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 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.
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.
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.
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.
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.
+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.
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!
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.
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 .... 🔥🤗
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.
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??
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?
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
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.
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.
Excellent presentation, and delivery of material content. Thoroughly explained in detailed information describing the devices characteristics and biasing approaches.
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
@@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
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?
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.
Do you mean Re from the example circuit? If so, I just arbitrarily picked values for the circuit resistors. Or do you mean re? That is a property of the transistor and it's operating point (re = 26mV/IE where 26mV is the thermal voltage).
@@ElectronXLab Reading back on my comment I feel it was a bit blunt, so apologies. I was frustrated after so much searching online on how to calculate bias resistor. I now understand that you have to make some assumptions and that is thanks to you. Subbed. Great videos
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.
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
More than 3 years after graduating, I am now understanding this. Thank you so much.
lmao
Wow, only 4 years to go then I guess I'll get it 😅
Why do you need to understand them now?
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.
Oh my god, you got your degree when I was born. I want to be a professional EE in the future.
@@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.
how do you get (beta+1) ? great video though! my professor cant explain shit!
Incredible video and explanations. I was able to get past a ton of homework problems thanks to you!
Please add more examples sir for this specific topic because we really understand your explanation. Thank You Sir :)
you made this so simple...thanks..
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.
thanks David ,you are my true teacher,keep going
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.
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.
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.
+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.
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!
Thank you, engineer, can we use it in the transmitter circuit for the of oscillation?
Hye. What is the software that you use to make this tutorial?
How you calculate the value of R1,R2,R3,R4?
divide the relative voltage with the relative current, use ohms law.
Brilliant Ember could you explain that different, not understanding
use this formula: Voltage = Current x Resistance
If i use exact method even if bre>>10r2 then is it right
Excellent explanation. Infinitely more helpful than a university professor
Thanks so much. Glad you think so!
this is a great example but how do you find Ib if RE is not given?
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
Great BUT why you say voltage divider R1 and R2 and immediately căLculate Rth as parralel ?/
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.
how if i want to find the R1 and R2 while i know the the Vce?
Can the resistor 'Re' be equal to zero in this circuit?
Sir how to get answer uA please tell me i don't uderstands
Could you please explain how you calculate R(th) as being 8.33 KΩ ?
Thanks for making this very *CLEAR* plain English video. I can definitely understand your Canadian English.
What is the equation to solve for ro? Anyone
Thanks so much . Your video has clarified all my doubt. Thanks once more
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 .... 🔥🤗
Thank you very very much Man!
Also is (Vth) the same as (Vb) the voltage looking into the base of the transistor?
I am confused about this too. Did you found out that Vth equal to Vb? Or is it Vth =Vb + Ib*Rth ?
Vth is equal to VR2, the voltage provide by the voltage divider. And so VR2 = VBE + VRE (equal to => (R2.IR2) = VBE + (RE.IRE)
thnx bro u r amazing
Great video! Addresses some of my longstanding questions. Thank you!
Thank you for this explanation, how do we define Beta here?
Best explanation
thank you sir
explained very clearly thankuu
How are those resistors in parallel? Having a little trouble getting my head around that
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.
This is an amazing video. The way you explained this made it very easy to understand. Thank you so much!
Awesome job :D
Thanks a lot sir
Thank youu..............clearly explained
8:16 could you explain why Rth +(b+1)*Re rather than Rth +Re. thanks
love you david, so glad for you
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??
Great Work!
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.
Thank you very much Sir..
thx a lot bro
Very useful methode
What is the use of RE in the circuit?how it provide stability?
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
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?
Congrats! You are antes excellent teacher!
Great
how to solve for IC if the beta is unknown? plss help
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
Is it Really the same ?? With the new circuit u drawn ,I1=Ic same generator wich is nit true ?
I find I'm more of a spice warrior then a scientific calculator nerd 😁
if you just want to target alaways use this magic resistors valors 50K, 10K, 3K and 1K. KKKKKKK
Great explanation sir! Now I understand what's going on there
4:16
What if there's a -VEE voltage source? How should I apply thevenin to it?
Thank you
Great
Thanks for the video. I've gained a headache.
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.
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.
Plugin Ic from where? 06:45
I'm not 100% sure what your question is, but if you've calculated IB based on the equation at 06:45, then you can calculate IC as (beta)xIB
This video helped me understand the biasing of BJT. Thanks a lot sir!
Glad it helped!
i have a similar problem with no mention on β, how can i solve it with-out it?
did you solve it? i have the same problem. no beta but i have all other values. how to calculate base current without it?
@@isuckatthisgame used loop theory and remember that at the grounded end V=0
Thank you for such a crystal clear explanation of a VDB-BJT.
Excellent presentation, and delivery of material content. Thoroughly explained in detailed information describing the devices characteristics and biasing approaches.
15:44 do you assume Rth is zero? in previously, Vb=Vth- Ib*Rth.
oh, i just realise that in this condition , Ib is nearly zero, the potential drop is near zero, so that Vth = Vb.
Thank you so much
Can anyone help with Deriving the equation at 4:44?
Great job simplifying this ckt. Thanks for the great job. It does help lots of ppl. Highly recommended
Thanks a lot
From iraq 🇮🇶
bless you sir. Great video
What would I do if R1 and R2 values ain't given?
Does it make a difference if the upper side is wired not the lower ?
Whats the value of i1 and i2?
Thankuuuu sooo muchhh for thiss😍 thankuu
Is the Thevenin Eq. Voltage the "Base Voltage"?
It's not the voltage at the base because it is applied through the Thevenin eq. resistance
Thanks so much!@@ElectronXLab
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
@@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
shouldn't it be Vce=Vcc-Ic[ Rc-(B+1)Re]
no
Wow!! Thanks..
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?
For future reference it's based on the transistor and the temperature of the device.
Very clear explanation!
How does the early voltage affect Ic in this type of calculations?
What would I do if R1 and R2 values ain't given?
R1 & R2 are must be given always, if its not given just use 1K
Is the Beta still in effect , even with the 10 times rule
Finally someone that made me understand this theory. Thanks a lot.
You're very welcome!
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?
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.
@@ElectronXLab thank you!
How to Calculate parallel 10k ll 50k ?
1/(1/10k + 1/50k) = 8.3k
wheres the part 2 of your AC Analysis of BJT?
+marhsall 27777 common emitter amplifier(part 6 and 7 of playlist)
Where did you get Re from?
Do you mean Re from the example circuit? If so, I just arbitrarily picked values for the circuit resistors. Or do you mean re? That is a property of the transistor and it's operating point (re = 26mV/IE where 26mV is the thermal voltage).
@@ElectronXLab Reading back on my comment I feel it was a bit blunt, so apologies. I was frustrated after so much searching online on how to calculate bias resistor. I now understand that you have to make some assumptions and that is thanks to you. Subbed. Great videos
The best video i've seen so far, thank you.
That's great to hear. Thanks for your comment.
Or is beta just the Ic / Ib?
Thanks for sharing...
at 9:34, why Vce is not as "VCE = Vcc-Ic*Rc-Ie*Re", since Ie=201/200Ic?
ua-cam.com/video/WY7WbGph_o4/v-deo.html
@@GSaiRakesh sir why Ie approximately equal to Ic ?
Sublime!
Making assumptions of previous knowledge.
Everything makes "assumptions of previous knowledge".
1+1 = 2 assumes you know what 1 is.
What if Current Gain is not given?
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.