Electrical Engineering: Ch 3: Circuit Analysis (34 of 37) Solving Basic Transistor Circuit (MESH) 1
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- เผยแพร่เมื่อ 26 ก.ย. 2024
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In this video I will used the MESH method to find the voltage from the collector to the emitter of a basic transistor circuit with a NPN transistor.
Next video in this series can be seen at:
• Electrical Engineering...
![[Solved]- Numerical for a BJT (Ib, Ic, Vce & Vcb =? )](/img/n.gif)
For the first time, I was able to solve a transistor circuit with actual confidence and understanding...
Probably the last time, too. Thanks anyway!
lol why the last time?
@@siiikebro1883 Things will start happening
Understandable speaking and writing, this helped massively. Looking forward to more videos.
Your the best man! Have a real bad Computer Technology teacher at UPV in Valencia, but thanks to you I'm learning everything he can't teach. Tons of love from Spain
Hi Michael ... Thanks for those mini lessons ... you make my day. I have to say something though. Two things dropping the quality of the presentations. 1) You have to turn, 2) you stand between the viewers and your wonderful and clear writing on the board.
My advice and I have to admit that I don't know how they do it.
Some you tubers like you, manage to stand behind a transparent board, so they always face the viewer, the viewers can see you and you never stand in between.
Thanks for the feedback. We are working on that problem.
What a beautiful piece. Very easy and straight forward. Thank you!
Thank you. Glad you found it helpful. 🙂
Sooo helpful! Thank you, you're doing an amazing job! The videos are well structured and well explained!
Thank you. Glad you found our videos. 🙂
Hallo, first thank you always for good videos. I have a question concerning the second loop. Why did you not includeVBE in the second look .does is it not include on that second mesh. (Loop) thank you so much. I am looking forward to your response
Vbe is part of Vo which is what we are solving for.
I liked your classes , best explanation about transistor I have ever seen.
Here in India, we learn all of this in high-school. Thanks a ton! I was having a lot of trouble wrapping my head around Kirchoff's law.
Welcome to the channel!
The greatest tutor of all time
Thank you. Glad you found our videos.
I like the way you give a lecture. Thanks for making videos of this kind though I have a little problem in understanding this and I hope you can help me. Why did you not include the current in the right loop (I2) when writing your KVL equation of the left loop? Also, when writing your KVL of the right loop , why did you not include the 0.7 volt ?
1) Since there are no voltage drops in the section of the 2 loops that are common, we don't need to worry about current 2 in loop 1 2) The 0.7 volts are included in the Vo
@@MichelvanBiezen Sir, thank you very much.
@@MichelvanBiezen 2) Isn't the Vce equal to 0 when in saturation mode and when in linear mode > 0.2V ?
This lecture series is phenomenal! Brilliant lecturer.
Thank you for making this video, first thing I've come across that makes good sense.
I really understand the method thank you sir
Great! Glad you found our videos. 🙂
@@MichelvanBiezen thank you sir
If my son's life wasn't gonna get impacted by it. I would've called him transistor
🙂
If I1=Ib and I2=Ic, shouldn't Ic=β(-IB) since Ib is in the opposite direction of Ic. In other words, shouldn't Ic=-4.3mA ??
Thank you.
The I2 current is drawn in the wrong direction so the whole thing is wrong
Ic and Ib are in right directions although I2 is in the opposite direction of Ic => which is wrong
But in spite of the huge mistake the calculations are supposed to be all right (not minding minuses)
just don`t get me wrong, I kinda like the guy :)
I love your bow tie! The video content was great too :-)
I really like your method but --at 2nd loop Vce [rather than V0 is clearer] is with minus ok - to +, but Ic being opposed to the loop direction should be minus too, and VCC +to - should be plus . I would write VCC = Vce - Ic R
If that method works for you, definitely go with your method.
how can I tell you ! 100% you help me and you make me clear from my confusion on this topic. thanks a lot.
You are welcome. Glad you found our videos.
Thanks a lot
Helped even for university
Glad you found our videos
Now i'm able to solve transistors ques confidently...
Thnx u so much sir...
Most welcome
Very good, thank you! Sadly though your body blocked your calculations.. if you could stand to one side while writing your calculations on the board your videos would be excellent! Please keep them coming!
Thank you for giving wonderful videos! But i have one question, sir. I don't undetstand why you put plus sign for Ic although its direction is opposited to second loop's direction.
When going across a resistor against the current, there will be a voltage rise.
@@MichelvanBiezen Thanks.. that means like E-(-IR)= E+IR ?
So voltage_sigma is the potential difference between the collector and emitter??
V sigma is the voltage drop across the entire transistor.
How is an opposite current always a voltage raise ? We are taking minus voltage according to KVL rotational loop choice , really disturbing
If you go across a resistor (against the current) then you will experience a voltage rise.
Where do you get the value of beta? Is it a fixed value or variable? ....from the philippines
Hi and welcome to the channel. Beta was a "given" in the problem.
Hey professor Biezen, one question. Why is there a 0.7V drop between the base and the emitter?
Across any p-n junction, there is a voltage drop due to two reasons. 1) there is an energy level difference between the two that must be overcome. 2) There is some resistance across the junction (the drop caused by this is current dependent)
@@MichelvanBiezen Will every npn transistor have the same voltage drop? If not, can it be calculated or found in the data sheet?
@colinmitchell7760What links?
Im sorry but on the very end when you calculated (50)(43miliamp) you converted the milliamps to amps right to solve ohm law and so it’s 0,215 shouldn’t it be 2,15 V? And then 8-2.15?
4.3 mA
This was a great video! would love if you covered MOSFETs as wel.
Thank you. We will get there eventually
This assumes we know the transistor is working in forward active mode. What if we don't know how the transistor is working? It could be in cut-off, saturation etc...
its actually in saturated since current gain*ib>ic
Hey, Michel! I am writing a program to simulate electric circuits and I am looking for generic formulas for transistors. You say that beta is Ic/Ib, but what if the 100kohm resistor was reduced to, say, a 30ohm resistor? The Ic/Ib ratio could not possibly be 100! Otherwise, there would be a negative v_sigma across the transistor, wouldn't there?
I have the same doubt, and if you change the value of the resistor on C, then the current at C changes without changing the current in B, so how the C current can depend on B current? I have tested this on real circuit and the formula to calculate C current using this beta value doesnt make sense. Surely i'm missing some critical information but i can't get it
For anyone that have the same doubt, the answer resides in the active mode and saturation mode of the transistors. When the voltage between C and E is more than 0.3 V the transistor is in active mode, if not then its in saturation. The formula to calculate C current with B current * Beta only works when transistor is in active mode, so if we increase C resistor value, there will be a point where voltage between C and E will be less than 0.3, and the formula will not work. For those cases we need to use Ohms and Kirchkoff laws to determine current and voltage in the circuit.
@@insertusername5737 Ebers molls model for BJT transistors is pretty good.
@@traiancoza5214 I didn't know that, thank you i will check it out
The thing we found in this question was not considered in the first loop,why?
shashi giddigam if you follow the circuit, V sub o is the voltage drop from collector to emittier. The first loop runs from the base to emitter, not collector to emitter
So it's a given..it will aways be.. 07 btw collector and emitter when calculating the mesh current. Love the videos btw. Using it to study for the FE. Thanks for responding and making sense.
+Christina Schumski Not 0.07 but 0.7 V. Yes there is very little difference between transistors. It is part of the property of the semi-conductor material.
nice lecture sir . but can u zoom it .. so that it ll be easier to see .. the board
y como se resuelve este mismo circuito, pero controlando las dos fuentes de poder ?
That would be hard to explain in a comment like this. (There are lots of examples in this playlist).
if knee voltage 0.3
how to solve this problems
In your 2nd loop equation why didn't you include the 0.7 v drop?
That is included in the Vo
Hello Professor... What happens when you have an input voltage but no resistor between it and the voltage drop between the Base and Emitter... Would you simply combine them?
You'd heat up the transistor until it burns!
Hi professor
thank u for this helpful lecture
but i was wondering why the V between C and E is equal to 0.7 ??
There usually is a small voltage drop across the transistor, depending on the amount of current flowing through it.
@@MichelvanBiezen Yes but you're still calculating the current so where did u get the 0.7 V from
@@MichelvanBiezen I wish you would have time to look around here, Sir :))
can you create one big loop for it? I want to solve for Vce
If you only use one loop, you may not be able to solve it.
that tie is epic
Thank you so much sir
From 🇳🇵 Nepal
Welcome to the channel!
Best ever explanation...
mr bizen at first loop you went throuhg positive terminal and you wrote -0.7 ?you should have done +0.7
The -0.7 V is the voltage drop from the base to the emitter.
@@MichelvanBiezen actually we have got diode characterisctic at vbe .you can think that 0.7 volt is thereshold voltage and first loop you went through positiv terminal ypu shpuld have wrote 0.7 not -0.7
@@MichelvanBiezen I am sorry mr. at first loop You drew the circle from left to right, but you created the opposite equation, my mistake.😅
No need :)
nice sir its great explaination
Excellent video! Thanks!
Yes, I understand now. Thanks for responding. Interesting. Your the best!
I've encountered a problem: This way of calculations seem to work well enough for situations where R1>R2*beta (R1 being base resistor, R2 the collector resistor). But when the collector resistor is higher, in this video's case it would be above 1k, the resulting Ic is no longer Ic=Ib*beta.
Is there a way around it?
Same here, I'm looking for the answer to that question as well. In my mind, I think that Ic is the *maximum* current that allows to float through the transistor. If the resistor doesn't cause the current to go below Ic, then the math adds up. Otherwise, it doesn't. It gets worse in the next video where an emitter resistor is added. If the collector resistor has high enough resistance, it will influence Ie, and if Ie is different, then it will change Ib as well. I find that quite hard to wrap my head around. I can reproduce this in simulators, but would really like to see some explanations on this topic.
Not sure if you're still interested, but Ic=Ib*beta is always true, as that is the definition. The value of beta changes, as it is the ratio of Ic and Ib.
@@Dogthedeadly Thank you. So, there is no solution, just an explanation, from a different view. As we, to solve this kind of problem, presume beta to be constant, to calculate Ic=Ib*beta, while it isn't, is there any way to predict the beta changing? What function of what it is?
@@nexusAa there are multiple betas depending on a few factors. What is shown in the video is a transistor in forward active mode using large dc signals, with the corresponding beta value for that mode. As you change Rc, you're going to change what mode it is in. The beta value for each mode isn't something we can really change for a given transistor from what I understand, it's a characteristic of the transistor.
@@Dogthedeadly The problem is that even in the same mode (for example, forward active) this beta value will constantly change for every resistor value in C. So, what is the point of using a beta value to calculate the current in C, if this beta value will change depending on the resistor value in C? Maybe i understand wrong and beta is only an estimate
Why we didn't use Ib=beta*Ic?for determining Ib
Very helpful video.. Thank you ❤🌹🙏
My pleasure 😊
❤
So there is a voltage drop of 7.785V on the transistor?
Yes, that is correct.
Thank you sir for this video.
I want to ask how you got 0.7v as the drop from the transistor to ground?
Is it the potential barrier voltage for silicon diodes?
Yes, it is the typical drop for transistors.
do you have also 3 phase power you are really good sir michel we are benefiting a lot from you
Not yet, that is the next chapter we have to cover.
Why isnt the Ic = beta * ( - Ib ) ?
Idk why but I enjoy watching him pressing the calculator buttons😆
th-cam.com/video/SEUtgpShfIs/w-d-xo.html
how are you michel ?do you have small and large equivalent circuıts with capacitors and four resistors ?
You may find what you are looking for here: PHYSICS 42 RESISTOR CIRCUITS and PHYSICS 39 CAPACITORS PHYSICS 48 RC AND RL CIRCUITS PHYSICS 49 RCL CIRCUITS
i mean analysis of bjt circuits michel
Those we haven't made yet. (something for the future)
this knowledge what I'm looking for!
Glad you found us.
@@MichelvanBiezen thank you,
But there is a little unclear to me
on designing a circuit what should I look first, I have BJT transistors to design a circuit, should I adjust on resistors to make the transistors work or I should look for transistors that can handle ng Current and Voltage of my power source?
How do we know that V(BE) is going to be 0.7V?
It is a property of semiconductors. The drop is always around 0.7 V for most ransistors and diodes
@@MichelvanBiezen I think we use a resistor of such ohms with it in order to have 0.7V drop. If we use a very very less resistance( with high DC voltage), then the drop will be very high and it will get damaged, I think.
The drop is inherent in semiconductors. The 0.7 volt drop is not due to resistors.
@@MichelvanBiezen I disagree with that. Because in ,for example, a PN JUNCTION DIODE of some type, you will have to increase external DC voltage to increase forward voltage of diode above 0.7V( below this, current will be negligible and too much above this, it will get damaged,,you may use some resistor with it, along with battery, to prevent damage). Now the same PN junction diode , when used in an NPN TRANSISTOR, again needs forward potential drop of nearly 0.7V. And again you need an external DC battery of 0.7V for that purpose. If you have 5V battery, for example, then you need to use a resistor with it.The exact value of resistor can be known by using manufacturer's data sheet in which the necessary current for that PN diode or transistor is already given.
Hi, 43mA isn’t 0.043?
It is hard to see, but the current is 4.3 ma (not 43 ma) Thanks for pointing that out.
Thanks for the help.. from a high school student in India. Your work is appreciated. :-)
i'm a primary school student in South Africa thank you :)
@@sashamuller9743 🤣😅💀
Lol
@@sashamuller9743 come once India and give JEE then u will understand boy 😂😂
@@GauravYadav-sl8vc🇮🇳💩🇮🇳💩🇮🇳💩🐷🇮🇳🐷🇮🇳🐷🇮🇳🐷🇮🇳🐷
The value you got for Vo is wrong because you made an error when you multiplied 50 Ohm by 43mA(0.043 not 0.0043)...The right answer is 5.85V
It is hard to see, but the current is 4.3 ma (not 43 ma) Thanks for pointing that out.
@@MichelvanBiezen I d'ont agree with that since it is 4.3/100kOhm for the value of Ib which will give 4.3×10^-5
Ah yes. I see it now. I was sloppy with the "100K ohms" . Thanks for pointing that out.
@@MichelvanBiezen No problem Sir
tA good example. Thank you.
Glad you liked it!
Is beta equal to hFe which can be measured in multimeter?
I am not familiar with "hFe".
Thank you sir
Welcome
Only thing I don't like is dropping all the units and suddenly intrducing them in the result again... that's bad style :) But nicely explained.
Always adding the units is not necessarily a good thing. It takes time, and clutters up the equations making it more difficult to flow through the problem. And when the units are known since you are typically solving for the current or the voltage it is often not necessary. However that said, there are times that it is advisable to incude the units
@@MichelvanBiezen I agree
Thanks a lot.....
You are welcome
Excellent, thank you!
I want to ask one question.which battery(5V or 8V) will the current I(E) go to? thank in advance
The actual current is directed according to the black arrows. The red circular arrows are just used to work out the problem using Kirchhoff's rules.
+Michel van Biezen I quite don't understand your answer. you mean I(E) will go to battery 8V only? right?
Ib is the base current ,it is driven by the 5V source. Ie is the emitter current, it is driven by the 8V source, but the amount of emitter current (Ie) is controlled by the amount of the base current.
Thank you very much sir.
@@MichelvanBiezen Doesn't that depend on if the transistor is on a specific operating mode. By the way, thx for the quality videos, Electrical Engineering is very tough and keep making more.
thank you
You're welcome 🙂
Thanks sir..
Welcome
How do you know the 100kohm resistor has a voltage drop?
If current flows through a resistor, there will be a voltage drop. V = IR
That is the drop for a P-N junction. It acts similar to a diode (diodes also a .7V drop typically)
The calculated V_0 is also the V_CE right?
Yes they are the same.
Great, thanks!
Excellent sir
Nice video, thank you for sharing :)
You are welcome. Glad you liked it. 🙂
thank you.
Thanks for the video.
You are welcome!
Thanks 😊👍🙏💯
You are welcome.
V + I*50 - 8 = 0.... Volts + Amps - Number... NaN :D Error
?
Awesome!
Is there a full DVD of all of these lectures combined?
No, only the individual examples and topics.
The IB(100k) is it related to V=IR??
I = V/R applies to all circuit branches
Thank you sir...from Bangladesh
Welcome to the channel!
I wish someday teachers across the world would start wearing cloaking clothes. All you would see then is their head, and their work on the board. Amazing presentation of information obscured by opaque fabrics. Leaving the student to press rewind 300 times, then it's time to move on. I am glad you've made this video and series, your presentation has helped me the most. Thank you.
How'd he get the. 07 voltage drop in first mesh
+Christina Schumski The drop is 0.7 Volts and it is due to the semi-conductor property of the base. Current will not flow from the collector to the emitter until there is a minimum of 0.7 Volts between the Base and the Emitter.
can you let us see the numbers while youre writing it?
Good suggestion. (working on it with the new videos)
Now-a-days Red Color Bow Tie ! What a....
the best
Glad you think so.
Best ever
What is V0? Thank you!
Vo represents the "output voltage".
থ্যাংকস
You are welcome
GREAT!!
🎉🎉🎉🎉🎉🎉
Thank you. Glad you liked it. 🙂
999
bro move out of the way
Yes, we changed our approach on the newer videos that will improve that.
The beauty of KVL is its assumed independence of potential . Potential can be 0 or 100 or x volts at any one point. The calculation result is the same. IOW potential at minus pole of Voltage Source can be voltage Y and not 0 V. It is none of your concern in that calculation. Imagine some entity creating a constant minus potential on 1 capacitor pin where obviously no DC current flows through.
It is a very powerful method indeed.
I really understand the concept very well be blessed
Glad the video helped.
Thank you
You're welcome
wtf where did that 0.7V come from?
Clemente Küchle He brought it from his house
It's the voltage drop from the transistor itself. It's a standard value.
For silicon .7V
For germanium .3V
VBE is the touching Voltage between 2 layers of semi-conductor, for Si : 0.7V and for Ge : 0.3V. This Transitor is from Si so it should be 0.7.
It is the drop of voltage in a junction in a simplifyed (and linearized) model.
thanks sir
Why do I automatically think bow tie wearers are possibly eccentric...?