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.
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.
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.
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.
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.
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 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
+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?
@@優さん-n7m 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
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.
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.
Why does a low reactance state of the capacitor though cause the signal to completely bypass the resistor? It's the path choice I'm getting confused about. Is it because they are in parallel? Apologies if you feel like you're repeating yourself and thanks for all your help thus far!
This video was somewhat helpful however I think it will fall short for most students studying the concepts. In the future could you make a video that describes the circuit in much more detail. For instance, the effects of the capacitors? What are the poles and zero's of each capacitor. How can one find the gain when taking them into consideration?
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".
The 600 ohm is the output impedance of the source. Without it, you can consider it an ideal source and there will be no voltage drop inside the source regardless of how much current is drawn from it.
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 ?
It allows the ac gain to be greater than the dc gain. The emitter resistor provides stability of the output voltage point against changes in temperature and supply voltage, as well as reducing the effects of variability of transistor parameters. The dc gain is -Rc/Re or about 2.67. The emitter bypass capacitor reduces the total emitter resistance at high frequencies where the reactance of the capacitor (1/2πjfC) is less than the dynamic emitter resistance of the transistor (re = 25mV/Ic). The ac gain is then Rc/re which can be considerably higher, although it will dramatically increase distortion for larger signals.
When finding Rout why do we replace the current source with an open circuit? I was taught that dependent sources should be untouched in such calculations
A current source is equivalent to a high voltage source with a high resistance in series such that the voltage divided by the resistance gives the required current. In other words, current sources are always going to have very high equivalent resistances, so they are effectively an open circuit in ac analysis.
First and foremost let me express my gratitude, this is absolutely crystal clear, many thanks Sir. Keep doing you. I have just one question, is the 26mV that you have divided by (Ic=Ie=DC Current), is that always that or you had calculated it?
The 26mV is the thermal voltage for a transistor = kT/q where k is Boltzmann's constant, q is the charge on an electron, and T is the absolute temperature. If you want to find out more detail, Google for "Shockley Diode Equation". The value clearly depends on temperature, but for normal use, we can take it to be around 25mV to 26mV for all transistors.
The output impedance at the collector of a transistor is very high (unless the transistor is in saturation), so it will always be many times greater than the collector resistor, with which it is in parallel for ac purposes, and we can ignore it.
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
For that final stretch of calculating the input & output voltages, how do you calculate it if there's no AC input resistor in front of the coupling capacitor and no load resistor / capacitor branching off the collector? This is the last bit of knowledge I need to understand this, but I'm having trouble finding the answer!
Whatever source you are connecting at the input will have an output impedance, so the input resistance will always be there. If it's such a low resistance that you can ignore it, then just assume that the full source voltage is applied at the input. As for the output, if there is no load resistor, then there will be no voltage drop across the CE amp's output impedance
your videos are beyond appreciation :).well i had a doubt..what is meant by output swing exactly?n it would be great if u could upload a video on clampers.
Thank you for such a nice video. Sir I have a question. what is the use/advantage of 600 ohm? why not we connect the input signal directly to the amplifier's input because it seems that some amount of input voltage will drop across 600 ohm resistor.
If you don't have the collector resistor, you will have some major problems with your circuit. Remember that from an AC point of view, capacitors are nearly shorts. Also, DC sources should be considered shorts as well. This means that from an AC point of view your output, which is at the collector is shorted to ground
Thank you for this informative video. Regarding the Zout resistance, if Rc (8k) and RL (12k) are in parallel in the ac circuit, shouldn't Zout be calculated as such to determine the correct voltage gain and ac Vout?
The 600 ohm resistor (from the output impedance of the source) and 2714 ohm resistor (the input impedance of the CE amp) are in series with each other, not in parallel. Also the output impedance (8k) of the CE amp and the load resistor (12k) are in series with each other too.
Hi, thank you for the tutorial i really like it. just a question. when you found Av gain which is equal to Vout / Vin. i believe that Vout should be Rc || RL because they are in parallel. am i right ? thank again sir
Thanks for the video. I got a doubt in between which goes..why can't we use a hybrid pi model in the AC equivalent of the transistor instead of the T model you have used? (using a Hie in b/w E and B and current source of Hfe(Ib) btw C and E..here the emitter is grounded due to the bipass. ) PS : can you make any videos on MOS amplifiers?? please, it'd be of a lot of help to guys like me :)
@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 8:43, in the calculation of r_e = 26.3 mV / 1.26 mA = 20.63 Ohms, where does the 26.3 mV voltage value come from? All I see in the initial circuit diagram is the function generator providing a 10 mV AC voltage. Thank you!
Okay, so I looked at your other video "Design a Simple Common Emitter Amplifier" and see someone else had the same question. Evidently, the 26mV value is a constant that represents the "thermal voltage" of a bi-polar junction transistor. Thanks for making these videos!
Mr. Williams, I believe I understand how to calculate most of the values in a common emitter amplifier but there still are a couple of things I'm not sure about. How do you calculate the value of the input and output coupling capacitors and the emitter bypass capacitor? I've seen two different explanations for the input and output capacitors. One says the input Xc and output Xc should match the Zin and Zout of the amplifier. The other says the higher the capacitance the better to assure that the capacitors act as short circuits and don't impede the input and output signal. Is either one of these explanations correct? I've also seen two different explanations for calculating the value of the emitter bypass capacitor. One says the Xc should match RE at the cutoff frequency. The other says Xc should be 1/10 RE at the cutoff frequency. Is either of these correct? The textbook I'm using, Grob Basic Electronics, contains a lot of good information but it doesn't go into much detail about these type of amplifiers. Any information you can provide will be greatly appreciated. Thank you.
You understand that the reactance of a capacitor, Xc depends on frequency? Xc = 1/2πjfC. For audio work, we want Xc to be less than the relevant resistance at a frequency somewhat less than 20Hz, so that's how we calculate each capacitor. The input impedance of the stage is about 2.7K, so we want Xc for the input capacitor to be 2.7K somewhere less than 20Hz. We calculate 1/2π.(20).(2.7K) = 3μF, so we use 3.3μF. The load resistance is 12K, so we want Xc for the output capacitor to be 12K somewhere less than 20Hz. We calculate 1/2π.(20).(12K) = 0.67μF, so we use 1μF. The dynamic emitter resistance of the transistor is about 20R, so we want Xc for the bypass capacitor to be 20R somewhere less than 20Hz. We calculate 1/2π.(20).(20) = 400μF, so we use 470μF. The answer to your question about the emitter bypass capacitor is that neither of the explanations you've been given is correct. The ac gain of the circuit is given by the ratio of the collector resistance to the total emitter resistance. If you think about it, that means you must make the bypass capacitor large enough that its reactance is negligible compared to the dynamic emitter resistor that it is series with. Otherwise the ac gain would be Rc/(re + Xc) and that will depend on frequency until Xc
Hello David ...great great video as usual..... when you reached to the part about finding the gain , input and output impedance , i got lost since i am missing basic theory about this matter . can you guide me to a link/links that i could learn the basic theory about how to calculate the gain , input and output impedance of any circuit ....thx alot
+Igor Mândru the 26mV is called the thermal voltage. It is constant when temperature is constant, but can be calculated as Vt = k⋅T ⁄ q. Where k is the Boltzmann constant, q is charge on an electron and T is temperature in Kelvins.
thanks for your time and for your response, if I got it right you compare if the voltage on the base is lower than the tenth of the voltage on the emitter of the transistor like this: 1 x R2 < 0.1 x beta x Re so 0.1 x beta is the current at the Emitter after amplification times Re gives the Voltage at the Emitter 1 x R2 is the current passes at R2 that gives the voltage at the base of that transistor
You're on the right track, but you don't need to compare voltages, you only need to compare resistances (i.e., the resistance seen looking in to the base which is (beta x Re) vs. the resistance of R2 (30k resistor)). You want to make sure the resistance of R2 is 1/10th (or less) of the resistance looking in to the base so that 90% or more of the current goes through R2
sure thing, we always seek calculations shortcuts, as far as I remember from my practical electronic knowledge, for me, this is a new kind of DC approximation technique (-: thanks a bunch, keep producing your concise lectures
just to confirm, in AC theory if you didn't connect a bypass cap on the emitter then i take it the re value calculated includes the resistor value 3k ohm connected to ground even though it would give a really low gain? just messing round with calculations
In calculating Z_inQ you used the normal relationship that I_E = (beta+1)*I_B. However, didn't we calculate the Q point with the assumption that I_B=0, thus making the input impedance infinite?
I did assume I_B=0, but there are two things about that assumption to point out. First, that was an approximation to make the Q-point calculation easier. Second that is for DC current while the Z_inQ that I calculate is for an AC signal
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.
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.
There are very few professors who can explain this with such simplicity. And thank god you're on the internet!
true
I know you made this vidio many years ago, but it helped me understand the effect of impedance today thank you!
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.
I studied electeonics but never developed such clarity. Thank you very much indeed.
Hay brother
Between your video and alot of back and forth in the book I'm much more confident about this material.
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.
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
I want to say this video has made me understand transistors more
I'm glad to hear it
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.
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?
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 you sir!! Clear explanation with the best audio and visual effect!! More easy to understand than the explanation of my lecturer at institute!!!
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 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
You are going to save me in my christmas exams... great tutorial, thank you!
best explanation, you made everything simple and clear. Thank you so much
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?
@@優さん-n7m 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.
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.
Great explanation - the best I've seen on TH-cam.
thank you very much sir williams, your explanations are very helpfull.Every video upload make me a happy student.greetings from Montreal
Great video and explanation. Very clear. I wish I’d found your videos weeks ago. Subscribed and bookmarked
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
Dear David Williams,
That's very useful video. Thanks much.
Peace be with you,
Best regards from Türkiye
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!!
you are a great teacher. these are wonderful tutorials
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.
Thank you for your video. Please how do you find the frequency of the output?
I think it might be open because ib is zero since there is no voltage source on the left when considering output impedance.
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.
Why does a low reactance state of the capacitor though cause the signal to completely bypass the resistor? It's the path choice I'm getting confused about. Is it because they are in parallel? Apologies if you feel like you're repeating yourself and thanks for all your help thus far!
What is the value of R in parallel with 0 ohms? The cap is assumed to be a short circuit at AC bypassing the R.
Great Tutorial. Finally understood how these things work
Thanks for these videos brother! I would be screwed on my exam if I didn't find these.
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
@tunicana thanks for the comments. I love hearing that these videos are useful
This video was somewhat helpful however I think it will fall short for most students studying the concepts. In the future could you make a video that describes the circuit in much more detail. For instance, the effects of the capacitors? What are the poles and zero's of each capacitor. How can one find the gain when taking them into consideration?
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".
Thank you so much David , i've Quiz Tomorrow and this video helps me a lot .
The 600 ohm is the output impedance of the source. Without it, you can consider it an ideal source and there will be no voltage drop inside the source regardless of how much current is drawn from it.
Hello sir
well done Mr.David !
Great video David. What's the correct starting point to calculate your resistor values and pick a suitable transistor?
Superb Tutorial! Very Very Helpful and Useful... Thanx for sharing the Knowledge!
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 ?
David, absolutely love your videos but can you explain the exact purpose of the bypass capacitor in parallel with emitter resistor?
It allows the ac gain to be greater than the dc gain.
The emitter resistor provides stability of the output voltage point against changes in temperature and supply voltage, as well as reducing the effects of variability of transistor parameters. The dc gain is -Rc/Re or about 2.67.
The emitter bypass capacitor reduces the total emitter resistance at high frequencies where the reactance of the capacitor (1/2πjfC) is less than the dynamic emitter resistance of the transistor (re = 25mV/Ic). The ac gain is then Rc/re which can be considerably higher, although it will dramatically increase distortion for larger signals.
When finding Rout why do we replace the current source with an open circuit? I was taught that dependent sources should be untouched in such calculations
A current source is equivalent to a high voltage source with a high resistance in series such that the voltage divided by the resistance gives the required current. In other words, current sources are always going to have very high equivalent resistances, so they are effectively an open circuit in ac analysis.
Thanks bro, i'll be better prepared for my exam now.
This made things so simple! thanks!
Thanks for this video! Very informative.
Just want to ask what will happen if there is no resistor 600 ohms.
one thing I didn't get is, why didn't you take the emitter resistance in series with re in T model ??
Appreciate Dave, diamond clear explaination
First and foremost let me express my gratitude, this is absolutely crystal clear, many thanks Sir. Keep doing you.
I have just one question, is the 26mV that you have divided by (Ic=Ie=DC Current), is that always that or you had calculated it?
The 26mV is the thermal voltage for a transistor = kT/q where k is Boltzmann's constant, q is the charge on an electron, and T is the absolute temperature. If you want to find out more detail, Google for "Shockley Diode Equation".
The value clearly depends on temperature, but for normal use, we can take it to be around 25mV to 26mV for all transistors.
I am going to write your name in my diploma man!!! Thank you :)
Davidica - Master Of Circuits
Very helpful and clear but what if you have the ro resistor as part of the model for the transistor? How can we analyze this?
The output impedance at the collector of a transistor is very high (unless the transistor is in saturation), so it will always be many times greater than the collector resistor, with which it is in parallel for ac purposes, and we can ignore it.
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
For that final stretch of calculating the input & output voltages, how do you calculate it if there's no AC input resistor in front of the coupling capacitor and no load resistor / capacitor branching off the collector? This is the last bit of knowledge I need to understand this, but I'm having trouble finding the answer!
Whatever source you are connecting at the input will have an output impedance, so the input resistance will always be there. If it's such a low resistance that you can ignore it, then just assume that the full source voltage is applied at the input. As for the output, if there is no load resistor, then there will be no voltage drop across the CE amp's output impedance
These are so helpful! Thank you so much!
Where can I learn more about BJT - common emitter, base and collector circuits?
your videos are beyond appreciation :).well i had a doubt..what is meant by output swing exactly?n it would be great if u could upload a video on clampers.
Thank you..You r clearing the concepts
Thank you for such a nice video. Sir I have a question. what is the use/advantage of 600 ohm? why not we connect the input signal directly to the amplifier's input because it seems that some amount of input voltage will drop across 600 ohm resistor.
600 ohm resistor represents the internal resistance of the source. Nothing you can do about it.
If you don't have the collector resistor, you will have some major problems with your circuit. Remember that from an AC point of view, capacitors are nearly shorts. Also, DC sources should be considered shorts as well. This means that from an AC point of view your output, which is at the collector is shorted to ground
Hello, can you give an explanation about the input and output impedances? What are those and when or why should someone take them in consideration?
great help .thanka for this fabulous video ,it clear all my doubts
Thank you for this informative video. Regarding the Zout resistance, if Rc (8k) and RL (12k) are in parallel in the ac circuit, shouldn't Zout be calculated as such to determine the correct voltage gain and ac Vout?
Disregard my comment...just did the math using my method, and I ended up with the same answer. Sheeeeeet.
@@RHCP08 ghnjhkıjoılö
The 600 ohm resistor (from the output impedance of the source) and 2714 ohm resistor (the input impedance of the CE amp) are in series with each other, not in parallel.
Also the output impedance (8k) of the CE amp and the load resistor (12k) are in series with each other too.
Very Good Explanation!!! This was an Enjoyable Video..Thanks For The Refresher!!!!( For myself)
Hi, thank you for the tutorial i really like it.
just a question. when you found Av gain which is equal to Vout / Vin. i believe that Vout should be Rc || RL because they are in parallel. am i right ?
thank again sir
Thanks for the video. I got a doubt in between which goes..why can't we use a hybrid pi model in the AC equivalent of the transistor instead of the T model you have used? (using a Hie in b/w E and B and current source of Hfe(Ib) btw C and E..here the emitter is grounded due to the bipass. )
PS : can you make any videos on MOS amplifiers?? please, it'd be of a lot of help to guys like me :)
@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 ...
+David Williams Why did put 600 ohms for the value of the input resistor instead of 600 Kohms in the r model?
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.
This was very helpful. Thank you so much!
At 8:43, in the calculation of r_e = 26.3 mV / 1.26 mA = 20.63 Ohms, where does the 26.3 mV voltage value come from? All I see in the initial circuit diagram is the function generator providing a 10 mV AC voltage. Thank you!
Okay, so I looked at your other video "Design a Simple Common Emitter Amplifier" and see someone else had the same question. Evidently, the 26mV value is a constant that represents the "thermal voltage" of a bi-polar junction transistor. Thanks for making these videos!
Great video! What program did you use to make this video? and what program are you using to draw with?
Brilliantly explained, thank you!
really useful sir..thank you
Very clear explanation. Thank you!
Spot On..! Glad to have Found out this Video. Amazing
fortnite
Amazing tutorial...you should do more tutorials
nice ..crystal clear explanation..
Mr. Williams,
I believe I understand how to calculate most of the values in a common emitter amplifier but there still are a couple of things I'm not sure about. How do you calculate the value of the input and output coupling capacitors and the emitter bypass capacitor?
I've seen two different explanations for the input and output capacitors. One says the input Xc and output Xc should match the Zin and Zout of the amplifier. The other says the higher the capacitance the better to assure that the capacitors act as short circuits and don't impede the input and output signal. Is either one of these explanations correct?
I've also seen two different explanations for calculating the value of the emitter bypass capacitor. One says the Xc should match RE at the cutoff frequency. The other says Xc should be 1/10 RE at the cutoff frequency. Is either of these correct?
The textbook I'm using, Grob Basic Electronics, contains a lot of good information but it doesn't go into much detail about these type of amplifiers. Any information you can provide will be greatly appreciated. Thank you.
You understand that the reactance of a capacitor, Xc depends on frequency? Xc = 1/2πjfC. For audio work, we want Xc to be less than the relevant resistance at a frequency somewhat less than 20Hz, so that's how we calculate each capacitor.
The input impedance of the stage is about 2.7K, so we want Xc for the input capacitor to be 2.7K somewhere less than 20Hz. We calculate 1/2π.(20).(2.7K) = 3μF, so we use 3.3μF.
The load resistance is 12K, so we want Xc for the output capacitor to be 12K somewhere less than 20Hz. We calculate 1/2π.(20).(12K) = 0.67μF, so we use 1μF.
The dynamic emitter resistance of the transistor is about 20R, so we want Xc for the bypass capacitor to be 20R somewhere less than 20Hz. We calculate 1/2π.(20).(20) = 400μF, so we use 470μF.
The answer to your question about the emitter bypass capacitor is that neither of the explanations you've been given is correct. The ac gain of the circuit is given by the ratio of the collector resistance to the total emitter resistance. If you think about it, that means you must make the bypass capacitor large enough that its reactance is negligible compared to the dynamic emitter resistor that it is series with. Otherwise the ac gain would be Rc/(re + Xc) and that will depend on frequency until Xc
Hello David ...great great video as usual.....
when you reached to the part about finding the gain , input and output impedance , i got lost since i am missing basic theory about this matter . can you guide me to a link/links that i could learn the basic theory about how to calculate the gain , input and output impedance of any circuit ....thx alot
Thanks for the compliment!
Genious, appreciate the explanation
I use OneNote for drawing and camtasia to capture and edit the screencast.
Nice video...just one question..8:50...when you calculate re=26mV/Ic......26mV is a constant number or it depends on something? ....thanks
+Igor Mândru the 26mV is called the thermal voltage. It is constant when temperature is constant, but can be calculated as Vt = k⋅T ⁄ q. Where k is the Boltzmann constant, q is charge on an electron and T is temperature in Kelvins.
+David Williams thankyou this information i loss ... good Luke
really helpful video
hi, this is a very good tutorial yet I could not understand your approximation at 2:27 with voltage divider where you said:
R2
The statement "R2
thanks for your time and for your response, if I got it right you compare if the voltage on the base is lower than the tenth of the voltage on the emitter of the transistor like this: 1 x R2 < 0.1 x beta x Re
so 0.1 x beta is the current at the Emitter after amplification times Re gives the Voltage at the Emitter
1 x R2 is the current passes at R2 that gives the voltage at the base of that transistor
You're on the right track, but you don't need to compare voltages, you only need to compare resistances (i.e., the resistance seen looking in to the base which is (beta x Re) vs. the resistance of R2 (30k resistor)). You want to make sure the resistance of R2 is 1/10th (or less) of the resistance looking in to the base so that 90% or more of the current goes through R2
sure thing, we always seek calculations shortcuts, as far as I remember from my practical electronic knowledge, for me, this is a new kind of DC approximation technique (-:
thanks a bunch, keep producing your concise lectures
just to confirm, in AC theory if you didn't connect a bypass cap on the emitter then i take it the re value calculated includes the resistor value 3k ohm connected to ground even though it would give a really low gain? just messing round with calculations
Hi! Thanks for the video! I just have a question. Why is V_E = V_B - 0.7?
V_E = V_B - 0.7 because the junction from base to emitter is a diode, and I'm assuming the voltage drop across the diode is 0.7V
sir thanks so much...so the RS doesnt contribute in Zin?
Yes, correct. Rs usually designates the output impedance of the source
You are a great teacher
In calculating Z_inQ you used the normal relationship that I_E = (beta+1)*I_B. However, didn't we calculate the Q point with the assumption that I_B=0, thus making the input impedance infinite?
I did assume I_B=0, but there are two things about that assumption to point out. First, that was an approximation to make the Q-point calculation easier. Second that is for DC current while the Z_inQ that I calculate is for an AC signal
Can you add a video without the values of the resistors and Vcc?
I wanted to learn the derivation only
very good tutorial! thanks so much!