How to design a single transistor amplifier with voltage divider bias
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- เผยแพร่เมื่อ 4 มี.ค. 2022
- This video simplifies the design of a small signal common emitter transistor amplifier that uses a voltage divider bias circuit on the base. Demonstration circuit is breadboarded after the design.
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Email: johnaudiotech@gmail.com
Thank You! - วิทยาศาสตร์และเทคโนโลยี
You are gifted when it comes to explaining stuff, I'm proud to announce I'm listening to this on a TDA1517p I built after watching your videos. Thanks for sharing your knowledge with the world John.
After absolutely scouring the internet for a straightforward explanation of how to chose resistor values for a voltage devider biased transistor amplifier, this video finally explained exactly what decisions need to be made and how to make them. Thank you so much for making these videos!
I attempted to do this (once!) "on paper" in order to re-design a 70's vintage pre-amp circuit. I ended up using a circuit simulator (TINA) to converge the circuit for lowest THD for the gain value (+9db / stage) I was shooting for. I used MPSA18 transistors (beta of 800+) for all three stages. The end result is still working quite well. Only a small amount of negative feedback is used from the follower (output) stage. The original design used un-obtainium FETS with a max output of 5 volts (!) less than .25% THD. I was able to achieve .11% THD at 3v rms using the mpsa18s WITHOUT changing the original board design. I just needed to add RB1. However it took quite a bit of "simulator" experimenting to home-in on the proper resistor values. It's two CE stages followed by an EF stage. The original design had NO RB1. I guess you didn't need them for the FETs. but I needed Rb1 at 10 meg and Rb2 at 1 meg. The simulation failed without Rb1. (even at 10 meg!) The original design also had Rb2 at 1 meg. I believe that's the only value that didn't change. Every board level component got replaced. This was originally done as a "re-capping" project back in 2006, but a few years later, I decided to replace the entire circuit. I had NO choice but to re-design the bias network even if I wanted to use modern J-FETS (the original plan) . But I eventually switched to the low noise (easily obtainable) NPN bi-polars. I used the same 36vdc that Sony was using off of their (re-capped) PSU board. The actual Pre-amp was a SONY TA-2000f (circa 1975). Except now it's got a re-built line amp. The line amp (20db) gain stage is a classic textbook design. Twin (daisy chained) CE stages followed by an EF stage, all cap-coupled. THD and transient response are much better than the original circuit.
I was thinking of getting Tina. I would like your opinion on Tina. ...Thanks
@@richardphillips2405 I'm (still) using on old version of Tina (6.0?) from 2006 on my "vintage" win-xp machine. It's one of the reasons I still keep that computer around. At the time it cost $50.00. I had already "played around" with some free SPICE Gui tools but TINA had S/N and THD simulation which is important for audio. So I purchased it. My only real gripe about this version (now unsupported) is that I can't update the device libraries. Last I looked...the latest version cost around $200.0. Is it worth 200? I guess if you design enough circuits it might be. I designed only three circuits that I actually physically built using TINA. The one described above (the SONY PreAmp), a phono stage re-build and an Op-amp based (2 to 6 ch) distribution circuit. I "fiddled around " with that pre-amp circuit for five years (!) before I actually got around to tearing up the working board and replacing ALL the parts! That circuit is also used for the tone controls, so it's four identical line amps on one board. I also have some audio design stuff that I haven't built (yet). I must say that the TINA simulation was quite accurate vs. the measurements that I did on the the devices I built. So I guess I got my $$'s worth. I've also used TINA to verify the working of some circuits that I found in design books. What I would REALLY like is a "budget" Linux version of TINA. I almost ended up purchasing some version of "WIN-SPICE" (PSPICE ?) but TINA seemed a better choice at the time. Hopefully I'll find a Linux equivalent someday.
@@johnnytoobad7785 I have the student version of Multisim. My version has a limited library. I would have to be enrolled as a student if I wanted to get a newer version. The commercial versions are way too expensive. I have seen several adds for Tina in the Practical Electronics magazine. So, I was wondering about Tina. My goal is to learn more about circuit design and also experiment with building circuits. I appreciate your review. I was wondering if Tina has a schematic capture similar to Multisim.
@@richardphillips2405 I know there is a student version. I doubt it captures schematics. Tina wants you to EXPORT your design from their platform. My version doesn't support EXPORTING to EAGLE CAD which drives me nuts. So I need to enter the schematic twice. Once for electronic design and once for board design. Other than that I like ALL the built in features of TINA.
Concise. Brief and to the point. Back up with protoboard hookups. Good job.
He’s good, but my knowledge is so limited I would need to watch this multiple times.
I built lm386 amplifier using your schematic and it works really well !
Its absolutely insane when I think about how much I've learned from you! Thank you John!
@@boonedockjourneyman7979 yet another person commenting when they don't even have any content of their own! 🤣🤣🤣 the only person flailing is you bro.
Same here, thank you so much, John!
The vbe reported by the transistor tester is probably at a 1ma bias current. The .65 v value is probably closer to correct at the 13 uA bias operating point of your circuit. Great intuitive explanation.
0:30 A small signal is not amplified. Rather a larger signal is regulated by a smaller signal. A minor distinction, but for the minds that think literally, a great insight.
Please make more electronics theory tutorials.
It was very interesting.
You explained that 100 times better than my electronics teacher .
Thank you !!
😀
Thank you!!! Simple explanations of elementary building blocks is something been looking for, for ages! If you could do an explanation of differences for mosfet (voltage input) bjp (current input). And npn pnp using the single transistor amp format like this that would be even better 😀.
Also I'm aware of differences between purely current output with the capacitor below the transistor and current and voltage output like you have on this video but am unsure of what those differences are.
Anyway. Thank you! Great stuff 😀
As ever, you make it look simple and easy peasy. And yet, I always learn something.
Well done, John. You hit all the right notes in this video. I have one small criticism and a couple of additions to make. The criticism is that the intrinsic emitter resistance is _not_ "26 ohms per milliamp" as that implies the resistance increases with emitter current. You corrected yourself by noting that it decreases with emitter current, but it's better not to have the confusion in the first place, Just say that the intrinsic emitter resistance is 26mV divided by the emitter (or collector) current.
My first addition would be to calculate the intrinsic emitter resistance (re), which is 26mV / 1.3mA = 20R. Knowing that value helps you decide on whether your other choices are good ones. Clearly the 470R emitter resistor is over 20x bigger, so you'll get a corresponding improvement in linearity, and It's worth making the point. Also, the voltage gain (Av) is better modelled as Rc / (Re + re) and that is then predicted to be 4700 / (470 + 20) = 9.6 which corresponds better to your measured voltage gain of 971mV / 103mV = 9.4. The remaining error is most likely, as you say, due to resistor tolerances.
Finally, note that the approximate gain is Rc / Re, and that the same current flows through both of them. That means that the quiescent voltage across the emitter resistor will be [voltage across collector resistor divided by gain]. If you set 6V across Rc, you will end up with (6V/Av) = (6V/10) = 0.6V across Re in this case. That gives a sort of upper practical limit to the achievable gain because a gain beyond 30 will leave less than 200mV across the emitter resistor and there comes a point where variations in base-emitter voltage between samples then become significant. Also at an approximate gain of 30, the emitter resistor has now become 150R and the 20R of the intrinsic emitter resistance is becoming a more noticeable factor. The only way to improve on that trade-off between linearity and gain is increase the supply voltage.
That made my brain explode
Yes all that makes sense. But John did say he was going to keep it simple. Like many people and myself have expressed, this is the BEST VIDEO explanation of how the process of choosing the components in this circuit have come across. This is a fantastic video. And could you post a link to your channel?
👍👍👍👍👍👍👍👍🙏🙏🙏🙏🙏
Nice job of presenting just enough detail to allow novices like me to understand how amplifiers are designed. You admit to glossing over some finer points, thus opening yourself up to accusations from the cognoscenti of either not telling or (worse) not knowing the full story. But your practical, simplified approach is worth far more to learners like me than a million hours of pontificating by "experts"!
Great video John! Thank you!
I’m happily hooked to this channel, just fantastic!!
Super explanation. Thank you!
Thank you! It's very informative video for hobbyist with simple explanation.
Howdy.
In another clip You demostrated that the output impedance is 2 x the collector resistor. Much gratitude. That set me on the track to figure out why I could not reach spec. power on my 4 X KT88 tube PA.
My design uses heavy neg. feedback from the end tube anodes to the grid. This reduces the tube stage input impedance. A lot actually. I had designed the end tube input imp. to equal the driver stage anode resistor. Too heavy loading of the driver. I need to redesign the end tube feedback for higher input. impedance.
I have seen many explanations of this common emitter designs. All make the false statement that the output imp. equals the collector resistor, just as I assumed earlier.
Another observation. Not many explain the value of the emitter resistor. The emitter resistor is elementary for thermal stabilization. It should be rated to produce more voltage drop than will be the base-emitter junction threshold voltage decrease when the transistor warms up. If not, there is a hazard of the current avalanching.
High Regards.
Thank you lecturer John. U explain it in simple easy applicable way.
Simple math.
I am a beginner hobbyist and this was immensely helpful! You get a like, comment, and subscribe good sir. Looking forward to more videos!
Great video! I would have been curious to see it’s performance at higher input levels. I would have liked to see where and how it clipped. I also would have liked to see how a transistor with a different measured hfe acted. Thanks for all your hard work. Like others have said, you are an asset to TH-cam and possess a gift for teaching.
Thanks John, really helpful tutorial. I know this is a tricky topic but nicely explained without it turning into an algebra lesson. 👍
This is a good lesson. Thanks!
Thank you so much dear human brother. You donated your knowledge.
Great, simple explanation
Great stuff very useful info, cheers Jhon.
Very clear explanation, thanks
Thank you for sharing your practical knowledge and posting this video. I specifically enjoy your talent for explaining how things work. It's a rare ability. Would you consider in a future video designing emitter follower? They are simple to build and work great in a wide range of frequencies, from audio to RF.
Enjoyable-relaxing&Educational,..thanks for sharing video(s) John.
ROAD TO 50K LESSSS GOOOO!!!!
I missed this videos
I've seen more complex examples on audio design books (mostly the ones recommend/shown by you and KISS analog) some to increase input impedance, linearity, etc. Even that most authors show different examples along the book, they always recommend using a differential pair instead since it can perform better.
Thanks for the upload Jhon!
Great refresher vid.
I've discovered some of the Beta tests on these meters and testers are pretty far off so if I need an accurate Beta measurement I will get the actual base and collector measurements and then use the formula B=Ic/Ib ...this always works out. Thanks !
Hey John just found you channel and it awesome.
Excellent discussion, thanks.
solid. Always helpful. Thanks
Great explanation. Thanks.
Absolutely love this explanation. Ohm's law always comes to the rescue
Thank you John!
you are 100 times better than my teacher
Thanks John!
Great tutorial...cheers.
Great video, thanks!
Fantastic, thank you
Outstanding!
Thanks. Great video.
Nice. BJT always were the most confusing active element to me, as they are current driven, rather than voltage driven like tubes, FETs, or opamps. So when messing with them it was more guesswork and trial + error for me than anything else.
This is a good tutorial on how to quickly calculate a class A stage though. Thanks.
This is verry helpfull - thx
Back to the technical school years in 1988 i remember for this circuit that hfe or beta does is not needed for the calculations because the base has a fixed bias voltage and collector current is calculated from (Vbase-0.7v) / RE.Also the circuit considered to be more stable to temperature variations.So calculate Vbase voltage higher than 0.7v is one important and the second is to choose RE resistor according the desired current and then calculate RC resistor to have half the power supply to the collector.
17:12
Question: should we always have the input signal connected to the negative terminal of the capacitor ? And have the positive terminal of the capacitor connected to the base of the transistor ?
As good as always!
Would something like this be practical to boost an electric guitar pickup signal to an input value approaching line level input needed for home audio amplifiers?
Guitar pickup need to have loading impedance more then 500k (this one have input impedance about 4.7k as mentioned in video) so you need an emitter follower circuit before this stage. But may be emitter follower is enough.
You should rather build a high input impedance curcuit with op-amp or Fet.
But be careful. You can very easily destroy home hifi speakers with a guitar. Use at least real guitar speakers.
@@westelaudio943 You are right, with TL071 or similar is the simplest solution.
Good morning God bless and much as normal thank for video
Awesome!
Thanks man this gives me a good place to start with my illegal drone jammer /s
Thks
Would you recommend the EEV Blog DMM?
EDIT: I just watched you review of the meter. The 5 mA current draw in auto-off is a deal breaker for me. Dave from Australia is a great guy, I'm surprised that he let his finished product have this kind of flaw; which completely defeats the purpose of auto-off.
re=.026 ohms per milliamp.The Emitter with NO cap.The gain is Collector resistor/Emitter Resister
So, when the transistor is on, does that mean the current through the load approaches zero (i.e. its off)? So a signal turning the transistor on seems like it would allow the current not to flow through the load? If there is no signal does that mean there would be 12V DC across the load? THanks!
What made you choose that Resistive value initally for R_c?
18:30
Question: Why the amplification is around 10 while the transistor Beta is 171 as measured by your component tester ?
Hi I wanted to make video amplifier circuit with one transistor wich transistor and which amplifier class are better? thanks
I wish I could get such a smooth sine wave out of my Feeltech. Everything below 500mv gets noise and jittering on my scope. Above 500mv (give or take) the generator works ok. I don't really know what could be the problem with it.
May I know how u get the Rc at 4.7K ohm? 😊 please advise thanks you
Howvoltage divider resistors will appear parallel to input signal and how we took 6 volt at collector
Excellent! Thank you. Do you know a source for the formulas and process that you used to determine all the values? Other than an overly complicated text book, or having to go back through the video and write it all down? Maybe a spreadsheet?
Here's how I'd recommend you do the design. First decide on the desired gain. Your supply voltage will be at least 0.5V times the gain. Then decide on your desired output impedance. Your collector current will be around half the supply voltage divided by that output impedance. Pick a transistor with a minimum β of at least 100 at that collector current.
Now set Rc = desired output impedance. Calculate Ic = Vcc / (2 * Rc).
Set Re = Rc / (desired voltage gain). Calculate voltage at emitter Ve = Re * Ic. Calculate voltage at base Vb = Ve + Vbe = Ve + 0.65V.
Calculate Ib = Ic / β. Set the current through the lower base bias resistor Irb2 = 10 * Ib (at least).
Calculate Rb2 = Vb / Irb2. Calculate Rb1 = (Vcc - Vb) / (Irb2 + Ib).
Now select nearest preferred resistor values and recalculate currents and voltages at base, emitter and collector. If the collector voltage is not close enough to mid-point, then adjust resistor values slightly to get the desired bias point.
The input impedance Rin will be the parallel combination of Rb1 || Rb2 || β*Re. If that's not high enough, either increase the output impedance or pick a transistor with higher β.
Choose an input capacitor whose reactance is at least equal to Rin at 20Hz. C = 1 / (40*π*Rin).
@@RexxSchneider Thanks, Rex! Good information!
Cool
Mr john, how to determine value of coupling capasitor in input and output, bypass capacitor in emitor to gnd?
You need to know the impedance of the circuit and low frequency cut off you want. Then you can use the equation to calculate for capacitance.
how do we know a collector current ?
I didn't see how you calculated the collector current
which transistor is used in this circuit?
What does Regeneration mean in a transistor? The Regeneration is doing what to the voltage and current?
when there is no signal, then how are 6 volts at collector after 4.7k resistor?
How must i calculate on + and - power supply people?
Hi John. I would like to see you do a test/ review on theTDA7498E DC 15-36V 2 Channel Audio Stereo Digital Power Amplifier Board 2*160W available on ebay. Is it bridgeable and applicable to drive 1 or 2 automotive subwoofers? This chip is a recent release from ST and data sheets are available from Mouser and Digi key.
I checked eBay and the boards are only around $20, so I bought one. Outputs from virtually all class D chip amps are already bridged, so I think they mean paralleling for driving lower impedance speakers.
what is the value of capacitor
John, @ 5:35 you say that Ib sets Ic. (Ib * beta=Ic). To me, that implies that the values of Rc and Re don't affect Ic (no input signal at this point). What am I missing since @ 8:34 you talk about the value of Rc setting the current?
And shouldn't adding the value of Re alter Ic?
Rc was chosen to set the output impedance and Ic was calculated using Rc to set the voltage on the collector at quiescent. Rc does not affect Ic. Re is used in the calculation of the voltage divider resistors on the base circuit. The simplification can be confusing to some people trying to understand what's going on in the circuit. This is where a more indepth look using the math might help in understanding.
@@JohnAudioTech Thank you John. After doing more research and testing and rewatching, I understand better. In your reply, you say "Rc does not affect Ic." but I am pretty sure you mean RE (not to be confused with "Re") does not affect Ic.
Anyway, I'm getting there and am thankful for your videos to help.
@@OIE82 Yes I do mean Rc does not affect Ic. With my demo circuit, short Rc out of the circuit by metering across it in milliamp mode on your multimeter and you'll find the current is about the same (slight change from the Early effect). Re (not to be confused with the intrinsic emitter resistance in this discussion) is a feedback mechanism which means it is intimately involved in the current, gain and linearity of the circuit. Try changing Re in the demo circuit to a different value and current, quiescent point and gain will all be affected.
@@JohnAudioTech Hey Mr. John. I did the experiment you suggested using a Class A setup similar to yours. You were right! Though I didn't really doubt you. So I measured 7.0mA between the collector resistor and the collector. And 7.1mA between the emitter and the emitter resistor. 0.1mA from the base current I guess? Then I shorted the collector resistor (would not have done that without you saying so) and the emitter current stayed the same!
Now, why? Every video I have watched on YT that mentions anything about the collector current says some of the following...
Rc and RE set the collector current. (I now know this is false but why do we start the design process by deciding what Ic should be and then calculating the values of Rc and RE around that current?
RC/RE set the gain
RC sets the output impedance
RE stabilizes the transistors variations due to thermal effects
I know there a lot of questions but you seem to know what you are talking about so I ask.
Thanks for all the videos, I have enjoyed learning.
@@OIE82 Recall that the base current * the current gain = Ic which in theory is constant no matter the load on the collector. In other words, the circuit is behaving as a constant current source (or sink in this case); therefore, Ic is independent of collector load.
I am trying to make an audio amp for a disk cutter, i keep chaing the schematic as i keep using parts for circuits, so am left with little to none to work with. I dont know where to start, Its bascially a phono amplifier, that ran on tubes, but it shocked me, so i figured id strip and rebuild, every wire or something was all dry rotted with age..Even a wire to the speaker was corroded,,I only need a 2.5 Watt amp. bypassing the transformer.. it has a mic input and phono pickup out.. Vintage 1950s switchraft Screw on type - mic in._ and Banna Jacks for line out.. Cutter head orgionally worked on 1200 volts D.C.. So Will go 8 ohm..4 volts...To cut. convert the cutter head to magnetic pickup...
Watching this video repeatedly because it's so good. But I don't understand how Rc and load resistor form a voltage divider for when there's a capacitor there.
In the pass band range of the amplifier, the capacitor is essentially a short and can be ignored. It is there only to block DC from entering the following circuit.
@@JohnAudioTech Yeah, I think I misunderstood what you were saying about the next stage. However, there is math voodoo going on in this circuit that is beyond my hobbyist ability to understand. I played with the value of Rb1 tonight and discovered it affects the measurement of Rc when trying to establish the 6v that is desired for getting half the supply voltage. I didn't have anything close to 40k - only had 33K and 47K. So I put in a pot, tuned it until the voltage at Rc equaled 6V. I then pulled it out of the circuit, and measured it with a DMM. Guess what? 41K was the number. Can you help me understand how the values of the base bias resistors are working to set Rc to half the supply voltage?
@@richardneel6953 The ratio of Rb1 to Rb2 sets the fraction of the supply voltage that appears at the base. The voltage at the emitter is always about 0.65V below the voltage at the base. The current through the emitter and collector is set to be the voltage at the emitter divided by Re. The voltage at the collector is set to the supply voltage less the voltage across the collector resistor (the collector current times the collector resistor).
So if you increase the ratio of Rb1 to Rb2 (either increase Rb1 or decrease Rb2), you increase the voltage at the base. That increases the voltage at the emitter. That increases the emitter and collector currents. That increases the voltage across the collector resistor. That decreases the voltage at the collector.
Decreasing the ratio of Rb1 to Rb2 (either decreasing Rb1 or increasing Rb2) will have the opposite effect, i.e. it will increase the voltage at the collector.
I think we need an in depth video about your component tester. There's a lot of lore about them. The original design and the multitude of Chinese knock offs. I'd like to see where yours fits in with it all. I have 3 myself and they're all different. One's almost an original clone. One's so horrible I don't use it. I didn't know as much about them when I bought it. Two I have I built from kits.
Trying to get some kind of answer so •••. How did 5170 ohms over 12 volts get to be 1.3 mA? It isn’t. It’s 2.3 mA. What is going on in this video?
I assume that the 5170 Ω you quote is meant to be the sum of Rc and Re = 4700 Ω + 470 Ω.
If all 12V were dropped across those two resistors, then you would be correct that the current would be 2.3mA.
However, we need to have some voltage dropped across the transistor, of course, in order for it to work.
You can see that since 1.3mA is flowing, then Rc and Re only drop 5170 Ω x 1.3mA = 6.7V. That leaves 12V - 6.7V = 5.3V across the transistor, which is just as we want it to be.
50k.....❤️
Thanks for the great content! Do you think you could find time to have like a 15 minute conversation with me? I have been struggling with a few things while trying to create some DIY speakers. I've done as much research as I can tolerate and I am at a stand still and can't seem to find some answers I need. I just need a diagram created for a simple bluetooth speaker design I made. I'd really appreciate it. Thanks!
Love your videos but you really need to focus on getting JAT done so I can feed my obsession. Let's regain focus, I am selfish and need to feed the audio beast inside of me. Love ya John...keep up the great work.
Almost there. Gotta do taxes first!
@@JohnAudioTech LOL...I am in no hurry to pay the government. I always wait until the very last day.
Keep it simple. Most of us are NOT designing missile guidance systems or satellites.
Keep it simple? It ain't simple dude, if you wanna simple go to do something else, we the engineers dont want the simple
You lost me at exactly 8:00. Where did the numbers come from? You say “acceptable” as though it’s a known thing. How on earth do we know what acceptable means? I’m not trolling. Please don’t deflect with that. I honestly don’t know where your numbers come from.
In all honesty you ate flipping back and forth from current to voltage so many times it looks like you don’t appreciate that BJTs are current dependent devices - exclusively. Talking about either bias or collector resister selection in terms of voltage is a freshmen EE error.
What are you talking about when you talk about “acceptable” values in impedeance matching of amplified stages.
I come here because you succeed. But your description is completely alien to me. I got my first Ph.D. In 1981. I may be dated but I lived the transistor age and I am telling you (as opposed to suggesting) you are making zero sense. I know you are a master at audio amplifier design but you are making no sense from a EE PoV.
Please help an old dog figure you out.
Just one example so maybe you understand. You talk about knowing collector current from collector resistor selection. That is simply not so. At this point you don’t even know the emitter resistance and you haven’t dealt with the impact of base-emitter current.
If you did know the emmiter resistor as you say, then 5170 ohms at 12 VDC is absolutely not 1.3 mA. It’s 2.3 mA. What •••
There is a reason designers communicate in specific ways. Your success is notable and significant. But trying to understand it based on your description is impossible.
When John used the word "acceptable" around 8:00 he was talking about the fact that you will lose some voltage gain because the input impedance of the next stage will load the output of this stage. As he said, if the Rin of the next stage is 22K and the Rout of this stage is also 22K, then you'll loose 1/2 of its voltage gain. That's because the 22K and 22K form a voltage divider and you only get 22/(22+22) = 0.5 or 50% of the signal passed on to the next stage.
However, if you designed the output impedance of this stage to be just 2K, then you would get 22/(22+2) = 0.92 or 92% of the signal passed on to the next stage. Unfortunately, this stage would then need 11x more current to operate. so there is always going to be a compromise between using lots of current and minimising the amount of gain lost when connecting to the next stage.
John suggested that an output impedance around 4.7k would be acceptable as that would only reduce the gain to 22/(22+4.7) = 82% when loaded by the next stage, while keeping the current down to just over 1mA.
The design process started by specifying the desired output impedance. That's something you will know because in real life, you will have an idea about what the input impedance of the next stage is. The output impedance of the common emitter stage is equal to the collector resistor, and so you have your first value, Rc.
Once you have Rc and know what supply voltage you have available, you can approximately calculate the target Ic so that roughly half the supply voltage will appear across Rc.
You now know Ic and hence Ie. From that you can calculate the intrinsic emitter resistance (re = 25mV/Ic). So to reduce distortion we make Re at least 10 times bigger. That means we set the voltage across Re to be at least 250mV.
The gain of the circuit is Rc/(Re + re) so you can get an idea of the maximum gain available from the stage before distortion becomes significant. If you need less gain, then just increase Re to get the gain you want.
Now you know Re and Ie, you can calculate the voltage at the emitter. The voltage at the base will need to be set about 0.65V higher because a small-signal silicon transistor has a Vbe drop of about 0.65V when conducting a few milliamps.
From the datasheet for your transistor, you can estimate the minimum value of β at the design collector current. That gives you the maximum possible base current, Ib = Ic/β. you then aim to pass 10 times that current through your base bias resistors Rb1 and Rb2, so that the current than goes into the base is negligible in comparison.
Now you know how roughly much current is passing through Rb1 and Rb2 and you know the voltage you want at their junction (the voltage at the base that you calculated earlier). That will allow you to calculate the values for Rb1 and Rb2.
That is effectively the path taken by John in his design, and I assure you that there's no errors, "freshman" or otherwise, involved in the process.
If there's something in this description of the design process that you still don't understand, then please feel free to let me know and I'll be happy to expand on my reasoning.
😇😇😇👍👍👍🙏🙏🙏 I like your class, really good.