Bruh, I've been scowering the internet for 5 days looking for someone to explain the meat and potatoes of how amplifiers work, not just, "well this voltage and this current goes here and creates this" Thank you bro, the universe needs more people like you
You rarely find a "how to" video which goes literally step by step. I'm arty not techy and have poor maths... but followed this without too much difficulty. The bonus is that I can replay the video to revise and review. I wish I had your way of teaching when I took the RAE exam in 83, I passed with double credits but didn't realise you could have AC and DC running on a circuit at the same time! Now I understand not only the "what" but the "why" as well. Thank-you Nick! 😄 👍
Thank you Louis. I have to admit that it has taken me a while to really get my head around all this stuff myself ... and I'm still learning I'm pleased to say. Glad it helped. 73, Nick
Very good information here to get those projects working, (and the grey matter!) Many people struggle with this very concept and it`s tempting to just grab an op-amp, until you need an amp at a high enough frequency to make op-amps too expensive or not suitable. Everything you need to get you going has been covered in this video. You do a great job to help keep this hobby alive. Chris, UK. (Gentlemen of 1977khz AM)
Thank you very much indeed Chris. I'm certainly not adverse to using an op-amp or two myself but I find there is something strangely empowering in understanding how the humble transistor amplifier works. Thanks again. 73, Nick
Good question. It is because of the way the transistor operates. The phase inversion happens because an increase in input signal causes a corresponding decrease in the output signal, and vice versa, due to the interaction between the base-emitter current and the collector voltage across the load resistor. When the input signal increases and makes the base-emitter junction conduct more (increasing Ic ), the voltage at the collector decreases. Conversely, when the input signal decreases, the base-emitter junction conducts less (decreasing Ic ), and the voltage at the collector increases. This relationship results in a 180-degree phase shift between the input signal at the base and the output signal at the collector. Hope this helps. Best wishes, Nick
What a great video Sr. Thank you very much for the simple explanation I really appreciate it . I have assembled the simple transistor amplifier from the video 56 using my breadboard and I am truly happy, is the first time that I build something that is not some kind of a blinking led =D, the amplifier works great. I don't need your notes, I have copied everything to my notebook while watching the video hehe (lots of pause and watching again =) I can't wait for the next. Thank you
Sweet! I ran my own hypothetical calculations with 10V and wanting 20ma/1v at Ve and built it on a breadboard. Took a few tries to go through the calculations and it was right as I wanted. I do find that calculating everything in the required units makes it a little easier than converting in my head on the fly, but I'm sure with some practice it would become 2nd nature.
I'm playing catch-up today mate, this is the second video, I'll save the next for tomorrow. Excellent explanation of the biasing theory, thanks for that. One note, clipping is a danger for the speakers - not at the power we're talking about here but when you move up in power it becomes so. When the output clips it becomes a DC signal, the voice coils start dissipating a lot more power than they're designed for and they release that magic smoke. There's an old saying - anything can be a smoke machine if you try hard enough.....
Thanks Al. That's a timely warning about clipping too so thanks for sharing that. Hope you are OK and that you both have a very Happy Christmas! 73, Nick
Appreciate the fact that you went through all the math. The only thing I didn't get really well was why use those condensers. I'm not an electronics engineer, just a computer engineer trying to learn a bit of electronics so I can make some DYI guitar pedals. Just came accross this and thought it was interesting, you can basically make a guitar power amp with this basic scheme.
Essentially the capacitors in this circuit all function so as to pass the frequency of interest and block DC. This means you need to have values of capacitance high enough to give a very low reactance at your lowest frequency of interest. I deliberately went over and above this in my design - as I explain. Just remember this is a small signal amplifier. Things are a bit different once you get into power amplifiers. Thanks for watching and commenting. Best wishes, Nick
Dear Nick, ... it could be, that I am the "number 7", looking your excellent 'show' here, ... ! Very interessting video, for all the 'newcomers'! This would be interested for all the 'new' people looking for the new 'beginners-licence' here in DL, since the last 3 month. Easy to understand ... and better to listen, I have ever heard before, ... ! Two bridges to go, in only one step! Learning 'English', (with your wonderfull and easy to understand speaking!) and doing the ground breaking 'electronic-knowlegement', (How does a small transistor amplifier work?) Side by side! - Wonderfull! - The new class is called 'N' in Germany, and so the 'beginners choise', for entering into HAM- radio! An 'up-date' from 'N', to class 'E' or even 'A', is possible and easy to do! Easier to work on, and not so 'technical', but also 'restricted' by the output oft the radio, and also the frequencies, that they could use. So Germany has a 'three class- system' again! From the 'N' (Newbie/Rookie/Entry- Class) over 'E' the 'in between' class, up to 'A', the top- notch class, where all is allowed, to the legal limits, of corse! (Hi Hi ! ! !) The class-'N' only restricted to 10m / 2m / 70cm with an EIRP of only 10W, ... but better than 'nothing'! This is, why it is called 'Newbie'- or 'Entry'- class! Better than doing 'illegale transmitting', on or near HAM- radio or CB-radio- frequencies! The easier the beginning of such new 'licence- courses' is, the better is the 'learning effect', and the number of 'new YL/OM', finishing these 'courses', and become an amateur- radio operator. I hope that a lot of interessted young people will up-date to higher classes, 'E' or 'A', ... so that HAM- radio ha a better chance to 'survive' in the jungle of internet and smart-phone euphoria! People are getting lazy, over the last three decades. ---> " One small step for a 'swl', but a giant leap into 'HAM- radio' ! " - something like that, ... !
Hi Markus, Thanks very much indeed for your kind words. Thanks also for filling me in on the Licensing changes in Germany - I wasn't aware this was happening. Hopefully it will encourage a few more folk to give ham radio a try. Please don't apologise for your English - it is MUCH better than my German! Nochmals vielen Dank. Machs gut, mein Freund. 73, Nick
Nice explanation Nick, It's a fairly easy process, but not that easy to explain to people coming into the task cold. This brings back so many memories of my first year electronics at college, with one exception it was ALWAYS hfe, never beta! which was reserved for those funny glass things with hot red leds in the bottom of them. 🙂 When you get to explaining the emitter bypass cap in the next episode, I will have a vision of my electronics lecturer (Mr Jones) jumping up and down on the spot to drive home the point of how that cap affects the gain. At the end of the day we also ended up just using the half supply rail voltage across Rc and 10 x the base current through R1 and R2 - It works. Andy
I agree with the previous accolades! I had been trying to figure out how the myriad of RF white noise generators and amps were designed but to no avail. This video was tremendously helpful because it showed the path and assumptions in the designs. I am anxious to watch the AC video! Thank you for taking the time to make this presentation!
Hi, over the past 12 months, I reckon I've watched every transistor video on UA-cam, trying to get the "penny to drop". With your detailed and very easy explanations and practical examples, I now get it. What a great video. It certainly wasn't too long as this interesting video seemed to be over in minutes. I was glued to the edge of my seat, as I was able to finally understand, put it all together and follow along with your examples with the transistor and resistor theory. Now for Part 2. Thank you for putting in so much time to put this video together, with your patience and easy to understand examples, you'd make a great teacher or instructor. I've also stumbled across another great channel to subscribe to. Cheers from Hobart, Tasmania, Australia. Darren
Hi Darren, thank you very much for your kind words of appreciation. I'm really pleased the video was helpful. It was a topic that I wanted to understand years ago but could never really get past all the maths. I suppose I try and make the videos that I wish I had discovered myself when I needed them! Best wishes for 2025!
Hi. I bought a few rf transistors and would like to design a rf amplifier of about 15db of gain but0 with 50 ohm input and output impedance. How to do that? I know that with only one stage it doesn't work. There may be no impedance matching stage in some place. The main frequency would be 915Mhz. Thanks
Hi Anderson, thanks for your question. I assume you mean a small signal RF amplifier (not a Power Amp - which is a whole different thing). If you watch the next video (No.58) on Small Signal Analysis you'll see the effect that the various components in the circuit have on input and output impedance. There are things you can do to try and design for certain input and output Z but failing that I'd go for a couple of impedance matching transformers. However, if you are building for 915 MHz then things get much more challenging in terms of the effects of stray capacitance etc. You're probably in the realms of striplines - where you create characteristic impedances by the length and width of the copper traces on your board. Honestly I can't really advise on this as all my experience is at much lower frequencies (HF) where homebrew is more forgiving. All the best to you. 73, Nick
@@M0NTVHomebrewing yes, I would like to amplify a small signal that leaves the SX1278 lora IC to an antenna. I know there are dedicated amplifier IC like Gali-5+, but I would like to use a transistor for that. I already have a NANOVNA and a Spectrum Analyzer to help me find an impedance match circuit. I'm learning how to make striplines with 50 ohm impedance characteristics from JLCPCB. These days I did a PCB in my CNC router that matched for a 915Mhz trace antenna for lora. Thank you very much.
Excellent.I am a beginner and this explained perfectly. However one query if anyone can help please. at around 09:50 you asser that we could get away with lowering Vcc power supply to 6v but wouldn't that then mean that rather than having 5.4 and 5.4 and 1.2 volts , the available voltage would be 6 volts minus 1.2 split over the output which means you would not have 5.4 you would have (6 - (1.2 /2))v = 2.4v. how then could you bias the RC output for 6 volts? furthermore if the supply is lowered to 6 volts , then how can the voltage be 6.6 volts at the collector side
Hi! I'm afraid I can't find any reference to lowering the VCC to 6V. If you did this you would have to recalculate everything. You need to start with your VCC value and then proceed from there in the way I describe. Thanks for watching. Best wishes, Nick
@M0NTVHomebrewing you said this at 42 mins (sorry I got the time wrong)m just play the vid from 42 mins and listen for 20 seconds you actually state quote "you could get away with halving the power supply unquote". as you correctly state in the previous comment this is wrong I just thought I'd bring it out
Thanks for the clarification. I see what you mean but I was talking about how to determine VC (the voltage at the collector). I spent some time accounting for the offset of 1.2V at VE which meant I set VC at 6.6V. But then I said you could "get away with halving the VCC and just going for 6V" i.e. there is no need to go into all that you could just divide the VCC by 2 and set THAT VALUE for VE = 6V. I was talking about how to calculate VE not changing the whole supply voltage. Sorry for any confusion. Hope that is clearer now. 73, Nick
@@M0NTVHomebrewing No probs I know you think you know what you were talking about , but it's just what you said was "half the power supply" exact words. clear what you said but anyway point made point taken and Keep up the good work making sure to be very clear and specific lol
Because we know we need 1.9V on the base (comprising VBE (0.7V) + VE (1.2V)). So that's why we've dropped 1.9V over R2. This means there is only Vcc - 1.9V remaining (i.e. 12 - 1.9 = 10.1V) and so this must all drop over R1. This is to satisfy Kirchhoff's Voltage Law about the the sum of all the voltage differences around a closed loop must be zero. Hope this helps. Best Wishes, Nick
Towards the end of your video, you review adding additional amplification stages. Are there any special design considerations for the final amp stage? I have seen multistage transistor amplifier designs online where stages 2-(n-1) were identical and the final stage used different resistor values. In my case, I am building a wideband RF noise generator that will pass the AC signal through 2 or 3 amplification stages. The final signal will be used with a Wheatstone bridge to tune an antenna. Thanks again for the great video!
In reality you'd probably want a more beefy power amplification stage - something like a push-pull with dedicated power transistors (TIP31C & TIP32C) in an audio amplifier. For RF it depends on how much power you need. I usually start at the end (PA) and work backwards i.e. what power does my PA need so what power does my driver need to produce? Then what power do I need to drive my driver etc. Hope this helps. All the best in your project. Best wishes, Nick.
@@M0NTVHomebrewing Thank you for your suggestions, Nick! I am building an RF noise bridge and I think your idea of working backward from the peripheral connections is spot-on. :). Happy New Year!
Hi there, it is because we want 10 x the base current running through the voltage divider to make it "stiff" i.e. it will swamp out the effects of Beta. You might need to go back and watch it again - which is what I regularly do when I'm trying to get my head round something. Hope it helps. Happy New Year! 73, Nick
Hi there! Good question. It was an arbitary choice. The proper way to establish the quiescent collector current is by Load Line Analysis which I really didn't want to get into in the video. In truth, my choice of 10mA is a bit higher than it should be. My own analysis shows that for my design, then 8.5mA is just about ideal. It'll still work with 10 though. If you are interested check out this excellent 4-part series on Load Line Analysis: ua-cam.com/video/dOv_nsBvYTM/v-deo.html. Thanks for watching and for your question. Best Wishes, Nick
Thanks very much. I'm running LTSpice (on a Mac). It's free and there is a huge user base which means plenty of people to ask when you get stuck! You can download it here: www.analog.com/en/resources/design-tools-and-calculators/ltspice-simulator.html?ADICID=PDSR_EMEA_Power-Management-LTSpice-NB_Google_Mult_202410&gad_source=1&gbraid=0AAAAACxqTx93ynCBN7P6ldtCDjBRBLK1q&gclid=Cj0KCQjwjY64BhCaARIsAIfc7YaCGr6t1mqlAvB6tgAb2VrqlOZJROvEP4T6aBsGwbSzZyeJz50zHy4aAkR4EALw_wcB
What a great video. Very clear. After watching many other videos on UA-cam, I think I now finally understand it. Or almost. I have one question. When you calculate RC, you want it to drop 5.4V, which I understand. But what I don't understand is, how can you assume that the transistor will drop the other 5.4 volts, and the emitter will stay fixed at 1.2V? In other words, what prevents RE from dropping part of that voltage and raise Ve?
Hi Willem, thanks very much. In answer to your question we are guided by Kirchhoff's Voltage Law which states that the sum of all the voltages in a closed loop must = 0. So we have the DC supply (= +12V) - Voltage drop over collector resistor (IC*RC = 10mA*540Ω = 5.4V) - Voltage dropped over transistor (VCE) - Voltage dropped over emitter resistor (IE*RE = 10mA*120Ω = 1.2V) = 0. Therefore VCE must = 12 - 5.4 - 1.2 = 5.4V. It will stay fixed from a DC point of view because what will change it? Once you add an AC signal then things get interesting but that is the topic of the next video - No.58 released yesterday. Hope this helps. Best wishes, Nick
@@M0NTVHomebrewing I think my expectation was that the resistance of RC would affect the current flowing through RE, and change the voltage dropped over RE. But I think Ve is actually fixed at 1.2V because it is necessarily 0.7V higher than Vb. And because the current flowing through the collector needs to be the same as the current flowing through the emitter, the voltage drop over RC can only be that current times its resistance (because ohms law). And the remaining 5.4 volts can then only be dropped by the transistor. Is that correct?
Hi Willem, I think so. The fact is that from a DC biasing perspective changing RC makes very little difference to the current in the emitter (IE) or even the collector (IC). Neither does it impact the voltage on the emitter (VE) or base ((VB) very much or the current through the base voltage divider. All it does is to drop more or less voltage over it and shift VC higher or lower. Just for fun, model it in LTSpice and set RC = 1Ω. Make a note of all the currents and voltages in the circuit. Then set RC = 1000Ω and do the same. The changes to everything except VC are tiny! The importance of RC comes through in the next video (58) where it not only sets the zero crossing line for our AC waveform but also pretty much singlehandedly sets the output impedance of the amplifier. Thanks again. 73, Nick
@@M0NTVHomebrewing I just tried it in LTSpice, and you're right, it has almost no impact! Not at all what I was expecting. I'll check out the followup video as well, thanks!
Well, I've watched a few times now to better understand and decided the best way to cement this topic is do my own presentation based on what you've done but using a beta I measured with one of my own transistors. Working out pretty well. Except for the current going through R1 and R2. I think I understand why there's a difference as you have explained, but so far my numbers there don't add up. I'm sure its just a math issue and I'll solve it without too much effort. Peter ve3poa
Hi Peter, well done for persevering! The current through the voltage divider just needs to be at least 10 times the base current. If you model it with LTSpice you can calculate what LTSpice thinks the Beta is (about 308 in my case). If you want the numbers to match up then you'll need to use this figure (not your measured one) in your calculations for R2 and R1. Real circuits with real components will also have some variations because of tolerance etc. This stuff can honestly take a while to get your head around. But once you've got it ...73, Nick
Thank you. This was very helpful :) > 3:19 Why polarized electrolytic instead of unpolarized electrolytic? (I see both used in some circuit diagrams and can't find an explanation) 49:40 where is the output C or E side? [Edit: I picked it up from video 56 :) ] > I am about to go back and watch 56. Trying to learn about creating some basic pre-amps, amps and filters for some home projects. This video explained it well combined with another by "Vocademy".
Hi there! The only reason I'm using electrolytic caps in this design is simply because of the higher value of capacitance required. At these values of hundreds of micro Farads then you're only really going to use electrolytics. 73, Nick
This particular design is an audio amplifier. Although you can put music through it I will only be using it in a Direct Conversion Radio Receiver so the range of interest will be 300 Hz - 3 kHz. You can of course calculate the capacitor values for other frequencies. If you are going to use it for RF though just be aware that you will need some impedance matching. Best wishes, Nick
Hey Nick, perhaps a simpler way to explain collector bias is simply halfway between Vcc and Ve... You got there, but you kind of went the long way around.
Hi Dean, thanks for watching and commenting. Yes, you can do (Vcc-Ve)/2. But when am I ever going to pass up the chance to go the scenic route?!! 73, Nick
1. The voltage gain of the circuit has nothing to do with β. The maximum voltage gain depends on the supply voltage (Vcc) and nothing else. You set the voltage across Rc to be about half of Vcc. Av(max) = Rc / re = (Vcc / 2.Ic) / (25mV / Ic) = Vcc / 50mV. However, the non-linearity of re creates distortion, so for a practical design, we "swamp" re by having an Re about 10 times bigger, showing that the highest practical gain from a swamped common emitter stage is when the emitter resistor has about 250mV across it, giving a gain of about Vcc / 500mV. With a supply voltage of 12V, you won't get a voltage gain of much more than about 12V/500mV = 24 without causing distortion. 2. The collector current is determined by the desired input and output resistances. The collector current is roughly half of the supply voltage divided by the output resistance (Rc). The resulting input impedance in kilohms is at most one tenth of β divided by the collector current in milliamps, assuming you pass ten times the base current (Ib) through R1 and R2 to swamp out variations in β. With a supply voltage of 12V and a collector current of 10mA, you expect an output impedance of 6V/10mA = 600R. If your 2N2904 has a β of 300, then you expect an input impedance of around 300/10 divided by 10mA = 3K. 3. The intrinsic emitter resistance (re) does not change from transistor to transistor. It is 25mV/Ic in every case for small signal transistors, where the bulk resistance is negligible. But remember that as the signal swings close to the positive rail, the collector current becomes small, increasing re, reducing the gain and causing distortion unless we use an unbypassed emitter resistor (Re) much bigger (say 10 times) the quiescent value of re. 4. You don't need 1V across Re. Since re = 25mV/Ic and Re can be 10 * re, it should be obvious that Re = 250mV/Ic. In other words, Ve = 250mV. You get diminishing returns by making it higher.
The thermal voltage is why CPUs need coolers! 😄 Without regulating the temperature, you risk something called thermal runaway-the main reason transistors release their magic smoke. 🔥 It's a self-reinforcing loop that grows exponentially. For the science nerds out there, the thermal voltage (Vt) is calculated using Boltzmann’s constant with this formula: Vt = kT/q. Here, k is Boltzmann's constant, T is the absolute temperature, and q is the elementary charge. The real trouble starts with the saturation current (Is), which depends on temperature like this: Is ∝ T³ * e^(-Eg/kT). Eg is the bandgap energy, and as T rises, the T³ factor increases fast, while the e^(-Eg/kT) part decreases slower. This means Is shoots up, which increases the emitter current (Ie). More current means more heat, which causes even more current-and pretty soon, _pop_ goes the junction! 😬
Thanks for the detailed explanation. I've only really troubled myself with thermal runaway when I'm building RF amplifiers which are handling more power. One of my favourite RF driver amp designs is by Eamon Skelton and can put out up to 1W of RF power. He helpfully includes a protection mechanism that will reduce the base current if the collector current goes too high. It does work pretty well and helps to keep the magic smoke inside! Thanks again. Best wishes, Nick
@@andymouseWith such resistor values, V of the base will be 2.05V (1.826 V LT Spice). This means that a signal with an amplitude higher than 2.05-0.7=1.35V will be cut off.
@@andymouseIn the scheme of the emitter repeater, the voltage at the base of the transistor is V=6.746 (LT Spice). Such transistors do not exist in nature. Greetings from Ukraine.
The intention is that you use that knowledge to design something (better) that you can use yourself. It was deliberately simple for that very reason. As my friend Pete Juliano (N6QW) often says, "When you know stuff you can do stuff". I wish you well in whatever you decide to build. 73, Nick
Bruh, I've been scowering the internet for 5 days looking for someone to explain the meat and potatoes of how amplifiers work, not just, "well this voltage and this current goes here and creates this" Thank you bro, the universe needs more people like you
Thank you very much. Glad it helped! Best wishes, Nick
Excellent video. Taught much better than my college electronics class. Thanks!
Thanks very much Dylon. I'm glad it made sense. Best wishes, Nick
You rarely find a "how to" video which goes literally step by step. I'm arty not techy and have poor maths... but followed this without too much difficulty. The bonus is that I can replay the video to revise and review.
I wish I had your way of teaching when I took the RAE exam in 83, I passed with double credits but didn't realise you could have AC and DC running on a circuit at the same time! Now I understand not only the "what" but the "why" as well. Thank-you Nick! 😄 👍
Thanks so much as always Ace. I'm pleased it was useful. Look after yourself. 73, Nick
Really great walk through of dc biasing Nick! Look forward to the second half.
Thanks Nate. Yes, I'll be starting work on the next one soon. Thanks for watching and commenting. Best wishes, Nick
Best explanation I've seen. Excellent teaching.
Thanks very much. Glad it was helpful. Best wishes, Nick
This is a wonderfully clear and easy to follow explanation.
Thanks very much. Best wishes, Nick
You are a natural teacher. Nice job explaining an often confusing topic.
Thank you Anthony! Best wishes, Nick
The first time I went through this was 1973. Now I think I have it! Thanks!
Thank you Louis. I have to admit that it has taken me a while to really get my head around all this stuff myself ... and I'm still learning I'm pleased to say. Glad it helped. 73, Nick
Yo dude, at it again with another amazing video!! You got me scouring YT for more, absolutely love it!!
Thanks Ed! Glad it was useful. Hope you're doing OK. Best wishes, Nick
Thanks Nick, I watched to the end, and learnt a few things!
Cheers Mike. Glad it was some help. 73, Nick
Many thanks for sharing and taking the time to explain it in great details!
Thanks Frank. Glad you found it helpful. Best Wishes, Nick
Very good information here to get those projects working, (and the grey matter!) Many people struggle with this very concept and it`s tempting to just grab an op-amp, until you need an amp at a high enough frequency to make op-amps too expensive or not suitable. Everything you need to get you going has been covered in this video. You do a great job to help keep this hobby alive. Chris, UK. (Gentlemen of 1977khz AM)
Thank you very much indeed Chris. I'm certainly not adverse to using an op-amp or two myself but I find there is something strangely empowering in understanding how the humble transistor amplifier works. Thanks again. 73, Nick
@@M0NTVHomebrewingCan you please explain why the Collector junction will be out of phase by 180 degrees?
Good question. It is because of the way the transistor operates. The phase inversion happens because an increase in input signal causes a corresponding decrease in the output signal, and vice versa, due to the interaction between the base-emitter current and the collector voltage across the load resistor. When the input signal increases and makes the base-emitter junction conduct more (increasing Ic ), the voltage at the collector decreases. Conversely, when the input signal decreases, the base-emitter junction conducts less (decreasing Ic ), and the voltage at the collector increases. This relationship results in a 180-degree phase shift between the input signal at the base and the output signal at the collector. Hope this helps. Best wishes, Nick
What a great video Sr. Thank you very much for the simple explanation I really appreciate it . I have assembled the simple transistor amplifier from the video 56 using my breadboard and I am truly happy, is the first time that I build something that is not some kind of a blinking led =D, the amplifier works great. I don't need your notes, I have copied everything to my notebook while watching the video hehe (lots of pause and watching again =)
I can't wait for the next.
Thank you
Hi Joao, thank you very much. I'm pleased that the video helped and very well done on breadboarding the amplifier. Take care. 73, Nick
The explanation starts at 3:12. Interesting video, well explained.
Thank you very much Heinz. 73, Nick
Put wonderfully simple, great. Thank you Nick.
Thanks very much. Best wishes, Nick
I like watching you vids, I am learning more Thank you
Thanks very much Jonny. I'm pleased the videos are helpful. Happy New Year! 73, Nick
Sweet! I ran my own hypothetical calculations with 10V and wanting 20ma/1v at Ve and built it on a breadboard. Took a few tries to go through the calculations and it was right as I wanted. I do find that calculating everything in the required units makes it a little easier than converting in my head on the fly, but I'm sure with some practice it would become 2nd nature.
Glad it helped. 73, Nick
I'm playing catch-up today mate, this is the second video, I'll save the next for tomorrow.
Excellent explanation of the biasing theory, thanks for that. One note, clipping is a danger for the speakers - not at the power we're talking about here but when you move up in power it becomes so. When the output clips it becomes a DC signal, the voice coils start dissipating a lot more power than they're designed for and they release that magic smoke.
There's an old saying - anything can be a smoke machine if you try hard enough.....
Thanks Al. That's a timely warning about clipping too so thanks for sharing that. Hope you are OK and that you both have a very Happy Christmas! 73, Nick
Appreciate the fact that you went through all the math.
The only thing I didn't get really well was why use those condensers.
I'm not an electronics engineer, just a computer engineer trying to learn a bit of electronics so I can make some DYI guitar pedals.
Just came accross this and thought it was interesting, you can basically make a guitar power amp with this basic scheme.
Essentially the capacitors in this circuit all function so as to pass the frequency of interest and block DC. This means you need to have values of capacitance high enough to give a very low reactance at your lowest frequency of interest. I deliberately went over and above this in my design - as I explain. Just remember this is a small signal amplifier. Things are a bit different once you get into power amplifiers. Thanks for watching and commenting. Best wishes, Nick
Long, but not long winded. Excellent ! 73 PE1LLA
Thanks very much Jacob. 73, Nick
Dear Nick, ...
it could be, that I am the "number 7", looking your excellent 'show' here, ... !
Very interessting video, for all the 'newcomers'!
This would be interested for all the 'new' people looking for the new 'beginners-licence' here in DL, since the last 3 month.
Easy to understand ... and better to listen, I have ever heard before, ... !
Two bridges to go, in only one step!
Learning 'English',
(with your wonderfull and easy to understand speaking!)
and doing the ground breaking 'electronic-knowlegement',
(How does a small transistor amplifier work?)
Side by side!
- Wonderfull! -
The new class is called 'N' in Germany, and so the 'beginners choise', for entering into HAM- radio!
An 'up-date' from 'N', to class 'E' or even 'A', is possible and easy to do!
Easier to work on, and not so 'technical', but also 'restricted' by the output oft the radio, and also the frequencies, that they could use.
So Germany has a 'three class- system' again!
From the 'N' (Newbie/Rookie/Entry- Class) over 'E' the 'in between' class, up to 'A', the top- notch class, where all is allowed, to the legal limits, of corse!
(Hi Hi ! ! !)
The class-'N' only restricted to 10m / 2m / 70cm with an EIRP of only 10W, ... but better than 'nothing'!
This is, why it is called 'Newbie'- or 'Entry'- class!
Better than doing 'illegale transmitting', on or near HAM- radio or CB-radio- frequencies!
The easier the beginning of such new 'licence- courses' is, the better is the 'learning effect', and the number of 'new YL/OM', finishing these 'courses', and become an amateur- radio operator. I hope that a lot of interessted young people will up-date to higher classes, 'E' or 'A', ... so that HAM- radio ha a better chance to 'survive' in the jungle of internet and smart-phone euphoria! People are getting lazy, over the last three decades.
---> " One small step for a 'swl', but a giant leap into 'HAM- radio' ! " - something like that, ... !
Hi Markus, Thanks very much indeed for your kind words. Thanks also for filling me in on the Licensing changes in Germany - I wasn't aware this was happening. Hopefully it will encourage a few more folk to give ham radio a try. Please don't apologise for your English - it is MUCH better than my German! Nochmals vielen Dank. Machs gut, mein Freund. 73, Nick
Very helpful - thanks for going through the calculations !
You are very welcome. Thank you for watching and commenting. 73, Nick
Nice explanation Nick,
It's a fairly easy process, but not that easy to explain to people coming into the task cold.
This brings back so many memories of my first year electronics at college, with one exception it was ALWAYS hfe, never beta! which was reserved for those funny glass things with hot red leds in the bottom of them. 🙂
When you get to explaining the emitter bypass cap in the next episode, I will have a vision of my electronics lecturer (Mr Jones) jumping up and down on the spot to drive home the point of how that cap affects the gain.
At the end of the day we also ended up just using the half supply rail voltage across Rc and 10 x the base current through R1 and R2 - It works.
Andy
Cheers Andy. Yes, it's true that sometimes the simple rule of thumb is more than sufficient! Thanks for watching and commenting. 73, Nick
Thank you! Greetings from Belarus.
Thank you very much. Glad it was of some interest. 73, Nick
I agree with the previous accolades! I had been trying to figure out how the myriad of RF white noise generators and amps were designed but to no avail. This video was tremendously helpful because it showed the path and assumptions in the designs. I am anxious to watch the AC video! Thank you for taking the time to make this presentation!
Thanks very much. Glad it was helpful. All the best for 2025! Nick
Hi, over the past 12 months, I reckon I've watched every transistor video on UA-cam, trying to get the "penny to drop".
With your detailed and very easy explanations and practical examples, I now get it. What a great video.
It certainly wasn't too long as this interesting video seemed to be over in minutes.
I was glued to the edge of my seat, as I was able to finally understand, put it all together and follow along with your examples with the transistor and resistor theory. Now for Part 2.
Thank you for putting in so much time to put this video together, with your patience and easy to understand examples, you'd make a great teacher or instructor.
I've also stumbled across another great channel to subscribe to.
Cheers from Hobart, Tasmania, Australia. Darren
Hi Darren, thank you very much for your kind words of appreciation. I'm really pleased the video was helpful. It was a topic that I wanted to understand years ago but could never really get past all the maths. I suppose I try and make the videos that I wish I had discovered myself when I needed them! Best wishes for 2025!
you did a better job than my physic teacher
Hi William, thanks very much for your kind words. I'm glad it made some sense! Look after yourself. 73, Nick
What makes you said that he did better job than your physics teacher? From the Philippines...
Hi. I bought a few rf transistors and would like to design a rf amplifier of about 15db of gain but0 with 50 ohm input and output impedance. How to do that? I know that with only one stage it doesn't work. There may be no impedance matching stage in some place. The main frequency would be 915Mhz. Thanks
Hi Anderson, thanks for your question. I assume you mean a small signal RF amplifier (not a Power Amp - which is a whole different thing). If you watch the next video (No.58) on Small Signal Analysis you'll see the effect that the various components in the circuit have on input and output impedance. There are things you can do to try and design for certain input and output Z but failing that I'd go for a couple of impedance matching transformers. However, if you are building for 915 MHz then things get much more challenging in terms of the effects of stray capacitance etc. You're probably in the realms of striplines - where you create characteristic impedances by the length and width of the copper traces on your board. Honestly I can't really advise on this as all my experience is at much lower frequencies (HF) where homebrew is more forgiving. All the best to you. 73, Nick
@@M0NTVHomebrewing yes, I would like to amplify a small signal that leaves the SX1278 lora IC to an antenna. I know there are dedicated amplifier IC like Gali-5+, but I would like to use a transistor for that. I already have a NANOVNA and a Spectrum Analyzer to help me find an impedance match circuit. I'm learning how to make striplines with 50 ohm impedance characteristics from JLCPCB. These days I did a PCB in my CNC router that matched for a 915Mhz trace antenna for lora. Thank you very much.
Sounds like you are on the right lines already. All the very best with your project. 73, Nick
@@M0NTVHomebrewing I'm doing a video to publish in my channel. I'm still learning how to make some editting.
Excellent tutorial, thank you.
Thank you very much Ronald. Best wishes, Nick
Thanks Nick. Enjoyed this video and looking forward to the on the AC portion.
Peter ve3poa
Cheers Peter. Thanks for watching and commenting. 73, Nick
Excellent.I am a beginner and this explained perfectly. However one query if anyone can help please. at around 09:50 you asser that we could get away with lowering Vcc power supply to 6v but wouldn't that then mean that rather than having 5.4 and 5.4 and 1.2 volts , the available voltage would be 6 volts minus 1.2 split over the output which means you would not have 5.4 you would have (6 - (1.2 /2))v = 2.4v. how then could you bias the RC output for 6 volts? furthermore if the supply is lowered to 6 volts , then how can the voltage be 6.6 volts at the collector side
Hi! I'm afraid I can't find any reference to lowering the VCC to 6V. If you did this you would have to recalculate everything. You need to start with your VCC value and then proceed from there in the way I describe. Thanks for watching. Best wishes, Nick
@M0NTVHomebrewing you said this at 42 mins (sorry I got the time wrong)m just play the vid from 42 mins and listen for 20 seconds you actually state quote "you could get away with halving the power supply unquote". as you correctly state in the previous comment this is wrong I just thought I'd bring it out
Thanks for the clarification. I see what you mean but I was talking about how to determine VC (the voltage at the collector). I spent some time accounting for the offset of 1.2V at VE which meant I set VC at 6.6V. But then I said you could "get away with halving the VCC and just going for 6V" i.e. there is no need to go into all that you could just divide the VCC by 2 and set THAT VALUE for VE = 6V. I was talking about how to calculate VE not changing the whole supply voltage. Sorry for any confusion. Hope that is clearer now. 73, Nick
@@M0NTVHomebrewing No probs I know you think you know what you were talking about , but it's just what you said was "half the power supply" exact words. clear what you said but anyway point made point taken and Keep up the good work making sure to be very clear and specific lol
35:00 why did you subtract the 1.9 Volts to calculate the R1?
Because we know we need 1.9V on the base (comprising VBE (0.7V) + VE (1.2V)). So that's why we've dropped 1.9V over R2. This means there is only Vcc - 1.9V remaining (i.e. 12 - 1.9 = 10.1V) and so this must all drop over R1. This is to satisfy Kirchhoff's Voltage Law about the the sum of all the voltage differences around a closed loop must be zero. Hope this helps. Best Wishes, Nick
@@M0NTVHomebrewing it helps thanks
Towards the end of your video, you review adding additional amplification stages. Are there any special design considerations for the final amp stage? I have seen multistage transistor amplifier designs online where stages 2-(n-1) were identical and the final stage used different resistor values. In my case, I am building a wideband RF noise generator that will pass the AC signal through 2 or 3 amplification stages. The final signal will be used with a Wheatstone bridge to tune an antenna. Thanks again for the great video!
In reality you'd probably want a more beefy power amplification stage - something like a push-pull with dedicated power transistors (TIP31C & TIP32C) in an audio amplifier. For RF it depends on how much power you need. I usually start at the end (PA) and work backwards i.e. what power does my PA need so what power does my driver need to produce? Then what power do I need to drive my driver etc. Hope this helps. All the best in your project. Best wishes, Nick.
@@M0NTVHomebrewing Thank you for your suggestions, Nick! I am building an RF noise bridge and I think your idea of working backward from the peripheral connections is spot-on. :). Happy New Year!
No worries. All the best with your building. 73, Nick
33:30 i don't get it why 10times ib, when find the R2
Hi there, it is because we want 10 x the base current running through the voltage divider to make it "stiff" i.e. it will swamp out the effects of Beta. You might need to go back and watch it again - which is what I regularly do when I'm trying to get my head round something. Hope it helps. Happy New Year! 73, Nick
i followed all the way but still a question , where did you get the 10 milliamp ?
Hi there! Good question. It was an arbitary choice. The proper way to establish the quiescent collector current is by Load Line Analysis which I really didn't want to get into in the video. In truth, my choice of 10mA is a bit higher than it should be. My own analysis shows that for my design, then 8.5mA is just about ideal. It'll still work with 10 though. If you are interested check out this excellent 4-part series on Load Line Analysis: ua-cam.com/video/dOv_nsBvYTM/v-deo.html. Thanks for watching and for your question. Best Wishes, Nick
What spice program are you using? Very useful information.
Thanks very much. I'm running LTSpice (on a Mac). It's free and there is a huge user base which means plenty of people to ask when you get stuck! You can download it here:
www.analog.com/en/resources/design-tools-and-calculators/ltspice-simulator.html?ADICID=PDSR_EMEA_Power-Management-LTSpice-NB_Google_Mult_202410&gad_source=1&gbraid=0AAAAACxqTx93ynCBN7P6ldtCDjBRBLK1q&gclid=Cj0KCQjwjY64BhCaARIsAIfc7YaCGr6t1mqlAvB6tgAb2VrqlOZJROvEP4T6aBsGwbSzZyeJz50zHy4aAkR4EALw_wcB
What a great video. Very clear. After watching many other videos on UA-cam, I think I now finally understand it. Or almost. I have one question. When you calculate RC, you want it to drop 5.4V, which I understand. But what I don't understand is, how can you assume that the transistor will drop the other 5.4 volts, and the emitter will stay fixed at 1.2V? In other words, what prevents RE from dropping part of that voltage and raise Ve?
Hi Willem, thanks very much. In answer to your question we are guided by Kirchhoff's Voltage Law which states that the sum of all the voltages in a closed loop must = 0. So we have the DC supply (= +12V) - Voltage drop over collector resistor (IC*RC = 10mA*540Ω = 5.4V) - Voltage dropped over transistor (VCE) - Voltage dropped over emitter resistor (IE*RE = 10mA*120Ω = 1.2V) = 0. Therefore VCE must = 12 - 5.4 - 1.2 = 5.4V. It will stay fixed from a DC point of view because what will change it? Once you add an AC signal then things get interesting but that is the topic of the next video - No.58 released yesterday. Hope this helps. Best wishes, Nick
@@M0NTVHomebrewing I think my expectation was that the resistance of RC would affect the current flowing through RE, and change the voltage dropped over RE. But I think Ve is actually fixed at 1.2V because it is necessarily 0.7V higher than Vb. And because the current flowing through the collector needs to be the same as the current flowing through the emitter, the voltage drop over RC can only be that current times its resistance (because ohms law). And the remaining 5.4 volts can then only be dropped by the transistor. Is that correct?
Hi Willem, I think so. The fact is that from a DC biasing perspective changing RC makes very little difference to the current in the emitter (IE) or even the collector (IC). Neither does it impact the voltage on the emitter (VE) or base ((VB) very much or the current through the base voltage divider. All it does is to drop more or less voltage over it and shift VC higher or lower. Just for fun, model it in LTSpice and set RC = 1Ω. Make a note of all the currents and voltages in the circuit. Then set RC = 1000Ω and do the same. The changes to everything except VC are tiny! The importance of RC comes through in the next video (58) where it not only sets the zero crossing line for our AC waveform but also pretty much singlehandedly sets the output impedance of the amplifier. Thanks again. 73, Nick
@@M0NTVHomebrewing I just tried it in LTSpice, and you're right, it has almost no impact! Not at all what I was expecting. I'll check out the followup video as well, thanks!
Well, I've watched a few times now to better understand and decided the best way to cement this topic is do my own presentation based on what you've done but using a beta I measured with one of my own transistors. Working out pretty well. Except for the current going through R1 and R2. I think I understand why there's a difference as you have explained, but so far my numbers there don't add up. I'm sure its just a math issue and I'll solve it without too much effort.
Peter ve3poa
Hi Peter, well done for persevering! The current through the voltage divider just needs to be at least 10 times the base current. If you model it with LTSpice you can calculate what LTSpice thinks the Beta is (about 308 in my case). If you want the numbers to match up then you'll need to use this figure (not your measured one) in your calculations for R2 and R1. Real circuits with real components will also have some variations because of tolerance etc. This stuff can honestly take a while to get your head around. But once you've got it ...73, Nick
Thank you. This was very helpful :)
>
3:19 Why polarized electrolytic instead of unpolarized electrolytic? (I see both used in some circuit diagrams and can't find an explanation)
49:40 where is the output C or E side? [Edit: I picked it up from video 56 :) ]
>
I am about to go back and watch 56.
Trying to learn about creating some basic pre-amps, amps and filters for some home projects.
This video explained it well combined with another by "Vocademy".
Hi there! The only reason I'm using electrolytic caps in this design is simply because of the higher value of capacitance required. At these values of hundreds of micro Farads then you're only really going to use electrolytics. 73, Nick
@@M0NTVHomebrewing Thank you. Much appreciate the explanation :)
👍Thank you sir.
You are welcome!
Range of frequencies?
This particular design is an audio amplifier. Although you can put music through it I will only be using it in a Direct Conversion Radio Receiver so the range of interest will be 300 Hz - 3 kHz. You can of course calculate the capacitor values for other frequencies. If you are going to use it for RF though just be aware that you will need some impedance matching. Best wishes, Nick
Hey Nick, perhaps a simpler way to explain collector bias is simply halfway between Vcc and Ve... You got there, but you kind of went the long way around.
Hi Dean, thanks for watching and commenting. Yes, you can do (Vcc-Ve)/2. But when am I ever going to pass up the chance to go the scenic route?!! 73, Nick
Superb
Thank you very much. Best wishes, Nick
Thankyou
You are welcome Simon. Best wishes, Nick
1. The voltage gain of the circuit has nothing to do with β. The maximum voltage gain depends on the supply voltage (Vcc) and nothing else.
You set the voltage across Rc to be about half of Vcc.
Av(max) = Rc / re = (Vcc / 2.Ic) / (25mV / Ic) = Vcc / 50mV.
However, the non-linearity of re creates distortion, so for a practical design, we "swamp" re by having an Re about 10 times bigger, showing that the highest practical gain from a swamped common emitter stage is when the emitter resistor has about 250mV across it, giving a gain of about Vcc / 500mV.
With a supply voltage of 12V, you won't get a voltage gain of much more than about 12V/500mV = 24 without causing distortion.
2. The collector current is determined by the desired input and output resistances.
The collector current is roughly half of the supply voltage divided by the output resistance (Rc). The resulting input impedance in kilohms is at most one tenth of β divided by the collector current in milliamps, assuming you pass ten times the base current (Ib) through R1 and R2 to swamp out variations in β. With a supply voltage of 12V and a collector current of 10mA, you expect an output impedance of 6V/10mA = 600R. If your 2N2904 has a β of 300, then you expect an input impedance of around 300/10 divided by 10mA = 3K.
3. The intrinsic emitter resistance (re) does not change from transistor to transistor. It is 25mV/Ic in every case for small signal transistors, where the bulk resistance is negligible.
But remember that as the signal swings close to the positive rail, the collector current becomes small, increasing re, reducing the gain and causing distortion unless we use an unbypassed emitter resistor (Re) much bigger (say 10 times) the quiescent value of re.
4. You don't need 1V across Re. Since re = 25mV/Ic and Re can be 10 * re, it should be obvious that Re = 250mV/Ic. In other words, Ve = 250mV. You get diminishing returns by making it higher.
Hi Rex, thanks very much for taking the time to give such detailed comments. I appreciate you doing that. Best wishes, Nick
The thermal voltage is why CPUs need coolers! 😄 Without regulating the temperature, you risk something called thermal runaway-the main reason transistors release their magic smoke. 🔥 It's a self-reinforcing loop that grows exponentially.
For the science nerds out there, the thermal voltage (Vt) is calculated using Boltzmann’s constant with this formula: Vt = kT/q. Here, k is Boltzmann's constant, T is the absolute temperature, and q is the elementary charge.
The real trouble starts with the saturation current (Is), which depends on temperature like this: Is ∝ T³ * e^(-Eg/kT). Eg is the bandgap energy, and as T rises, the T³ factor increases fast, while the e^(-Eg/kT) part decreases slower. This means Is shoots up, which increases the emitter current (Ie). More current means more heat, which causes even more current-and pretty soon, _pop_ goes the junction! 😬
Thanks for the detailed explanation. I've only really troubled myself with thermal runaway when I'm building RF amplifiers which are handling more power. One of my favourite RF driver amp designs is by Eamon Skelton and can put out up to 1W of RF power. He helpfully includes a protection mechanism that will reduce the base current if the collector current goes too high. It does work pretty well and helps to keep the magic smoke inside! Thanks again. Best wishes, Nick
@@M0NTVHomebrewing True, it's not such a concern with a properly designed small signal amplifier, especially in the pre-amp stage.
It’s entertainment for the “nerds”…. I wouldn’t have it any other way ….
Thanks Cesar! Best wishes, Nick
It takes long time to explain a smaal item
It certainly does ... or it takes ME a long time anyway! Happy Christmas! 73, Nick
(12:(16+3.3))×3.3=2.05 V; 2.05-0.7=1.35V;
Not enough for class A amplifier.
(12:(8+12))×12=7.2V.
2n3904 will be burned.
Hi Mike, thanks very much for watching and for sharing your thoughts. Best wishes, Nick
Your notation is odd to me please explain further.
@@andymouseWith such resistor values, V of the base will be 2.05V (1.826 V LT Spice). This means that a signal with an amplitude higher than 2.05-0.7=1.35V will be cut off.
@@andymouseIn the scheme of the emitter repeater, the voltage at the base of the transistor is V=6.746 (LT Spice). Such transistors do not exist in nature.
Greetings from Ukraine.
@@mikepaul2831 Thanks :)
Now its designed, what can I do with it, chuck in a bin. Pretty useless thing.
The intention is that you use that knowledge to design something (better) that you can use yourself. It was deliberately simple for that very reason. As my friend Pete Juliano (N6QW) often says, "When you know stuff you can do stuff". I wish you well in whatever you decide to build. 73, Nick
Top Nick! Iu2paq
Thanks very much Alfredo! 73, Nick