@@SightCentralVideos It was a joke. Half of electronic videos are in Hindi although they have an English title. Which ofcourse doesn't matter if you understand it, but sucks if you don't.
I wish you were my teacher... Mine doesn't teach us any of this because its easy for him. Since he knows it his students should already know it... Thanks for the video, it really helped out and im pretty sure im going to watch plenty more!
Great teaching and very logical. The heading is AC analysis so I did not mind that you skipped the DC. Love how you used the box to get me thinking of how to replace that box with a model.
Which is the output amplified signal? Is the Collector or the Emitter? Because I am confused when I am creating circuit but I don't know where will the amplified signal flows.
Is the following correct? the base of the transistor is getting the AC current PLUS the DC current (the AC is riding on the DC). So the transistor is always on. Then the amplified result has the DC removed by a capacitor. If that is right, then a DC battery between C1 and the base of Q1 would eliminate a lot of resistors and capacitors(?). I'm trying to get an Arduino microprocessor to register a fluctuating analog (not alternating) signal from a weak piezo crystal at one of its Analog pins. Any suggestions?
Hi, great tutorial and great refresher for me. I was wondering where can I find the next part, or if it was created? I can't seem to find it in the playlist. Thanks again
The capacitor act as Open circuit in the DC circuit and as short in the AC circuit, that’s why the Re is cancelled in AC circuit but we will consider it in the DC circuit, Is that correct
hi sir , great job you have done , where is the video which shows how to find the voltage gain and impedance , can you reply me with the link of that video thank you :)
The current through a junction goes up exponentially as the voltage increases. At normal small-signal emitter currents (say 100μA to 10mA), the voltages needed to produce those currents lie in a small range from about 0.6 to 0.7 volts. If you drew a curve of Ie against Vbe it would turn upwards quite rapidly around that voltage range. At any point on that graph, the slope of the graph would represent the change in current for a small change in voltage. That is the reciprocal of the dynamic resistance of the junction at that point. If you do the maths (see "Shockley's diode equation), you find that the dynamic resistance is around 25 or 26mV divided by the emitter current at that point (or the collector current as they are virtually the same). So the higher the quiescent collector current you use, the lower the dynamic resistance of the base-emitter junction. You'll need that when calculating the gain and input impedance of the amplifier stage.
I think there's a mistake in here. While calculating Vce, you had used the formula...Vce=Vcc - Ic(Rc + Re), whereas I'd would use the formula Vce=Vcc - Ic(Rc) . I fail to understand why is the resistance at the emitter even coming into picture? Please clarify this doubt of mine :)
Your calculation of Vcc - Ic(Rc) Would give you the voltage at the collector with respect to ground. To get the voltage between the collector and the emitter, you have to calculate how much voltage difference there is between ground and the collector - this calculation requires the IcRe (i.e., the voltage at the emitter with respect to ground) be subtracted from the voltage at the collector with respect to ground.
This is VT, the thermal voltage of the PN junction. It's not actually constant, but is temperature dependent. VT = kT/q. k is Boltzmann's constant (1.38x10^-23 J/Kelvin). T is the temperature in Kelvins and q is the charge on an electron (1.6x10^-19C)
I didn't use a specific Beta value; I just assumed that it was high enough that the resistance looking into the base was high enough that I could ignore the base current.
Those equations would of course give different values for re. The correct one is re = 26mV/ie. I think that what you are referring to in the other equation is what the equivalent resistance would be if we modeled the resistance in the base instead of in the emitter. in that case, the resistance would be re(beta + 1) which is equal to 26mV/ib
Why would you use a model for the transistor that makes it so easy to jump to the wrong conclusions? If the dynamic emitter resistance is really seen between the base and the emitter (i.e. between base and ground in your circuit) as far as the ac analysis is concerned, then you're giving the impression that the input impedance of the transistor is re. Now, we know that the effect of the current source is to provide an effective resistance that is β times bigger than re in the base-emitter circuit, but you have to derive that part of the analysis every time, and if you have have students, you'll confuse them because they will forget the effect of the current source and think that the input impedance at the base is about 8 ohms (26mV/3mA). It is far easier (and kinder to students) to use a model where the base and collector are separated; the collector-emitter resistance is re; and the base-emitter resistance is β.re. You only have to derive that model once, and then everyone has an easy time calculating the input impedance (which btw, depends far too strongly on β with the values you've used), as well as the gain and the values needed for the capacitors.
i'd solve for Ic i got 2.63mA i did find first the Vb then loop from ground to ground in Vb to Ve. so this is my equations Vb-Vbe-Ie(Re)=0 then its 2.63mA
Hi Steve With all the respect, that circuit of yours does not work, it's nice on paper, but not in the real world. The AC gain will vary like crazy, especially if you have good swing at the load. ie the AC gain depends on re , which also depends on IE , and sense IE varies as the output changes, the AC gain will vary.
If that capacitor connected to the emitter is big enough, to Short the emitter to ground, then this circuit is doomed to fail. The dynamic emitter resister re , will keep changing in value, in simple words AC gain =-Rc/re , re =26mv/IE , when Vc swings by 10v pp , you will see magic. Sorry to say that.
I was trying to split up the topics into individual videos. This one shows how to convert the circuit into AC a format that you can do AC analysis on. Check out my BJT playlist for more: ua-cam.com/play/PL4651816D92AB6B2B.html for
Wow. An electrical engineering video with reasonable ok sound quality , not filled with noise. Unusual. Thanks !
racist, hindi isn't noise, it's a language...
@@costbart who said anything about hindi
@@SightCentralVideos It was a joke. Half of electronic videos are in Hindi although they have an English title. Which ofcourse doesn't matter if you understand it, but sucks if you don't.
@@costbart true. I admire the effort, but as a Dutch guy it’s very difficult to understand.
I wish you were my teacher... Mine doesn't teach us any of this because its easy for him. Since he knows it his students should already know it... Thanks for the video, it really helped out and im pretty sure im going to watch plenty more!
I can't thank you enough. This is the best playlist about BJT analysis that I found on UA-cam!
Same, starting my analogue electronics course
@@ggripen Same
@@AdityaKumar2128 i finished my EE degree :D
Great teaching and very logical. The heading is AC analysis so I did not mind that you skipped the DC. Love how you used the box to get me thinking of how to replace that box with a model.
Explained it so well, just doing the diagram step by step made much more sense.
Thank you so much David. So easy to understand when you're teaching. What a gift.
Thank you!! Really helpful. Where is the other video ? did you solve Ac Analysis?
Thanks for making this introductory video! I learned more from this video than from my first two weeks of analog class. ._. Liked and subscribed.
You're welcome!
You're the second best teacher on youtube, after me!
Fantastic...very well conceptualized and explained
Which is the output amplified signal? Is the Collector or the Emitter?
Because I am confused when I am creating circuit but I don't know where will the amplified signal flows.
Hey David Thank for share your knowledge and help me understand this concepts. I hope you continue posting more videos. again thank for your time.
why does DC source become ground? Instead of simply shorting?
Is it because negative portion of AC makes positive DC to become ground?
"I'm not going to do all the calculations" ??? So much for the analysis.
+Digger D
Video title DOES say "AC analysis", not DC analysis. DC analysis is probably covered in one of his other videos. Have you looked?
+SixSixSix exactly.... he does the DC, not the AC
your video is great, it clarified everything about ac analysis for me, thanks a lot!
Very nice analysis. Intuitive. Thank you!
Is the following correct? the base of the transistor is getting the AC current PLUS the DC current (the AC is riding on the DC). So the transistor is always on. Then the amplified result has the DC removed by a capacitor. If that is right, then a DC battery between C1 and the base of Q1 would eliminate a lot of resistors and capacitors(?).
I'm trying to get an Arduino microprocessor to register a fluctuating analog (not alternating) signal from a weak piezo crystal at one of its Analog pins. Any suggestions?
Why u can connect VCC to the ground when doing AC analysis?
Shouldn't VRC be 8.9 Volts instead of 13.1 Volts on the DC analysis?
Hi, great tutorial and great refresher for me. I was wondering where can I find the next part, or if it was created? I can't seem to find it in the playlist.
Thanks again
So what is the actual numerical value of (ac + dc)? Is it the DC level + AC pk? Or is it ACrms + DC level?
The capacitor act as Open circuit in the DC circuit and as short in the AC circuit, that’s why the Re is cancelled in AC circuit but we will consider it in the DC circuit, Is that correct
In the AC analysis, what is the reasoning behind turning the DC source in a ground? That does not make sense to me. Thank you.
just wondering, for the first DC bias point analysis, are you assuming that the beta = 100 for the transistor? Thanks!
It's been a while since I made this, but I think I just assumed that it is big enough that we can ignore it.
How did you determine Vce at 2.97V. Also Vc and Ve
good explanation...need more videos
Is it even possible to find Ib when the Beta is not given like in the example?
Same Transistor simulation used a diode for BE - It was understandable too and reached the same result.
we take beta dc in transistor circuit? means beta dc already given in question? how?
YOU ARE AN AMAIZNG TEACHER PLEASE MAKE MORE VIDEOS!!!
Why not using p equivalent bjt model?
When you calculate the collector current, do we use the DC part of the input as our Vbe? I guess I should say Vb since there is an emitter resistor.
how did you find VCE if it's not given?
Please for the love of god do a PI model video! PLEASE PLEASE PLEASE
Also do one where IB is not neglected
Great Explanation thanks a lot
hi sir , great job you have done ,
where is the video which shows how to find the voltage gain and impedance , can you reply me with the link of that video
thank you :)
Great tutorial, well explained!
best explanation ever.. Thanks a lot
Thank you for the good explanation.
wouldn't your current source be a dependent source?
Good explanation but can you clear up one question. How does the value of re given relate to a 0.7v drop across the be junction. Many thanks.
The current through a junction goes up exponentially as the voltage increases. At normal small-signal emitter currents (say 100μA to 10mA), the voltages needed to produce those currents lie in a small range from about 0.6 to 0.7 volts. If you drew a curve of Ie against Vbe it would turn upwards quite rapidly around that voltage range. At any point on that graph, the slope of the graph would represent the change in current for a small change in voltage. That is the reciprocal of the dynamic resistance of the junction at that point. If you do the maths (see "Shockley's diode equation), you find that the dynamic resistance is around 25 or 26mV divided by the emitter current at that point (or the collector current as they are virtually the same). So the higher the quiescent collector current you use, the lower the dynamic resistance of the base-emitter junction. You'll need that when calculating the gain and input impedance of the amplifier stage.
great introduction sir......
Could you please give me the link to your DC analysis video?
Thank you
How to find voltage drop across resiator?
What would be an equivalent circuit of an ideal current source/?
I think there's a mistake in here. While calculating Vce, you had used the formula...Vce=Vcc - Ic(Rc + Re), whereas I'd would use the formula
Vce=Vcc - Ic(Rc) . I fail to understand why is the resistance at the emitter even coming into picture? Please clarify this doubt of mine :)
Your calculation of Vcc - Ic(Rc) Would give you the voltage at the collector with respect to ground. To get the voltage between the collector and the emitter, you have to calculate how much voltage difference there is between ground and the collector - this calculation requires the IcRe (i.e., the voltage at the emitter with respect to ground) be subtracted from the voltage at the collector with respect to ground.
Oh yeah! Vce = (Vc - Ve) . Okay I got your point. Thanks for clafifying. :)
+Jdeep Shiva
Good question , and good explanation ! I learned things.
I love the way you teach! thank you so much!
in the T model , where does the 26mV come from?
This is VT, the thermal voltage of the PN junction. It's not actually constant, but is temperature dependent. VT = kT/q. k is Boltzmann's constant (1.38x10^-23 J/Kelvin). T is the temperature in Kelvins and q is the charge on an electron (1.6x10^-19C)
which program do you use to write?
I just use OneNote and an old Wacom tablet
Awesome explanation thx
You're welcome!
Thank you for this video, it helped me a lot.
With what value of Beta did you get those DC analysis values?
I didn't use a specific Beta value; I just assumed that it was high enough that the resistance looking into the base was high enough that I could ignore the base current.
Excellent intro video, thanks!
Hello Dave!
Thanks! I never understood this in class!
what is the differentiate between re=26mv/ie........... and re=26mv/ib
Those equations would of course give different values for re. The correct one is re = 26mV/ie. I think that what you are referring to in the other equation is what the equivalent resistance would be if we modeled the resistance in the base instead of in the emitter. in that case, the resistance would be re(beta + 1) which is equal to 26mV/ib
Where did 26mV come from?
There are a few. If you look on My Channel, there is a BJT playlist. The first few are DC analysis (Characteristic curves, base biased, fixed bias)
Where's the continuation of this video?
Very clear. I thank you.
Excellant session
YOU ARE THE MAN!!!!!!!!!!!!!!!!
Very helpful after 11:50, before this is a bit long winded.
what is the use of c2 capacitor
Makes the circuit more stable while allowing it to amplify input at higher frequencies.
Thanks for sharing.
Why would you use a model for the transistor that makes it so easy to jump to the wrong conclusions? If the dynamic emitter resistance is really seen between the base and the emitter (i.e. between base and ground in your circuit) as far as the ac analysis is concerned, then you're giving the impression that the input impedance of the transistor is re. Now, we know that the effect of the current source is to provide an effective resistance that is β times bigger than re in the base-emitter circuit, but you have to derive that part of the analysis every time, and if you have have students, you'll confuse them because they will forget the effect of the current source and think that the input impedance at the base is about 8 ohms (26mV/3mA).
It is far easier (and kinder to students) to use a model where the base and collector are separated; the collector-emitter resistance is re; and the base-emitter resistance is β.re. You only have to derive that model once, and then everyone has an easy time calculating the input impedance (which btw, depends far too strongly on β with the values you've used), as well as the gain and the values needed for the capacitors.
thank you mr david from egypt :)
WERE IS ANOTHER VIDEO >>>>>>>>>>>>>...NEXT VID
Good video thank you
Thank you great video
well explained
thanks a lot sir
You are most welcome. Thanks for the comments
Best not to teach the supply is shorted to ground for AC analysis....you simply consider it a ground to AC signals
where is the next part?
All of my BJT circuit videos are in this playlist: ua-cam.com/play/PL4651816D92AB6B2B.html
+David Williams
Sir answer my question? please
Muhammad Abid Khan
Man but I needed all the calculations for everything :(
Thanks Prof dude.
No, because 13.1 is the remainder after the voltage drop at RC,
i'd solve for Ic i got 2.63mA
i did find first the Vb then loop from ground to ground in Vb to Ve.
so this is my equations Vb-Vbe-Ie(Re)=0
then its 2.63mA
I think you missed the part where he changed the supply voltage from 20V to 22V. If he had not done that, your answer would have been correct.
Thank You
Start watching at 11:50.
Thank you so much :)
good video ty bro
thanks!
Thanks alot
subscribed!
Hi Steve
With all the respect, that circuit of yours does not work, it's nice on paper, but not in the real world. The AC gain will vary like crazy, especially if you have good swing at the load. ie the AC gain depends on re , which also depends on IE , and sense IE varies as the output changes, the AC gain will vary.
wow!!
Thanks for the "wow"...I think
can't hear you at all.....
If that capacitor connected to the emitter is big enough, to Short the emitter to ground, then this circuit is doomed to fail. The dynamic emitter resister re , will keep changing in value, in simple words AC gain =-Rc/re , re =26mv/IE , when Vc swings by 10v pp , you will see magic. Sorry to say that.
sorry can you give me your contact ,i am student from Tanzania??
Lots of info not needed,, just do the formulas and complete then explain.. WOW I went back to the my book..
You set all this up, then did nothing with it.
I was trying to split up the topics into individual videos. This one shows how to convert the circuit into AC a format that you can do AC analysis on. Check out my BJT playlist for more: ua-cam.com/play/PL4651816D92AB6B2B.html for
really poorly taught. skipped alot of stuff like how the dc values were gotten, and how re was gotten too.
you can check the previous videos for DC analysis. there, you can find really good stuff
"Replace caps with shorts"
The double entendre lol