BJT - Voltage Divider Bias Circuit

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ElectronX Lab

ElectronX Lab

Күн бұрын

Learn how a BJT voltage divider bias circuit works and why it is the preferred method of biasing a BJT. Two different analysis methods are shown; a more precise one and one where you assume the base current is effectively 0.
David Williams
www.elen.ca

Пікірлер: 183
@gideonsaina5199
@gideonsaina5199 4 жыл бұрын
More than 3 years after graduating, I am now understanding this. Thank you so much.
@thepianist817
@thepianist817 2 жыл бұрын
lmao
@iluvyunie
@iluvyunie Жыл бұрын
Wow, only 4 years to go then I guess I'll get it 😅
@AdityaYadav-lv8zd
@AdityaYadav-lv8zd 8 ай бұрын
Why do you need to understand them now?
@mrkattm
@mrkattm 7 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_007
@homerodaniel_007 4 жыл бұрын
Oh my god, you got your degree when I was born. I want to be a professional EE in the future.
@guntherzoldy1342
@guntherzoldy1342 3 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.
@spinloki
@spinloki 5 жыл бұрын
Incredible video and explanations. I was able to get past a ton of homework problems thanks to you!
@xMrJanuaryx
@xMrJanuaryx 7 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!
@carlgradolph9676
@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.
@yvesbouchard5197
@yvesbouchard5197 3 жыл бұрын
Thank you for such a crystal clear explanation of a VDB-BJT.
@montezeminor2854
@montezeminor2854 4 жыл бұрын
Excellent presentation, and delivery of material content. Thoroughly explained in detailed information describing the devices characteristics and biasing approaches.
@blalmal10a
@blalmal10a 8 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
@zacktakahashi468
@zacktakahashi468 5 жыл бұрын
thanks David ,you are my true teacher,keep going
@monke12354
@monke12354 5 жыл бұрын
Hey man, as a student with their electronics final in 3 days, this is the only video I could find clearly explaining where everything came from and then how to use it in an example. Subscribed!
@williamgray884
@williamgray884 5 жыл бұрын
what if r2 is greater than beta * re
@TomLeg
@TomLeg Жыл бұрын
​@@williamgray884 You mean "greater than 1/10 Beta * Re" ... then you have to do the calculations the long way because the approximation is outside the tolerance on your resistors
@democracyforall
@democracyforall Жыл бұрын
The problem with this divider stuff is that it takes too many resisters and you want less component in your circuit so the simple method of operating is by far the best one.
@eepower
@eepower 7 жыл бұрын
Great job simplifying this ckt. Thanks for the great job. It does help lots of ppl. Highly recommended
@zackkassner3374
@zackkassner3374 3 ай бұрын
This is an amazing video. The way you explained this made it very easy to understand. Thank you so much!
@mileslegend
@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 .... 🔥🤗
@orpheneh
@orpheneh 6 жыл бұрын
love you david, so glad for you
@shashankbiswas4130
@shashankbiswas4130 8 жыл бұрын
you made this so simple...thanks..
@JCooperAudioSystems
@JCooperAudioSystems 3 жыл бұрын
Thanks for making this very *CLEAR* plain English video. I can definitely understand your Canadian English.
@alfredoshamed3254
@alfredoshamed3254 7 жыл бұрын
Congrats! You are antes excellent teacher!
@dingbop963
@dingbop963 2 жыл бұрын
Excellent explanation. Infinitely more helpful than a university professor
@ElectronXLab
@ElectronXLab 2 жыл бұрын
Thanks so much. Glad you think so!
@Festus2022
@Festus2022 5 ай бұрын
Great video! Addresses some of my longstanding questions. Thank you!
@arthuryeung671
@arthuryeung671 3 жыл бұрын
Great explanation sir! Now I understand what's going on there
@bryanbuildmindchanger1415
@bryanbuildmindchanger1415 2 жыл бұрын
Thanks so much . Your video has clarified all my doubt. Thanks once more
@diazolguinoscar2629
@diazolguinoscar2629 2 жыл бұрын
The best video i've seen so far, thank you.
@ElectronXLab
@ElectronXLab 2 жыл бұрын
That's great to hear. Thanks for your comment.
@refaymohamed2960
@refaymohamed2960 5 жыл бұрын
Thank you, engineer, can we use it in the transmitter circuit for the of oscillation?
@RexxSchneider
@RexxSchneider 2 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.
@maramaeolasiman8237
@maramaeolasiman8237 7 жыл бұрын
Please add more examples sir for this specific topic because we really understand your explanation. Thank You Sir :)
@puspamadak
@puspamadak Жыл бұрын
This video helped me understand the biasing of BJT. Thanks a lot sir!
@ElectronXLab
@ElectronXLab Жыл бұрын
Glad it helped!
@sudarshankj
@sudarshankj 4 жыл бұрын
Very clear explanation!
@warrickramnand9437
@warrickramnand9437 Жыл бұрын
Finally someone that made me understand this theory. Thanks a lot.
@ElectronXLab
@ElectronXLab Жыл бұрын
You're very welcome!
@chintumali2935
@chintumali2935 8 жыл бұрын
explained very clearly thankuu
@Avionics1958
@Avionics1958 8 жыл бұрын
very comprehensive and well explained.
@paulgazey2899
@paulgazey2899 2 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
@gauravjhamnani8915
@gauravjhamnani8915 7 жыл бұрын
Thank youu..............clearly explained
@elektroguz14
@elektroguz14 5 жыл бұрын
Great Work!
@Kanterneit
@Kanterneit 7 жыл бұрын
Best explanation
@atankent
@atankent 6 жыл бұрын
Hye. What is the software that you use to make this tutorial?
@yakubumshelia1668
@yakubumshelia1668 Жыл бұрын
bless you sir. Great video
@Amrhossam96
@Amrhossam96 7 жыл бұрын
Awesome job :D
@sadi3230
@sadi3230 8 жыл бұрын
Thank you very much Sir..
@mustafaghanim9597
@mustafaghanim9597 7 жыл бұрын
How does the early voltage affect Ic in this type of calculations?
@yeluntalan
@yeluntalan Жыл бұрын
Thank you for this explanation, how do we define Beta here?
@claudioadam164
@claudioadam164 5 жыл бұрын
how if i want to find the R1 and R2 while i know the the Vce?
@Ashish-_-
@Ashish-_- 7 жыл бұрын
Thank you very very much Man!
@lailakc9288
@lailakc9288 4 жыл бұрын
I think you are one of the best teachers ever! Please make more videos related to EE courses. Thanks. GOD BLESS YOU!
@justadreamerforgood69
@justadreamerforgood69 4 жыл бұрын
What's your Instagram baby?
@michaelcostello6991
@michaelcostello6991 3 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.
@mathushekmathu6735
@mathushekmathu6735 3 жыл бұрын
Very useful methode
@shaonahamed1667
@shaonahamed1667 3 жыл бұрын
Thank you so much
@moadibnasir2526
@moadibnasir2526 5 жыл бұрын
Thankuuuu sooo muchhh for thiss😍 thankuu
@tedtate57
@tedtate57 6 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?
@ElectronXLab
@ElectronXLab 6 жыл бұрын
I do have a number of videos on logic circuits. Some of them are in a Boolean Algebra playlist: kzfaq.info/get/bejne/q6d_otSX1Nndlnk.html
@tedtate57
@tedtate57 5 жыл бұрын
Thanks a million David. Your teaching skills and content are incredible.
@CheckThisOut380
@CheckThisOut380 6 жыл бұрын
Does it make a difference if the upper side is wired not the lower ?
@nauraizsubhan01
@nauraizsubhan01 6 жыл бұрын
Thanks a lot sir
@samsammy4220
@samsammy4220 7 жыл бұрын
You're a legend
@santoshdebakikrishna381
@santoshdebakikrishna381 7 жыл бұрын
Thanks for sharing...
@lem09able
@lem09able 6 жыл бұрын
What if there's a -VEE voltage source? How should I apply thevenin to it?
@seankennedy8506
@seankennedy8506 4 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?
@esuelle
@esuelle 6 жыл бұрын
Thank you!
@mba2ceo
@mba2ceo 5 жыл бұрын
for Vth why can U disconnect the Rc ?
@homamalbakry7067
@homamalbakry7067 7 жыл бұрын
thnx bro u r amazing
@knowall5792
@knowall5792 6 жыл бұрын
Good video. How do you draw on the screen?
@ElectronXLab
@ElectronXLab 6 жыл бұрын
Thanks. I use OneNote and a tablet and use Camtasia for screen capture.
@slipnorris5882
@slipnorris5882 6 жыл бұрын
Is the Beta still in effect , even with the 10 times rule
@balandraken4838
@balandraken4838 Жыл бұрын
Thank you
@waelfathe9909
@waelfathe9909 7 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??
@ha.w.s4992
@ha.w.s4992 4 жыл бұрын
Thanks a lot From iraq 🇮🇶
@meth5475
@meth5475 7 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.
@kameeltv9126
@kameeltv9126 7 жыл бұрын
What would I do if R1 and R2 values ain't given?
@hagegeabay4542
@hagegeabay4542 7 жыл бұрын
this is a great example but how do you find Ib if RE is not given?
@ElectronXLab
@ElectronXLab 7 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
@sandeepkamboj7675
@sandeepkamboj7675 7 жыл бұрын
How you calculate the value of R1,R2,R3,R4?
@FireSymphoney
@FireSymphoney 6 жыл бұрын
divide the relative voltage with the relative current, use ohms law.
@darrenallen1387
@darrenallen1387 6 жыл бұрын
Brilliant Ember could you explain that different, not understanding
@FireSymphoney
@FireSymphoney 6 жыл бұрын
use this formula: Voltage = Current x Resistance
@poetas84
@poetas84 2 жыл бұрын
Could you please explain how you calculate R(th) as being 8.33 KΩ ?
@0Bariq0
@0Bariq0 8 жыл бұрын
thank you
@athulvinayak3328
@athulvinayak3328 Жыл бұрын
What is the use of RE in the circuit?how it provide stability?
@TheRicardihino
@TheRicardihino 8 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?
@animasaurus2007
@animasaurus2007 5 жыл бұрын
For future reference it's based on the transistor and the temperature of the device.
@simoncheung2427
@simoncheung2427 3 жыл бұрын
8:16 could you explain why Rth +(b+1)*Re rather than Rth +Re. thanks
@MrPinkster1
@MrPinkster1 7 жыл бұрын
Can anyone help with Deriving the equation at 4:44?
@CheckThisOut380
@CheckThisOut380 6 жыл бұрын
Is it Really the same ?? With the new circuit u drawn ,I1=Ic same generator wich is nit true ?
@simozy4475
@simozy4475 3 жыл бұрын
thx a lot bro
@AkshayKumar-jq9ph
@AkshayKumar-jq9ph Жыл бұрын
If i use exact method even if bre>>10r2 then is it right
@hossamfadeel
@hossamfadeel 5 жыл бұрын
Thanks Alot
@rakeshdutta623
@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
@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.
@eljefe6852
@eljefe6852 4 жыл бұрын
Whats the value of i1 and i2?
@goranvuletic8873
@goranvuletic8873 3 жыл бұрын
Sublime!
@GomesTugaPT
@GomesTugaPT 5 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?
@ElectronXLab
@ElectronXLab 5 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.
@GomesTugaPT
@GomesTugaPT 5 жыл бұрын
@@ElectronXLab thank you!
@seankennedy8506
@seankennedy8506 4 жыл бұрын
Also is (Vth) the same as (Vb) the voltage looking into the base of the transistor?
@age04125
@age04125 4 жыл бұрын
I am confused about this too. Did you found out that Vth equal to Vb? Or is it Vth =Vb + Ib*Rth ?
@gwenhurel5125
@gwenhurel5125 4 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)
@victord3867
@victord3867 Жыл бұрын
How are those resistors in parallel? Having a little trouble getting my head around that
@ElectronXLab
@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.
@chickennugget3362
@chickennugget3362 3 жыл бұрын
Thanks for the video. I've gained a headache.
@wnk25
@wnk25 10 ай бұрын
Can the resistor 'Re' be equal to zero in this circuit?
@ASDERTORETRY
@ASDERTORETRY Күн бұрын
how to find Vc and Vbc
@DMES-wp2fj
@DMES-wp2fj 3 жыл бұрын
What is the equation to solve for ro? Anyone
@TheRicardihino
@TheRicardihino 8 жыл бұрын
Or is beta just the Ic / Ib?
@sonimuthoni9200
@sonimuthoni9200 7 жыл бұрын
how did u get the value of vbe
@JoMaFjErynMaLaZOr
@JoMaFjErynMaLaZOr 7 жыл бұрын
we just assume it's 0.7 V
@michaelzlab
@michaelzlab 4 жыл бұрын
Wow!! Thanks..
@homayoonfhrvnd1470
@homayoonfhrvnd1470 7 ай бұрын
Great
@harrisppm2575
@harrisppm2575 8 ай бұрын
i have a similar problem with no mention on β, how can i solve it with-out it?
@isuckatthisgame
@isuckatthisgame 7 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?
@harrisppm2575
@harrisppm2575 7 ай бұрын
@@isuckatthisgame used loop theory and remember that at the grounded end V=0
@smartpk9534
@smartpk9534 3 жыл бұрын
Sir how to get answer uA please tell me i don't uderstands
@fahemzoldyck1729
@fahemzoldyck1729 6 жыл бұрын
how did you calculate Rth ! 7:20
@mustafaalsamarray5027
@mustafaalsamarray5027 6 жыл бұрын
R1//R2 = (R1 * R2) / (R1 + R2)
@marhsall-bw5kv
@marhsall-bw5kv 8 жыл бұрын
wheres the part 2 of your AC Analysis of BJT?
@WoloW1zard-
@WoloW1zard- 8 жыл бұрын
+marhsall 27777 common emitter amplifier(part 6 and 7 of playlist)
@jenadolfo9358
@jenadolfo9358 4 жыл бұрын
how to solve for IC if the beta is unknown? plss help
@ElectronXLab
@ElectronXLab 4 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
@Festus2022
@Festus2022 5 ай бұрын
Is the Thevenin Eq. Voltage the "Base Voltage"?
@ElectronXLab
@ElectronXLab 4 ай бұрын
It's not the voltage at the base because it is applied through the Thevenin eq. resistance
@Festus2022
@Festus2022 4 ай бұрын
Thanks so much!@@ElectronXLab
@Festus2022
@Festus2022 4 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
@ElectronXLab
@ElectronXLab 4 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
@engenhariaeletricaprojetoi7629
@engenhariaeletricaprojetoi7629 5 жыл бұрын
good guy
@akshayaugustine9741
@akshayaugustine9741 6 жыл бұрын
hey, how do you reach the assumption that beta*RE>=10R2
@ElectronXLab
@ElectronXLab 6 жыл бұрын
If beta*RE >= 10R2, then almost all of the current will flow through R2 and very little will flow in to the base. The assumption that you can then make is that essentially no current flows in to the base.
@akshayaugustine9741
@akshayaugustine9741 6 жыл бұрын
David Williams but why particularly 10R2, why not some other value like 5R2 or 20R2?
@ElectronXLab
@ElectronXLab 6 жыл бұрын
It is somewhat arbitrary, but 10R2 ensures that more than 90% of the current goes through R2.
@akshayaugustine9741
@akshayaugustine9741 6 жыл бұрын
David Williams Thanks man. Really great video
@alexstone691
@alexstone691 6 жыл бұрын
what is that || at RTH ?? how did you get 8.33k ?
@tvptr9302
@tvptr9302 6 жыл бұрын
Alex Stone( || )means parallel
@louies6914
@louies6914 5 жыл бұрын
Why not R1 and R2 are in series?
@mks9111
@mks9111 8 жыл бұрын
why u add 1 to beta in 8:14 ?
@RJWaynerium
@RJWaynerium 7 жыл бұрын
Mohammed Sowwan it's another way of stating this: Ie=Ic+Ib
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