43 BJT Circuits at DC

This is the 43rd video in a series of lecture videos by Prof. Tony Chan Carusone, author of Microelectronic Circuits, 8th Edition, covering chapters 1 - 7 of the text: Devices and Basic Circuits. Find more resources at analogicdesign.com.
In this video, we will look at practical techniques for the analysis of BJT circuits at DC.

[@mugiwara-no-luffy]

I hope you know how valuable these videos are! Cheers from Canada.

[@ahyungrocks5509]

This is super helpful. 30 years ago when I took microelectronics while learning English, I was not able to comprehend despite I passed the course. Because of that I stayed away from the dark side (analog). Didn't realize this is so much more fun than 1s and 0s.

[@mohamedqasim7787]

Thank you Prof. for this precious explanation 🙏

[@ramirosuarez9537]

I am so grateful for your explanation. Thanks

[@jxcjxuan]

clear and smooth explanation, suggesting to insert the special formula for reference

[@gilbertgonzalez7890]

Thank you! Superb explanation

[@tonykimani9933]

perfect for what I have been looking for

[@asarekofikwaasi8010]

This playlist is damn good!

[@labanidharnaik1103]

Very clear.

[@GOODBYE.SHAHZAIBNAJAF]

Great man

[@niccettajames6072]

Good job explaining

[@raccoonlord9646]

Big W, U saved my ahh for this exam bro u the best

[@AllenSA_Lily]

Thank you 👍

[@angelabryan7529]

I hope you will consider adding videos for the second half of the textbook: chapters 9 -14

[@sipoVvids]

Great!!

[@jonelberdejo9446]

In your final analyis I think the transistor in saturation mode. Because the base current is approximately 129 micro amps and then the current gain is 100 so the collector current should be(100*129microA), which is 12.9 mA. So to find the collector voltage we have to subtract 15-(12.9mA*5000) and we get an impossible answer that is -49.5 which indicates that the transistor is operating in the saturation mode. A hard saturation

[@gwav1a]

Yes it turns out like that but in actual fact you apply 0.7 across the base emmitter diode and the large part of the majority carriers from the emmitter cross the base collector junction to constitute the collector current. The smaller portion comes out through the base. The proportionality of collector current to base is beta. Viewing it this way shows causation - what results in what. To me base current is the result, not the cause of collector current. I would say the base current is a function of collector current not the other way

[@Lexyvil]

For the circuit at 9:30, why is there a current I_c when the diode at the collector point is against I_c? (Reversed)? Since it's reversed bias, shouldn't I_c be 0?

[@tinaquayum8910]

How did you got .2v

[@babatundetaiwo2817]

So let me get this straight: If I apply a 0.7 Voltage bias from Base with respect to the emitter while also supplying current in the base of the NPN, I will get an output current from Collector to emitter that is a multilocation beta of the Base current. This the emitter is what I'm taking everything in respect to : Voltage form Base to emitter with a current applied to the base outputs a BETA times the input current flowing from collector to emitter.

[@sibilakshman9575]

In saturation why is Vce = 0.2 V ?

[@makokakao6782]

which books is you took the problem examples ?

[@nelsonkh001]

Microelectronic Circuits by Sedra & Smith

[@thokozaninhlakahh9255]

how much is alpha ,please show steps

[@ikmalhakimi2957]

TQ OIIIIII !!!🤣

[@shantnukumar3982]

What is this book name

[@austinpetersen6899]

Microelectronic circuits

[@worldgalaxyuniversemilkiwa1321]

Negative and positive 😊

[@VeritasEtAequitas]

Thanks, but in the final analysis around 22:00 - 23:00 you dropped the k on the resistor. I think it's 33.3k(IB) plus (beta+1)IB. That's 33,300 +303, so then Ib = 4.7/(33,603) ≈ 139.87 uA.

[@ohbaewan3147]

not quite, he factors out the k at first because it is part of the units, so its (beta+1)IB*3 which is why it simplifies to 303 but is still in units of k, then in the end he adds them up, 303 + 33.3 = 336.3 and adds the k back during unit evaluation.

[@malekhaboosh476]

Use BJT transistor to design an amplifier circuit that operates in the frequency range 10kHz100kHz. Verify your design by simulating the design using the circuitmaker software. Use standard resistor values (10% tolerance). The design should meet the following criteria:

[@VeritasEtAequitas]

Lol

[@eckee]

ok

[@arashyusefi1889]

Thanks Microelectronics 💖✨💫🙏🌹💖