I spent 4 years at the University of Architecture, but didn't like it, then moved onto the University of Engineering and quit that too. Why couldn't you have been one of my lecturers??? I would have stayed!!! Thank you for this video, I have looked all over for it! Thank you! Can't wait for part two :)
You remind me of a college a college CAD professor I had. Learning from him was a pleasure. Not intimadating, very patient and really wanted the student to grasp concepts and not regurgitate information. Thank you sir. I hope you make many more videos.
What software do you use now? Im testing differen programs and trying to get into 3d Modeling. What are some good free software u found? Also whats a good laptop for 3d modeling?
Thank you, Sir. I wanted to homebrew my first CW transmitter and I came across this circuit but instead of just following the schematics I wanted to understand a little bit more about linear transistor amplification. I watched many videos and read many articles on the subject but your video is by far the most clear and simple. It really helped me a lot, thanks.
Every now and then I come back and watch this video for a refresher. So clear and concise and non-intimidating. I wish all EE professors were as skilled
Thank you so much for being kind enough to explain everything patiently and taking the time to do it even though you're retired.. Your explanations are very enlightening. Wish you well :)
A really informative and easy to follow video, for which many thanks, Tom. I look forward to watching the other parts of this soon. You are a great teacher!
Thank you very much!! I am new to electronic hobby and searched for months about the transistor working circuit but found nothing usefull.... You came like an angel to me
Wonderful! I've been reading text after text, for years looking for this exact information. All I've ever seen was, use transistor A with register 1, 2 and 3 at such-and-such ohms, Yada Yada Yada. Now, finally you give me the GREAT SECRETS of how to determine the values of every part in a circuit, including the source voltage. With a bonus feature of why I really shoild buy a meter for testing transistors. Awesome! And thank you.
This was just great and reasonably simplest explanation I've ever seen. Without complex maths and with focus on fast and effective design. Thanks for that and respect.
I've done plenty of digital but I've recently wanted to design something that needs an amplifier and this video is perfect for helping me work out how to do the calculations. Thanks so much. I'm off to do part 2 now.
Hm, I would usually design a transitor amplifier stage in another way: At first I would define the collector current and calculate the collector resistor in a way that the output voltage is half of the power supply voltage in order to get the best linearity over a wide voltage range. In the second step I would set the emitter resistor according to the desired voltage gain and calculate the voltage across it when the amplifier sits idle. At third I would calculate the base voltage divider considering a bias current Ib through it of 0.05 to 0.1 of the collector current. R2 = (V(Re) + V(be)) / Ib; R1 = (+V - V(Re) - V(be)) / Ib.
An excellent and informative video. The instructional techniques employed by tomtektest reminds me of my ECE professor, Dr. Nicholas Sloan, at the George Washington University almost 50 years ago.
This was very helpful to me. Even though I watched part 2 first. Will be watching these again. I'm learning a lot. I wish you were my next door neighbour. You would have to change me tuition. :)
I liked the clear dervation of value of resistant, hfe of transistor, volt applied,and data sheet of transistor charateristic. I am glad to know u ways of clear UNDERSTANDING CIRCUIT DESIGN.
really struggling with this because my professor will just go rom step 1 to 5 and assume we can all follow. His notes are really difficult but you have explained it so well. Love how u explain
A good introduction and a very usable circuit with decent linearity and a sensible voltage gain for a single stage. However, when biasing the emitter close to the negative rail (i.e. comparable to the Vbe drop of the transistor), you'll get an operating point for the collector that is sensitive to changes in resistor tolerances and temperature. The dc gain of the circuit is the same as the ac gain, approximately -10. When you calculate a base bias point as low as 1.7 volt and then pick standard resistor values for R1 and R2, it's always a good idea to recalculate the bias point and see what effect that has. If the transistor beta is high, then we should expect that (15-Vb)/R1 = Vb/R2, so Vb = 15.R2/(R1+R2) = 1.36V, rather lower than the 1.7V target. That will give an emitter current of about (1.36-0.65)V/100R = about 7 mA, but that will depend quite strongly on the exact Vbe for that transistor. The voltage on the collector would then be 15-7 = 8V. Those are quite close to the observed values in the actual circuit. However, the actual circuit shows a collector voltage of 8.6V, which implies a collector current of (15-8.6)/1K = 6.4mA; while the emitter voltage shows 0.687V implying an emitter current of 6.87mA. That's a rather large discrepancy and I'd be looking at whether your resistors are too far away from their marked values - they look like 5% tolerance. Finally, the small-signal input impedance will be the parallel combination of R1 || R2 || beta*Re = 15K || 1.5K || 250*100R (assuming beta=250, but it could well be lower) = 1.3K and the output impedance is Rc (for audio frequencies) = 1K.
Absolutely fantastic. I have spent months trying to build my skills and learnt more from this than just about everything else put together. Most everything else I have seen is either to simple leaving me wondering but what is the math to calculate that or way above my skill and assume I know it all already. Please please put out more videos like this for other circuit building blocks.
What I dont get how did he get the initial Ure voltage, just used the Vce value from datasheet and and the end of video explained that Vce is 4V for some unknown reason.
those transistor testers on meters are really only good for letting know the component is working!! can they be used for matching transistors??? I mean it should be a common V and current each time you use it! I enjoyed that very much Tom! going to watch part two now...
Thanks for this informative lecture. It seems that you may have chosen to simplify the presentation by glossing over some of the transistor's characteristics. This choice may annoy the engineering types, but it may allow know-nothings like me to understand the basic principles. Some educated people have a hard time explaining things to those with less knowledge. I for one greatly appreciate your efforts!
@tomtektest Hi, at about 8 minutes you say for the gain to be x 10 the top resistor (Rc) has to be about 10x the value of the bottom resistor (Re 100 ohms),is that logic based on the fact that 1k (1000 ohms) is x10 larger than 100 ohms?, does it always work out like this, so for example if i wanted the gain to be x5would i make (Rc) 500 ohms instead of 1K (1000 ohms), is that how it works? basically both Rc and Re setup is a voltage divider
You have the principle right. Remember though that this is an approximation and only works over a normal range. For example, if you put a 1 Meg resistor in Rc and a 1 Ohm resistor in Re, you will not get a gain of a million. Like all approximations, they work to get you started, which is the real idea I was trying to get across. Also, as you see in Part 3, if you bypass the emitter resistor, then the gain goes up, regardless of the initial ratio of Rc/Re. There are always factors that limit the useful range of approximations, but they are very useful in getting to a starting point.
The meter is not measuring an external voltage. It is not on volts measuring mode. The meter is on diode test mode and he is testing the emitter-base junction, which is effectively a diode junction. In diode test mode, the voltage is supplied by the meter itself, often from an internal PP3 9V battery. This way, semi-conductor devices can be tested without being in circuit,
I still don't get it how did you use 1V as voltage across emiter resistor, although datasheet says that 1V is voltage on Vce? Vce and Ve are different things...
Data sheets give you approximations for one design based on a given hFE. He made his own design for the desired hFE or DC gain. If the numbers on the data sheet were static this wouldn’t be designable, only buildable.
That's just a value that the designer is free to choose. It is not connected to any value on the transistor datasheet. If you try to make that voltage too small, then it becomes difficult to establish stable bias points for the circuit. If you make it too large, then you need a very large supply voltage to allow a decent gain while allowing the output to swing over a useful range. Those are the consequences of changing that 1V, but you might want to raise it a bit to improve stability, or lower it a bit if you've only got a 9V battery as a supply. Unfortunately, you have to design a lot of circuits to get a feel for what is going to be the best choice for parameters like the emitter bias point or the collector current (which sets the input and output impedances), given constraints like supply voltage, available supply current and required input impedance.
Yes it is. But that voltage appears across a 100ohm resistor and (assuming the value is accurate), we then know that the current flowing through that resistor is 0.687V/100R = 6.87mA, all of which flows from the emitter, so we know the emitter current. When measuring currents in circuit, it is usual to measure the voltage across a known resistance, as you don't need to interrupt the circuit to put an ammeter inline.
The circuit as designed isn't going to give an output swing of more than about 12 to 13V peak-to-peak _at best_. If the gain is 10, then I assume that you're not trying to drive an input signal of more than 1.2V pk-pk (i.e. about 400mV rms). I assume you're connecting your input signal through a blocking capacitor - try a 1μF capacitor (if it's polar, then + goes to the circuit input) With those conditions, If it is clipped at the top, then the collector bias point is too high. Measure the voltage on the collector with no signal - it should ideally be around 8.5V above ground. If it is higher, then you need to reduce R1, say to 12K (or connect a 56K resistor across it for the same effect). That will increase the base voltage to 1.7V from 1.4V and drop the collector bias point by around 3V (the DC gain is -10). You could use a 100K potentiometer across R1 and trim it around the high end to give you a fine tuning of that bias point. Hopefully that will allow you to set the clipping symmetrically and then reduce the input amplitude to something that the circuit can cope with.
Seems like your picking the supply and not designing based on the supply? Also on the data sheet for the nte123 there is just a minimum gain 75 at 10volts vce and 10ma
Can anybody help me understand the relationship between the desired gain and the minimum/maximum gain values found in the datasheet. The DS says minimum gain is 100 but we enter this project with a desired gain of 10. Are these 2 different gains that we're talking about or is he running the circuit at one tenth the minimum gain the Datasheet calls for?
he is running the circuit at one tenth the minimum gain the Datasheet calls for. The gain is determined by Rc/Re. If you add a shunt capacitor (10uF) at Re, the gain will be ~250 (as the tester showed). Datasheet is only telling: if you buy 10 transistors and check gain in the tester, it should be within this range 100-300. It was clearly 250 for this transistor.
Those are two different gains. The datasheet is showing the *current* gain of the transistor, called Hfe or beta, and that appears to be anywhere between 40 and 300 depending on collector current and individual transistor variations. The circuit is designed to produce an inverting _voltage_ gain of 10 , and good design aims to make that as independent of the transistor's parameters as possible. In this case, we arrange for the voltage gain to be set by the ratio of two resistors Rc/Re. The only thing which slightly depends on the transistor's current gain is the tiny base current that will slightly lower the bias points for the base and emitter, which is otherwise also set by the ratio of two resistors, R2/(R1+R2).
It's a parameter that the designer is free to choose within certain constraints. If that 1V is reduced much, variations in the transistor Vbe between samples and with temperature will become significant so the circuit becomes difficult to bias stably and distortion will increase. Because the voltage across the collector resistor is 10 times as that across the emitter resistor (for a gain of ten), if you increase that 1V by much, you'll need a much higher supply voltage to keep the transistor from clipping the output signal. You'll need to know how much temperature stability is needed, how much distortion can be tolerated, and how much voltage swing at the output is needed to be able to optimise that 1V that was pulled from the sky. Usually 1V isn't a bad value to go with.
The label is wrong. As the audio explains, the assumed current (approximately) through R1 is .001 Amperes or 1 miliAmpere. I apologize for any confusion.
The gain will be the same: Rc/Re = 10 (not 100 or 250). With a shunt capacitor on Re, the gain for the signal will be - what the tester shows (in this case 250)
@@urkosh No, The ac gain using a shunt capacitor will be Rc/re where re is the transistor's intrinsic emitter resistance, which is equal to 25mV/Ic. In other words, for a collector current of 10mA, the effective emitter resistance will be 2.5 ohms, and the gain will be 1K/2.5 = 400. Unfortunately, it will be temperature dependant and highly non-linear for anything more than tiny signals.
A transistor with a higher current gain (beta) will shift the base bias voltage up by a few tens of millivolts, increasing the emitter and collector currents by a few percent and moving the voltage on the collector down by a few hundred millivolts. A transistor with a lower current gain will have the opposite effect, but the effect will only remain small in the given circuit if the beta stays much bigger than 10 (say 50 minimum).
From what I've learned, it is not an assumption but rather a value that we control. We usually want VCE to be around half the value of the supply voltage Vcc. This is because VCE determines what we call the Q-point of the transistor, which is a point located at a certain coordinate in the load line of the transistor. By having VCE as half of the supply voltage, thus centering the Q point in the middle of the load line, we ensure that the AC signal that we introduce into the amplifier is not clipped at the output. I don't know exactly why he chose a value of 4 instead of 6 or 7... maybe because the AC signal isn't too big so it doesn't matter that much.
@@Lijericosas You can go for 7V across the transistor (Vce) if you want, but you'll still want around another 7V across the collector resistor. Given the approximate 1V across the emitter resistor, that's fine for a 15V supply. That implies using a 7mA collector current (which is what the actual circuit turned out to be using). If you want more voltage swing with the same gain, you'll need to design with a higher supply voltage.
Very good except we seem to assume quite a lot. That way I have no idea how to design my own class A amplifier since I don't intend to assume random numbers.
Nice, basic, understandable lecture. I got a bit lost with how you got 1V, but anyway what do you expect of a lowly electronics technician and not an engineer. I think you should have told your viewers that if the collector resistor (1K ohm one) is not in parallel with a diode (positive side toward V+), which is also in series with a 0.001 microfarad CD capacitor, then a voltage spike from V+ would destroy the transistor.
Why 4 V across the transistor? it is not the way this thing works, just assuming. OMG man you made a complicated thing be. more complicated. Gain Rc/Re is only when you use Ve = 1V with a lower value of Rc, and that is never a pratical value, in a real world a value is usually under 0,06V. You should start with a given power supply value and estimate the courrent will be used on the coletor depending what source of signal will be applied. Your explanaition is very twisted and confused but you are not the only one. Every time I watch videos on youtube they try to confuse things that I know a litle bit.
I was thinking the same, if Vcc = 15V, so Vc = Vcc - Ic(Rc+Re) then Vc = 4V. If we follow the ecuation Vce = Vc - Ve, we see that Vce (voltage across the transistor) it must be 3V. It is more easier and accurate stablish Vcc first, then the current thru Rc, caculate Vc an Ve following the commom values, Vc = 40 to 50% of Vcc, and Ve = 10 to 25% of Vcc, then calculate Vce an so on.
@@GrayeLara I think he wants to find resistor values based on a gain he wants, and that's wrong. Exact gain is very complicated, involves transconductance GAv = - RC / (1 / gm + RE) Aproximately Ve around 0.5v gain 6 to 10 times Ve = 0.05v gain 41 to 70 times Ve = 1V gain around 5 times or less suppose Vcc is 15V make Vc = any valour between 7.5 - 0.65 (or 0.6 or 0.7), swing will be smaller than +3.5 -3.5V estimate Ic based on the signal applied in the base, high voltage 1mA to 5mA, small voltage < 500mV smaller than 1mA Rc = Vc/Ic pick a closest manufactored resistor and reset the current before you find Re set Ibase first Ibase = Ic/Hfe (don`t pick the minimun, 1/2 or more over minimun) Re = Ve(choose the gain)/ Ic + Ibase (Ibase is very small could be omitted) for currents IR1 and IR2 there are two proportions, IR1 use Ibase x 10 and IR2 Ibase x 9 or IR1= Ibase x 6.5 and IR2= Ibase x 6 (i like this one) , depends the input impedance you want VR2 = 0.65V + Ve(chose) then you find R2, discount VR2 from Vcc and find R1 To change the input impedance R1/R2 = factor , pick any R2 valour and find R1 keeping the factor
@@magicguitarpedal You'll find that the gain actually is determined by the ratio of Rc/Re as long as Re is much greater than the intrinsic base resistance of the transistor, (re) which is always 25mV/Ic = 1/gm. We don't need an exact gain, but if you want to calculate Rc/(Re + 1/gm) = Rc/(Re + re) = Rc/(Re + 25mV/Ic) for this circuit, you'll see it is close enough to Rc/Re to be within the resistor tolerances. Good design aims to make the circuit performance independent of the transistor characteristics as far as possible. Of course you set Rc/Re to get the gain you want. Since in this circuit, the voltage across Rc is 10x the voltage across Re, you'll want the voltage across Re to be about 1/20 of the supply voltage. That gives about 0.75V with a 15V supply. Then the voltage at the base is about 1.4V and that determines the ratio between R1 and R2. The current through R2 has to be much bigger than the base current Ib which will be Ie/beta. So the collector current along with the minimum beta determines the input impedance and you're free to choose that current within limits to get the input impedance you want. With this design the input impedance is a little less than R2. A collector current of 1mA in this design will require R2 to be much less than 1.4V/0.01mA if the minimum beta is 100. that makes R2
You talke about a gain of 10 because Rc is 10x bigger then Re and we saw it on the oscilloscope. What has this gain to do with the hFE gain? You should be more clear about that to not confuse the targeted viewers.
The _voltage_ gain has nothing to do with the HFE _current_ gain. The whole point of this particular circuit design is to make its operation independent of the transistor's characteristics.
