Chapter 5: Setting Up An Emitter Follower Circuit

[@miketony2069]

Watching videos like this makes me realize how horrible my college professors were... the teacher makes all the difference in the world. Thank you for a clear video

[@nyambemundopu1168]

The professor makes life easier than some of my former lecturers in Analogy Electronics. Very refreshing, I must say. I think I'II get addicted, if no hooked already, to Professor Curtis Meyer videos. Great job professor. Keep up the good work.

[@davidwisbey943]

This man is a rock star! Thank you for a clear concise lecture! I am using what you do in my classes. Thank you so much!

[@toxicore1190]

one of the best lectures on electronics

[@Bishka100]

Thanks for the clear and concise vid'.

[@user-su5sq5ib3i]

I love watching your lectures

[@mikesfarmmikesfarm3977]

like your teaching style. really simplified it

[@EdwinFairchild]

that was a beautiful explination.

[@coby1kenobii]

This was very nice. thank you

[@gabrielrodrigo5626]

Very good!!!!!! thank u very much!!!

[@DwiNoviantoUntidar]

excuse me Prof, how to derivate Output Resistance on Emitter Follower with Voltage Divider Bias? Thank you.

[@maxbarnett5924]

How can the input impedance be high if the first thing the signal sees when exiting the input capacitor is the 1K to ground resistor?

[@JurekPrzezdziecki]

Super clear and simple explanation. One question: where is a load resistor? Normally 4 resistors are used for this type of circuits?

[@bart99gt]

JurekPrzezdziecki It is an emitter follower, the output is developed across the emitter resistor.

[@ElectronicsLearning]

Hey, great, great video. One question I have, why do you apply the 20Hz to the angular frequency equation, ω =2πf? What's the significance of using this equation for the calculation?

[@WMTeWu]

+ElectronicsLearning The cutoff frequency of high pass filter is 1/(2 pi R C1) where R is 500 (just like shown in 9:20) and C1 is value for the capacitor you are looking for. So you just need to solve 1 --------------------------- = 20 2 * pi * 500 * C1 for C1, and you get C1 = 0000159155, approximately 16uF.

[@CarlosHenrique-ve9nf]

Salvou os irmao do brasil thank u

[@danish6192]

for dc input and output the emitter follower does not follow the rule (Ve = Vb - Vdiode) until about (Vin > 1/2 * Vc) , where can I get explanation of that, please somebody help me on this

[@psych.fenatic]

0:00 but now I don' wanna learn anymore

[@perceptrongaming4290]

sir my voltage drop when putting a 8 ohm speaker at the output side .

[@satadalsen2178]

how did omega(Rc) became 125 s inverse? plz explain

[@nogrend]

w = 2*pi*f [rad/s] or [rad*s^-1] and f = 20 [Hz] or [s^-1] w = 2*pi*20 = 125 rad/s Note that because the radian is a dimensionless unit, the radian per second is dimensionally equivalent to the hertz-both are defined as one s−1. This means that great care must be taken to avoid confusing angular frequency ω and frequency f. Source URL: en.wikipedia.org/wiki/Radian_per_second

[@HalfVccTronYente]

👍👍👍🎖️🏅🎖️

[@smallyericho8037]

at 9:18 .. why did u x100 times the 560 ohm to 56k ohm ???

[@danish6192]

100 is beta (the hFE) and according to formula for input impedance Zin = b*Rload

[@fernandohood5542]

You will lose a few people with 125. Also is that input impedance useful 500 ohms? What about small re?

[@martovify]

Great!, im not sure why most professors choose to solve a circuit that already has resistance values instead of designing from specs to find the resistor values needed.

[@wilmdrdo1228]

why cant you have large voltage divider resistor

[@sudipdhara8752]

+William Madrideo That will limit the base current I guess! Theoretically input to the base is current, not voltage.

[@Kronkulus9282]

Lol this is from 4 years ago but im bored so: the output impedance of this circuit is related to the input impedance by a factor of Beta (current gain of transistor), which he was using 100 for Beta in the video. If you choose too high of bias resistors, even dividing by beta still leaves you with a high output impedance and you can't properly drive heavy loads (which is what this circuit is sort of meant for).