As a child, I could never do maths, as a geriatric, it takes me a while to absorb anything new but I followed this and understood the main principles straight away. If I view it a few times, I will nail it... and it's down to how you explained and taught this! Thank-you! I wish you had been my school maths/physics teacher. I actually enjoyed doing this!
Dude, you're reading my mind. Love the cumulative progression through each of these concepts. I've been trying to understand the schematics of microphone preamps I own and I'm finally getting it. Thanks for your efforts here.
You are an excellent teacher. You have a gift of patience of being able to explain the very details involved in analyzing these circuits, and keeping our interest in following along.
when i have time I watch your video's over and over, always learning more each time!! I thought it would be cool to trying and build a amplifier for an ipod music player putting together the various stages to drive an 8 ohm speaker.
It's the intrinsic emitter resistance of the transistor. As the current through the emitter changes, there is a very small change in the voltage between the base and collector, the change in voltage divided by the change in current is effectively the dynamic or intrinsic resistance inside the transistor's emitter. You can look up "Shockley's diode formula" to get a more in-depth treatment, but one result is that this resistance is approximately equal to 25mV divided by the emitter (or collector) current at room temperature. So with a collector current of 2.5mA, the transistor behaves as if it had a 25mV/2.5mA = 10 ohm resistance in its emitter.
Question: I was wondering, WHY: You never talked about how to figure out by calculations or rules of thumb about how those COUPLING and BYPASS 'CAPACITORS' are 'selected'.
When used 47uF or 100uF for the capacitor the output wave form is clipped at the top. Whats if the fix for this ? Where the values of capacitor u used different or are they all equal ?
Hi what value of risistor would you recomend for RB1and RB3 for useing the circuit with a nine volt battery as 56k seems not to work with nine volts as the risistance is to high thanks 🤔 great vidio im learning a lot great bit on voltage testing different inputs and output voltages that the circuit should have at each stage i think you learn more if you can test voltages that the circuit uses st what point great work keep it up any help with this would be great cheers 👍
So if say a transistor has a continuous collector current of 700mA would that be the maximum current that can flow in the transistor when it is in saturation?
For CE stages shouldn't the V gains have been -11.31 and -10.58? Obviously you still come up with the same positive number (in phase with input) once you multiply for the total "composite" gain, but I feel it's important to note the 180° phase shift of both stages.
Hi, great explanation, but I got lost at 9.00. 1. How did you get to the formula for Av2. I thought that the formula for gain had to include Alpha (Av = Alpha * RL/Rin). Or did you assume that Ic and Ie are almost equal and so Alpha is 1? 2. What is rej? Where did the value of 25mV came from? Thank you in advance. Great tutorials.
Because Beta is at least 100 in this transistor, alpha has to be .99 at a minimum so I make IC and IE equal. The 25mV is a given for a transistor at room temperature using Boltzmann's constant.
If I only do 1 stage to my transistor amplifier is the sound will be as loud as a two stage or more transistor amplifier? It's a dumb question but I just want to prove it.
so if i make a three stage amplifier the first stage is for v gain and 2 to 3rd stage are all current gain right? and where is the input and output connected? thanks :)
Driving speakers at correct levels is a complicated issue with chapters and books being written on the topic. A small, two stage amplifier is a good compromise. The first stage can be a common emitter, input to the base and output on the collector. The second stage could be a common collector with its input (the bases) taken from the collector of the previous circuit. This will give you a voltage gain on stage 1 (CE) and a current gain on stage 2 (CC). Small signal transistors will work for both stages. Multiple common emitters can give you more voltage gain. If you need some serious current on the final stage, use with a TO-3 power transistor.
I understand that it's not the Earth ground, just wondering if you're talking about electron current flow, or was that to show the "negative" cycle of the AC signal?
In this circuit, the emitter resistor is split. The bypass capacitor (around 100μF) is connected in parallel with the lower half of the emitter resistor and so only bypasses that part. The remaining 120R resistor still presents a resistance to the ac signal and therefore must be part of the gain calculation.
When we work with our circuits, we use a resistor in series with the whole circuit that sits before the first capacitor (they call it Rg). Does this effect any of the equations other than just adding that resistor in series to the parallel combination of resistors for Zin?
I had to stop at 4:30. The 1.3 v is not the voltage at the based but the Thevenin voltage that should be in series with the Thevenin resistance. The calculations might need revision.
Hello - I make all of my own drawings. I'll see if I can dig up the original but unfortunately, some got lost in transfer and this may have been one of them. Thanks for watching.
Yes. The gain equation should be Rc / (Re + re), so setting Re=0 leaves Rc/re which is the maximum possible gain of the stage. It's also a nonlinear gain since re varies with the instantaneous collector current. That produces very unpleasant distortion as the output signal gets larger.
Hello Farhad, I am not sure exactly how to answer without a little more information. The input voltage should have no effect on gain at frequencies of a few KHz. Gain will decrease as the frequency goes up because of things like the Miller effect. As long as the voltage doesn't saturate the output of go to high in frequency (check the hfe on the data sheet) you should be okay.
That would work if the base-emitter voltage of the transistor also scaled down, but it doesn't. However, it's not a bad starting point to experiment with. You'll find that Rb2 doesn't need to change much, but Rb1 now has to drop about 3V less, so needs to be smaller. I would recommend aiming to keep the collector current around 2.5mA and use Rb1 = 47K; Rb2 = 6.8K; Rc = 1.8K and Re = 100R + 100R which would give roughly the same performance.
I have using 2 channel amplifier.i would like to very punchy and depth smooth bass.but in my amplifier Voice only loudly Come out please give me any solution.
Hello, Without knowing more about your circuit, I can't really suggest much. Are there caps in series with the output, inductors in parallel? There are to many variables for me to give you answer.
+ The Offset Volt To set something straight for me; The single transistor amp, with a single rail supply should by rights only swing between 0v to lets say 12v rail when the emitter voltage is at roughly 6 volts in that simple configuration of amp, (and through a DC decoupling cap of course). The load, e.g. the coil/cone of a loudspeaker should then move ONLY from 0v to 12v, and back to 0v, (if you measure it load side and when a signal is present), yes? To get a full 'swing' from the speaker coil/cone where the output also becomes negative (-12v), and an so an inward coil/cone movement you must need a split rail supply i.e. -12v, 0v, and +12v, (with a decoupling cap), then the coil/cone movement would follow 0v, +12v, 0v, -12v, 0v, +12v when a signal is present, as in the case of a multi transistor npn/ pnp push pull amp. Correct?So the -ve cycle of the cone/coil is missing in the first case common emitter configuration?
@@mcwolfus4718 If you have a single 12V supply, then you bias your output stage to be at +6V when there is no signal. Then you use a large capacitor between the output and your speaker to block the dc offset of +6V, but allow ac signals to pass through. So without a signal, the loudspeaker has 0V at the end connected to the capacitor and the other end is grounded (also 0V). The amplifier can then pas a maximum signal between 0V and 12V. That is an ac signal of ±6V around the +6V bias point. The ac signal of ±6V will pass through the capacitor driving the speaker to +6V and -6V relative to ground. That's how you get negative voltages onto your speaker with a single rail power supply.
Adjusting the emitter resistors would be easiest but then the Q points change as well. You could use two resistors on the emitter that equal the same value as the original and put a capacitor between them to set the gain. Kinda complicated to explain here. Hope it helps a little!
@@TheOffsetVolt If you keep the sum of the two emitter resistors constant, you won't change the bias points. I'd simply change the first stage emitter resistors to Rb1=100R; Rb2=140R (=100R+39R) and call that close enough. 🙂
hello sir can i know what is the exact value of the capacitor ? i would highly appreciate it if u could provide me the value of the capacitor for my simulation test
I don't know why I never put the values down for those ??? I don't remember the exact values but 10uF electrolytics are more than enough. Thanks for watching!
Hello, The current from the collector of the second transistor would go through RC2 and the load. Since the current split into two paths at the point where RC2 and the load connect, these are parallel to one another.
The Offset Volt thank you for replying, it's the way that you've written it for the calculations, I've not seen the || used before and I was baffled as to how you got your answer. Forgive me for sounding a bit dense
Not a problem. I do the videos for people new to the field and I always forget two or three points that I should include after the video is published. Your question was one of those dis-remembered points. Thanks for watching.
I believe I was using 10uF although you will get pretty good results with 2.2uF as well. The idea is to have an Xc that is, at it's maximum, 1/10th of the resistance you are bypassing.
Thank you so much, btw can you share with me a tested and working circuit diagram of a common emitter - common collector two stage amplifier if you have any. I am working on a project at college and struggling with it. Yours was greatly helpful.
The emitter bypass capacitors need to have a reactance less than Re1+re' (i.e. about 130ohms) at the lowest frequency of interest. For audio, we usually work down to 20Hz. So the emitter bypass capacitors should be more than 1/2πfR = 1 / (2π x 20 x 130) = 60μF. Use 100μF. The input and interstage coupling capacitors need a reactance less than about 6K (the input impedance of each stage), so you can use 2.2μF. That will be okay for the output cap as well.
At frequencies where the capacitors have negligible impedance, the input impedance is the parallel combination of Rb1, Rb2, and β x (Re1 + re'), which is 56K || 6.8K || 90 x (120+10) in the worst case. That's about 6K.
I have been trying to implement a bifet amplifier but with a load resistance of 10 ohms and Vcc as 5V to get a gain of 20 V/V but it seems not to work. im working with LTSpice. i keep thinking probably my load resistance is too low or the Vcc. What could be done please. Ineed help
Hello, Without knowing more about the circuit I can only offer some general ideas. A 10 ohm load on a common emitter stage, if built using the circuit values in the video will not give a gain at all. Av=(Rc||RL)/RE. What is the voltage in and the desired voltage out? Remember Vout can only go between Vcc and ground. With a 5V supply, the biggest AC voltage out, if perfectly centered on 2.5V DC is 5Vpp.
Your supply voltage is too low for any comfortable design. When working with FET input stages, you have to account for the large range of important parameters. Discrete JFETs and MOSFETs really need 15V supplies to allow you to compensate for gate turn-on or turn-off voltages of several volts and still get a usable voltage gain. If you're using a single-ended FET first stage, you'll want a split source resistor with most of it bypassed for ac with a capacitor to increase the ac gain if you want to get x20 out of it. Then your problem will be that you need considerable current gain from the bipolar output stage and you may need a Darlington in common collector to achieve that with a 10R load.
It's the intrinsic emitter resistance of the transistor. As the current through the emitter changes, there is a very small change in the voltage between the base and collector. The change in voltage divided by the change in current is effectively the dynamic or intrinsic resistance inside the transistor's emitter. You can look up "Shockley's diode formula" to get a more in-depth treatment, but one result is that this resistance is approximately equal to 25mV divided by the emitter (or collector) current at room temperature. So with a collector current of 2.5mA, the transistor behaves as if it had a 25mV/2.5mA = 10 ohm resistance in its emitter.
Hello Hena, There is a callout in the video that corrects the calculation. The resistance of Re was 240 ohms so the current is 2.5mA, Thanks for watching.
Hello, A good source would be Electronic Devices (Electron Flow Version) Hardcover: 976 pages Publisher: Pearson; 9 edition (February 18, 2011) Language: English ISBN-10: 0132549859 ISBN-13: 978-0132549851 Good luck in your classes.
As a child, I could never do maths, as a geriatric, it takes me a while to absorb anything new but I followed this and understood the main principles straight away. If I view it a few times, I will nail it... and it's down to how you explained and taught this!
Thank-you! I wish you had been my school maths/physics teacher. I actually enjoyed doing this!
Dude, you're reading my mind. Love the cumulative progression through each of these concepts. I've been trying to understand the schematics of microphone preamps I own and I'm finally getting it. Thanks for your efforts here.
He's not your dude bruhhh stfu
You are an excellent teacher. You have a gift of patience of being able to explain the very details involved in analyzing these circuits, and keeping our interest in following along.
Wow, thank you and thanks for watching!
Could not figure out how to do this when explained in my amplifiers class, this came in clutch, Thank you
What a gem of a channel
Thanks for sharing these tips. even it was already 4 years ago, still helping students a lot. Appreciate it very much
Your teaching style is great! More videos please
Thank you! Will do!
Man where you was before 😢. I just found you on the last hour before the exam.
Hello again, It was to show the movement of the electrons.
You solidified my knowledge for my exam tomorrow. Thank you
Good luck on that exam!!!
Can you explain (at5.05) why you ignored Re2 from the emitter current calculations?Is it not 0.6/240 =2.5 ma?
your video gave me enough explanation to handle my mini project, thanks
when i have time I watch your video's over and over, always learning more each time!! I thought it would be cool to trying and build a amplifier for an ipod music player putting together the various stages to drive an 8 ohm speaker.
GOOD PRESENTATION!
What is rej ? Amazing videos !!!!
I would like to know that,too!
Great video and explantion except rej. What that is and where it comes from remains a mystery.
Please give some hint.
It's the intrinsic emitter resistance of the transistor. As the current through the emitter changes, there is a very small change in the voltage between the base and collector, the change in voltage divided by the change in current is effectively the dynamic or intrinsic resistance inside the transistor's emitter. You can look up "Shockley's diode formula" to get a more in-depth treatment, but one result is that this resistance is approximately equal to 25mV divided by the emitter (or collector) current at room temperature.
So with a collector current of 2.5mA, the transistor behaves as if it had a 25mV/2.5mA = 10 ohm resistance in its emitter.
Question: I was wondering, WHY: You never talked about how to figure out by calculations or rules of thumb about how those COUPLING and BYPASS 'CAPACITORS' are 'selected'.
Should Ie be lower? You divided by 120 instead of 240?
He wrote 120, but the did the calculation using 240, so the answer is correct.
When used 47uF or 100uF for the capacitor the output wave form is clipped at the top. Whats if the fix for this ? Where the values of capacitor u used different or are they all equal ?
How to calculate the minimum alternating signal voltage that can be applied to transistor after setting DC biasing voltage?
Should try changing the frequency and observing effect of the capacitors.
Hi what value of risistor would you recomend for RB1and RB3 for useing the circuit with a nine volt battery as 56k seems not to work with nine volts as the risistance is to high thanks 🤔 great vidio im learning a lot great bit on voltage testing different inputs and output voltages that the circuit should have at each stage i think you learn more if you can test voltages that the circuit uses st what point great work keep it up any help with this would be great cheers 👍
So if say a transistor has a continuous collector current of 700mA would that be the maximum current that can flow in the transistor when it is in saturation?
For CE stages shouldn't the V gains have been -11.31 and -10.58? Obviously you still come up with the same positive number (in phase with input) once you multiply for the total "composite" gain, but I feel it's important to note the 180° phase shift of both stages.
Yup
Hi, great explanation, but I got lost at 9.00.
1. How did you get to the formula for Av2. I thought that the formula for gain had to include Alpha (Av = Alpha * RL/Rin). Or did you assume that Ic and Ie are almost equal and so Alpha is 1?
2. What is rej? Where did the value of 25mV came from?
Thank you in advance. Great tutorials.
Because Beta is at least 100 in this transistor, alpha has to be .99 at a minimum so I make IC and IE equal. The 25mV is a given for a transistor at room temperature using Boltzmann's constant.
Much appreciated.
You have polarized capacitor at c1 I thought it would destroy a polarized capacitor?
so this amplifier on first transistor amplifi the negative and then on second transistor positive ? we will have 10x increase on both ?
whats the values of the capacitors ?
Thanks for your time. I learnt a lot ...
Would the same formulas be used under different amplifier configurations (common base, common collector, etc.)?
Hello, I am not sure to which formulas you would be referring. Gain calculations would be different for each species.
If I only do 1 stage to my transistor amplifier is the sound will be as loud as a two stage or more transistor amplifier? It's a dumb question but I just want to prove it.
Same question here sir.
so if i make a three stage amplifier the first stage is for v gain and 2 to 3rd stage are all current gain right? and where is the input and output connected? thanks :)
Driving speakers at correct levels is a complicated issue with chapters and books being written on the topic. A small, two stage amplifier is a good compromise. The first stage can be a common emitter, input to the base and output on the collector. The second stage could be a common collector with its input (the bases) taken from the collector of the previous circuit. This will give you a voltage gain on stage 1 (CE) and a current gain on stage 2 (CC). Small signal transistors will work for both stages. Multiple common emitters can give you more voltage gain. If you need some serious current on the final stage, use with a TO-3 power transistor.
Hi, if we were to connect a speaker, say an 8 ohm speaker where would we connect it. Is it where we have the RL of 5.6k ?, If yes how?
you are good teacher thank you!
Its very good video which include theory as well as practical .......thanks
Do this also apply to FET common source configuration?
Excellent video Thank you for your time and effort!!! I can finally finish my project....
Hey sir, did you finish your project? I'm just wondering.
He did not draw the hybrid-pi model for the AC analysis. One star :)
Can you do an example of multistage amplifiers with FET configurations? Thank you very much! Very helpful tutorial.
Thank you - I'll see what I can do.
How much capacitor you used?
Please explain what you said at 9 minutes in. Why does the AC current flow backwards from ground, into the emitter, and out the collector of T2?
I understand that it's not the Earth ground, just wondering if you're talking about electron current flow, or was that to show the "negative" cycle of the AC signal?
Great video man!
I would kindly like some clarification I thought if you have a capacitor at Re then you ignore Re when calculating the voltage gain
In this circuit, the emitter resistor is split. The bypass capacitor (around 100μF) is connected in parallel with the lower half of the emitter resistor and so only bypasses that part. The remaining 120R resistor still presents a resistance to the ac signal and therefore must be part of the gain calculation.
At 17:30 i think the equa. Of ic should contain VC instead of VCE because in the re equivelant circuit Vout equal to VC not VCE
Book name please?
When we work with our circuits, we use a resistor in series with the whole circuit that sits before the first capacitor (they call it Rg). Does this effect any of the equations other than just adding that resistor in series to the parallel combination of resistors for Zin?
Hello and thanks for watching. It shouldn't be an issue. It's probably done to match the test setup to the real world implementation.
you saved my life thank you very much ❤❤
First time my video saved a life. Glad it was of help.
How he is saved your life??? Tell about that?
Great video!
Glad you enjoyed it
Do you have a pdf document for your videos ? If you have could you upload a link to it please
Unfortunately, no. It is something I should consider doing though. Thanks for watching.
Does this work as it is on ltspice or the must be any editing??
By the way great vid and great explanation
Thank you
Hi, It should work, it's basic components.
I had to stop at 4:30. The 1.3 v is not the voltage at the based but the Thevenin voltage that should be in series with the Thevenin resistance. The calculations might need revision.
Where do you get the Schematic for this, I wana practice and print some of these off.
Hello - I make all of my own drawings. I'll see if I can dig up the original but unfortunately, some got lost in transfer and this may have been one of them. Thanks for watching.
If R_E1 was a short, how would that change the gain A_V1? Would I just divide by r_ej in the A_V1 equation?
Thank You Very Much
Yes. The gain equation should be Rc / (Re + re), so setting Re=0 leaves Rc/re which is the maximum possible gain of the stage. It's also a nonlinear gain since re varies with the instantaneous collector current. That produces very unpleasant distortion as the output signal gets larger.
Sir, how can we calculate the effect of input voltage on Voltage gain? Please guide me.. Is there any use of input voltage in formulas?
Hello Farhad,
I am not sure exactly how to answer without a little more information. The input voltage should have no effect on gain at frequencies of a few KHz. Gain will decrease as the frequency goes up because of things like the Miller effect. As long as the voltage doesn't saturate the output of go to high in frequency (check the hfe on the data sheet) you should be okay.
great vid
Excellent! thanks a lot!
Glad you liked it!
what if want the voltage source to be 9 volts rather than 12 should i just take percentages of every resistor or it's more complex?
That would work if the base-emitter voltage of the transistor also scaled down, but it doesn't. However, it's not a bad starting point to experiment with.
You'll find that Rb2 doesn't need to change much, but Rb1 now has to drop about 3V less, so needs to be smaller.
I would recommend aiming to keep the collector current around 2.5mA and use Rb1 = 47K; Rb2 = 6.8K; Rc = 1.8K and Re = 100R + 100R which would give roughly the same performance.
what would be Av2 if there where no Rc and RL on the second stage BJT, would it be a case that there is no loading effect?
Why does it bother you bro?
Why does the equation for Rin((base) have Beta instead of Beta+ 1? Doesn't ie=ib(Beta +1)
Your numbers are correct. When we rearrange to find beta we get,
B=(ie/ib)-1
It is such a small variation that it is negligible in most calculations.
How do you figure how many ohms can that amplifier handle ?
Please, can anyone reply?
Very useful...cheers.
Cheers mate
If i have a multistage with 2 bjt couple with capacitor, but at q1 output there's no signal what could cause that
That could be a lot of things, no AC input, no base bias voltage. Hard to say with the info given.
Thanks for sharing.
thanks a lot,sir
do you have full derivation of DC bjt analysis for cascade and cascode?thanks
I haven't done a Cascode amplifier yet. I hope to in the near future.
I have using 2 channel amplifier.i would like to very punchy and depth smooth bass.but in my amplifier Voice only loudly Come out please give me any solution.
Hello, Without knowing more about your circuit, I can't really suggest much. Are there caps in series with the output, inductors in parallel? There are to many variables for me to give you answer.
whats the name of the transistors being used ?
Hello,
I used 2N3904s. 2.2uF caps will work fine.
thank you!!
Grazie
What's the Application of CE Amplifier ..
Anytime you need amplification of a low frequency signal, the common
emitter is a good choice. It is a general "go to" type of circuit.
+ The Offset Volt To set something straight for me; The single transistor amp, with a single rail supply should by rights only swing between 0v to lets say 12v rail when the emitter voltage is at roughly 6 volts in that simple configuration of amp, (and through a DC decoupling cap of course). The load, e.g. the coil/cone of a loudspeaker should then move ONLY from 0v to 12v, and back to 0v, (if you measure it load side and when a signal is present), yes? To get a full 'swing' from the speaker coil/cone where the output also becomes negative (-12v), and an so an inward coil/cone movement you must need a split rail supply i.e. -12v, 0v, and +12v, (with a decoupling cap), then the coil/cone movement would follow 0v, +12v, 0v, -12v, 0v, +12v when a signal is present, as in the case of a multi transistor npn/ pnp push pull amp. Correct?So the -ve cycle of the cone/coil is missing in the first case common emitter configuration?
@@mcwolfus4718 If you have a single 12V supply, then you bias your output stage to be at +6V when there is no signal. Then you use a large capacitor between the output and your speaker to block the dc offset of +6V, but allow ac signals to pass through. So without a signal, the loudspeaker has 0V at the end connected to the capacitor and the other end is grounded (also 0V). The amplifier can then pas a maximum signal between 0V and 12V. That is an ac signal of ±6V around the +6V bias point. The ac signal of ±6V will pass through the capacitor driving the speaker to +6V and -6V relative to ground. That's how you get negative voltages onto your speaker with a single rail power supply.
hai can i ask you which resistor i need to incrase or decrease value because i want get voltage gain total 140
Adjusting the emitter resistors would be easiest but then the Q points change as well. You could use two resistors on the emitter that equal the same value as the original and put a capacitor between them to set the gain. Kinda complicated to explain here. Hope it helps a little!
@@TheOffsetVolt If you keep the sum of the two emitter resistors constant, you won't change the bias points. I'd simply change the first stage emitter resistors to Rb1=100R; Rb2=140R (=100R+39R) and call that close enough. 🙂
hello sir can i know what is the exact value of the capacitor ? i would highly appreciate it if u could provide me the value of the capacitor for my simulation test
I don't know why I never put the values down for those ??? I don't remember the exact values but 10uF electrolytics are more than enough. Thanks for watching!
@@TheOffsetVolt oh ok i see ..thanks for ur kindness sir highly appreciate it
As I commented above, for audio work down to 20Hz, you need bypass capacitors of 100μF and coupling capacitors of 2.2μF.
when you say RC2 is parallel to RL, what do you mean by that?
Hello, The current from the collector of the second transistor would go through RC2 and the load. Since the current split into two paths at the point where RC2 and the load connect, these are parallel to one another.
The Offset Volt thank you for replying, it's the way that you've written it for the calculations, I've not seen the || used before and I was baffled as to how you got your answer. Forgive me for sounding a bit dense
Not a problem. I do the videos for people new to the field and I always forget two or three points that I should include after the video is published. Your question was one of those dis-remembered points. Thanks for watching.
What size caps did you use?
I believe I was using 10uF although you will get pretty good results with 2.2uF as well. The idea is to have an Xc that is, at it's maximum, 1/10th of the resistance you are bypassing.
what is the merit of the capacitor in the circuit? i need brief explanation
to bypass or decoupling
In brief, it doubles the ac gain of the circuit without affecting its dc stability.
Would you call this a RC Coupling Amplifier?
yes
Sir , please make for single stage amplifier also
I thought that for the DC biasing that RE was to be sum of RE1 and RE2 which would be 240 ohms?
bigbread9000000 It is indeed. The answer is right I just wrote 120 instead of 240.
Great tutorials, you are very thorough!
Wasn't that gain has negative sign? Av= -(Rc2IIRL)/(Re3IIrej)
Hi, Yes, any inversion should have a negative sign. I have the habit of ignoring for sine waves.
Do you mind giving the values of the capacitors you used?
Hi, I used 10uF - what I had on hand at the time. Lower values would work but for the caps to be electrically transparent, bigger is better.
Thank you so much, btw can you share with me a tested and working circuit diagram of a common emitter - common collector two stage amplifier if you have any. I am working on a project at college and struggling with it. Yours was greatly helpful.
Thank you. I don't have anything on hand but might come up with something.
The emitter bypass capacitors need to have a reactance less than Re1+re' (i.e. about 130ohms) at the lowest frequency of interest. For audio, we usually work down to 20Hz. So the emitter bypass capacitors should be more than 1/2πfR = 1 / (2π x 20 x 130) = 60μF. Use 100μF.
The input and interstage coupling capacitors need a reactance less than about 6K (the input impedance of each stage), so you can use 2.2μF. That will be okay for the output cap as well.
@@RexxSchneider thank you for this...
What is the value of your capacitor ???
10uF anything larger will work. I think I used 47uF. Thanks for watching.
how can i find the total input resistance?
At frequencies where the capacitors have negligible impedance, the input impedance is the parallel combination of Rb1, Rb2, and β x (Re1 + re'), which is 56K || 6.8K || 90 x (120+10) in the worst case. That's about 6K.
How do you come up with an answer if there's Rb1||Rb2?...I'm confused by the parallel symbol
I have been trying to implement a bifet amplifier but with a load resistance of 10 ohms and Vcc as 5V to get a gain of 20 V/V but it seems not to work. im working with LTSpice. i keep thinking probably my load resistance is too low or the Vcc. What could be done please. Ineed help
Hello,
Without knowing more about the circuit I can only offer some general ideas. A 10 ohm load on a common emitter stage, if built using the circuit values in the video will not give a gain at all. Av=(Rc||RL)/RE. What is the voltage in and the desired voltage out? Remember Vout can only go between Vcc and ground. With a 5V supply, the biggest AC voltage out, if perfectly centered on 2.5V DC is 5Vpp.
Your supply voltage is too low for any comfortable design. When working with FET input stages, you have to account for the large range of important parameters. Discrete JFETs and MOSFETs really need 15V supplies to allow you to compensate for gate turn-on or turn-off voltages of several volts and still get a usable voltage gain. If you're using a single-ended FET first stage, you'll want a split source resistor with most of it bypassed for ac with a capacitor to increase the ac gain if you want to get x20 out of it. Then your problem will be that you need considerable current gain from the bipolar output stage and you may need a Darlington in common collector to achieve that with a 10R load.
About of rej parameters??
I am going to do a video on this soon. A lit of peaple have the same question.
It's the intrinsic emitter resistance of the transistor. As the current through the emitter changes, there is a very small change in the voltage between the base and collector. The change in voltage divided by the change in current is effectively the dynamic or intrinsic resistance inside the transistor's emitter. You can look up "Shockley's diode formula" to get a more in-depth treatment, but one result is that this resistance is approximately equal to 25mV divided by the emitter (or collector) current at room temperature.
So with a collector current of 2.5mA, the transistor behaves as if it had a 25mV/2.5mA = 10 ohm resistance in its emitter.
👍
Really good tutorials!! Mind doing one on feedback. Negative feedback. To minimize distorion, THD, etc? :)
Thank you. I'll see what I can do but it might be a while.
No rush! With the quality of your tutorials, I am looking forward to it! :D
Great
3Q
Sir please help
some one helpe me plaese😢😞
Why 0.6/120=25?
This equation is wrong it is like this 0.6/120=5mA
Hello Hena, There is a callout in the video that corrects the calculation. The resistance of Re was 240 ohms so the current is 2.5mA, Thanks for watching.
@@TheOffsetVolt Thanks Sir
Hello Sir my name is Tripti Das , 2nd Year Jadavpur University B.E ETCE student , and need your help.
Your video is good and I just need a book where I get your explanation ..
I have books but the explanation is not good and so I need a book from where I get all the formula deduction you used in this video.
Hello, A good source would be
Electronic Devices (Electron Flow Version)
Hardcover: 976 pages
Publisher: Pearson; 9 edition (February 18, 2011)
Language: English
ISBN-10: 0132549859
ISBN-13: 978-0132549851
Good luck in your classes.
The Offset Volt
*please write the writer/editor/author name ..*
There are lot of Pearson publisher books in the library.
Thank you for your reply.
Thomas Floyd, Prentice Hall
Video searching for
Hagalo en español.
خويا تقدر تعوني
S ok
aslen karak
Rc1|| Zin
2k||4.13
= 1.35
I would like to know how did you come up with that answer? I'm confused by the parallel symbol
what are these capacitors values?
Hello Tahir, how are you?
what was the capacitors value?
They were 100uF.