A video showing how to do AC analysis of a common emitter amplifier. AC analysis involves figuring out the voltage gain, the input impedance and the output impedance of the amplifier.
Good video, confirmed a different method I learned for solving these beasts. Highly satisfying when you make your calculations, build it, and your in a close ballpark to actual values.
If anyone is wondering why the output impedance is found by treating Rc as if its in series with the load when its clearly in parallel in the diagram, its because in the simplified amplifier model (the triangle) we model the output as a thevinin equivalent circuit with the output impedance being the thevinin equivalent resistance. Its kind of unexplained in the video but him opening up the current source is just the usual step you take to finding thevinen resistance, which ends up being just Rc. And so when you put together the output as a thevinin equivalent circuit in the final model you have Zout as Rc in series with the load. Very helpful video by the way i just wanted to clear this up because it stumped me for a while
you what's even more treacherous, the fact that a forward baised diode on a voltage divider circuit gives of a .637mV out put that means the whole time Ve is just .637mV, since base to emitter junction is a forward biased.
+David Williams I did not understand the Rin part, why not do Vin/Iin for the 24.4k ohm resistor at the input too but do it for the 20.36k ohm??? what is going on here dude?
@@user-ww2lc1yo9c It's a mistake. The ac input impedance of the transistor is β x re or 150 x 20.36 = 3K. Put that in parallel with the 24K and you'll get sensible answers.
BRO!!! I have learned a lot more from you than from my whole electronics semester at school. I FUCKING LOVE YOU! So clear and you explain every detail... keep it up man, I wish I could take back that tuition money and give it to you LOL
You've got a great voice for tutorials. My circuit professors are Indian, and they always sound like they're shouting and angry when they explain these things.
To those asking where the RE went in the AC analysis, it was shorted by the bypass capacitor. Capacitors are shorted in AC and therefore current passe through the shorted component to avoid the resistance.
Howdy again. I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly. Regards.
The quick answer is that at high enough frequencies a capacitor looks like a short circuit. At low enough frequencies, a capacitor looks like an open circuit. The deeper reason is that a capacitor presents a reactance to the signal which is an opposition to a change in voltage. The reactance of a capacitor is equal to 1/(2*pi*f) where f is the frequency. As the frequency increases, the reactance of the capacitor decreases.
Thank you for your efforts .. If we applied this circuit c emmiter to the radio frequency via a resonant circuit, we know that the signal will be inverted 180 degrees. Can we receive the signal in the reception circuit and be mirrored, or should it be the same as the phase ?
Dear David,I really like this video, as well as your video on the emitter follower circuit. I just had one question on this circuit: Could you please tell me how to determine the peak AC current leaving the amplifier? The reason I am asking is I am trying to combine an emitter follower with a common emitter. I see now how to find the output impedance of the amplifier, and how to get Vout. Would the outward current just be Vout/amplifier output impedance?Thanks,Tim
@David Williams why is the current source replaced by an open ?? If the idea is to obtain the output impedance by using a thevenin equivalent, dependent sources shouldn't be removed if I remember correctly ...
Output swing refers to the maximum and minimum voltage that the output reaches. I have a list of video ideas that I would like to create. I'll add clampers to the list, but don't expect it too soon.
At 4:30 you were able to calcualte VE and VB but not VC why is that? Also, if *R2 > (0.1*B*RE)* how do you calculate? & At the end you calculated Vout but not Vin so is Vin automatically the 10mVp?
The midband AC gain of a common emitter amplifier is equal to about -Rc/(re+RE). When you bypass the emitter resistor (RE) with a capacitor, you are making RE invisible from an AC point of view (i.e you are bypassing it). This makes the gain equal to about -Rc/re
That's only true when Xc (the reactance of the capacitor) is negligible compared to re. For slightly lower frequencies, the ac gain is -Rc/(re + Xc) as long as Xc
Howdy. Absolutely Great. However. The output impedance is 2 x the collector resistor. JohnAudioTech corrected me on that point. And I verified it yesterday. I used T1 = BC109, Rc = 10 k, Re = 1 k, Rb1 = 47 k and Rb2 = 12 k. B+ was 9 V. DC work point was about 4,2 V on the collector. Loading the collector with Rload = 22 k the Uload dropped to 1/sqrt2 of the unloaded value (-3dB). 1 kHz sine wave. With a fixed Rc this is the most energy efficient transfer of power. So. I say. Zout = 2 x Rc. But of course. Designing the stage for a smaller output impedance will increase the load power. Yes. But the stage will consume way more power that what is gained over the load. Regards.
Load impedances for audio systems are usually 4 or 8 ohms, so common emitter amplifiers are not used for driving speakers. Amplifiers are used in many different types of applications that have higher load impedances. Also common emitter amplifiers are single stage amplifiers and are often found as part of a multistage amplifier system.
On the contrary, a pure Class-A amplifier will have a common emitter stage as its output, and many high-quality audio amplifiers use that, although they may use an active collector load to improve the open loop gain. There are plenty of examples on the web.
+Sylwester Kogowski. The beta of the transistor is the current gain while the gain of 193 is the voltage gain of the amplifier circuit that uses the transistor
Nicely done! I would prefer to get out a general-purpose equation for gain in terms of source and load impedances, and the resistors and source voltages being used, but perhaps that’s one of those exercises better left to the viewer. In any case, CE-amplifiers are a lot more demystified as far as I’m concerned.
Sure. If Rs is the source impedance, Rl is the load impedance, Zin is the amplifier input impedance, Zout is the amplifier output impedance, and Av is the amplifier open-loop gain, then the overall gain is given by: Zin/(Zin+Rs) x Av x (Rl/(Rl + Zout). For a well-designed common emitter stage, Av will equal -Rc/Re; Zin will approximately equal the lower base bias resistance, and Zout will equal the collector resistance. Unfortunately, the circuit in the video doesn't fit my definition of "well-designed".
Eleven years later I come across this masterpiece.
Thank you so much
my god, you explained this much more clear than my microelectronics professor. bless your soul.
I want to say this video has made me understand transistors more
I'm glad to hear it
Good video, confirmed a different method I learned for solving these beasts. Highly satisfying when you make your calculations, build it, and your in a close ballpark to actual values.
best series on BJT's possible, i've learned alot and it all makes sense now, thank you very much
Great explanation - the best I've seen on UA-cam.
Thank you so much David, this cleared my doubt for long time and thanks to people who asked the question about Re, since I got the same questions!!
Great video and explanation. Very clear. I wish I’d found your videos weeks ago. Subscribed and bookmarked
I know you made this vidio many years ago, but it helped me understand the effect of impedance today thank you!
Thank you very much ! Your videos helped me tremendously in understanding this topic, and the use of "t-model" is a lot more easier to work with.
If anyone is wondering why the output impedance is found by treating Rc as if its in series with the load when its clearly in parallel in the diagram, its because in the simplified amplifier model (the triangle) we model the output as a thevinin equivalent circuit with the output impedance being the thevinin equivalent resistance. Its kind of unexplained in the video but him opening up the current source is just the usual step you take to finding thevinen resistance, which ends up being just Rc. And so when you put together the output as a thevinin equivalent circuit in the final model you have Zout as Rc in series with the load.
Very helpful video by the way i just wanted to clear this up because it stumped me for a while
Can you illustrate more. And what do you mean by thevinin equivalent circuit ?
you what's even more treacherous, the fact that a forward baised diode on a voltage divider circuit gives of a .637mV out put that means the whole time Ve is just .637mV, since base to emitter junction is a forward biased.
You are going to save me in my christmas exams... great tutorial, thank you!
If only my prof could explain as well as you did! Thanks!
You are very good at explaining electronics. Thank you for posting this video.
Very clean and well done...thank you!
Ancient lawz thanks for the comment. I appreciate the feed back.
+David Williams I did not understand the Rin part, why not do Vin/Iin for the 24.4k ohm resistor at the input too but do it for the 20.36k ohm??? what is going on here dude?
@@user-ww2lc1yo9c It's a mistake. The ac input impedance of the transistor is β x re or 150 x 20.36 = 3K. Put that in parallel with the 24K and you'll get sensible answers.
Great Tutorial. Finally understood how these things work
Brilliantly explained, thank you!
Between your video and alot of back and forth in the book I'm much more confident about this material.
you are a great teacher. these are wonderful tutorials
best explanation, you made everything simple and clear. Thank you so much
This was very helpful. Thank you so much!
Appreciate Dave, diamond clear explaination
This made things so simple! thanks!
These are so helpful! Thank you so much!
Thanks you sir!! Clear explanation with the best audio and visual effect!! More easy to understand than the explanation of my lecturer at institute!!!
There are very few professors who can explain this with such simplicity. And thank god you're on the internet!
true
You are a great teacher
Spot On..! Glad to have Found out this Video. Amazing
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Thank you so much David , i've Quiz Tomorrow and this video helps me a lot .
Very Good Explanation!!! This was an Enjoyable Video..Thanks For The Refresher!!!!( For myself)
Very clear explanation. Thank you!
great help .thanka for this fabulous video ,it clear all my doubts
BRO!!! I have learned a lot more from you than from my whole electronics semester at school. I FUCKING LOVE YOU! So clear and you explain every detail... keep it up man, I wish I could take back that tuition money and give it to you LOL
Man this is the best tutorial i have ever seen . thanks a lot. i have exam after an hour.I knew nothing before watching this :D
haha how did that go?
I studied electeonics but never developed such clarity. Thank you very much indeed.
Hay brother
Thanks for these videos brother! I would be screwed on my exam if I didn't find these.
Dear David Williams,
That's very useful video. Thanks much.
Peace be with you,
Best regards from Türkiye
Thank you..You r clearing the concepts
thank you very much sir williams, your explanations are very helpfull.Every video upload make me a happy student.greetings from Montreal
well done Mr.David !
You've got a great voice for tutorials. My circuit professors are Indian, and they always sound like they're shouting and angry when they explain these things.
Thank you for providing this to everyone :)
Nice videos by the way, keep them coming!!!
Genious, appreciate the explanation
To those asking where the RE went in the AC analysis, it was shorted by the bypass capacitor. Capacitors are shorted in AC and therefore current passe through the shorted component to avoid the resistance.
very good tutorial! thanks so much!
You're the best, thank you.
Superb Tutorial! Very Very Helpful and Useful... Thanx for sharing the Knowledge!
Howdy again.
I was wrong. You and JohnAudioTech are correct. Rc is the output impedance. Then the output signal is ½ the unloaded and most power is transferred. I remembered JAT's reply wrongly.
Regards.
Thank you so much! Subscribed.
thank u my good sir . this was amazing
This is excellent thank you
thank you sir ..well explained!!
nice ..crystal clear explanation..
Well explained..Thanks!
@tunicana thanks for the comments. I love hearing that these videos are useful
The quick answer is that at high enough frequencies a capacitor looks like a short circuit. At low enough frequencies, a capacitor looks like an open circuit.
The deeper reason is that a capacitor presents a reactance to the signal which is an opposition to a change in voltage. The reactance of a capacitor is equal to 1/(2*pi*f) where f is the frequency. As the frequency increases, the reactance of the capacitor decreases.
Thanks bro, i'll be better prepared for my exam now.
Thanks for the compliment!
really useful sir..thank you
thank you so much for the video!
Thank you for your efforts .. If we applied this circuit c emmiter to the radio frequency via a resonant circuit, we know that the signal will be inverted 180 degrees. Can we receive the signal in the reception circuit and be mirrored, or should it be the same as the phase ?
Great video David. What's the correct starting point to calculate your resistor values and pick a suitable transistor?
great explanation, thanks
Thanks for the comment. I appreciate it
one thing I didn't get is, why didn't you take the emitter resistance in series with re in T model ??
I am going to write your name in my diploma man!!! Thank you :)
really helpful video
+David Williams Why did put 600 ohms for the value of the input resistor instead of 600 Kohms in the r model?
Great video! What program did you use to make this video? and what program are you using to draw with?
Thank you for your video. Please how do you find the frequency of the output?
Dear David,I really like this video, as well as your video on the emitter follower circuit. I just had one question on this circuit: Could you please tell me how to determine the peak AC current leaving the amplifier? The reason I am asking is I am trying to combine an emitter follower with a common emitter. I see now how to find the output impedance of the amplifier, and how to get Vout. Would the outward current just be Vout/amplifier output impedance?Thanks,Tim
perfect explaination
very good, thanks
Amazing tutorial...you should do more tutorials
@David Williams SIr? why do we ignore 600 ohms when we calculate? would you describe this please?
@David Williams
why is the current source replaced by an open ?? If the idea is to obtain the output impedance by using a thevenin equivalent, dependent sources shouldn't be removed if I remember correctly ...
Where can I learn more about BJT - common emitter, base and collector circuits?
I use OneNote for drawing and camtasia to capture and edit the screencast.
Does this apply to Self Bias and Fixed Bias which are CE configs? Thanks in advance.
thank you so so so much
Helpful! Thx!
hello do you mind explaining how to bias a voltage divider common emtter configuration used for making an oscillator(With a gain of 1)
Output swing refers to the maximum and minimum voltage that the output reaches. I have a list of video ideas that I would like to create. I'll add clampers to the list, but don't expect it too soon.
At 4:30 you were able to calcualte VE and VB but not VC why is that? Also, if *R2 > (0.1*B*RE)* how do you calculate? & At the end you calculated Vout but not Vin so is Vin automatically the 10mVp?
what circuit schematic software did you build this on?
Thanks for this video! Very informative.
Just want to ask what will happen if there is no resistor 600 ohms.
Thanks a lot!
The midband AC gain of a common emitter amplifier is equal to about -Rc/(re+RE). When you bypass the emitter resistor (RE) with a capacitor, you are making RE invisible from an AC point of view (i.e you are bypassing it). This makes the gain equal to about -Rc/re
That's only true when Xc (the reactance of the capacitor) is negligible compared to re. For slightly lower frequencies, the ac gain is -Rc/(re + Xc) as long as Xc
Hey I have a trouble with my circuit analysis D: my base is being short circuit to the output....
Howdy. Absolutely Great.
However. The output impedance is 2 x the collector resistor. JohnAudioTech corrected me on that point. And I verified it yesterday.
I used T1 = BC109, Rc = 10 k, Re = 1 k, Rb1 = 47 k and Rb2 = 12 k.
B+ was 9 V. DC work point was about 4,2 V on the collector.
Loading the collector with Rload = 22 k the Uload dropped to 1/sqrt2 of the unloaded value (-3dB). 1 kHz sine wave. With a fixed Rc this is the most energy efficient transfer of power.
So. I say. Zout = 2 x Rc.
But of course. Designing the stage for a smaller output impedance will increase the load power. Yes. But the stage will consume way more power that what is gained over the load.
Regards.
Load impedances for audio systems are usually 4 or 8 ohms, so common emitter amplifiers are not used for driving speakers. Amplifiers are used in many different types of applications that have higher load impedances. Also common emitter amplifiers are single stage amplifiers and are often found as part of a multistage amplifier system.
On the contrary, a pure Class-A amplifier will have a common emitter stage as its output, and many high-quality audio amplifiers use that, although they may use an active collector load to improve the open loop gain. There are plenty of examples on the web.
+David Williams
This transistor had beta of 150, but the result is 193 times amplification.
Is that ok?
+Sylwester Kogowski. The beta of the transistor is the current gain while the gain of 193 is the voltage gain of the amplifier circuit that uses the transistor
Nicely done! I would prefer to get out a general-purpose equation for gain in terms of source and load impedances, and the resistors and source voltages being used, but perhaps that’s one of those exercises better left to the viewer. In any case, CE-amplifiers are a lot more demystified as far as I’m concerned.
Sure. If Rs is the source impedance, Rl is the load impedance, Zin is the amplifier input impedance, Zout is the amplifier output impedance, and Av is the amplifier open-loop gain, then the overall gain is given by: Zin/(Zin+Rs) x Av x (Rl/(Rl + Zout).
For a well-designed common emitter stage, Av will equal -Rc/Re; Zin will approximately equal the lower base bias resistance, and Zout will equal the collector resistance. Unfortunately, the circuit in the video doesn't fit my definition of "well-designed".
What if you didn't have β there. How would you go on fidning the gain?
how did you find voltage at the eimmiter
thank you for clearing my (silly)doubt!
very important tutorial
why is this different from the voltage divider bias circuit for dc
thanks bro...................
I believe Zin is not 2.714 kilo ohms but only ohms , i apologise if i am wrong
Video is , aside this detail, very good
Thank you from brussels