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.
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!
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 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
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.
Excellent presentation, and delivery of material content. Thoroughly explained in detailed information describing the devices characteristics and biasing approaches.
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.
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 .... 🔥🤗
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.
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?
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.
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?
I do have a number of videos on logic circuits. Some of them are in a Boolean Algebra playlist: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-uAOoqfosyfE.html
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.
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.
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.
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.
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
