The transistor as an amplifier (Part 1)

[@jeffpereira8891]

it's one of the best tutorial I 've come across.

[@malcolmhodgson7540]

The clearest explanation on how to calculate the values. And I already have the Art of Electronics book!!!! Thank you for helping me understand. You are a star!!!!

[@michaelfuchs441]

That is too sweet. Thank you!

[@michaelbauers8800]

This is the best explanation I have ever experienced in my quest to understand transistor circuits. I still failed to fully understand it, but I feel that's my problem not the videos. I especially appreciated computing all the values.

[@patvax532]

Never have I ever understood this concepts better then after this video. I had all of this in my electronics class in my studies but some things weren't exactly clear up to this point. Of course it was probably easier in understanding as I already got to work with an transistor. Nevertheless a great video and even better explanations. Thank you and keep up the good work.

[@marytennis7884]

You saved our homework. Thank you so much!

[@carlgradolph9676]

This is the best beginner-level explanation of transistor amplifier design I have found! Clear, concise and well illustrated. Explains the necessary (and sometimes scary) calculations without unnecessary chatter. I'm afraid I don't know much German, but I can say: "Vielen Dank" and "Ausgezeichnet"! I have subscribed to your channel and look forward to learning more.

[@Palamatar]

I've been watching a dozen videos on transistors trying to find the best way to calculate values for biasing. Your way to explain it really got me to understand it. Thanks a lot.

[@michaelfuchs441]

Thank's a lot! I'm always glad to hear that.

[@bestbytest1974]

Finally, I understood how these values are calculated and their meaning. Thanks, great video!

[@raphaeldpaiva]

That was the most amazing lesson I've ever had on transistors! Thank you very much!

[@harperkody466]

you probably dont give a shit but if you guys are stoned like me atm you can watch pretty much all of the new series on Instaflixxer. Have been binge watching with my gf for the last couple of days xD

[@casenquentin5879]

@Harper Kody yup, I've been watching on InstaFlixxer for years myself :)

[@eeemrky]

Best explanation ever. İt doesn't matter that the video is not in my mother tongue but that guy lectures ß times better than my teachers 😅

[@skullfuxxed4462]

I just wanted to thank you for the great lesson

[@michaelfuchs441]

Thank you for this great compliment! Always nice to read. ;)

[@rayhanizzat2311]

Thank you for uploading this video, god bless you with helping thousands of science and engineering students worldwide.

[@a.nelprober4971]

the best video ever period

[@ife.tugraz]

Thank you so much!

[@johnyjli]

10:16 I wonder if the equation is correct? I would think it should be r1//r2

[@michaelfuchs441]

I guess you're right. We need 750k/10=75k. Thank you for noticing.

[@user-jt4me1jj3b]

best video about NPN in the internet ! thank you

[@stepiankalunga752]

Please continue with the program am enjoying, I really want to start designing my own circuits

[@michaelfuchs1607]

We continued on kzbin.info

[@1Reevee]

Thank you, this is the best video on this topic I’ve seen

[@rajagohar4319]

Thank you very much for this easy and understandable explanation. Vielen Dank.

[@valiant3917]

Excellent video! that stuff finally makes sense.

[@Zombi-uw9ml]

Thank you!

[@johnlong5676]

Great Video ! This one is just CLEAR, easy to understand and with a usable example. Thank a lot. Keep up !

[@MrSeimson]

This video sets a benchmark! Very good explanation and nice editing.

[@fernandodecastro1783]

Good understanding of transistor principle as presented here will make troubleshooting of electronic equipment easier. Simple presentation but it "rocks". Kudos to the presentor. I am one of your subscriber and follower. Regards and keep safe😀

[@user-st2pw7ex6b]

At last . best explanation how to choose components sizes. Very informative . keep up.

[@rebsguevarra7508]

thanks sir.best explanation ever.

[@ver64]

Excellent explanation 👍🏻👍🏻👍🏻 and subscribed.

[@arkanjo7509]

very cool, thanks.

[@pyotrleflegin7255]

A fine video - good explanation and easy analogies to follow. Thank you very much.

[@jamesgilmore2389]

Excellent video. Thanks you.

[@kwekker]

why didn't I find this sooner I don't get everything yet but I just learned a heck of a lot more in 13 minutes than in the 20 hours I've listened to the teachers of my high school

[@manvirauppal]

Very Nicely explained everything . I'll be greatful if you suggest me what should I do to learn to build my own amplifier and low pass filter. I am a layman in electronics. But have gained some knowledge about basics of electronics components by watching KZbin videos learned a lot from video. Thank you in advance

[@michaelfuchs441]

I think the key in electronics is that you build stuff by yourself. Take an old radio, tear it appart and try to figure out how it works, watch videos on KZbin and try to understand the basic circuits that are used in electronics. Sooner or later you will need some measurement equipment, like a multimeter and an oscillsocope. You don't have to spend a lot of money on them at first. There are also videos on how to get affordable lab equipment for a good price. To learn basic circutis I can also recommend our new channel. We will publish a new video on electronics every week! kzbin.info

[@debanjanchakraborty9946]

The sound example is very practicle and seems like i am experimenting good explanation

[@goldendaffodils8085]

Terminology demonstration overall representation is awsome dear one U R assets. Thnx a lot sir May God bless you.

[@danish6192]

I guess there's a problem in expression at 07:30 it should be Z1out * 10

[@imooxat]

You're absolutely right. Thank you for bringing the mistake to our attention.

[@williamhunter4562]

This error also appears at 10:27. It should read "(R_1 || R_2) ≤ R_B/10" . Thanks for the video.

[@user-ce6xe1kf3q]

Danke schön, dein Englisch ist perfekt!

[@michaelfuchs441]

That's so nice of you! I know it's a lie, but it's very nice to hear.

[@poiuwnwang7109]

nice analogy to handle-valve at 3:44. Thanks.

[@Brian-mf3ry]

shouldnt the equation be Zout

[@michaelfuchs441]

You're right. Unfortunately we made this mistake.

[@RexxSchneider]

To refine it a little, once you have chosen standard resistor values, you quite rightly recalculated the base bias point to be 8.0V. However, that assumes an infinite beta, and it's worth noting that a transistor beta of 100 and a collector current of 1mA would draw an extra 0.01mA from that bias point, lowering it to about 7.3V. The effect of the rule of thumb (using R1 || R2 < 10 x emitter resistor) is bigger than you might anticipate, and considerably bigger than the effect of using standard resistors that you did calculate. The lowering of the output bias point by 0.7V isn't a bad thing as it gives a bit more headroom to keep the transistor out of saturation on positive swings. Obviously, you can never get the full rail-to-rail swing with bipolar devices. Finally - and it's only a small point - once you have fixed R1 and R2 as standard values, you should recalculate their parallel resistance which turns out to be 70K, not too far different from the 75K you used to calculate the input impedance, which is about 64K in this case. It close enough not to alter the value for the input capacitance calculated.

[@ife.tugraz]

That's a very smart observation, thank you. Looks like we've gotten a little sloppy with our rules of thumb here.

[@user-ub5su9vj6d]

Best of the best. 👍👍👍🙏🙏🙏

[@EdwinCloud]

Finally I understood transistor as amplifier with this video..

[@kwekker]

thanks for existing :)

[@peerappel2012]

Super danke!

[@cosminbel79]

Brilliant explanation!

[@james77011]

so pretty if u want to amplify a signal u have to put a voltage to the base emitter to forward bias the diode and than a larger voltage will flow from collector to emitter?

[@ramonluquin2612]

@11:08 I might need a refresher. what equation is used to derive 75k x (2.17/1.17). how do you multiply the equivalent resistance to the equation for parallel resistance, inverted, with the ratio values of a voltage divider to find R1 and R2.

[@imooxat]

For this calculation you need to know how parallel resistances are calculated. You can find a good explanation here: bit.ly/2Dr0Yl9. In Short: R_1//R_2=(R_1*R_2)/(R_1+R_2)=75k. If you now put in the ratio from above (R_2=1.17*R_1) you get 75k=(1.17*R_1^2)/(R_1+1.17*R_1) and if you solve this equation you will end up with R_1=75k*2.17/1.17. I hope this helps!

[@bogdankhamelyuk3857]

danke Michael und schöne Grüße aus Vorarlberg!

[@ingvarpekelny1281]

Hallo and thanks for the video. I cannot unfortunately understand why the output voltage does not depend on the supply voltage. Vcc is also applied to R making VR bigger when Ic starts growing. Could you please explain ho it works? Thank you in advance!

[@michaelfuchs441]

But the output voltage is dependent on the supply. The supply limits the maximum voltage at the output. With 15V supply and correct biasing you can never reach more than ~7.5V at the output.

[@abdelazeemhassan5629]

Great discussion , awesome perfect , thanks a lot for you for simplifying the knowledge

[@Enigma758]

8:58 I believe the capacitor is coupling capacitor to remove any DC offset. I don't believe it is there as an intentional filter, although it's cutoff frequency must be dealt with.

[@michaelfuchs441]

Yes, that's true, but I don't see why you cannot use it as such when it's there.

[@Enigma758]

@@michaelfuchs441 I think my point is that the goal is to remove DC offset but as a side effect, you end up with a filter as well. I think you want to size the capacitor so as to minimize the impact of the filter on your desired signal.

[@michaelfuchs441]

@@Enigma758 I can't argue with this conclusion. You're right.

[@Enigma758]

@@michaelfuchs441 Thanks for your response, just trying to be precise...

[@PhysicsNinja]

Very nice video with clear animations and explanations. Can i ask what software do you use for animations? Thanks

[@imooxat]

Thank you for your nice feedback. We were using Premiere Pro for the animations.

[@douglasstrother6584]

You beat me to it: I was going to recommend H&H as well! Look up Bob Pease's "What's all this about ..." for a ton (909kg) of insights on analog electronics.

[@iqraalvi4327]

👍👍👍👍👍👍👍👍

[@zacharytaylor190]

may someone please explain to me what the 2 vertical parallel lines mean in these equations?

[@michaelfuchs441]

Very good question! Two vertical lines between two resistor values mean that they are switched in parallel. If for instance the equation sais R1 || R2, this means that the total resistance of the two parallel resistors is calculated like this: R_total = R1 || R2 = 1 / (1 / R1 + 1 / R2) Look here for further explanation: www.sengpielaudio.com/calculator-paralresist.htm

[@abdul-alimbalarabedawud9628]

Hi, thanks for your efforts in enlighten us about the operation of a transistor. i need help. i have a very basic knowledge with regard to circuit building. i have been given the task of building an RF power amplifier for a small FM transmitter. Do i have to start the design from the biasing of the transistor? i am confuse because this is an RF amplifier not just audio amp. i want the output of the FM transmitter to be about 30 - 40W. can i use a 2N3632 transistor as the RF amplifier?

[@amitanaudiophile]

Really great upload....

[@incxxxx]

nice!

[@peterwiley706]

Very good. One question: where does the value 1.17 come from when talking about R1/R2 @9:08?

[@imooxat]

The supply voltage of 15V is divided by the two resistors R1 and R2. If you want an 8.1V voltage drop at R1 and therefore a 6.9V drop at R2, you need a divider ratio of 6.9 to 8.1, which is equal to 1 to 1.17.

[@billybob-392]

yes .. so its just 8.1 divided by 6.9

[@omarmostafa8385]

Great job,we Waite more more successful video in electronics

[@sandlertossone1813]

Don't you want the 6.9V to drop across R1?

[@ajitnandakumar]

@@imooxat Or can use the classic formula for Voltage divider... 15* ( R2/(R1+R2) = 8.1. If you solve for R1/R2 you will get 0.85. Which is equivalent to 1/1.17

[@ats89117]

I don't think the equation at 7:31 is correct. You really want Zout to be

[@chnithinsai601]

In electronics how to select parts bro for specific specifications like I have 4ohm 6w 60v speakers and I don't know what type of capacitor,transistor and resistor to use for making an amplifier

[@michaelfuchs441]

That's a hard question, because you need experience to answer this correctly. First of all you need to know if these 6 watts are RMS or some peak value. In any case, if you want to play it safe, you overdimension your amplifier. So your transistors must at least dig a collector current of about 1.5A (sqrt(P/R)=sqrt(6W/4ohm)=1.23A), your capacitors a voltage of 60V (which is deffinitly enough) and also the resistor at the collector must be able to sustain 1.5A, which leads to a pretty big resistor for high values of R (calculate over P=I^2*R). But as I said you are on the very safe side here. If you want to make better assumptions you can calculate even better or you can use simulation tools. Unfortunately most of all you will need practice and you will blow stuff up in getting there. That's the fun part of learning. ;)

[@chnithinsai601]

Tq for your information bro and anyway nice explanation

[@chnithinsai601]

Bro I was not able to get tip112 transistor and 60v capasitor can I use tip31c and 63v capacitor

[@michaelfuchs441]

@@chnithinsai601 ​Yes, I don't see why not.

[@chnithinsai601]

Bro I for got to say I have 2speakers each one with specifications of 60v, 6w and 4ohms now should I + these two speakers specifications and do an amplifier or should I take only one speaker specifications to do an amplifier

[@chitaramshathasivam4586]

Great.. Thanks for the good work. Subscribed

[@palavirajgude5717]

Thank god you exist, or else I would have keep Vc always lower than Ve

[@michaelfuchs1607]

I laughed way too hard at that comment. Thank you very much!

[@andyhubbard7374]

Finally, I get it! Thanks for the great video

[@zerta1681]

Thank you so much!

[@hamidbesharati3688]

Hi .is it pussible to amplify a fast edge squre wave (rise time=30ns) with 10 Vp-p to 100 Vp-p and with rise time being less than 60ns?

[@michaelfuchs441]

That's a tricky question, but I think in this case this is not the right circuit for you. If you want to amplify only square wave signals (so no sinusoidals or any other form), you don't need to use this kind of amplifier. You would then rather build a switching circuit with a high speed switching transistor. 30ns rise time will be possible with high power MOSFETs is this case. Also if you want negative voltages (because you wrote Vp-p) you will need two transistors (in push-pull configuration) for this endeavor.

[@hamidbesharati3688]

@@michaelfuchs441 thanks for the respond. Yes i want to build a high voltage squre wave genrator. So its pussible to reach 100 volt with power mosfet and have a fast edge still? I built a fast edge squre wave with schmitt triger & another one with ic555. But i could not amplify those more than 20 Vp-p.

[@mohammedhamid3715]

awesome

[@lillyzegarra8025]

can you explain the voltage divider a little and how you got r1/r2=1/1.17. The resistors r1 and r2 don't have values. please explain carefully. thank you.

[@michaelfuchs1607]

This question comes up a lot. Here is my previous explanation. Feel free to ask again, if it's still unclear. In short you get the 1/1.17 from the voltage divider at the input. If the biasing voltage at the base would be half of the suplly voltage, R1/R2 would be exactly 1. But since we need 0.6V more at the base, we get R1/R2 = 1/1.17. Now for small AC signals at the input the DC voltage source looks like a short and the input resistance looks like a parallel connection of R1 and R2. This is what is troubeling most of our viewers, which is why we will shortly do a video about small signal analysis. But from there on it's simple: You can calculate resistors R1 and R2 by knowing the ratio from above, and knowing, that they both in parallel need to be at least 10*R_E=10*7,5k=75k. From there on you can calculate the parallel resistance R_1//R_2=(R_1*R_2)/(R_1+R_2)=75k. If you now put in the ratio from above (R_2=1.17*R_1) you get 75k=(1.17*R_1^2)/(R_1+1.17*R_1) and if you solve this equation you will end up with R_1=75k*2.17/1.17. I hope this helps.

[@justadreamerforgood69]

You have to make more videos on inductors, filters MOSFETs. Also add a Patreon to your page so people can donate money for your brilliant work

[@michaelfuchs441]

Thank you very much! We will definitally continue to make videos. However, since I'm currently involved in research and teaching at the technical university I don't have time to put more focus on producing such videos. So there will be videos, but probably never more than 8 per year. Fortunately there is no need for a Patreon account, since I get paied by the university. ;)

[@kamleshchavan7451]

Sir Please 🙏 explain R1/R2 = 1/ 1.17 HOW did you get it

[@BlindRunner09]

@iMooX at @10:24 you said: the impedance looking into the voltage divider; must be at least 10 times smaller than the 'INPUT' impedance looking into the base. But it confused my mind; I think it shouldn't the INPUT impedance but the OUTPUT. So the correction is input to output. The sentence turned to this. The impedance looking into the voltage divider; must be at least 10 times smaller than the 'OUTPUT' impedance looking into the base. Otherwise the input impedance looking into the base is the same as is? :D

[@michaelfuchs441]

There is a mistake regarding the directions. Thank you for noticing! I should have said: "The OUTPUT impedance of the voltage divider must be at least 10 times smaller than the INPUT impedance looking into the base."

[@reynolds087]

How is the impedance of the follower calculated to 750K at 11:35? The value of Rb is not explained, and I can't figure out how to do the high pass filter without that value...

[@michaelfuchs441]

Rb is simply beta*R_E, as shown in 9:57. Still there seems to be an error in the calculation of R1 parallel to R2, which is more like 70k instead of 75k.

[@valiant3917]

@2:20, what is ment by Vc being 'more positive' than Vb?

[@michaelfuchs441]

It means that V_CE must be a positive voltage (with + on the collector and - on the emitter. Otherwise a positive input voltage would not lead to conduction of the transistor.

[@akashwa7969]

Amazing!! Please keep it going guys

[@BluetekFREEGAMES]

i undrestood everything till the thumb rules came , how in earth did you get that R1/R2 = ratio is 1/1.17

[@sauravmalpani1153]

So many things clear in video , I have some questions Can you please derive the total expression for video at 10:51 min. how R1 is llel to R2 and how 1.17 came .

[@michaelfuchs441]

Good question! The supply voltage of 15V is divided by the two resistors R1 and R2. You want to have 8.1V voltage drop at R2 and therefore a 6.9V drop at R1. So what you need a divider ratio of R1/R2 = 6.9/8.1, which is equal to R1/R2 = 1/1.17 .

[@justadreamerforgood69]

@@michaelfuchs441 Ok, finding R1 and R2 are easy, but how is Rin=(R1||R2)||RB?? Shouldn't Rin=R1||R2

[@sauravmalpani1153]

@@michaelfuchs441 ok thanks .

[@michaelfuchs441]

@@justadreamerforgood69 I admit, that is a little tricky to understand, since we are looking at a small-signal model of a bipolar transistor. You can find a pretty good explanation of what is going on here: wiki.analog.com/university/courses/electronics/text/chapter-9 Look at the small-signal model of the BJT in chapter 9.4 and following to get what's going on.

[@Dannydawson537]

Please can anybody help me I need to make a LF pre amplifier to amplify a 125khz am signal which is then going to be fed into another power amplifier then transmitted at 5.2ghz

[@TheXuism]

7:32 Is the inequation wrong？ should it be Zout

[@imooxat]

Yes, you are right. That is a mistake, which @Danish ALI already stated in the comments. We're sorry to confuse you.

[@michaell4527]

Very helpful.

[@j.w.2093]

The link to Wikipedia/KZbin no longer is active.

[@ife.tugraz]

Thank you for noticing. Here is a link that should work: kzbin.info/www/bejne/babOdX6MfdOUmZY

[@alcoyot]

I don’t understand this circuit. Why is the output voltage lower than the input. How does that make it an amplifier if it is less than what it started ?

[@michaelfuchs441]

Legitimate question! The reason is that the transistor is a current-amplifier. It only amplifies currents, not voltages. If you want to learn how to build a voltage amplifier, you will be interested in our next video on transistor amplifiers: kzbin.info/www/bejne/jZ2umZ6grt-ZhKs

[@johanariff1460]

The voltage amplification factor of this circuit is not defined as the ratio of the dc output voltage to dc input voltage as you seem to think. Rather it is defined as the ratio of peak-to-peak output voltage to peak-to-peak input voltage. Detailed circuit analysis will show that the voltage gain for this common collector amplifier circuit is slightly less than 1 (about 0.98). It's not employed for its amplification property but for its high input impedance and low output impedance property, i.e. as a buffer between two circuits with "incompatible" impedance levels.

[@cemkarakarcayildiz5462]

thank you for video good job. but i dont understand at 10:41 where that 2.17 coming from????

[@michaelfuchs441]

We know the ratio R_1/R_2 = 1/1.17, or in other words R_2 = 1.17*R_1. Because we do a "small signal analysis" R_1 is considered to be in parallel with R_2 (written as R_1 || R_2), which should be 75 kOhm. If you do the math and calculate the total value of R_1 || R_2, you get: 75 kOhm = (R_1*R_2) / (R_1 + R_2) Now plug in the ratio from before, you get: 75 kOhm = (R_1* 1.17*R_1) / (R_1 + 1.17*R_1) = (1.17*R_1^2) / (2.17*R_1) = 1.17*R_1 / 2.17 In the last step you solve for R_1 and get: R_1 = 75 kOhm * 2.17/1.17

[@eggxecution]

why is the output sometimes at the collector? pls help, also your vids are more clearer than the others

[@michaelfuchs441]

For the emitter follower the output is at the emitter. The idea is to amplifiy a current. If you want to build a voltage amplifier you need to add another resistor between Vcc and the collector and then the output is also at the collector. Just watch our second video on the subject: kzbin.info/www/bejne/jZ2umZ6grt-ZhKs

[@tsundzukanileonmashamba3458]

It was great thank you

[@mohamadnajm1465]

Hi Thanks for the video Question How R1||R2 = 1.17??

[@jibrealkhan3525]

Great Explanation but my doubt is that why are resistors R1 and R2 considered as parallel for the purpose of choosing their values ? As according to me they are clearly in series.

[@michaelfuchs441]

Good point! The explanation for this mystery is something like this: What we want to do is a so-called "small signal analysis". We only want to look at the AC signals, not the DC signals. In the case of an AC signal with high frequency, any DC supply (such as the 15 V) can be considered a short circuit. Also the capacitor C_1 at the input can be considered as a short circuit for high frequency AC signals. If you now look from the input side, you see a resistor R_2 going to ground and a resistor R_1 going to the 15 V DC supply, which is considered a short circuit to ground. Therefore the two resistors are considered to be parallel. I know this is a little confusing at first, but you get used to it. There are a lot of videos covering small-signal analysis. We are also working on one right now. I hope this explanation will help in the mean time.

[@jibrealkhan3525]

@@michaelfuchs441 thank you for the quick reply and yes it definetly helped.

[@bixenbaer]

Nice Video. Ist seems That you just took the explanation from the Book the art of Electronics and put it in a Video with simple Language. Thanks. Edit: ok u mention it at the end ;)

[@attarpattar]

at 10:12 why Rb=beta × Re ??

[@michaelfuchs441]

As it is explaned in 5:04 , I_C=beta*I_B and I_C is approximately the same as I_E. It is actually I_C+I_B, but I_B is very small and can be neglected. You will also often see in literature, that I_E=(1+beta)*I_B and then (becaus beta is much higher than 1) the approximation I_E=beta*I_B. Because the currents are indirectly proportional to the resistances (the higher the current, the lower the resistance and vice versa), consequential R_B=beta*R_E.

[@HDESM]

If this is amplifying the current but you are assuming only 1 mA at the output (9:24) and no current flowing through a load, then where is the current all going ?

[@imooxat]

The current will flow solely through R to ground (and back to the power supply, which provides the 15 volts).

[@HDESM]

In the diagram at 6:52 it appears that all of the amplified current IC is just flowing to ground through the resistor R. Is R the load ?

[@imooxat]

No, R is not the load in this case. A load could be connected in parallel to R. Usually you will also find an additional capacitor at the output acting as a high pass filter. Here you can find an example of such a circuit here: bit.ly/2SsrJxw

[@mystwalker479]

How do i choose a transistor? I'm sure there are transistors rated for high voltage applications. I needed to buy and explore one rather than listening and reading..

[@michaelfuchs441]

Typically you choose a transistor according to the most important parameters for your application in the datasheet. But it's not easy to tell you what you need for a specific application. You might want to amplify high currents (collector current)), or you might want to have a very fast transistor for switching (turn on time) or for high freuency amplification (gain bandwidth product), or you might want to switch high voltages (max. collector-emitter voltage or better look for MOSFETs, Thyristors, JFETs,...) maybe with high efficiency (on resistance) or maybe you need very low leakage currents, or you might be more concerned about linearity, or temperature behavior. It all depends...

[@namename8986]

9:35 that is 2 mA ?

[@michaelfuchs441]

No, 7.5 volts at the output divided by the 7.5 k resistor is 1mA.

[@fc2p]

Nice job Mike!

[@joeenglert7105]

Thank you very much for the great videos. I'm watching this after enjoying your ADC series. I am confused by the modeling of a transistor as two diodes. In the NPN example, the collector diode would stop current from flowing to the emitter. Not sure what I'm missing about the analogy.

[@michaelfuchs441]

This analogy of course does not represent the real behaviour of a transistor on the output side. It is just to remind you that - seen from the base - we are dealing with 2 pn junctions here. So we definitely have a voltage drop from the base to the emitter. By the way, the analogy to the two diodes also points to another circumstance that is often forgotten. Theoretically, the collector and emitter of the transistor could be swapped. It would still work. In reality, of course, it won't do that very efficiently, but it does.

[@ahmetkoraysonal5841]

Sir, you said that it should be 750 k, you said that the input resistance should be 10 times bigger than it should be 75k, according to what did you choose it? First you said it would be B times bigger, then you said it as 10 times.

[@ife.tugraz]

I see your confusion. I'm talking about two different impedance values here, which an AC current would see. - The first one I'm talking about is the input impedance of the transistor. Imagine you are an AC signal, standing at the base connection of the transistor. If you look to the right (the direction of the base), you will see an input impedance of beta*R_E. Since beta is approximately 100, the base impedance is about 750k. - The second impedance is the one of the voltage divider. Again imagine standing at the base connection, but this time you look to the left and see the parallel impedance of the voltage divider (R1 and R2), which should be at least 10 times lower than the base impedance (on the right side) --> therefore 75k. Finally you can imagine yourself standing at the input looking to the right. If you neglect the capacitor for a moment you will see the two impedances we just calculated in parallel (75k || 750k), which is about 68k. With this value you can calculate the necessary value vor the capacitor C_1. I know it's a little tricky to understand the AC signal point of view at first. If you are struggling with it, it's worth to google "small signal analysis" for further reading. I hope this helps!

[@ahmetkoraysonal5841]

@@ife.tugraz Sir.Thank you very very much for detail explanation.I understood now .Thnak you

[@ahmetkoraysonal5841]

@@ife.tugraz sir, you explained it very well, I have a request from you, would you make such a video about mosfet? (how should input, output impedance etc.)

[@ife.tugraz]

@@ahmetkoraysonal5841 Yes, we probably will and publish it on our new channel: kzbin.info/door/LrXYcSgmlRdFM3ob8Q4Wnw

[@attilarivera]

👏👏👏👏👏👏👏

[@joshuasangabriel1088]

Why R1/R2 is 1/1.17? Why is that 1? In numerator and after that you use 2.17 in numerator?

[@michaelfuchs441]

The supply voltage of 15V is divided by the two resistors R1 and R2. You want to have 8.1V voltage drop at R2 and therefore a 6.9V drop at R1. So what you need a divider ratio of R1/R2 = 6.9/8.1, which is equal to R1/R2 = 1/1.17 . Does this help?

[@edwardv4546]

THANK YOU

[@nerijusvilcinskas7851]

beta ratio is also sometimes written as hfe

[@ektro-johnny573]

I have a question for anyone here: I'm trying to make a small amplifier to drive an analog VU meter. It's a stereo meter which means I have to have two amplifier circuits for each side L/R. I'm using a C945 transistor and some spare resistors. I have a dual potentiometer as well. Now for a couple questions. In the schematic at 11:12, are the leads going to ground meant to go to, for example, minus on a battery? I'm guessing the 15V at the collector is for the plus side of a battery at least. And if I wanted to be able to adjust the max output of the VU meter, where would I add the potentiometer?

[@michaelfuchs441]

To answer the first question: Yes, the ground connection is made at the minus side of the battery. I'm afraid though, that the answer to the second question is like an advertisement for our own videos. Maybe the problem is easier to tackle if you would watch our second video on transistor amplifiers (kzbin.info/www/bejne/jZ2umZ6grt-ZhKs). As you will see, you can build a voltage amplifier quite easy by taking a collector resistor, which can also be a potentiometer. I hope this helps.

[@sanjaygsn2439]

6:41 is output current is ic ?

[@michaelfuchs441]

In most literature you will find I_C as output current, as this allows a simple definition of the gain beta (I_C = beta * I_B). But you have a point there. One could argue that the actual output current must be I_E, since this is the current flowing into the load. In this case, you must add the base current to I_C to obtain the new output current. With the previously defined gain beta you get I_E = (beta + 1) * I_B.

[@sanjaygsn2439]

@@michaelfuchs441 ohh ! thank you for your reply! ..............understood.

[@styzr]

The impedance calculations were confusing at best. Care to elaborate on that?

[@michaelfuchs441]

Since this question is rather vague, I do not have a good answer. But I can provide you with some additional material for a better understanding. A good example can be found on pages 13-15 in this document: www.ee.ic.ac.uk/pcheung/teaching/aero2_signals&systems/transistor%20circuit%20notes.pdf Go through it, and if you have additional, more specific questions, I will be happy to answer them.