He's not your dude bruhhh stfu

Wow, thank you and thanks for watching!

Thank you! Will do!

Good luck on that exam!!!

Hey sir, did you finish your project? I'm just wondering.

Thank you - I'll see what I can do.

Glad you enjoyed it

Hello, I am not sure to which formulas you would be referring. Gain calculations would be different for each species.

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.

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?

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.

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.

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.

Hi, It should work, it's basic components.

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.

Glad you liked it!

I haven't done a Cascode amplifier yet. I hope to in the near future.

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.

Why does it bother you bro?

First time my video saved a life. Glad it was of help.

How he is saved your life??? Tell about that?

Yup

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 could be a lot of things, no AC input, no base bias voltage. Hard to say with the info given.

Cheers mate

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.

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.

He wrote 120, but the did the calculation using 240, so the answer is correct.

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.

Unfortunately, no. It is something I should consider doing though. Thanks for watching.

bigbread9000000 It is indeed. The answer is right I just wrote 120 instead of 240.

Great tutorials, you are very thorough!

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.

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.

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.

Hi, Yes, any inversion should have a negative sign. I have the habit of ignoring for sine waves.

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.

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. 🙂

yes

Hello Tahir, how are you?

Grazie

Hello, I used 2N3904s. 2.2uF caps will work fine.

They were 100uF.

to bypass or decoupling

In brief, it doubles the ac gain of the circuit without affecting its dc stability.

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.

10uF anything larger will work. I think I used 47uF. Thanks for watching.

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...

3Q

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.

Please, can anyone reply?

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

Thanks for the great compliment. I appreciate your time watching.

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

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.

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

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

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