Hi from South Africa l am 80 now and love your channel it brings back memories from when l started way back in 1965 fixing national portable radios in a radio factory Barlow's in Durban and from there started my Industrial electronics company in 1970 that l still run today. And still use my old Simpson that l serviced a few times but never missed a beat in all that time. ( it has left a few modern style stuff now in the out box ) So thanks again reg Lee 😉
I can't tell you how much I enjoy watching your Video's. I recently retired from my job as a Electronics Controls Technician , a civilian employee of the Department of Defense. My knowledge is in the Industrial World and you've given me a new interest in Ham Radio. Something I needed to prevent my brain from turning to mush. Much to learn and Ham Equipment peaks my interest. I live in Alabama so when I get my new ICOM-7300 online, perhaps we can talk on the radio soon. I began my career in 1973 repairing TV's and CB radio's. Keep those Video's coming !!
Good stuff! I've never done the experiment to determine beta sensitivity with high/low transistor values. It varied a bit more than I was expecting even for the standard divider bias. Thanks
Thanks for providing these examples of transistor biasing. I'm constructing some of the circuits in "The Art of Electronics" and seeing this live puts a new light on the subject.
hello, do you have some pictures of those circuits from the book? it will be very useful to see diagrams translate in to real. it will be very useful for non-experts who make mistakes in the beginning. thank you.
These videos are amazing, its exactly what I wanted in learning material because you can build the circuits and go through the experiments in full and gain a full understanding of whats going on as for people just starting Electronics it just accelerates understanding of Electronics.
Thank you for all the time and effort you've put into these videos; I'm currently studying side-by-side for my intermediate and full amateur radio license here in the UK, and *all* your material has been so useful and superbly clearly explained. I can't thank you enough! 73 good sir.
I really enjoy your videos😁. My instructors at college have got me scratching my head way to often and wondering🤔. Most of our lab assignmentd are done at home because of COVID-19 😷. What really impressed me about your videos are how easy they are to understand the principles of each circuit. Thank You!
I love revisiting some topics that I not quite understood earlier. Recently I was designing a simple CE preamp and was researching the biasing stuff and suddenly something clicked and I get it! And it even worked with my prototype in the real world! And now watching this was so much better than years ago 😁
Very good explanation of this subject. From the practical point of view, relating to your 3rd circuit, I usually do not exceed a bias current factor of 10 in order to keep the imput impedance of the transistor stage high enough. The emitter voltage across Re is also quite limited because the ratio of Rc / Re determines the dc voltage amplification of the circuit. For ac you can keep it high when you short Re with a capacitor but you have to keep your low frequency roll-off in mind.
I completely agree with keeping the bias current to ten times the maximum base current to retain a usable input impedance. However, if you bypass the emitter resistor, you introduce distortion because the remaining internal emitter resistance (re) is non-linear (equal to 25mV/Ic) and you need 330μF to get a 20Hz roll-off with a 1mA emitter current (re=25R). In addition, you then reduce the input impedance again because the base impedance of β times 25R is now in parallel with the input and is probably the dominant term. Practically, it's better to set no less than 250mV across the emitter resistor (so 270R for 1mA emitter current) and dispense with any bypass capacitor. That will represent the maximum practical gain of a single transistor stage if you want to reduce third harmonic distortion to a tolerable level.
Fantastic video. Before I have seen it I struggled a lot trying to understand the reasons of the advantages and disadvantages of the various configurations for making biasing in a transistor. Now im ready for the exams, thank you a lot!!!!
Very good demo of different transistor biasing methods. Resistors are cheap. I learned this at university and usually design transistor stages with a voltage divider base bias. I never actually did this experiment you demonstrate here, but I have done the math using the Ebers Moll model. This would be a good lab for sophomore EE students
A very educational video on this great topic. So very cleverly presented. I really appreciate the practical experiments. It really helps drive the points home. True learning! Thank you very much for your hard work.
Not with a 9V supply, unless you are amplifying very small signals. Also you'll have a gain of about 2, unless you split the 2K emitter resistor and bypass most of it with a large capacitor to increase the ac gain. Personally, I'd recommend using half the value of R1 and taking it to the collector instead of the positive rail. This gives the benefits of both emitter and collector feedback to stabilise the bias point. In that case, Ve can be as low as 250mV with good stability, allowing a gain around 16 with minimal distortion.
This would have been SO useful when I was wading thru the transistor section of the "Art of Electronics" the first time. As with many such books, I almost had to understand the device before I could figure out what the text was about. Of course they also offered a student guide, which was a book to help you understand the book. Here you dispense with all that...
I'm exactly in that predicament. Trying to understand the examples in "Art of Electronics", but first after watching +w2aew videos, the coin drops. I've said thanks in other videos, but I don't think I can thank you enough. It's nice to see that there still are people out there helping for altruistic reasons. Kudos Sir!
Nice demonstration of what can be gained by negative feedback. But it is also important to know what you pay, in power consumption increase, voltage gain reduction, input impedance drop and output impedance rise. Comparing those effects would make for a good second video with this setup.
jpopelish Yes, there are a LOT of topics that could be given their own video when it comes to the good and bad affects of feedback and bias arrangements.
Another great video, well done. If I may could I request that you continue on with this topic and discuss cascading BJT amplifies stages and the effects of input and output impedance specifically as it pertains to RF frequencies and Maximum Power transfer. I know that there is a lot of confusion out there when it comes to this topic, at least there is with me, and I have never really understood why in some applications you design for high input impedance and low output impedance and in other applications you match the input impedances to achieve maximum power transfer.
For the collector feedback circuit, the equation for Vrc can be re-written by dividing top and bottom by Rc.(β+1) to give this: Vrc = (Vcc - Vbe) / (1 + Rb/(Rc.(β+1))) Since you want Vrc to be roughly 1/2 (Vcc-Vbe), that means you want Rb roughly equal to Rc.(β+1). If you make (β+1),Rc large with respect to Rb, then Vrc becomes roughly equal to (Vcc-Vbe). That's not the bias point you want with the transistor close to saturation. If you make the current through R1 fifty times bigger than Ib. then you'll have a stiff Vb, almost independent of β of course, but you sacrifice input impedance quite a lot. It's much more usual to set the current through R1 to about ten times Ib, which is a compromise that results in a small β dependence, but with a significantly higher input impedance.. The most stable biasing is a mixture of voltage divider and collector feedback, by halving the value of R1 and taking it from the collector instead of from the supply rail. That produces a very stable collector bias point, which then allows you to increase the stage gain, if desired, by reducing the emitter resistor to no less than about 270R with a corresponding change in the bias resistor ratio to maintain 1mA emitter current. Lower value emitter resistors will tend to introduce distortion as the non-linear intrinsic emitter resistance of about 25Ω starts to become significant.
Very interesting videos. Now I know that When I replace a transistor with an equivalent, if the biasing scheme is the voltage divider I haven't to worry about the beta. Thank you
As usual, very impressive. I have a question though. In the last section of voltage divider bias, how did you determine Base current? Isn't that beta dependent? So which beta is to be used? The highest or the lowest? Thanks.
@@w2aew Thanks. Just used the technique to stabilise my 2MHz crystal oscillator circuit on breadboard. There is no alternative to learning by doing under guidance of experts. ❤
Excellent video. ...the third example with voltage dividers resistor is best way to work it. Next time teach us the way how power supply a transistor NPN or PNP. Thanks.. Kostas
I watch all your videos very carefully. Well done! I'd vote for a step up in complexity, to, idk, start designing a pre-amp for 2m or HF. 73's de WB0WHD.
I am going through these old video's because of the wealth of knowledge contained in them, Because I can't just "wing it" anymore on other peoples designs.... I have to learn this stuff.... One question I have is, (and if you do have a video for this please point me to it, I learn a lot better by seeing) how is adjustable bias set up? Like in old valve and transistor amps you need to set up or re-adjust the bias current after replacing said valve and even transistors? Thanks in advance..
Great video that really shows why circuits should be made independent of Beta. You should have probably mentioned that the voltage divider bias you had in this particular example was not very good because of the very low input impedance resulting in higher current drawn from the power supply which in turn makes it a bad circuit to couple with another input signal.
Yes, there's always a trade off. Of course, and input impedance of >2kohms isn't something that I'd consider very low in many applications. You could always increase the input bias resistors if a high input Z and lower quiescent current is desired, at the expense of increased dependence on Beta. For cases where input impedance needs to be much higher, then a FET may be more appropriate.
I love your notebooks. They look so clean and to the point. You should make a video on how to write great notes. I know this sounds stupid, but few people know how to write notes clearly like you. (oh and BTW, would it be interesting to simulate the circuits on LTSpice, after the notebook explanation ?)
In the voltage divider bias calculations, you have said choose r1 and r2 to give you a Vb of 1.1V, and that r1 r2 current should be 50 times Ib and that Ie is 1mA. But given such a variation of beta, how do you chose a suitable r1 r2 value? You don't actually know Ib at this stage surely? I'm a bit confused as you could chose values in MegOhms which would give you 1.1V but might break the 50x rule of thumb. Thanks for your video, very helpful, been binge watching alot of your content recently, keep it up!
The 50x rule of thumb is a very approximate figure - the idea is to make the R1-R2 current significantly higher than the base current so that it can be somewhat ignored. If the R1-R2 current is close to the base current, then the base current variations (and the variations in Beta) will cause a change in the bias voltage. You can choose to use the minimum beta listed in the datasheet if you like.
@@w2aew ahhh OK, so If I have a bjt with gain of 20 to 200, as long as I work with the 50x guideline and a min beta of 20, then it'll probably be OK? Thanks again:)
I have an audio preamplifier that has a three transistor gain stage. A while back I wanted to figure out what the dc operating points should be because it had gone wrong. I had (have) the circuit, which tells me the Vcc and all the resistor values etc. The first transistor stage, had what looks like a potential divider bias, but there is an extra resistor shared by both the base bias resistors and the emitter resistor. In other words, the emitter current and bias current go through this 390K resistor, then they divide into the emitter resistor and the potential divider, so an equilibrium is reached. This is a stage that has a +ve supply, but the first tranisistor is PNP, so the collector of this transistor is on the 0V side of the supply. Because the only voltage on the circuit is the 50V above this extra resistor, it meant I couldn't calculate the base voltage easily, in the way it is shown in your video and all the textbooks and therefore none of the other voltages. The only way I could do it was by simulating the circuit in SPICE. But this seemed a lot of hassle, there mustve been a more clever way.
It's difficult to tell from your description, but it sounds like there was some feedback around the transistor(s). That always makes the bias calculations a bit tricky.
There is other interesting methods for biasing, like using a diode for thermal stability, or providing a resistor from a voltage divider and feedback from emitter to base to increase input resistance. Do you have videos about it?
Thank you. This explains it better for me but I am still having difficulty. I plugged in the values on a breadboard and it doesn't add up at the moment for the emitter resistor. Try again after rest. KC9RDM out.
Thank you. Very good video and clear instructions. It would be interesting to re-do this experiment with the first 2 circuits but simply adding and Re, like you did in stage 3. Adding Re to the first 2 designs would further reduce the dependence on Beta, wouldn't it?
The output impedance is just the collector resistor, or 3K9 in these examples. For the voltage divider case, the input impedance is the parallel combination of R1, R2 and β.Re. If R1=20K, R2=2K7 and β.Re = 100 x 470R = 47K, then the input impedance is 1 / (1 / 20K + 1 / 2.7K + 1 / 47K) ≈ 2.3K. That won't change significantly with β.
The V+ lead of the DMM is connected to the power supply voltage, and the COM lead is connected to my handheld probe, so that I can probe the voltage drop across the collector resistor.
W2aew, Some Technicians say that Biasing is "Balancing" the voltage and current of the vacuum tube, FET, Transistor. You can bias a Vacuum tube Hot by over biasing it by biasing the current higher or you can bias the vacuum tube cold by under biasing it by biasing the current lower. So Biasing is Balancing the Voltage & Current Ratios it seems for Vacuum tubes and transistors? but FETS is something different because of the compression point and saturation point which is a different type of biasing/balancing of the voltage and current ratios?
For transistor circuits, the supply voltages are typically much lower. Thus, there isn't as much "headroom" for the signal swing. In these cases, the concerns for biasing is typically centered on ensuring there is enough headroom for the desired signal amplitude and ensuring it doesn't get clipped or saturated.
@@w2aewThanks for the help, yes biasing is about headroom so the desired amplitude does clip or saturate. I haven't seen a youtube video going over vacuum tube and transistor 1dB compression point biasing. What would be your method of biasing a tube and a transistor to have a desired 1dB compression point biasing? what I mean is when you're adjusting the biasing its adjusting the tubes or transistor 1dB compression point?
Some heat sink compound is non-conductive and other heat sink compound is conductive. I'm confused which one to use and why would you want non-conducitive or conductive heat sink compound? I'm guessing microprocessor chips use "conductive" heat sink compound?
I have always thought that R2 is for temperature compensation and Re is negative feedback that reduce sensitivity of the bias point to the value of hfe.
rapsod1911 R2 by itself doesn't provide any temperature compensation per se, but putting a diode in series with it will provide a first order compensation of the temperature variation of Vbe. Re does provide negative feedback, which reduces gain and helps to stabilize the operating point.
Transistors and Op amps when getting close to the Rails will clip off and chop off the positive and negative cycle compared to Tubes & JFETS/MOSFETS when getting close to the Rails it will have a voltage drop and round off the edges, any reasons why? Transistors and Opamps need to add "Soft clipping diodes" in the feedback loop to round off the edges of the clipping. You should do a video lesson about soft clipping didoes and what happens when getting close to the Rails. The Older Op amps in the 70's when the output signal would get close to the rails would Snap to either the positive rail or negative rail or invert, they did unusual things which the newer modern op amps don't do that when the output signal gets close to the rails. Older Op amps with the same part numbers would have a Harder time trying to keep up with the input signal because they had slower slew rate so if you drive the input harder it took time for the op amps output to catch up with the input signal which they had a harder time trying to keep up. I'm not sure the slower slew rate was the only problem but other parameters in the datasheets or it could be just the Op amp wafer processing materials they used back in the 70's.
very good video! i understand the r1 and r2 u choose to get 1.1V at Vb. but what about the base current( 50x base current), dont quite getting calculation for the voltage divider resistor to get the 50x base current.
There are many (infinite) combinations of R1 and R2 that will give you the desired base voltage at the junction of R1 and R2. It is the ratio that determines the base voltage. Choosing to make the standing current going through R1 and R2 approximately equal to 50x the base current adds another constraint - which says that VCC over the sum of R1+R2 should be approx equal to 50x the base current (and the base current is already approximated as the collector current divided by beta). Thus, with these two conditions, you can arrive at good values for R1 and R2. In the example in the video, assume that beta is 100. Collector current is 1mA, so the base current is 10uA. We want the current in the divider to be about 50x10uA, or about 500uA. With VCC of 9V, this says I want R1+R2 approx equal to 9/500uA or 18k. I had a limited selection of resistors in my junk box, so I started with 20k as R1. We want 7.9V across R1, and 1.1V across R2. 7.9V/20k gives me 395uA. Thus R2 = 1.1V/395uA = 2.78k.
I love your videos. Is there a quick and dirty way to approximate the resistors values? Like how can I just start building a circuit and know to use a low resistance or a high resistance 1k or less or 10k, 100k . If i am designing a circuit and I just want to amplify a signal that is in the nano amp range do I start with high values and as each additional stage progresses use lower values? I am just trying to see if there is a rule of thumb before doing the actual numbers.
The design goals will usually dictate where you start with the values. These goals might include things such as desired power dissipation, required input/output impedance, circuit bandwidth required, noise performance, etc. Some of these are obvious (lower power dissipation means higher resistor values, higher bandwidth means lower resistor values, etc.). Maybe you require a specific bias current to meet a certain performance parameter of the transistor, etc. So, unfortunately there isn't a general rule of thumb. However, there are shortcuts that you can often take. This video describes some of these design shortcuts for a common-emitter amplifier: th-cam.com/video/VWY2WQcKJgk/w-d-xo.html
For voltage divider bias, its not clear why you chose 20k for R1. My guess is that you assume beta to be between about 100-200 so that you can estimate Ib to be about Ie/beta or about Ib=10uA. Then, with I(R1) = 50*Ib, you get R1=(Vcc-1.1)/I(R1)=15.8k. R2 can be adjusted to give correct values for Vb=1.1. Is that right?
That's basically right. The desired target current through *both* R1 and R2 should be much greater than the base current - the 50x factor is an example, but it it not a precise number by any means - so 9V/22.7k is about 400uA, well above the expected base current. You could use a factor of 5 or 10x if the desire is to keep current consumption at a minimum (at the expense of a little more Beta dependence), or 75-100x or more if you want very good Beta independence at the expense of current consumption. So, the 20K+2.7K were chosen to give me two things - a nice Beta independent bias point, and the desired 1.1V Vb (to result in about 0.5V across Re).
Why does the voltage divider bias require 50x Ib of current across R1 and R2? Wouldn’t that increase the current through the base-emitter junction? Therefore the transistor would no longer be in the active region?
The current flowing through R1 and R2 will not affect the current drawn by the base. That will be determined by the voltage at the base and the emitter resistor value. Choosing to make the current through R1 and R2 much larger than the base current will ensure that the base current doesn't influence the voltage at the base - making it determined only by the resistor values (and power supply value).
@@w2aew thank you for the response. I always forget that the current at some point is determined by the voltage at that point and not anything else. Awesome clarification!
Sir at 15:10 where your talking about beta dependence vs the ratio of RC to RB.The IB equation IB= Vcc-Vbe/ Rb + [B+1]Rc Doesn"t this imply to reduce dependence onBeta you would want to swamp out [B+1]Rc by making Rb much greater than BRc?
We want to minimise the dependency of Vrc (the voltage across Rc) on β. If β >>1 then we can write β for β+1. Then Ib = (Vcc - Vbe) / (Rb + β.Rc). Now Ic= β.Ib, so we have: Ic = β.(Vcc - Vbe) / (Rb + β.Rc) and we know that Vrc = Ic.Rc = β.Rc.(Vcc - Vbe) / (Rb + β.Rc). this simplifies to: Vrc = (Vcc - Vbe) / (Rb/β.Rc + 1) If you now make Rb >> β.Rc, then Rb/β.Rc >> 1 and we get Vrc = β.Rc.(Vcc - Vbe)/Rb which is directly dependent on β and is very small. Neither of those is desirable. However, if you make Rb
When OverBiasing a Transistor why does it increase the gain? normal operating biasing is HALF of the supply voltage which is 4.5vdc biasing point for +Vcc 9vdc but Overbiasing is set to be higher than 4.5vdc from 5 vdc to 8vdc. What happens to the transistor when its overbiased? if you apply a function signal generator sinewave or squarewave input signal the overbiased transistor will changing the pulsewidth length and symmetry any reasons why?
I"ve never heard the term 'overbiasing'. If the transistor isn't biased to mid supply, then it will be driving into saturation or cutoff earlier in the cycle. This is what causes the symmetry issues.
@@w2aew Overbiasing a transistor is also called biasing HOT or class A. Its biasing the transistor ABOVE mid supply. Underbiasing is called COLD biasing because its BELOW mid supply. You hear these terms in biasing tube amplifiers.
I liked the video. But could someone explain things a little more simply for a noob to electronics like me like how to choose the resistors you need to get the Q point and why 1 ma across the collector
w2aew, if you plot the frequency response of different silicon transistor part numbers. Why does different silicon or germanium transistor part numbers have different frequency responses from 20hz to 20khz the frequency response curve profile are all different profiles, any reasons why?
Yes I know but what I mean is that its not the transistors HFE gain that changes the frequency response profile, its the miller effect capacitance, depletion regions, or the doping is what shapes the frequency response profile curve?
Unfortunately, no - that's as simple as it gets - even this calculation isn't terribly precise because beta is not a very tightly controlled parameter. This is precisely why you want to design circuits that are largely independent of beta and base current variations.
W2aew, how can you make a circuit with an LED light that tells you when the germanium transistors are Biased correctly? If you have a comparator circuit set to a referenced bias voltage that the germanium transistor should be "bias correctly" and the LED light will turn on & light. But when the germanium transistor temperature will drift up and down from the cold and hot temperature the comparators reference Correctly bias voltage will not be good anymore? because the correct bias voltage has drifted also so the LED will turn on and Light at the wrong bias point because the transistors have drifted in temperature? I'm trying to get the LED to turn on at the correct calculated bias voltage even if the germanium transistors have drifted in temperate up and down the Bias circuit comparator will adjust and know what the correct bias voltage should be. Not sure how to do this, do you know what I mean?
This is not really possible, because the "correct" bias for a given transistor depends ENTIRELY on the specific circuit it is designed into. The bias condition for a Class A common emitter amplifier running off of 12V is entirely different from the proper bias condition for a Class C amp, which is different from a common collector amp, which is different from a circuit where the transistor is being used as a switch, etc. There is no single "right" answer.
@@w2aew this engineer calms his bias circuit will only turn on the LED light when the germanium transistors are correctly bias, check out the video link th-cam.com/video/cN76y_Ybs04/w-d-xo.html I'm not sure how this biasing circuit works because the germanium transistors will drift with temperature so the bias circuit has to drift also to keep the bias voltage correct.
W2AEW DE KD5SMF Let's say you want to design a HF transceiver that had a true output of 5 Watts minimum & 10 Watts Max output at the Antenna. Most qrp is actually less than 5 watts & no one actually even gets a true full output of the actual designed circuit. Which transistors and MOSFETS would you use to build a Superhet transciever?
8 years later... still one of the best videos on YT about transistor biasing and Beta coefficient. Cudos!
There's a certain elegance to your explanations. I'm going around a BJT learning cycle again and find your videos excellent.
Hi from South Africa l am 80 now and love your channel it brings back memories from when l started way back in 1965 fixing national portable radios in a radio factory Barlow's in Durban and from there started my Industrial electronics company in 1970 that l still run today.
And still use my old Simpson that l serviced a few times but never missed a beat in all that time. ( it has left a few modern style stuff now in the out box )
So thanks again reg Lee 😉
I can't tell you how much I enjoy watching your Video's. I recently retired from my job as a Electronics Controls Technician , a civilian employee of the Department of Defense. My knowledge is in the Industrial World and you've given me a new interest in Ham Radio. Something I needed to prevent my brain from turning to mush. Much to learn and Ham Equipment peaks my interest. I live in Alabama so when I get my new ICOM-7300 online, perhaps we can talk on the radio soon. I began my career in 1973 repairing TV's and CB radio's. Keep those Video's coming !!
Good stuff! I've never done the experiment to determine beta sensitivity with high/low transistor values. It varied a bit more than I was expecting even for the standard divider bias. Thanks
How far we've come
To See is to Believe and to Remember forever. Thanks.
Thanks for all the time that you put into making these videos , a great aid to revision when the books get relegated to reference sources .
Thanks for providing these examples of transistor biasing. I'm constructing some of the circuits in "The Art of Electronics" and seeing this live puts a new light on the subject.
hello, do you have some pictures of those circuits from the book? it will be very useful to see diagrams translate in to real. it will be very useful for non-experts who make mistakes in the beginning. thank you.
A nice practical, yet thorough explanation of a tricky topic. Thank you.
This is one of the most clear and helpful videos I’ve ever watched. Thank you for sharing your expertise! - Jim
These videos are amazing, its exactly what I wanted in learning material because you can build the circuits and go through the experiments in full and gain a full understanding of whats going on as for people just starting Electronics it just accelerates understanding of Electronics.
Thank you for all the time and effort you've put into these videos; I'm currently studying side-by-side for my intermediate and full amateur radio license here in the UK, and *all* your material has been so useful and superbly clearly explained. I can't thank you enough! 73 good sir.
I really enjoy your videos😁. My instructors at college have got me scratching my head way to often and wondering🤔. Most of our lab assignmentd are done at home because of COVID-19 😷. What really impressed me about your videos are how easy they are to understand the principles of each circuit. Thank You!
I love revisiting some topics that I not quite understood earlier. Recently I was designing a simple CE preamp and was researching the biasing stuff and suddenly something clicked and I get it! And it even worked with my prototype in the real world! And now watching this was so much better than years ago 😁
Great complement on my last video on BIAS adjustment on amateur radio. Will link your video in my video..
Thanks Alan, great as always.
God bless what a clear explanation. Thank you soo much
Best explanation I've ever seen about different transistor configurations for beta dependence.
I'm just starting in electronics as a hobby and your videos on transistor really have helped me out. Really good stuff. Thanks Alan
Very good explanation of this subject. From the practical point of view, relating to your 3rd circuit, I usually do not exceed a bias current factor of 10 in order to keep the imput impedance of the transistor stage high enough. The emitter voltage across Re is also quite limited because the ratio of Rc / Re determines the dc voltage amplification of the circuit. For ac you can keep it high when you short Re with a capacitor but you have to keep your low frequency roll-off in mind.
I completely agree with keeping the bias current to ten times the maximum base current to retain a usable input impedance.
However, if you bypass the emitter resistor, you introduce distortion because the remaining internal emitter resistance (re) is non-linear (equal to 25mV/Ic) and you need 330μF to get a 20Hz roll-off with a 1mA emitter current (re=25R). In addition, you then reduce the input impedance again because the base impedance of β times 25R is now in parallel with the input and is probably the dominant term. Practically, it's better to set no less than 250mV across the emitter resistor (so 270R for 1mA emitter current) and dispense with any bypass capacitor. That will represent the maximum practical gain of a single transistor stage if you want to reduce third harmonic distortion to a tolerable level.
Thank you Alan for a great explanation. Better than most classes I had in university.
Incredible quality education. Thank you sir!
I lift my hat off and say: -"Thank you for educating us, professor"
Keep up your good work.
Fantastic video. Before I have seen it I struggled a lot trying to understand the reasons of the advantages and disadvantages of the various configurations for making biasing in a transistor. Now im ready for the exams, thank you a lot!!!!
Very good demo of different transistor biasing methods. Resistors are cheap. I learned this at university and usually design transistor stages with a voltage divider base bias. I never actually did this experiment you demonstrate here, but I have done the math using the Ebers Moll model. This would be a good lab for sophomore EE students
A very educational video on this great topic. So very cleverly presented. I really appreciate the practical experiments. It really helps drive the points home. True learning! Thank you very much for your hard work.
500mV for Ve may be a good starting point to have Beta tolerance, but to have also a good thermal stability I'd recommend to increase it up to 2Volts.
Not with a 9V supply, unless you are amplifying very small signals. Also you'll have a gain of about 2, unless you split the 2K emitter resistor and bypass most of it with a large capacitor to increase the ac gain.
Personally, I'd recommend using half the value of R1 and taking it to the collector instead of the positive rail. This gives the benefits of both emitter and collector feedback to stabilise the bias point. In that case, Ve can be as low as 250mV with good stability, allowing a gain around 16 with minimal distortion.
This would have been SO useful when I was wading thru the transistor section of the "Art of Electronics" the first time. As with many such books, I almost had to understand the device before I could figure out what the text was about. Of course they also offered a student guide, which was a book to help you understand the book. Here you dispense with all that...
+Chris Gonzales Glad you liked it!
Chris Gonzales I am going through art of electronics now and I am lost on the transistor section. This video helped a lot!
I'm exactly in that predicament. Trying to understand the examples in "Art of Electronics", but first after watching +w2aew videos, the coin drops. I've said thanks in other videos, but I don't think I can thank you enough. It's nice to see that there still are people out there helping for altruistic reasons. Kudos Sir!
That was explained better than I got from the lecturers at college way back when.....
Nice video Alan.
This video gave me my aha moment with transistor biasing. Thank you so much!
I know right....!!! Someone is finally making it make sense for me
Great video. I wished having you as my professor when I was at the university. Greetings from La Paz, Bolivia.
Makes transistor biasing so easy to understand. thumbs up.
Your explanations are really best I was able to find and now finally I understand lot of thing that I could not get before - thank you!
Great explanation! I didn't realize the differences could be so significant.
Two weeks of EE202 in 19 minutes. Very helpful!
That was an excellent presentation sir, you’re a very knowledgeable teacher and I thank you kindly!
Nice demonstration of what can be gained by negative feedback. But it is also important to know what you pay, in power consumption increase, voltage gain reduction, input impedance drop and output impedance rise. Comparing those effects would make for a good second video with this setup.
jpopelish Yes, there are a LOT of topics that could be given their own video when it comes to the good and bad affects of feedback and bias arrangements.
Another great video, well done. If I may could I request that you continue on with this topic and discuss cascading BJT amplifies stages and the effects of input and output impedance specifically as it pertains to RF frequencies and Maximum Power transfer. I know that there is a lot of confusion out there when it comes to this topic, at least there is with me, and I have never really understood why in some applications you design for high input impedance and low output impedance and in other applications you match the input impedances to achieve maximum power transfer.
L.G. Cowles says:
the collector feedback circuit needs to be Vcc > 8 volts,
and thanks again well done
Not 7V? I thought 6.5-7 on collector, no? Or you mean something different?
Thank you very much for your instructional videos, bring me good memories of me EE studies!
Thank you from Sweden
You are awesome. Thanks so much for your videos, now you are my electronics professor
Thanks - be sure to send me your tuition! (just kidding!)
@@w2aew hahaha
lovely video dude, very clear and concise explanation.
:D
Out standing as usual, one to reference again and again.
Thank You Alan For such a Great Video on one of my most Interesting Topics ...Transistor Biasing Circuits!!!!
For the collector feedback circuit, the equation for Vrc can be re-written by dividing top and bottom by Rc.(β+1) to give this:
Vrc = (Vcc - Vbe) / (1 + Rb/(Rc.(β+1)))
Since you want Vrc to be roughly 1/2 (Vcc-Vbe), that means you want Rb roughly equal to Rc.(β+1).
If you make (β+1),Rc large with respect to Rb, then Vrc becomes roughly equal to (Vcc-Vbe). That's not the bias point you want with the transistor close to saturation.
If you make the current through R1 fifty times bigger than Ib. then you'll have a stiff Vb, almost independent of β of course, but you sacrifice input impedance quite a lot. It's much more usual to set the current through R1 to about ten times Ib, which is a compromise that results in a small β dependence, but with a significantly higher input impedance..
The most stable biasing is a mixture of voltage divider and collector feedback, by halving the value of R1 and taking it from the collector instead of from the supply rail. That produces a very stable collector bias point, which then allows you to increase the stage gain, if desired, by reducing the emitter resistor to no less than about 270R with a corresponding change in the bias resistor ratio to maintain 1mA emitter current. Lower value emitter resistors will tend to introduce distortion as the non-linear intrinsic emitter resistance of about 25Ω starts to become significant.
Very interesting videos. Now I know that
When I replace a transistor with an equivalent, if the biasing scheme is the voltage divider I haven't to worry about the beta. Thank you
Gran explicación,gracias.
Great tuto. Thanks a lot!
very useful this video to bias two AC128, which are very unstable with the temperature. my proyect is a fuzz pedal for the guitar.Thanks for sharing
Excellent demonstration and video! Learned a lot.
Thanks for the very informative video. Enjoy the scope of your videos. Take care.
Thank you for all the work you put into making these videos highly informative and understandable!
Another great back to basic tutorial sir.
Love the clarity!
As usual, very impressive. I have a question though. In the last section of voltage divider bias, how did you determine Base current? Isn't that beta dependent? So which beta is to be used? The highest or the lowest? Thanks.
To consider worst case, use the lowest Beta. To consider typical operation, use the typical Beta.
@@w2aew Thanks. Just used the technique to stabilise my 2MHz crystal oscillator circuit on breadboard. There is no alternative to learning by doing under guidance of experts. ❤
A very useful refresher. Thanks Alan.
Thank you. Clear explanation.
Excellent video. ...the third example with voltage dividers resistor is best way to work it.
Next time teach us the way how power supply a transistor NPN or PNP.
Thanks..
Kostas
I watch all your videos very carefully. Well done! I'd vote for a step up in complexity, to, idk, start designing a pre-amp for 2m or HF. 73's de WB0WHD.
I am going through these old video's because of the wealth of knowledge contained in them, Because I can't just "wing it" anymore on other peoples designs.... I have to learn this stuff.... One question I have is, (and if you do have a video for this please point me to it, I learn a lot better by seeing) how is adjustable bias set up? Like in old valve and transistor amps you need to set up or re-adjust the bias current after replacing said valve and even transistors? Thanks in advance..
There really isn't one universal way to adjust bias - it depends entirely on the specific circuit design.
Another great video. Thanks for taking the time to make it.
Great video!
Very nice! Thanks for sharing!
Really good stuff.
Thanks
Carl
Great, video!
The IT-18 is a very nice instrument, and extremely sensitive as well- I have found the beta adjust to be very touchy.
Agreed on all counts
Great video that really shows why circuits should be made independent of Beta. You should have probably mentioned that the voltage divider bias you had in this particular example was not very good because of the very low input impedance resulting in higher current drawn from the power supply which in turn makes it a bad circuit to couple with another input signal.
Yes, there's always a trade off. Of course, and input impedance of >2kohms isn't something that I'd consider very low in many applications. You could always increase the input bias resistors if a high input Z and lower quiescent current is desired, at the expense of increased dependence on Beta. For cases where input impedance needs to be much higher, then a FET may be more appropriate.
Thanks for a great tutorial
Great video with examples. it was very helpful.
I love your notebooks. They look so clean and to the point. You should make a video on how to write great notes. I know this sounds stupid, but few people know how to write notes clearly like you.
(oh and BTW, would it be interesting to simulate the circuits on LTSpice, after the notebook explanation ?)
I learned how to make clean and clear notes by example - starting with the Notebook series by Forrest Mimms.
***** Thank you, I didn't know this interesting fellow. His notebooks are a thing to behold !
Very good video !! you are great Mr. w2aew.
In the voltage divider bias calculations, you have said choose r1 and r2 to give you a Vb of 1.1V, and that r1 r2 current should be 50 times Ib and that Ie is 1mA.
But given such a variation of beta, how do you chose a suitable r1 r2 value? You don't actually know Ib at this stage surely?
I'm a bit confused as you could chose values in MegOhms which would give you 1.1V but might break the 50x rule of thumb.
Thanks for your video, very helpful, been binge watching alot of your content recently, keep it up!
The 50x rule of thumb is a very approximate figure - the idea is to make the R1-R2 current significantly higher than the base current so that it can be somewhat ignored. If the R1-R2 current is close to the base current, then the base current variations (and the variations in Beta) will cause a change in the bias voltage. You can choose to use the minimum beta listed in the datasheet if you like.
@@w2aew ahhh OK, so If I have a bjt with gain of 20 to 200, as long as I work with the 50x guideline and a min beta of 20, then it'll probably be OK?
Thanks again:)
I have an audio preamplifier that has a three transistor gain stage. A while back I wanted to figure out what the dc operating points should be because it had gone wrong. I had (have) the circuit, which tells me the Vcc and all the resistor values etc. The first transistor stage, had what looks like a potential divider bias, but there is an extra resistor shared by both the base bias resistors and the emitter resistor. In other words, the emitter current and bias current go through this 390K resistor, then they divide into the emitter resistor and the potential divider, so an equilibrium is reached.
This is a stage that has a +ve supply, but the first tranisistor is PNP, so the collector of this transistor is on the 0V side of the supply.
Because the only voltage on the circuit is the 50V above this extra resistor, it meant I couldn't calculate the base voltage easily, in the way it is shown in your video and all the textbooks and therefore none of the other voltages. The only way I could do it was by simulating the circuit in SPICE.
But this seemed a lot of hassle, there mustve been a more clever way.
It's difficult to tell from your description, but it sounds like there was some feedback around the transistor(s). That always makes the bias calculations a bit tricky.
Sounds like they were biasing with a constant current source
There is other interesting methods for biasing, like using a diode for thermal stability, or providing a resistor from a voltage divider and feedback from emitter to base to increase input resistance.
Do you have videos about it?
Not at this time - just this "basic" video.
well done and demonstrated. Thanks again
I missed this video earlier. Watching it now.
Thank you. This explains it better for me but I am still having difficulty. I plugged in the values on a breadboard and it doesn't add up at the moment for the emitter resistor. Try again after rest. KC9RDM out.
Great Video as always. Thanks a million .
Thank you. Very good video and clear instructions. It would be interesting to re-do this experiment with the first 2 circuits but simply adding and Re, like you did in stage 3. Adding Re to the first 2 designs would further reduce the dependence on Beta, wouldn't it?
Yes it would - not to the extent as in the 3rd example though.
Hi Alan,
Could you explain how to calculate the input and output impedance of the circuit ?
Thank you
The output impedance is just the collector resistor, or 3K9 in these examples.
For the voltage divider case, the input impedance is the parallel combination of R1, R2 and β.Re. If R1=20K, R2=2K7 and β.Re = 100 x 470R = 47K, then the input impedance is 1 / (1 / 20K + 1 / 2.7K + 1 / 47K) ≈ 2.3K. That won't change significantly with β.
Great explanation many thanks... Just a question please .. Why you are connecting the COM lead of DMM to the positive of power supply??
The V+ lead of the DMM is connected to the power supply voltage, and the COM lead is connected to my handheld probe, so that I can probe the voltage drop across the collector resistor.
@@w2aew Thanks.. understood
W2aew, Some Technicians say that Biasing is "Balancing" the voltage and current of the vacuum tube, FET, Transistor. You can bias a Vacuum tube Hot by over biasing it by biasing the current higher or you can bias the vacuum tube cold by under biasing it by biasing the current lower. So Biasing is Balancing the Voltage & Current Ratios it seems for Vacuum tubes and transistors? but FETS is something different because of the compression point and saturation point which is a different type of biasing/balancing of the voltage and current ratios?
For transistor circuits, the supply voltages are typically much lower. Thus, there isn't as much "headroom" for the signal swing. In these cases, the concerns for biasing is typically centered on ensuring there is enough headroom for the desired signal amplitude and ensuring it doesn't get clipped or saturated.
@@w2aewThanks for the help, yes biasing is about headroom so the desired amplitude does clip or saturate. I haven't seen a youtube video going over vacuum tube and transistor 1dB compression point biasing. What would be your method of biasing a tube and a transistor to have a desired 1dB compression point biasing? what I mean is when you're adjusting the biasing its adjusting the tubes or transistor 1dB compression point?
Some heat sink compound is non-conductive and other heat sink compound is conductive. I'm confused which one to use and why would you want non-conducitive or conductive heat sink compound? I'm guessing microprocessor chips use "conductive" heat sink compound?
Thank you for this amazing explanation.. l hope that I can get more useful crisis like this:😊
I have always thought that R2 is for temperature compensation and Re is negative feedback that reduce sensitivity of the bias point to the value of hfe.
rapsod1911 R2 by itself doesn't provide any temperature compensation per se, but putting a diode in series with it will provide a first order compensation of the temperature variation of Vbe. Re does provide negative feedback, which reduces gain and helps to stabilize the operating point.
Transistors and Op amps when getting close to the Rails will clip off and chop off the positive and negative cycle compared to Tubes & JFETS/MOSFETS when getting close to the Rails it will have a voltage drop and round off the edges, any reasons why? Transistors and Opamps need to add "Soft clipping diodes" in the feedback loop to round off the edges of the clipping. You should do a video lesson about soft clipping didoes and what happens when getting close to the Rails. The Older Op amps in the 70's when the output signal would get close to the rails would Snap to either the positive rail or negative rail or invert, they did unusual things which the newer modern op amps don't do that when the output signal gets close to the rails. Older Op amps with the same part numbers would have a Harder time trying to keep up with the input signal because they had slower slew rate so if you drive the input harder it took time for the op amps output to catch up with the input signal which they had a harder time trying to keep up. I'm not sure the slower slew rate was the only problem but other parameters in the datasheets or it could be just the Op amp wafer processing materials they used back in the 70's.
Great video, Can you do a video on emitter feedback bias and emitter bias with the 2 dc supplys.those are the two I don't understand.
Excellent! Thank you!
very good video! i understand the r1 and r2 u choose to get 1.1V at Vb. but what about the base current( 50x base current), dont quite getting calculation for the voltage divider resistor to get the 50x base current.
There are many (infinite) combinations of R1 and R2 that will give you the desired base voltage at the junction of R1 and R2. It is the ratio that determines the base voltage. Choosing to make the standing current going through R1 and R2 approximately equal to 50x the base current adds another constraint - which says that VCC over the sum of R1+R2 should be approx equal to 50x the base current (and the base current is already approximated as the collector current divided by beta). Thus, with these two conditions, you can arrive at good values for R1 and R2. In the example in the video, assume that beta is 100. Collector current is 1mA, so the base current is 10uA. We want the current in the divider to be about 50x10uA, or about 500uA. With VCC of 9V, this says I want R1+R2 approx equal to 9/500uA or 18k. I had a limited selection of resistors in my junk box, so I started with 20k as R1. We want 7.9V across R1, and 1.1V across R2. 7.9V/20k gives me 395uA. Thus R2 = 1.1V/395uA = 2.78k.
@@w2aew thanks sir, very well explanation!
I love your videos. Is there a quick and dirty way to approximate the resistors values? Like how can I just start building a circuit and know to use a low resistance or a high resistance 1k or less or 10k, 100k . If i am designing a circuit and I just want to amplify a signal that is in the nano amp range do I start with high values and as each additional stage progresses use lower values? I am just trying to see if there is a rule of thumb before doing the actual numbers.
The design goals will usually dictate where you start with the values. These goals might include things such as desired power dissipation, required input/output impedance, circuit bandwidth required, noise performance, etc. Some of these are obvious (lower power dissipation means higher resistor values, higher bandwidth means lower resistor values, etc.). Maybe you require a specific bias current to meet a certain performance parameter of the transistor, etc. So, unfortunately there isn't a general rule of thumb. However, there are shortcuts that you can often take. This video describes some of these design shortcuts for a common-emitter amplifier: th-cam.com/video/VWY2WQcKJgk/w-d-xo.html
you do help me a lot, thank you!
I think you need a Back To Basics video for this Back to Basics video
For voltage divider bias, its not clear why you chose 20k for R1. My guess is that you assume beta to be between about 100-200 so that you can estimate Ib to be about Ie/beta or about Ib=10uA. Then, with I(R1) = 50*Ib, you get R1=(Vcc-1.1)/I(R1)=15.8k. R2 can be adjusted to give correct values for Vb=1.1. Is that right?
That's basically right. The desired target current through *both* R1 and R2 should be much greater than the base current - the 50x factor is an example, but it it not a precise number by any means - so 9V/22.7k is about 400uA, well above the expected base current. You could use a factor of 5 or 10x if the desire is to keep current consumption at a minimum (at the expense of a little more Beta dependence), or 75-100x or more if you want very good Beta independence at the expense of current consumption. So, the 20K+2.7K were chosen to give me two things - a nice Beta independent bias point, and the desired 1.1V Vb (to result in about 0.5V across Re).
Why does the voltage divider bias require 50x Ib of current across R1 and R2? Wouldn’t that increase the current through the base-emitter junction? Therefore the transistor would no longer be in the active region?
The current flowing through R1 and R2 will not affect the current drawn by the base. That will be determined by the voltage at the base and the emitter resistor value. Choosing to make the current through R1 and R2 much larger than the base current will ensure that the base current doesn't influence the voltage at the base - making it determined only by the resistor values (and power supply value).
@@w2aew thank you for the response. I always forget that the current at some point is determined by the voltage at that point and not anything else. Awesome clarification!
Sir at 15:10 where your talking about beta dependence vs the ratio of RC to RB.The IB equation IB= Vcc-Vbe/ Rb + [B+1]Rc Doesn"t this imply to reduce dependence onBeta you would want to swamp out [B+1]Rc by making Rb much greater than BRc?
We want to minimise the dependency of Vrc (the voltage across Rc) on β. If β >>1 then we can write β for β+1. Then Ib = (Vcc - Vbe) / (Rb + β.Rc). Now Ic= β.Ib, so we have:
Ic = β.(Vcc - Vbe) / (Rb + β.Rc) and we know that Vrc = Ic.Rc = β.Rc.(Vcc - Vbe) / (Rb + β.Rc). this simplifies to:
Vrc = (Vcc - Vbe) / (Rb/β.Rc + 1)
If you now make Rb >> β.Rc, then Rb/β.Rc >> 1 and we get Vrc = β.Rc.(Vcc - Vbe)/Rb which is directly dependent on β and is very small. Neither of those is desirable.
However, if you make Rb
When OverBiasing a Transistor why does it increase the gain? normal operating biasing is HALF of the supply voltage which is 4.5vdc biasing point for +Vcc 9vdc but Overbiasing is set to be higher than 4.5vdc from 5 vdc to 8vdc. What happens to the transistor when its overbiased? if you apply a function signal generator sinewave or squarewave input signal the overbiased transistor will changing the pulsewidth length and symmetry any reasons why?
I"ve never heard the term 'overbiasing'. If the transistor isn't biased to mid supply, then it will be driving into saturation or cutoff earlier in the cycle. This is what causes the symmetry issues.
@@w2aew Overbiasing a transistor is also called biasing HOT or class A. Its biasing the transistor ABOVE mid supply. Underbiasing is called COLD biasing because its BELOW mid supply. You hear these terms in biasing tube amplifiers.
I liked the video. But could someone explain things a little more simply for a noob to electronics like me like how to choose the resistors you need to get the Q point and why 1 ma across the collector
If and only if my teacher explained as clear as your explanation...
w2aew, if you plot the frequency response of different silicon transistor part numbers. Why does different silicon or germanium transistor part numbers have different frequency responses from 20hz to 20khz the frequency response curve profile are all different profiles, any reasons why?
Different transistors are designed differently, so they behave differently - just as different cars behave differently (handling, speed, etc).
Yes I know but what I mean is that its not the transistors HFE gain that changes the frequency response profile, its the miller effect capacitance, depletion regions, or the doping is what shapes the frequency response profile curve?
Hi Alan,
Is there a simple formula to calculate Ib in this circuit else than Ib=Ic/beta ?
Thank you !
Unfortunately, no - that's as simple as it gets - even this calculation isn't terribly precise because beta is not a very tightly controlled parameter. This is precisely why you want to design circuits that are largely independent of beta and base current variations.
W2aew, how can you make a circuit with an LED light that tells you when the germanium transistors are Biased correctly? If you have a comparator circuit set to a referenced bias voltage that the germanium transistor should be "bias correctly" and the LED light will turn on & light. But when the germanium transistor temperature will drift up and down from the cold and hot temperature the comparators reference Correctly bias voltage will not be good anymore? because the correct bias voltage has drifted also so the LED will turn on and Light at the wrong bias point because the transistors have drifted in temperature? I'm trying to get the LED to turn on at the correct calculated bias voltage even if the germanium transistors have drifted in temperate up and down the Bias circuit comparator will adjust and know what the correct bias voltage should be. Not sure how to do this, do you know what I mean?
This is not really possible, because the "correct" bias for a given transistor depends ENTIRELY on the specific circuit it is designed into. The bias condition for a Class A common emitter amplifier running off of 12V is entirely different from the proper bias condition for a Class C amp, which is different from a common collector amp, which is different from a circuit where the transistor is being used as a switch, etc. There is no single "right" answer.
@@w2aew this engineer calms his bias circuit will only turn on the LED light when the germanium transistors are correctly bias, check out the video link th-cam.com/video/cN76y_Ybs04/w-d-xo.html I'm not sure how this biasing circuit works because the germanium transistors will drift with temperature so the bias circuit has to drift also to keep the bias voltage correct.
Thank you sir!
W2AEW DE KD5SMF
Let's say you want to design a HF transceiver that had a true output of 5 Watts minimum & 10 Watts Max output at the Antenna. Most qrp is actually less than 5 watts & no one actually even gets a true full output of the actual designed circuit. Which transistors and MOSFETS would you use to build a Superhet transciever?
I really don't know, off the top of my head - it would take a bit of research to even begin to narrow it down.
Try looking at QST December 2002, No Excuses QRP transceiver.