers etc. Otherwise great content!
100% agree. Amazing videos but the clipping pretty much makes me decrease the volume below 50%.
10:08
why ?
i did not understood this part
and why the transistor will get off
A great video overall, thank you! But Where did alpha come from? @5:28 you state i_c=i_e which makes perfect sense. But then 10 seconds later @5:38 essential states i_c = alpha*i_e. Would appreciate help here, thanks!
Such a clear and awesome video. I am confused about the Amop in 555 chips. Hope someday you can post an analysis video about it. Thank you.
This video is very... ...informative.
Thanks sir for such a awesome explanation
Not sure if I missed it but it would be nice if it was explained that the input itself is assumed to be differential. This amp doesn't create a differential signal although I realize is referred to that in industry. It only maintains the differential. Otherwise it just looks like to npns in parallel. 6:26 is especially misleading?
It is a bit confusing on equations between 5:00 and 8:00. If you substitute Vcm - Vbe to Vp in 8:00, the polarity of Vbe is different from equation 5:00.
I didn't get what you mean to say. But at 5:00 I was referring to the DC analysis, while at 8:00, it is how differential amplifier will behave to the common-mode signal, where the common-mode signal is present at both inputs of the differential amplifier. At 5:00, VCM is 0.
@@ALLABOUTELECTRONICS yes you are right. But if you substitue Vcm - Vbe to Vp, the formula in 8:00 will become IEE = ( VEE - VCM + VBE ) / RE. So if Vcm = 0, what's left is IEE = ( VEE + VBE ) / RE . the sign of vbe is wrong.
why can't we use ic=beta*Ib instead of ic=gm*del*VBE?
Thanks for video.
Only problem is that u are using way too much equations instead of just simply explaining how it works. This video with explaining could have last 2 minutes instead of 17. People come here to fastly and easy understand how something works, not to look at formulas. For formulas and equations we have our professors. You should just in most simple way describe how something works, if you can ofc.
Anyways, this is just my opinion about it. Your work is great. Keep it up. :)
The man is volunteering, in his own time to show how these circuits work by equations and is brilliant at it. And you have contributed? Precisely nothing. You'll probably understand the requirement for formulas when you pass into adulthood.
@@neilcullimore5798 chillout boomer. Sliceurthroaz
We don't even have professors for formula, so it is good to explain whole concept
Nice presentation. One doubt is that while analyzing DC, you say that the emitter current from each emitter terminal is IEE/2; why not IE and the total current to the RE is 2IE?
Very Nice video Sir
Sir please make a video on operational amplifier plz
There is entire series on opamp. Please check the playlist page of the channel. You will find the playlist
Excellent sir
is this the dimensioning of Diffrential amplifier? If yes, can you tell me which part it is? I need it for writing my report. Thank you !
Can you please make a video about discrete opamp made with transistor or jfets?
I will cover the internal structure of the op-amp soon. Once I cover all the relevant topic related to it, I will make a video on it.
@@ALLABOUTELECTRONICS great
Thanks bro! great material:)
What happens if we use two different emitter resistors..?
Which text book we have to refer for this?
Microelectronic Circuits by Sedra and Smith.
1.07 diagram, there must a phase difference between input and output
Hello Sir,
I'm student of Diploma and I am studying in Nashik. I have request regarding to the videos . Can you make the special playlist for Audio Video engineering Basics and advance. I have got very less videos of this course. So can you make it ?. And I have also one request that can you make it in hindi?
What's your branch name??
Super .. Keet up it
I have an important question though.
What happens if we want to delete a BG music in my signal, if I put it on the other side it will change as time passes, so if we apply the same locic, to cancel it from the output the voltage at Vp will be equal to that BG music, right?
If the both transistor are not same than it will manage with resistance
Nice...cheers.
Tell us something about Battery BMS in next video.
भाई जी मुझे इलेक्ट्रॉनिक सीखना है बताओ मैं क्या करूं आईटीआई करूं या कोई और कोर्स
nice
أحسنت حياك الله ^_^
Thanks!
make an app, please
Thanks
Make an app
Placement mili
Confused
28th October 2024
This is why i hate electrical engineering. You EEs NEVER WRITE WHAT YOUR VARIABLES ARE REFERENCING! WRITE DOWN WHAT CURRENT, VOLTAGE, RESISTANCE YOU ARE REFERRING TO. It makes so damn difficult to follow what you are talking about when I have to guess which voltage or current you mean.
Also they didnt explain from where current flowing
Some of the words you slaughter are very simple "OP" "AMP". 2 words and very simple to say. I know English isn't your first language. But, these very simple mistakes undermine the listeners' belief that you know what you're talking about. The second thing that is somewhat helpful is that you should not write on the bottom of the screen -- because we are watching the Closed Captioned English down there and it covers anything your write on the bottom 15% of the screen. Keep your writing above that point please.