Thanks Charlie … very useful for reviving my half forgotten radio amateur’s exam studies. I need to do more work on designing the collector load for RF amplifiers.
Charlie, one of my fav electronics books is by Malvino, picked it up used, "Transistor Circuit Approximations", that teaches that the resistor divider common emitter amps are beta independent. It uses such simple equations to determine bias, the book is really cool.
I think that "Semiconductor Circuit Approximations," by Malvino, is an updated edition. It was published in 1985 and includes some information on IC circuits. I picked up a used copy ("very good" condition) from Abebooks for $7.57, shipped. Thanks, edbeckerich3737, for the tip!
Hi Ram. I'm pleased you found it useful. Please note that I am now using the geometric (SQRT[min HFE x max HFE]) mean for determining beta off the spec sheet. That approach seems to work well.
This looks like a excel spread sheet is needed with the rules of thumb being the calculations and just drop in transistor info with voltages. I am surprised this is not already available along with many commonly used devices for ham radio. Certainty the info is inside simulation modeling software but it would be useful to have it unmasked and simply available. Thanks for the video!
Thanks Kenneth. I expect this is somewhere on the web. I quite like breaking out the ol-HP48G and running the numbers from scratch. Thanks for the input.
Bit of a late reply - only just came across the video and thanks for posting. The approach you have taken reminds of methods I was taught years ago at "school" and techniques I have seen in many text books. I too have wondered why few datasheets contain the characteristics graphs. What I have found from limited homebrew experience is that the HFE values for transistors can vary considerably over the specified range. Maybe that's why manufacturers are reluctant to put the graphs on the datasheets? Also, some of the datasheets values will be valid only for very specific test conditions (e.g. specific Ic and Vce). So, unless I measure the HFE of a transistor I am going to use, I would normally go with the specified minimum HFE for calculating the dc base bias resistors. Also, especially for transistors like the 2N3904 that don't have a super high transition frequency, you can of course expect the hfe to drop off as frequency increases. So, some expirementation, reliance on previous experience and reference to tried and tested designs is what I would go for.
CHARLIE MORRIS, Does the AC/DC load line mean the maximum output wattage power of the transistor or does the AC/DC load line mean the power dissipation of the transistor? Because the AC/DC load line is selected by the EE designer of the operating voltage and operating current for the transistor and the DC bias Q point which will determine the maximum output wattage power out of the transistor and the power dissipation of the transistor. I'm not sure what the LOAD means in the AC load line and DC load line what they are calling the LOAD.
Hi Charlie Been playing with 'traditional' calculations for this specific biasing technique... I too come from the 'old school' days of electronics teaching. What my own calculations seem to suggest, is that your rule of thumb is pretty spot on. That rule of thumb also appears to be confirmed when running modern tools such as LT Spice (XVII). The only variation I did come across was Rc. In both the 'traditional' calculations, online biasing calculators (yes, they do exist) and, LT Spice, the best overall gain came with an Rc (purely resistive) value of between 450 and 650 ohms. Playing out that value, in LT Spice, I replaced that purely resistive load (Rc) with a 9:1 transformer bringing, in this case, 450 ohms down to 50 ohms modeling an FT-37-43 with 9T:3T in LT Spice. The gain measured into a 50 ohm resistive load was identical to that measured at the collector output with a purely resistive Rc. As I have never, in real world experimenting, encountered an Rc larger than ~300 ohms with a 2N3904 based amplifier, indeed, I typically reference 200-250 ohms for Rc, I attribute this to a couple of things. First is that LT Spice, as good as it is, is still just a simulation. Second, there is the issue of the internal collector resistance. Lastly, I could be totally wrong! :p One last thing is the debate over beta vs hFE. It would seem hFE (DC current gain) is the proper reference. I've seen beta and hFE used interchangeably and, I've seen arguments that they are not necessarily related. Given that datasheets today rarely list a beta value and do list hFE, using the latter seems to make more sense, as does your choosing a design hFE of 200 with the 2N2904. In any event, as I stated previously, it seems your rule of thumb is spot on, in my opinion. The real world operational results of your methodology would seem to support that it is correct as well! Great video! 73, Robin - AC7LX
HI Robin. Firstly thank you very much for the reply, it was very useful. So, from your experience, do you just set a value of Rc to be-say-450ohm or do you determine that each time by finding the open circuit Vo, then Vo/2 across a variable resistors (as Lar's suggested a video or two back)? Also, for determining hFE from the spec sheet, have you across geometric mean before? I seem to recall way back working out the hFE by SQRT(min hFE * max hFE), and not (min hFE + max hFE)/2 . I'm about to start the mic amp so I might as well use a refreshed design approach and 'rules of thumb'. Thanks again Robin. Charlie
Hi Charlie, Actually, until about a month ago, I had basically assumed an Rc between 200-300 Ohm. That came from looking at other designs and, a datasheets (I don't recall which one), where there was a reference to some level using an Rc of 275 Ohm. Later, I found and watched a video by W2AEW on determining an unknown output impedance from a signal source. It was then I had started experimenting with different values of Rc (or RL) to discover at what point the signal was half of the unloaded collector. I also discovered there was a lot about amplifier design I needed a refresher on! After your video, I tried something similar to see what Rc got the idle voltage at the collector to Vcc/2 for a given design and then, depending on what the amplifier end use, would adjust to feed a given impedance or, in the case of RF, calculate an impedance ratio to match to, in the case of RF, 50 ohm. I kind of like LT Spice for that because you can try things without destroying a device. Pretty sure I'd have some issues with simply tying the collector to Vcc in this design. In actual experiments I have started with some ridiculously high values for Rc to attempt to obtain something like an open/unloaded collector and then see what value for Rc brings the signal to 1/2 of unloaded. The apparent difference between the two methods is that one is looking for half of the unloaded signal and the other is looking for half of Vcc at the collector with no signal. I kind of prefer the latter because it can be calculated before attempting. Honestly, I should be better at this stuff, but it's been 40+ years since school and 10 years since I retired and took a break from radio, only renewing my interest a couple of years ago. As far as the hFE calculations, I've never done the sqrt calculation, but it seems to end up with similar/close values. ~172 vs 200. However, one never knows which is better until one sees the results. It is certainly interesting. Most of the time I use Spice to be close and then work with an actual circuit. Probably doesn't help much, but there it is. :D
Hi Charlie, Just a followup on my last comment. Simply put, I screwed up! LOL! I think I know how LT Spice came up with that high Rc value. I (think), that it was because I was looking at Rc as 'the load', when what I was apparently doing was looking at an open load condition. through a capacitor. I replaced the high value resistor with a reasonable inductance (~10uh), measured the collector output again and got the same value as with the high value resistor. I then coupled the collector (in that configuration) via a 100n cap to a variable resistor (load) and discovered what I had always assumed. The output impedance of the 2N3904 is indeed, around 190-240 ohms, at least in LT Spice. Interesting brain fog on my part, but now the world seems to make sense again. Well, at least where this stuff is concerned! LOL! 73, Robin - AC7LX
Cheers Robin. That 190-240ohm falls nicely around the 200ohm rule of thumb in the SSDRA and EMRFD texts. I do like Lar's process of determining the actual output resistance through test. Thanks again for the update! Charlie
Hi Charlie, great video as always. I think your idea of the quiescent voltage on the emitter is not right. At 3:44 you explain the 1V rule of thumb because the working line should not enter the cutoff region. I think this (your way of explaining) is not right, because Ve is completely external to the transistor. The diagram you use shows Vce, which is the emitter-collector voltage, completely inside the transistor. You could lift the voltages around the transistor any amount you wish, as long as you keep the Vbe and Vce as designed. I always design Ve to give me the ability for some swing on the emitter, although this will not be very much. But, this may not be the best way. Cheers, Koen
You are dead right Koen. I was lying in bed last night reflecting on the video and realised what i had said was incorrect. I will update the video description to make that clear. In regards to your designs, do you run with approx. 1V on the emitter or 10% Vcc? Charlie
As far as I know, the external emitter resistor is only needed to swamp the internal emitter resistor (little r sub e), which is temperature dependant. Without external emitter resistor, the amplification of the common emitter transistor amplifier would be highly dependant on temperature. Also, the external emitter resistor gives some negative feedback, which helps to minimise distortion. I just choose Ve to be somewhere between .5 and 1 Volts.
@@MrKoenvandijken You're right that the emitter resistor is there to dampen the effects of temperature and variations in the transistor's parameters from sample to sample. The real problem with omitting an emitter resistor (or bypassing it completely) is that re depends on collector current and for large output swings, the gain for the positive half of the signal will decrease, while the gain for the negative half will increase, producing serious amounts of distortion. Since the internal emitter resistance (re) is 25mV/Ic, we can choose Re to be at least ten times that (as a rule of thumb) in order to swamp out variations in re. But that means its value is at least 250mV/Ic, which immediately leads to the conclusion that it requires 250mV across it. If you can ensure that the base bias point is "stiff" (i.e. doesn't change much with changes in the transistor's β or its Vbe from sample to sample) and repeatable (i.e. you use decent low tolerance resistors whose value won't change), then you can afford to bias the emitter as low as 250mV and do away with the bypass capacitor completely. Otherwise, you have to split the emitter resistor and have 250mV across the unbypassed part and several times that in the bypassed part, depending on how "floppy" your base bias point is. Maybe a volt or so at the emitter turns out to be needed, but using 1% resistors and something like a BC547C, which has a very high β with a narrow variation, might allow you to get down to a few hundred millivolts at the emitter without problems.
@@CharlieMorrisZL2CTM As I commented to Koen, modern, high-β transistors and 1% resistors will allow you to push down the voltage at the emitter. It doesn't depend on Vcc. By that I mean, the lowest bias points you can get away with are the same whether you have a 6V supply or a 30V supply. Higher supply voltages allow larger gains and larger output swings, but don't affect other design parameters. The "knee" in the Ic vs. Vce curves is just the saturation voltage of the transistor, which you can get from the data sheets, typically about 0.2V for general purpose transistors. Then the lowest voltage that the collector can swing to is simply Ve + Vcesat (the voltage you set on the emitter + the saturation voltage). For maximum output swing, you have to bias the collector to be at (Vcc + Ve + Vcesat) / 2 volts.
This is an older video but I would like to offer some perspective that I feel is being lost by the video and the comment section, and would like to offer ideas that few use. First: All of these 'hallowed'-calculations that are taught, are for designing for production NOT performance! Different transistors will perform differently (even within the same-type and even same-lot!) at different Icq's, frequencies and output impedances. Second: This is driven home, by transistor Beta variations that can exceed 100% for a given transistor even within the same manufacture datasheet! If you want to properly quantify performance, it is a fool's errand to try and use the widest geometric-mean or the lowest value etc. for beta. Another example is the 10X rule for the input DC divider, this is really only a guarantee for lousy transistors within a lot. Third: There is this magical Vcc (in this case 13.8V) that is not real world unless you are never going to use it off that perfect power-supply. Many times, you will be using a battery that can be a 14.3V down to 11.7V (or even worse) say in an emergency. So, a circuit worth its salt has to perform at any one of these to be practical. Some of the major ham equipment manufactures are terrible, requiring boost circuits even when using a partially charged battery! Fourth: External component tolerance variations (especially when all their statistical variations conspire to give the worst outcome!) will play a significant role in the exact circuit performance. Lastly, if you design using one of these 'production'-calculated/derived circuits methodologies, what do you have. Well you can claim that the circuit will work with the crappiest transistor out of the lot; the worst resistor tolerance variation/degradation. Does that sound like designing for performance--No! Moral-of-the story: The real answer for the hobbiest (not-production) is to think in terms of greater precision and to set ourselves up for success. How? First off, buy a bunch of your favorite transistor and grade them for good performance and tight similarity on a unit-to-unit basis. Constructing a high-frequency amplifier or oscillator will allow you to more accurately quantify performance, than calculating geometric means and worst-case values from a datasheet. So, out of a hundred transistors you will usually find 75% of them are really tight with each other--use the rest to drive LED's or relays! Then come up with your own specifications as if you were in charge of doing so for your sweetheart lot. Finally, the most important concept, is empirical testing of your circuit. Every engineer should sit down and vary all the circuit parameters just to learn and observe EXACTLY what your circuit REALLY does. When I was playing around with QRP gear I quickly found that using 10mA quiescent-points quickly added up and did NOT add ANYTHING to the NEEDED circuit performance. You will find that you can use much less current whether quiescent or input divider (because you don't have wild beta-variations!). Many times a free-lunch can be had with higher-gain transistors but make sure everything stays stable! Make a habit of measuring ALL of your external components for tolerance. Most importantly, lower the operating voltage and make sure the circuit still performs (oscilloscope patterns etc.). You will find it fun to push the circuit elements to the max when you think in terms of performance NOT production. 73...
In a battery powered environment 10 ma in a multistage amplifier can add up pretty quickly. What about the input impedance which will be dependent on R1 and R2?
That's certainly true. I'm going to design all of the amps at 1mA for the Go QRP rig and see how that works. I chose 10mA for this one just for the exercise.
In another reply, I noted that the input impedance also depended on (β x the total emitter resistance), and showed that if you bypass the emitter resistor, it could be as low as 250 ohms. Reducing the collector current chosen by a factor of X will increase the input impedance (and also the output impedance) by the same factor. Adding an extra emitter resistor that is not bypassed can also significantly increase the input impedance (as well as reducing the distortion and gain).
It seem that ALL similar tutorials (class A amp) start form the "middle" - electing Ic. The amp "amplifies" - controls output POWER using input POWER. Hence the design objective should be HOW much amplification is desired. Perhaps that would improve on some of those "rules of thumb" .
You are correct. The PA videos later on take that approach. Please note that these are not tutorials, but simply a video log of what I have done. I am not an expert. The whole aim of the videos is to encourage others to give homebrew a go. Thanks again.
Thanks. I am no expert, and this uses a number of rules of thumb. The sim is to make it simple and hopefully attractive for other to give homebrew a go.
@@CharlieMorrisZL2CTM No, you really made it easy for me. Thank a lot. I was stuck somewhere at collector, and needed to match the impedance of Rl with Rc. Your approach to biasing seems a lot more easier than reading through several books, which actually complicate things. Also, I just wanted to know more about constructing RF transformers to match impedance between the stages. Coz, this is actually the device that effects the overall efficiency of the circuit. So, requesting your next video, please. And thank you once again for your reply.
Hi Charlie -thanks for video The approach I have used in the past is to cheat ! Go to old service manuals find the device in question and a circuit close to the one you want ,copy it and alter the value's you need That way you will be quite close to what's needed Cheating -yes,but it works ! What do you think?, or should I be drummed out of the Ham community ! Thanks again
Charlie was talking about designing an amplifier stage for use on a radio frequency band that is normally called the "80-metre band", which has been allocated for amateur use. It extends from 3.5MHz to 3.9MHz or 4.0Mhz, depending on which region of the world you are in. There's plenty of information on the web if you want to search for it.
Hi Bill. I don't know the answer to that off the top of my head. Current would be a function of the load hanging off the cap once it is charged up. Full charge will be 5x CR (capacitance x resistance). Same for discharge.
@@CharlieMorrisZL2CTM without the LOAD, how much current in amps can a 4700uf hold store with a power supply DC voltage at +450vdc? the 4700uf filter cap will charge up to +450vdc but how much current is the 4700uf holding storing? What formula can I use to calculate how much current the filter cap is storing in amps?
Bill. I hope I have interpreted your question correctly. In a DC system current does not flow through the capacitor. When a voltage is applied, the capacitor will charge up to that voltage. Once it is charged, the capacitor will hold a charge (in coulombs) equal to Capacitance x Voltage (recall, charge a coulomb is the charge (Q) transported by a constant current of one ampere (A) in one second (S), i.e., 1 Coulomb = 1 Ampere for 1 Second). Now for your capacitor, the charge will be 4700uF x 450V = 2.115 coulombs. Similar to a battery, that capacitor could provide, once a suitable load is applied, 2.115 amps for 1 second, or any other combination (e.g., 1.058 amps for 2 seconds). Note too that the time to fully discharge will be approx. 5x (Capacitance x Resistance) seconds.Not sure if this answers your question. Apologies if I have misunderstood your question.
@@CharlieMorrisZL2CTM yes for the formula is Capacitance X Voltage = Current Stored? Why I ask is because the AC ripple voltage will increase higher when the load is drawing more current. Any reasons why the AC ripple will increase when the Load draws more current? The Load is drawing the current OUT of the filter cap 4700uF
Hi Bill. I'm not expert but if I recall the filter caps should be providing adequate current to the load during the negative portion the supply cycle. If the ripple is increasing under load, then that suggest you don't have adequate capacitance. Have you tried another capacitor in parallel?
Good Topic and a well done video, I recently discovered your channel and I left a comment regarding your casecod amp which was also a good video. Good enough that it got me thinking and a little bit motivated. May I ask what your education is? I have a BSEE from an ABET accredit school (SVSU), graduated in 92 and I am also an extra class amateur licensee My concentration of study was control systems so I am almost useless when it comes to RF design but I have recently become very interested in the subject and I have been devouring every text book, web site and you tube video that I come across. The best text, thus far, that I have come across is "RF circuit design by Chris Bowick". I referenced that text in regards to this video and like you he really stressed the importance of setting the proper operating point. His analysis focused on maintaining a stable bias point vs temperature and while doing so made some assumptions (regarding hFE / hfe ) while I think that understand why, he makes no mention why he did it. His example uses a 2N5179 and he just stated it to be 50 (its range is 25 to 250) and you did a similar thing choosing your Beta but using characteristic curves and a load line. As I recall from school that parameter is very hard to control in manufacturing and can vary wildly, there are design technics one can use to minimize the affects, I will have to see if I can dig up my notes from my college days. Aside from picking that value, which can be arbitrary, you did everything else by the book. I can't wait until the next video, thanks for sharing. 73
Hi Michael. Firstly thanks for the feedback. I did 30 years in the Air Force, where I started as an avionics technician. All of my basis electronics theory was taught there. Later on I did a BE in electronics, but never worked as a design engineer. It's playing around at home where I try and put those lessons from years ago to good use. I must look up that RF circuit design by Chris Bowick book too. As for beta, I took a simple mean between the min and max values out of the spec sheet. If i recall, taking the geometric mean (square-root of min*max) maybe a better option. As an aside, I don't use characteristic curves, but tend to start just with the spec sheet. I think i will start to use 10mA as a quiescent Ic, as that seems to be a good option. Again, thanks to you and the others for taking the time to leave a comment. I really do appreciate it. 73s Charlie, ZL2CTM
Charlie, thanks for getting back to me, I think I have just what you need. Are you familiar with w2aew youtube channel ? He has a few videos on this topic. Video # 182, and #272, here is a link to his Bias video # 182 th-cam.com/users/results?search_query=w2aew+185
@@CharlieMorrisZL2CTM Because β does vary in manufacturing, it makes sense for the hobbyist to look for transistors that are gain-selected. I currently prefer BC547C for general use which has been selected to have a β between 420 and 800. That's a tight range and typically remains independent of collector current from 1mA up to 100mA. You can design with those and assume a minimum β of 420 safe in the knowledge that the calculated base current won't be any less than half the value calculated. If you use 10mA as the quiescent collector current with a transistor like the 2N3904 (min =100) and bypass the emitter resistor, you can end up with an input impedance looking into the base of 100 x re = 100 x 25mV/10mA = 250R. That will be the dominant value in determining the stage input impedance since R1 and R2 were found to be 22K and 3.9K, and all three are in parallel. It should be obvious, I hope, that reducing the collector current by a factor of 10, i.e. to 1mA will reduce all of the currents by the same factor, and hence all of the resistances and the input impedance will be increased by a factor of 10. For RF use, you can match stage impedances with transformers, so it makes sense to select a Ic that makes the parasitic capacitances negligible and 10mA might be sensible, or maybe a lower Ic might be just as good. For audio use, an input impedance of 250R would not be a good idea because of the size of coupling and bypass capacitors needed to get a flat response down to 20Hz. For audio, an Ic somewhere between 100μ and 2mA would be a more common choice
Hi. Sorry for the late reply. The main aim I was going for by matching the load to the collector circuit was to achieve maximum power transfer, as opposed to maximum voltage gain (as you point out). Charlie
Hi. I don't have time to make a video sorry. Here are some generic biasing calcs for a 12VDC system, with 10mA quiescent current through the BC547A. See if you can follow the logic. From the BC547 spec sheet the hFE range for the BC547A is 110-240. I now use the geometric mean for biaising calculations, so Beta-DC = SQRT(110 x 240) = 162. Let Ve = 1/10Vcc = 1.2VDC. Thus Re = 1.2 / 10mA = 120ohm. R2 = (1.2V + 0.7V) / (10 x (10mA / 162)) = 3078ohm (try 3k). R1 = (12V - (1.2V + 0.7V)) / (11 x (10mA / 162)) = 14874ohm (try 15k). If you are trying to determine input impedance etc, then you will need Beta-AC. From the spec sheet Ft = 300MHz. So Beta-AC = 300MHz / the highest freq of your amplifier operation. The input/output coupling capacitors, emitter bypass capacitor etc are a function od what frequency you intend to run the amplifer at (i.e., audio versus RF). A lot of the more recent videos and the blog (zl2ctm.blogspot.com/) have examples of that. Hope that helps. Charlie
Hi Madalin. You are right that those transistors would be more suitable, but the approach I have taken (right or wrong) for the video series is to readily available parts in an effort to encourage others to give homebrew a go. I guess I'm trying to demonstrate that a functional radio can be build from common parts. I can see another build coming on using the parts you identified as a way of showing performance differences. Thanks for the input
If my recall my theory correctly, the low freq performance is set by the input coupling capacitor. I typically use a 100nF cap which works well for the frequencies I'm using. Otherwise, set the capacitive reactance to be 1/10th Rin.
@@CharlieMorrisZL2CTM Thanks. I will look into both of those. Currently using an AC supply of 0.5v at a frequency of 10K Hz then have an Rin if 300. So I should either have a capacitive reactance of 30 or have a 100nF. Thanks for the great video and fast reply man.
Thank you for the video! I've put together an interactive notebook that can calculate the circuit using that method given the input values. Might be useful to someone: observablehq.com/@flpvsk/calculating-a-common-emitter-amplifier-circuit Someone has asked that already, but I didn't quite get the answer. How do we know the operating frequency? Would that be 20Hz - 20kHz for audio? And so for the Ce we'd use 20Hz, is that right?
The collector current determines the input and output impedances, nothing else. The high limit occurs when the transistor's maximum power dissipation of 625mW is approached. The low limit occurs when the parasitic capacitances' reactances at the frequencies of interest become significant compared to the resistors used. Around 13:30 you should realise that if you fully bypass the emitter resistor, the ac gain of the stage becomes RL/ re, where re is the internal emitter resistance (the slope of the Vbe/Ie curve at that collector current and equal to 25mV / Ic). But you make the mistake of assuming that Re determines the value of the bypass capacitor. It doesn't. Until the reactance of the bypass capacitor is less than re (which is just 2.5 ohms at 10mA collector current), the gain will increase with frequency because it becomes RL / (re + Xc). The capacitance would actually have to be C = 1 / 2π x 3.5MHz x 2.5R = 18nF, rather than the 3.3nF your calculation would indicate. if we used your value, the gain at 3.5MHz would be RL / (2.5 + 13.8j) which is phase-shifted, highly frequency dependant, and 15dB down on what it could be. If you split the emitter resistor so that a part of it is unbypassed, then as a rule of thumb, that part should be ten times re (25mV/Ic), say 27R. That will remove much of the non-linear distortion and make the ac gain more predictable. Then you use the 27R to calculate your bypass capacitor, of course.
That sounds interesting Madalin. Do you have a recommendation in mind? It would be interesting to see how much an actual device differs from the spec sheet based mean.
Yes a beta-meter with difrent collector current test condition like this sound.whsites.net/project31.htm.Of course your designs must be dependent on beta to other viewers to replicate schematics,not everyone have a complex betameter to design the best pesformance of the transistor which they have at their disposal.
![Gain block RF Amplifiers - Theory and Design [1/2]](/img/n.gif)
Sir, you explained all things even better than what learn in an university! Thank you sir
Thanks! I know there will be errors, but it works for me.
Thanks Charlie … very useful for reviving my half forgotten radio amateur’s exam studies.
I need to do more work on designing the collector load for RF amplifiers.
Charlie, one of my fav electronics books is by Malvino, picked it up used, "Transistor Circuit Approximations", that teaches that the resistor divider common emitter amps are beta independent. It uses such simple equations to determine bias, the book is really cool.
Cheers Ed. I'll jot that down and take a look.
Charlie
I think that "Semiconductor Circuit Approximations," by Malvino, is an updated edition. It was published in 1985 and includes some information on IC circuits. I picked up a used copy ("very good" condition) from Abebooks for $7.57, shipped.
Thanks, edbeckerich3737, for the tip!
thanks for a great explination so far i have not seen such a clean explination and thanks for it....
Hi Ram. I'm pleased you found it useful. Please note that I am now using the geometric (SQRT[min HFE x max HFE]) mean for determining beta off the spec sheet. That approach seems to work well.
FOR GENERAL CIRCUIT DESIGN YOU DID A GOOD JOB.
Thanks David. I'm sure there will be errors, but it works for me.
This video helped me out a lot Charlie.
Thanks. 🙂
Excellent. I'm pleased you found it useful Fredrik.
@@CharlieMorrisZL2CTM again and again. 🙂 im back to refresh a bit
This looks like a excel spread sheet is needed with the rules of thumb being the calculations and just drop in transistor info with voltages. I am surprised this is not already available along with many commonly used devices for ham radio. Certainty the info is inside simulation modeling software but it would be useful to have it unmasked and simply available. Thanks for the video!
Thanks Kenneth. I expect this is somewhere on the web. I quite like breaking out the ol-HP48G and running the numbers from scratch. Thanks for the input.
Bit of a late reply - only just came across the video and thanks for posting.
The approach you have taken reminds of methods I was taught years ago at "school" and techniques I have seen in many text books. I too have wondered why few datasheets contain the characteristics graphs. What I have found from limited homebrew experience is that the HFE values for transistors can vary considerably over the specified range. Maybe that's why manufacturers are reluctant to put the graphs on the datasheets? Also, some of the datasheets values will be valid only for very specific test conditions (e.g. specific Ic and Vce). So, unless I measure the HFE of a transistor I am going to use, I would normally go with the specified minimum HFE for calculating the dc base bias resistors. Also, especially for transistors like the 2N3904 that don't have a super high transition frequency, you can of course expect the hfe to drop off as frequency increases. So, some expirementation, reliance on previous experience and reference to tried and tested designs is what I would go for.
GOOD ADVISE, I TOO REMEMBER LEARNING THOSE METHODS.
I'm afraid I don't know enough to add anything constructive but appreciated the learning curve.
My pleasure. Keep at it.
Charlie
CHARLIE MORRIS,
Does the AC/DC load line mean the maximum output wattage power of the transistor or does the AC/DC load line mean the power dissipation of the transistor?
Because the AC/DC load line is selected by the EE designer of the operating voltage and operating current for the transistor and the DC bias Q point which will determine the maximum output wattage power out of the transistor and the power dissipation of the transistor. I'm not sure what the LOAD means in the AC load line and DC load line what they are calling the LOAD.
Hi Charlie
Been playing with 'traditional' calculations for this specific biasing technique... I too come from the 'old school' days of electronics teaching.
What my own calculations seem to suggest, is that your rule of thumb is pretty spot on. That rule of thumb also appears to be confirmed when running modern tools such as LT Spice (XVII).
The only variation I did come across was Rc. In both the 'traditional' calculations, online biasing calculators (yes, they do exist) and, LT Spice, the best overall gain came with an Rc (purely resistive) value of between 450 and 650 ohms.
Playing out that value, in LT Spice, I replaced that purely resistive load (Rc) with a 9:1 transformer bringing, in this case, 450 ohms down to 50 ohms modeling an FT-37-43 with 9T:3T in LT Spice.
The gain measured into a 50 ohm resistive load was identical to that measured at the collector output with a purely resistive Rc.
As I have never, in real world experimenting, encountered an Rc larger than ~300 ohms with a 2N3904 based amplifier, indeed, I typically reference 200-250 ohms for Rc, I attribute this to a couple of things.
First is that LT Spice, as good as it is, is still just a simulation.
Second, there is the issue of the internal collector resistance.
Lastly, I could be totally wrong! :p
One last thing is the debate over beta vs hFE.
It would seem hFE (DC current gain) is the proper reference. I've seen beta and hFE used interchangeably and, I've seen arguments that they are not necessarily related.
Given that datasheets today rarely list a beta value and do list hFE, using the latter seems to make more sense, as does your choosing a design hFE of 200 with the 2N2904.
In any event, as I stated previously, it seems your rule of thumb is spot on, in my opinion. The real world operational results of your methodology would seem to support that it is correct as well!
Great video!
73, Robin - AC7LX
HI Robin. Firstly thank you very much for the reply, it was very useful. So, from your experience, do you just set a value of Rc to be-say-450ohm or do you determine that each time by finding the open circuit Vo, then Vo/2 across a variable resistors (as Lar's suggested a video or two back)? Also, for determining hFE from the spec sheet, have you across geometric mean before? I seem to recall way back working out the hFE by SQRT(min hFE * max hFE), and not (min hFE + max hFE)/2 .
I'm about to start the mic amp so I might as well use a refreshed design approach and 'rules of thumb'.
Thanks again Robin.
Charlie
Hi Charlie,
Actually, until about a month ago, I had basically assumed an Rc between 200-300 Ohm. That came from looking at other designs and, a datasheets (I don't recall which one), where there was a reference to some level using an Rc of 275 Ohm.
Later, I found and watched a video by W2AEW on determining an unknown output impedance from a signal source.
It was then I had started experimenting with different values of Rc (or RL) to discover at what point the signal was half of the unloaded collector. I also discovered there was a lot about amplifier design I needed a refresher on!
After your video, I tried something similar to see what Rc got the idle voltage at the collector to Vcc/2 for a given design and then, depending on what the amplifier end use, would adjust to feed a given impedance or, in the case of RF, calculate an impedance ratio to match to, in the case of RF, 50 ohm.
I kind of like LT Spice for that because you can try things without destroying a device. Pretty sure I'd have some issues with simply tying the collector to Vcc in this design.
In actual experiments I have started with some ridiculously high values for Rc to attempt to obtain something like an open/unloaded collector and then see what value for Rc brings the signal to 1/2 of unloaded.
The apparent difference between the two methods is that one is looking for half of the unloaded signal and the other is looking for half of Vcc at the collector with no signal. I kind of prefer the latter because it can be calculated before attempting.
Honestly, I should be better at this stuff, but it's been 40+ years since school and 10 years since I retired and took a break from radio, only renewing my interest a couple of years ago.
As far as the hFE calculations, I've never done the sqrt calculation, but it seems to end up with similar/close values. ~172 vs 200. However, one never knows which is better until one sees the results. It is certainly interesting.
Most of the time I use Spice to be close and then work with an actual circuit.
Probably doesn't help much, but there it is. :D
Hi Charlie,
Just a followup on my last comment.
Simply put, I screwed up! LOL!
I think I know how LT Spice came up with that high Rc value. I (think), that it was because I was looking at Rc as 'the load', when what I was apparently doing was looking at an open load condition. through a capacitor.
I replaced the high value resistor with a reasonable inductance (~10uh), measured the collector output again and got the same value as with the high value resistor. I then coupled the collector (in that configuration) via a 100n cap to a variable resistor (load) and discovered what I had always assumed. The output impedance of the 2N3904 is indeed, around 190-240 ohms, at least in LT Spice.
Interesting brain fog on my part, but now the world seems to make sense again. Well, at least where this stuff is concerned! LOL!
73, Robin - AC7LX
Cheers Robin. That 190-240ohm falls nicely around the 200ohm rule of thumb in the SSDRA and EMRFD texts. I do like Lar's process of determining the actual output resistance through test.
Thanks again for the update!
Charlie
Thanks man, very useful video.
Great explanation 👍
My pleasure. I'm not an expert, but it works for me.
Hi Charlie, great video as always. I think your idea of the quiescent voltage on the emitter is not right. At 3:44 you explain the 1V rule of thumb because the working line should not enter the cutoff region. I think this (your way of explaining) is not right, because Ve is completely external to the transistor. The diagram you use shows Vce, which is the emitter-collector voltage, completely inside the transistor. You could lift the voltages around the transistor any amount you wish, as long as you keep the Vbe and Vce as designed. I always design Ve to give me the ability for some swing on the emitter, although this will not be very much. But, this may not be the best way. Cheers, Koen
You are dead right Koen. I was lying in bed last night reflecting on the video and realised what i had said was incorrect. I will update the video description to make that clear. In regards to your designs, do you run with approx. 1V on the emitter or 10% Vcc?
Charlie
As far as I know, the external emitter resistor is only needed to swamp the internal emitter resistor (little r sub e), which is temperature dependant. Without external emitter resistor, the amplification of the common emitter transistor amplifier would be highly dependant on temperature. Also, the external emitter resistor gives some negative feedback, which helps to minimise distortion. I just choose Ve to be somewhere between .5 and 1 Volts.
@@MrKoenvandijken You're right that the emitter resistor is there to dampen the effects of temperature and variations in the transistor's parameters from sample to sample. The real problem with omitting an emitter resistor (or bypassing it completely) is that re depends on collector current and for large output swings, the gain for the positive half of the signal will decrease, while the gain for the negative half will increase, producing serious amounts of distortion.
Since the internal emitter resistance (re) is 25mV/Ic, we can choose Re to be at least ten times that (as a rule of thumb) in order to swamp out variations in re. But that means its value is at least 250mV/Ic, which immediately leads to the conclusion that it requires 250mV across it. If you can ensure that the base bias point is "stiff" (i.e. doesn't change much with changes in the transistor's β or its Vbe from sample to sample) and repeatable (i.e. you use decent low tolerance resistors whose value won't change), then you can afford to bias the emitter as low as 250mV and do away with the bypass capacitor completely. Otherwise, you have to split the emitter resistor and have 250mV across the unbypassed part and several times that in the bypassed part, depending on how "floppy" your base bias point is. Maybe a volt or so at the emitter turns out to be needed, but using 1% resistors and something like a BC547C, which has a very high β with a narrow variation, might allow you to get down to a few hundred millivolts at the emitter without problems.
@@CharlieMorrisZL2CTM As I commented to Koen, modern, high-β transistors and 1% resistors will allow you to push down the voltage at the emitter. It doesn't depend on Vcc. By that I mean, the lowest bias points you can get away with are the same whether you have a 6V supply or a 30V supply. Higher supply voltages allow larger gains and larger output swings, but don't affect other design parameters.
The "knee" in the Ic vs. Vce curves is just the saturation voltage of the transistor, which you can get from the data sheets, typically about 0.2V for general purpose transistors. Then the lowest voltage that the collector can swing to is simply Ve + Vcesat (the voltage you set on the emitter + the saturation voltage). For maximum output swing, you have to bias the collector to be at (Vcc + Ve + Vcesat) / 2 volts.
This is a good explanation
Excellent. Pleased to hear that.
This is an older video but I would like to offer some perspective that I feel is being lost by the video and the comment section, and would like to offer ideas that few use.
First: All of these 'hallowed'-calculations that are taught, are for designing for production NOT performance! Different transistors will perform differently (even within the same-type and even same-lot!) at different Icq's, frequencies and output impedances.
Second: This is driven home, by transistor Beta variations that can exceed 100% for a given transistor even within the same manufacture datasheet! If you want to properly quantify performance, it is a fool's errand to try and use the widest geometric-mean or the lowest value etc. for beta. Another example is the 10X rule for the input DC divider, this is really only a guarantee for lousy transistors within a lot.
Third: There is this magical Vcc (in this case 13.8V) that is not real world unless you are never going to use it off that perfect power-supply. Many times, you will be using a battery that can be a 14.3V down to 11.7V (or even worse) say in an emergency. So, a circuit worth its salt has to perform at any one of these to be practical. Some of the major ham equipment manufactures are terrible, requiring boost circuits even when using a partially charged battery!
Fourth: External component tolerance variations (especially when all their statistical variations conspire to give the worst outcome!) will play a significant role in the exact circuit performance.
Lastly, if you design using one of these 'production'-calculated/derived circuits methodologies, what do you have. Well you can claim that the circuit will work with the crappiest transistor out of the lot; the worst resistor tolerance variation/degradation. Does that sound like designing for performance--No!
Moral-of-the story: The real answer for the hobbiest (not-production) is to think in terms of greater precision and to set ourselves up for success. How?
First off, buy a bunch of your favorite transistor and grade them for good performance and tight similarity on a unit-to-unit basis. Constructing a high-frequency amplifier or oscillator will allow you to more accurately quantify performance, than calculating geometric means and worst-case values from a datasheet. So, out of a hundred transistors you will usually find 75% of them are really tight with each other--use the rest to drive LED's or relays! Then come up with your own specifications as if you were in charge of doing so for your sweetheart lot.
Finally, the most important concept, is empirical testing of your circuit. Every engineer should sit down and vary all the circuit parameters just to learn and observe EXACTLY what your circuit REALLY does.
When I was playing around with QRP gear I quickly found that using 10mA quiescent-points quickly added up and did NOT add ANYTHING to the NEEDED circuit performance. You will find that you can use much less current whether quiescent or input divider (because you don't have wild beta-variations!). Many times a free-lunch can be had with higher-gain transistors but make sure everything stays stable! Make a habit of measuring ALL of your external components for tolerance. Most importantly, lower the operating voltage and make sure the circuit still performs (oscilloscope patterns etc.). You will find it fun to push the circuit elements to the max when you think in terms of performance NOT production. 73...
Great input. Thanks for taking the time.
PICKY,PICKY!
In a battery powered environment 10 ma in a multistage amplifier can add up pretty quickly. What about the input impedance which will be dependent on R1 and R2?
That's certainly true. I'm going to design all of the amps at 1mA for the Go QRP rig and see how that works. I chose 10mA for this one just for the exercise.
In another reply, I noted that the input impedance also depended on (β x the total emitter resistance), and showed that if you bypass the emitter resistor, it could be as low as 250 ohms. Reducing the collector current chosen by a factor of X will increase the input impedance (and also the output impedance) by the same factor. Adding an extra emitter resistor that is not bypassed can also significantly increase the input impedance (as well as reducing the distortion and gain).
It seem that ALL similar tutorials (class A amp) start form the "middle" - electing Ic. The amp "amplifies" - controls output POWER using input POWER. Hence the design objective should be HOW much amplification is desired. Perhaps that would improve on some of those "rules of thumb" .
You are correct. The PA videos later on take that approach. Please note that these are not tutorials, but simply a video log of what I have done. I am not an expert. The whole aim of the videos is to encourage others to give homebrew a go. Thanks again.
What is that swing you mention in the minute 3:08? Thank you
Is that the maximum sinusoidal wave amplitude you can drive the transistor without distortion? No?
Sir, how can i magnetostiction oscillator amplifier? Which make ultrasonic vibration.
The practical way of biasing BJT transistors needs to be taught in universities.
You are the best!
Thanks. I am no expert, and this uses a number of rules of thumb. The sim is to make it simple and hopefully attractive for other to give homebrew a go.
@@CharlieMorrisZL2CTM No, you really made it easy for me. Thank a lot. I was stuck somewhere at collector, and needed to match the impedance of Rl with Rc. Your approach to biasing seems a lot more easier than reading through several books, which actually complicate things. Also, I just wanted to know more about constructing RF transformers to match impedance between the stages. Coz, this is actually the device that effects the overall efficiency of the circuit. So, requesting your next video, please. And thank you once again for your reply.
Hi Charlie -thanks for video
The approach I have used in the past is to cheat !
Go to old service manuals find the device in question and a circuit close
to the one you want ,copy it and alter the value's you need
That way you will be quite close to what's needed
Cheating -yes,but it works !
What do you think?, or should I be drummed out of the Ham community !
Thanks again
Whatever works. So long as you are having fun!
From where u got that frequency value 3.5 Mhz while calculating capacitor value plz give repley
Charlie was talking about designing an amplifier stage for use on a radio frequency band that is normally called the "80-metre band", which has been allocated for amateur use. It extends from 3.5MHz to 3.9MHz or 4.0Mhz, depending on which region of the world you are in. There's plenty of information on the web if you want to search for it.
thank you for explain... helpful...
My pleasure. Please note that this is not a tutorial. It's simply what I do and it works well for me.
if its possible can you please make a video on biasing a BC547 transistor
its just same dude values of resistor could be different
CHARLIE MORRIS< what is the formula to get "how much current in amps" can a 4700uf cap can hold from B+450vdc from a linear power supply?
Hi Bill. I don't know the answer to that off the top of my head. Current would be a function of the load hanging off the cap once it is charged up. Full charge will be 5x CR (capacitance x resistance). Same for discharge.
@@CharlieMorrisZL2CTM without the LOAD, how much current in amps can a 4700uf hold store with a power supply DC voltage at +450vdc? the 4700uf filter cap will charge up to +450vdc but how much current is the 4700uf holding storing? What formula can I use to calculate how much current the filter cap is storing in amps?
Bill. I hope I have interpreted your question correctly. In a DC system current does not flow through the capacitor. When a voltage is applied, the capacitor will charge up to that voltage. Once it is charged, the capacitor
will hold a charge (in coulombs) equal to Capacitance x Voltage (recall, charge a coulomb is the charge (Q) transported by a constant current of one ampere (A) in one second (S), i.e., 1 Coulomb = 1 Ampere for 1 Second). Now for your capacitor, the charge will be 4700uF x 450V = 2.115 coulombs. Similar to a battery, that capacitor could provide, once a suitable load is applied, 2.115 amps for 1 second, or any other combination (e.g., 1.058 amps for 2 seconds). Note too that the time to fully discharge will be approx. 5x (Capacitance x Resistance) seconds.Not sure if this answers your question. Apologies if I have misunderstood
your question.
@@CharlieMorrisZL2CTM yes for the formula is Capacitance X Voltage = Current Stored?
Why I ask is because the AC ripple voltage will increase higher when the load is drawing more current. Any reasons why the AC ripple will increase when the Load draws more current? The Load is drawing the current OUT of the filter cap 4700uF
Hi Bill. I'm not expert but if I recall the filter caps should be providing adequate current to the load during the negative portion the supply cycle. If the ripple is increasing under load, then that suggest you don't have adequate capacitance. Have you tried another capacitor in parallel?
What about the other capacitors?
Good Topic and a well done video, I recently discovered your channel and I left a comment regarding your casecod amp which was also a good video. Good enough that it got me thinking and a little bit motivated. May I ask what your education is? I have a BSEE from an ABET accredit school (SVSU), graduated in 92 and I am also an extra class amateur licensee My concentration of study was control systems so I am almost useless when it comes to RF design but I have recently become very interested in the subject and I have been devouring every text book, web site and you tube video that I come across. The best text, thus far, that I have come across is "RF circuit design by Chris Bowick". I referenced that text in regards to this video and like you he really stressed the importance of setting the proper operating point. His analysis focused on maintaining a stable bias point vs temperature and while doing so made some assumptions (regarding hFE / hfe ) while I think that understand why, he makes no mention why he did it. His example uses a 2N5179 and he just stated it to be 50 (its range is 25 to 250) and you did a similar thing choosing your Beta but using characteristic curves and a load line. As I recall from school that parameter is very hard to control in manufacturing and can vary wildly, there are design technics one can use to minimize the affects, I will have to see if I can dig up my notes from my college days. Aside from picking that value, which can be arbitrary, you did everything else by the book. I can't wait until the next video, thanks for sharing. 73
Hi Michael. Firstly thanks for the feedback. I did 30 years in the Air Force, where I started as an avionics technician. All of my basis electronics theory was taught there. Later on I did a BE in electronics, but never worked as a design engineer. It's playing around at home where I try and put those lessons from years ago to good use. I must look up that RF circuit design by Chris Bowick book too. As for beta, I took a simple mean between the min and max values out of the spec sheet. If i recall, taking the geometric mean (square-root of min*max) maybe a better option. As an aside, I don't use characteristic curves, but tend to start just with the spec sheet. I think i will start to use 10mA as a quiescent Ic, as that seems to be a good option.
Again, thanks to you and the others for taking the time to leave a comment. I really do appreciate it.
73s
Charlie, ZL2CTM
Charlie, thanks for getting back to me, I think I have just what you need. Are you familiar with w2aew youtube channel ? He has a few videos on this topic. Video # 182, and #272, here is a link to his Bias video # 182 th-cam.com/users/results?search_query=w2aew+185
Charlie Morris Wow you were in Army. That explains it.you are fantastic.
@@CharlieMorrisZL2CTM Because β does vary in manufacturing, it makes sense for the hobbyist to look for transistors that are gain-selected. I currently prefer BC547C for general use which has been selected to have a β between 420 and 800. That's a tight range and typically remains independent of collector current from 1mA up to 100mA. You can design with those and assume a minimum β of 420 safe in the knowledge that the calculated base current won't be any less than half the value calculated.
If you use 10mA as the quiescent collector current with a transistor like the 2N3904 (min =100) and bypass the emitter resistor, you can end up with an input impedance looking into the base of 100 x re = 100 x 25mV/10mA = 250R. That will be the dominant value in determining the stage input impedance since R1 and R2 were found to be 22K and 3.9K, and all three are in parallel.
It should be obvious, I hope, that reducing the collector current by a factor of 10, i.e. to 1mA will reduce all of the currents by the same factor, and hence all of the resistances and the input impedance will be increased by a factor of 10. For RF use, you can match stage impedances with transformers, so it makes sense to select a Ic that makes the parasitic capacitances negligible and 10mA might be sensible, or maybe a lower Ic might be just as good. For audio use, an input impedance of 250R would not be a good idea because of the size of coupling and bypass capacitors needed to get a flat response down to 20Hz. For audio, an Ic somewhere between 100μ and 2mA would be a more common choice
Excellent video.
Another design consideration...
Doesn't using a 2:1 transformer effectively cut your gain in half ?
Hi. Sorry for the late reply. The main aim I was going for by matching the load to the collector circuit was to achieve maximum power transfer, as opposed to maximum voltage gain (as you point out).
Charlie
Can this work with an H biased amplifier design.
I'm not so sure Kaseem. I'd have to look into it.
How do you get the lowest operating frequency ? If it's an audio amplifier should the lowest operating frequency should be 20 Hz right
Hi. I use standard communications audio bandwidth of 300-3000Hz.
i have tried but i couldn't get the exact results so i thought you might help me in biasing a BC547 transistor
Hi. I don't have time to make a video sorry. Here are some generic biasing calcs for a 12VDC system, with 10mA quiescent current through the BC547A. See if you can follow the logic. From the BC547 spec sheet the hFE range for the BC547A is 110-240. I now use the geometric mean for biaising calculations, so Beta-DC = SQRT(110 x 240) = 162. Let Ve = 1/10Vcc = 1.2VDC. Thus Re = 1.2 / 10mA = 120ohm. R2 = (1.2V + 0.7V) / (10 x (10mA / 162)) = 3078ohm (try 3k). R1 = (12V - (1.2V + 0.7V)) / (11 x (10mA / 162)) = 14874ohm (try 15k). If you are trying to determine input impedance etc, then you will need Beta-AC. From the spec sheet Ft = 300MHz. So Beta-AC = 300MHz / the highest freq of your amplifier operation. The input/output coupling capacitors, emitter bypass capacitor etc are a function od what frequency you intend to run the amplifer at (i.e., audio versus RF). A lot of the more recent videos and the blog (zl2ctm.blogspot.com/) have examples of that. Hope that helps.
Charlie
why don't use a rf transistor such as BF199,BF254,BF214,BFY90 series or japanese .The 2N3904 it not design for rf and it very noisy
Hi Madalin. You are right that those transistors would be more suitable, but the approach I have taken (right or wrong) for the video series is to readily available parts in an effort to encourage others to give homebrew a go. I guess I'm trying to demonstrate that a functional radio can be build from common parts. I can see another build coming on using the parts you identified as a way of showing performance differences. Thanks for the input
Good
Thanks. Is this in line with your designs?
Charlie Morris yes
Great. Thanks for that. If you don't mind me asking, is Almohtarif your first or family name?
Charlie
what about inpu and aoutput capacitor values?
I typically set those to have a Xc < 100 ohms at the lowest freq of operation. More often than not I set RF amp caps to be 100nF.
Anyone know where the lowest operating frequency comes from?
If my recall my theory correctly, the low freq performance is set by the input coupling capacitor. I typically use a 100nF cap which works well for the frequencies I'm using. Otherwise, set the capacitive reactance to be 1/10th Rin.
@@CharlieMorrisZL2CTM Thanks. I will look into both of those. Currently using an AC supply of 0.5v at a frequency of 10K Hz then have an Rin if 300. So I should either have a capacitive reactance of 30 or have a 100nF. Thanks for the great video and fast reply man.
Thank you for the video! I've put together an interactive notebook that can calculate the circuit using that method given the input values. Might be useful to someone: observablehq.com/@flpvsk/calculating-a-common-emitter-amplifier-circuit
Someone has asked that already, but I didn't quite get the answer. How do we know the operating frequency? Would that be 20Hz - 20kHz for audio? And so for the Ce we'd use 20Hz, is that right?
Nice work Andrey. That will be useful.
most of video out there says R2
9:42 0.1A = 100mA
what if av id given how to calculate correctly
Hi. I would use Av=-Rc/Re, then once Re and Ve are determined, re-calc R1 and R2.
@@CharlieMorrisZL2CTM so in ce voltage divider bypass AV = -RC/re i can determine RC , but if my RE> RC that have no problem right?
Can you please make a calculator for a power amplifier with some "base value" suk as 11*10^x 22*10^x 39*10^x 47*10^x they are anywhere
The collector current determines the input and output impedances, nothing else. The high limit occurs when the transistor's maximum power dissipation of 625mW is approached. The low limit occurs when the parasitic capacitances' reactances at the frequencies of interest become significant compared to the resistors used.
Around 13:30 you should realise that if you fully bypass the emitter resistor, the ac gain of the stage becomes RL/ re, where re is the internal emitter resistance (the slope of the Vbe/Ie curve at that collector current and equal to 25mV / Ic).
But you make the mistake of assuming that Re determines the value of the bypass capacitor. It doesn't. Until the reactance of the bypass capacitor is less than re (which is just 2.5 ohms at 10mA collector current), the gain will increase with frequency because it becomes RL / (re + Xc). The capacitance would actually have to be C = 1 / 2π x 3.5MHz x 2.5R = 18nF, rather than the 3.3nF your calculation would indicate. if we used your value, the gain at 3.5MHz would be RL / (2.5 + 13.8j) which is phase-shifted, highly frequency dependant, and 15dB down on what it could be.
If you split the emitter resistor so that a part of it is unbypassed, then as a rule of thumb, that part should be ten times re (25mV/Ic), say 27R. That will remove much of the non-linear distortion and make the ac gain more predictable. Then you use the 27R to calculate your bypass capacitor, of course.
Thanks Rex. I'll take a closer look at this.
Rex i need the text book that you're taking these infos from pls 🥺
@@NoName-yy1jx Sorry, none of this is from a text book. It's just the result of spending over 50 years designing and making transistor circuits.
"Clickter current" 😂
You must have a ''beta-meter '' !!!
That sounds interesting Madalin. Do you have a recommendation in mind? It would be interesting to see how much an actual device differs from the spec sheet based mean.
Yes a beta-meter with difrent collector current test condition like this sound.whsites.net/project31.htm.Of course your designs
must be dependent on beta to other viewers to replicate schematics,not everyone have a complex betameter to design the best pesformance of the transistor which they have at their disposal.
sound.whsites.net/project31.htm
@@madalinbetea9871 Great. Thanks Madalin.