More than 3 years after graduating, I am now understanding this. Thank you so much.
@thepianist8172 жыл бұрын
lmao
@iluvyunie Жыл бұрын
Wow, only 4 years to go then I guess I'll get it 😅
@AdityaYadav-lv8zd Жыл бұрын
Why do you need to understand them now?
@mrkattm8 жыл бұрын
Awesome job, I am EE, graduated in 1992 from SVSU, I was feeling a little nostalgic for the those simpler college days when I came across your video and decided to watch. All I can say is that I wished that I had these types of resources available to me when I was learning this material, you did an excellent job, keep up the great work.
@homerodaniel_0074 жыл бұрын
Oh my god, you got your degree when I was born. I want to be a professional EE in the future.
@guntherzoldy13424 жыл бұрын
@@homerodaniel_007 Just started EE in Uni, a tough ride so far and these videos are a lifesaving resource. I was born 2002, so I'm really lucky to have grown up with videos like these.
@blalmal10a8 жыл бұрын
thanks man.. why am i not searching for vids at the first time?? i'm so glad... all the books and pdf.s provide nonsense derivations
@carlgradolph9676 Жыл бұрын
Thanks for a very clear and helpful explanation. I now have a much better understanding of how to set up a bias circuit for a BJT, and why a certain amount of "slop" necessarily occurs when calculating actual voltage, current, and resistance values.
@spinloki5 жыл бұрын
Incredible video and explanations. I was able to get past a ton of homework problems thanks to you!
@xMrJanuaryx8 жыл бұрын
Now this was a great example thank you very much! Bonus I asked my professor and he said he is fine with me making the assumption that Ib is zero if I make that argument you showed me!
@mileslegend Жыл бұрын
As I was hovering looking for a perfect solution .. I just saw your the cover photo and I knew my problem is solved.... And guess what am just 4mins I to the video and I have understood everything... Thanks man .... 🔥🤗
@zackkassner33746 ай бұрын
This is an amazing video. The way you explained this made it very easy to understand. Thank you so much!
@ding96332 жыл бұрын
Excellent explanation. Infinitely more helpful than a university professor
@ElectronXLab2 жыл бұрын
Thanks so much. Glad you think so!
@JCooperAudioSystems3 жыл бұрын
Thanks for making this very *CLEAR* plain English video. I can definitely understand your Canadian English.
@yvesbouchard51973 жыл бұрын
Thank you for such a crystal clear explanation of a VDB-BJT.
@montezeminor28544 жыл бұрын
Excellent presentation, and delivery of material content. Thoroughly explained in detailed information describing the devices characteristics and biasing approaches.
@zacktakahashi4685 жыл бұрын
thanks David ,you are my true teacher,keep going
@Festus20228 ай бұрын
Great video! Addresses some of my longstanding questions. Thank you!
@puspamadak2 жыл бұрын
This video helped me understand the biasing of BJT. Thanks a lot sir!
@ElectronXLab2 жыл бұрын
Glad it helped!
@diazolguinoscar26292 жыл бұрын
The best video i've seen so far, thank you.
@ElectronXLab2 жыл бұрын
That's great to hear. Thanks for your comment.
@eepower7 жыл бұрын
Great job simplifying this ckt. Thanks for the great job. It does help lots of ppl. Highly recommended
@arthuryeung6713 жыл бұрын
Great explanation sir! Now I understand what's going on there
@warrickramnand94372 жыл бұрын
Finally someone that made me understand this theory. Thanks a lot.
@ElectronXLab2 жыл бұрын
You're very welcome!
@RexxSchneider2 жыл бұрын
The "short-cut" method calculates the quiescent collector current and dc bias points of a transistor that has infinite β. The stability of the dc bias against changes in β depends purely on the relative size of the emitter resistor Re. That resistor provides negative feedback that stabilises the operating points as well as reducing the distortion of the stage. It comes at the cost of lowering the voltage gain of the stage, which is 3 in this case. It also "absorbs" around 2.5V of the potential output swing. You can improve the performance of a common emitter by removing some of the negative feedback from the emitter resistance by lowering it, while providing a feedback path from the collector to the base. In other words, derive your base voltage bias (the positive end of R1) from the collector rather than the positive supply rail. Since the emitter resistor only needs to be more than 10 times the dynamic emitter resistance of the transistor (25mV/Ic) to provide reasonable linearity, the emitter can be usefully biased to as little as 250mV above ground, putting the base at about 900mV above ground. If you assume that the collector bias point is around half the supply voltage, you can quickly calculate the ratio of R1 to R2. Knowing the β will allow you to make R1 || R2 less than 1/10 of β.Re for any value of Re. That shows you that the input impedance of the stage is inversely proportional to the collector current, and (all else being equal) you should choose Ic to obtain whatever Zin is required. Somewhere between 1mA and 2mA with a β > 250 or so will result in an input impedance around 5K to 10K. That solution will allow reasonable gains of around 20 to 30 with good stability and linearity, while maintaining independence of transistor parameters. I recommend it to you.
@shashankbiswas41308 жыл бұрын
you made this so simple...thanks..
@bryanfoot2 жыл бұрын
Thanks so much . Your video has clarified all my doubt. Thanks once more
@lailakc92884 жыл бұрын
I think you are one of the best teachers ever! Please make more videos related to EE courses. Thanks. GOD BLESS YOU!
@justadreamerforgood694 жыл бұрын
What's your Instagram baby?
@maramaeolasiman82378 жыл бұрын
Please add more examples sir for this specific topic because we really understand your explanation. Thank You Sir :)
@alfredoshamed32547 жыл бұрын
Congrats! You are antes excellent teacher!
@orpheneh6 жыл бұрын
love you david, so glad for you
@Kanterneit7 жыл бұрын
Best explanation
@sudarshankj4 жыл бұрын
Very clear explanation!
@Avionics19588 жыл бұрын
very comprehensive and well explained.
@chintumali29358 жыл бұрын
explained very clearly thankuu
@elektroguz145 жыл бұрын
Great Work!
@gauravjhamnani89158 жыл бұрын
Thank youu..............clearly explained
@ha.w.s49924 жыл бұрын
Thanks a lot From iraq 🇮🇶
@kubumshel Жыл бұрын
bless you sir. Great video
@mathushekmathu67353 жыл бұрын
Very useful methode
@meth54758 жыл бұрын
I think you should clarify that the Thévenin resistance and voltage are taken from Base to datum node (GND). Otherwise it seems like you are doing 2-port analysis on just one point, the base, which obviously doesn't make sense. Other than that, great vid.
@sadi32308 жыл бұрын
Thank you very much Sir..
@michaelcostello69914 жыл бұрын
Only complaint is there is not a download link for a pdf document of this lecture. The information is so good there should be a link.
@Ashish-_-7 жыл бұрын
Thank you very very much Man!
@Amrhossam967 жыл бұрын
Awesome job :D
@nauraizsubhan017 жыл бұрын
Thanks a lot sir
@yeluntalan Жыл бұрын
Thank you for this explanation, how do we define Beta here?
@govarthananbaskar72246 жыл бұрын
thank you sir
@chickennugget33624 жыл бұрын
Thanks for the video. I've gained a headache.
@refaymohamed29605 жыл бұрын
Thank you, engineer, can we use it in the transmitter circuit for the of oscillation?
@homamalbakry70677 жыл бұрын
thnx bro u r amazing
@paulgazey28992 жыл бұрын
Thanks for the good video. How would you solve the question if there is also a resistor between base of transistor and opposite node? Thanks
@shaonahamed16674 жыл бұрын
Thank you so much
@balandraken4838 Жыл бұрын
Thank you
@AaJaY-px3mh4 күн бұрын
R1 AND R2 are in series why are they considered to be in paralell??
@samsammy42208 жыл бұрын
You're a legend
@moadibnasir25265 жыл бұрын
Thankuuuu sooo muchhh for thiss😍 thankuu
@hagegeabay45427 жыл бұрын
this is a great example but how do you find Ib if RE is not given?
@ElectronXLab7 жыл бұрын
I'm not totally sure what you mean about RE not being given, but if there is no RE, then IB = (VTH-0.7)/RB
@sandeepkamboj76757 жыл бұрын
How you calculate the value of R1,R2,R3,R4?
@FireSymphoney6 жыл бұрын
divide the relative voltage with the relative current, use ohms law.
@darrenallen13876 жыл бұрын
Brilliant Ember could you explain that different, not understanding
@FireSymphoney6 жыл бұрын
use this formula: Voltage = Current x Resistance
@tedtate576 жыл бұрын
Great information. I'm in the process of refreshing my basic knowledge of electronics so I might be able to go further in depth. Q. Do you have videos on logic circuits?
@ElectronXLab6 жыл бұрын
I do have a number of videos on logic circuits. Some of them are in a Boolean Algebra playlist: kzbin.info/www/bejne/q3KyoKScpNisnKc
@tedtate576 жыл бұрын
Thanks a million David. Your teaching skills and content are incredible.
@santoshdebakikrishna3817 жыл бұрын
Thanks for sharing...
@thatsreallyshort Жыл бұрын
Can the resistor 'Re' be equal to zero in this circuit?
@rakeshdutta623 Жыл бұрын
Just awesome explanation sir, thank you so much for that video. I have one doubt about beta, how can we take the beta value as constant because in any transistor data sheet the beta value have a range. For example BC548 has a beta range from 110 to 500. So which value should i take while using this transistor for amplification (for a perfect active region). Please explain sir, if possible.
@ElectronXLab Жыл бұрын
You make a good point. What value of beta do you use? When my students are first learning about transistors, we make the assumption that it is constant and this simplifies analysis a lot. Then we measure beta in the lab for a particular transistor and even though everyone has the same part number, the values of beta change and to make it worse the value changes when temperature changes. Real circuits are designed so that the value of beta doesn't matter as much (e.g., voltage divider bias is less sensitive to changes in beta than a fixed bias circuit). So I think the progression of learning is assume that beta is constant at first because then there is less cognitive load, then recognize that beta can change, then look at ways to make circuits less sensitive to changes in beta. It's probably not the answer you are looking for, but I hope it helps a bit.
@athulvinayak33282 жыл бұрын
What is the use of RE in the circuit?how it provide stability?
@waelfathe99097 жыл бұрын
the bais is stable ...but what about the ac signal to be amplified later...re usually bypassed by cap ..so the ac does not see re ...which means that amplification depends on the beta which may enlarge the amplification till saturation of cut off??
@simozy44753 жыл бұрын
thx a lot bro
@esuelle7 жыл бұрын
Thank you!
@claudioadam1646 жыл бұрын
how if i want to find the R1 and R2 while i know the the Vce?
@atankent7 жыл бұрын
Hye. What is the software that you use to make this tutorial?
@seankennedy85064 жыл бұрын
Also is (Vth) the same as (Vb) the voltage looking into the base of the transistor?
@age041254 жыл бұрын
I am confused about this too. Did you found out that Vth equal to Vb? Or is it Vth =Vb + Ib*Rth ?
@gwenhurel51254 жыл бұрын
Vth is equal to VR2, the voltage provide by the voltage divider. And so VR2 = VBE + VRE (equal to => (R2.IR2) = VBE + (RE.IRE)
@poetas843 жыл бұрын
Could you please explain how you calculate R(th) as being 8.33 KΩ ?
@AkshayKumar-jq9ph Жыл бұрын
If i use exact method even if bre>>10r2 then is it right
@CheckThisOut3806 жыл бұрын
Does it make a difference if the upper side is wired not the lower ?
@smartpk95344 жыл бұрын
Sir how to get answer uA please tell me i don't uderstands
@kailm.51077 жыл бұрын
I understand how you derived the equation for IB using KVL, but the intuition behind the equation itself is not clear to me. So why is it not IB=VB/RB and then IB=VB/Rth? Additionally my textbook frequently uses VB=VE+.7V, so if I combine this with your method, specifically where Vth-VBE, then I am taking the .7V into account twice. This does not make sense to me.
@ElectronXLab7 жыл бұрын
+Kail M. I'm not totally sure I understand your question, but I think your confusion arises from determining the IB from the Thevenin equivalent circuit. The Thevenin equivalent does not physically exist, I am just using it as a trick to help me calculate IB (which is not the current that flows through either one of R1 and R2, it is the current into the base only). I hope that helps...if not, let me know.
@lem09able7 жыл бұрын
What if there's a -VEE voltage source? How should I apply thevenin to it?
@simoncheung24274 жыл бұрын
8:16 could you explain why Rth +(b+1)*Re rather than Rth +Re. thanks
@mustafaghanim95977 жыл бұрын
How does the early voltage affect Ic in this type of calculations?
@michelebianchi78614 жыл бұрын
how do you get (beta+1) ? great video though! my professor cant explain shit!
@seankennedy85064 жыл бұрын
So (Rth) is the same as the resistance at the base of the transistor so (Ib) times (Rth) will give me the voltage at the base of the transistor and therefore will give me the voltage at the resistor going to the ground (Rb2) does that sound right?
@johnjcb46902 жыл бұрын
Great BUT why you say voltage divider R1 and R2 and immediately căLculate Rth as parralel ?/
@ElectronXLab2 жыл бұрын
Because I wanted to Thevenize the voltage source and resistance that the base "sees". When calculating the Rth, you short out the voltage source, so in that step both R1 and R2 are connected to GND, and are therefore in parallel.
@hossamfadeel6 жыл бұрын
Thanks Alot
@CheckThisOut3806 жыл бұрын
Is it Really the same ?? With the new circuit u drawn ,I1=Ic same generator wich is nit true ?
@slipnorris58827 жыл бұрын
Is the Beta still in effect , even with the 10 times rule
@victord38672 жыл бұрын
How are those resistors in parallel? Having a little trouble getting my head around that
@ElectronXLab Жыл бұрын
I believe the resistors that you are referring to are the ones connected to the base. One goes to Vcc and one goes to ground. They are only in parallel from an AC point of view. If you are doing AC analysis, you ignore the DC source. In other words you replace the DC source with a short and it therefore becomes a short to ground. So from an AC point of view those two resistors are connected to the base on one side and to ground on the other side.
@harrisppm2575 Жыл бұрын
i have a similar problem with no mention on β, how can i solve it with-out it?
@isuckatthisgame11 ай бұрын
did you solve it? i have the same problem. no beta but i have all other values. how to calculate base current without it?
@harrisppm257511 ай бұрын
@@isuckatthisgame used loop theory and remember that at the grounded end V=0
@goranvuletic88733 жыл бұрын
Sublime!
@DMES-wp2fj3 жыл бұрын
What is the equation to solve for ro? Anyone
@eljefe68524 жыл бұрын
Whats the value of i1 and i2?
@TheRicardihino8 жыл бұрын
Or is beta just the Ic / Ib?
@nonenone45396 жыл бұрын
Is it always 1/10 ?
@GomesTugaPT5 жыл бұрын
How do you apply the voltage divider rule there? We should considerate both resistor are subjected to the same current because Ib is very small? Or should we assume that transistor is operating in the cut-off region?
@ElectronXLab5 жыл бұрын
Assuming that you meet the condition that (beta+1)RE > 10*R2, then you can assume that IB is small enough to ignore and apply voltage divider rule to R1 and R2.
@GomesTugaPT5 жыл бұрын
@@ElectronXLab thank you!
@jenadolfo93584 жыл бұрын
how to solve for IC if the beta is unknown? plss help
@ElectronXLab4 жыл бұрын
With the Voltage Divider Bias circuit, as long as it is "big enough", the value of beta can be ignored. You can say the IE and IC are equal. IE can be calculated by first determining VE (the voltage at the emitter): VE = VB-0.7. Then IE = VE/RE
@TheRicardihino8 жыл бұрын
How do you determine the beta value? From my understand is this an assigned value within the circuit or is it derived from certain equations?
@animasaurus20075 жыл бұрын
For future reference it's based on the transistor and the temperature of the device.
@Festus20228 ай бұрын
Is the Thevenin Eq. Voltage the "Base Voltage"?
@ElectronXLab8 ай бұрын
It's not the voltage at the base because it is applied through the Thevenin eq. resistance
@Festus20228 ай бұрын
Thanks so much!@@ElectronXLab
@Festus20228 ай бұрын
BUT, at 13:30 (in the shortcut) you give the same formula for the Base-Voltage as you had for the Thevenin Voltage previously. What has changed with respect to the base voltage form the long version calculation? Thanks
@ElectronXLab7 ай бұрын
@@Festus2022. Oh, yeah, good question...It is more accurate to apply the thevenin voltage through the thevenin resistance...this is the analysis in the first half of the video. In certain conditions (i.e., if the base current is small enough (R2
@qamarrashid5795Ай бұрын
Great
@michaelzlab4 жыл бұрын
Wow!! Thanks..
@knowall57926 жыл бұрын
Good video. How do you draw on the screen?
@ElectronXLab6 жыл бұрын
Thanks. I use OneNote and a tablet and use Camtasia for screen capture.
@MrPinkster18 жыл бұрын
Can anyone help with Deriving the equation at 4:44?
@simoncheung24274 жыл бұрын
15:44 do you assume Rth is zero? in previously, Vb=Vth- Ib*Rth.
@simoncheung24274 жыл бұрын
oh, i just realise that in this condition , Ib is nearly zero, the potential drop is near zero, so that Vth = Vb.
@marhsall-bw5kv8 жыл бұрын
wheres the part 2 of your AC Analysis of BJT?
@WoloW1zard-8 жыл бұрын
+marhsall 27777 common emitter amplifier(part 6 and 7 of playlist)
@engenhariaeletricaprojetoi76296 жыл бұрын
good guy
@rollandjeffersonmejia59835 жыл бұрын
What if Current Gain is not given?
@ElectronXLab5 жыл бұрын
This biasing circuit is not affected much by big changes in beta. Watch the video from about 12:20 on to see a method that doesn't rely on beta (except for an initial check). The beta value of real transistors can vary quite a lot, so it is valuable to use circuits like this where the beta value doesn't affect the operation too much.
@jeffm27873 жыл бұрын
I find I'm more of a spice warrior then a scientific calculator nerd 😁