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.
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?
How did you derived 0.3 v value? How can electron pass base collector junction, when it is in reverse bias? Generally, in PN junction, in reverse bias, election can not cross junction in reverse bias.
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
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
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.
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:
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.
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.
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.
I hope you know how valuable these videos are! Cheers from Canada.
Thank you Prof. for this precious explanation 🙏
clear and smooth explanation, suggesting to insert the special formula for reference
Big W, U saved my ahh for this exam bro u the best
perfect for what I have been looking for
I am so grateful for your explanation. Thanks
thank you for your video
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?
I hope you will consider adding videos for the second half of the textbook: chapters 9 -14
Thank you! Superb explanation
19:10 isn't it 10V for Rth because 15V*(100/150) = 10 not 5
No because Vb is the voltage drop across the 50k resistor not the 100k resistor
How did you derived 0.3 v value?
How can electron pass base collector junction, when it is in reverse bias? Generally, in PN junction, in reverse bias, election can not cross junction in reverse bias.
Thanks Microelectronics 💖✨💫🙏🌹💖
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
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
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.
Good job explaining
Thank you 👍
In saturation why is Vce = 0.2 V ?
Very clear.
which books is you took the problem examples ?
Microelectronic Circuits by Sedra & Smith
Great man
How did you got .2v
Great!!
how much is alpha ,please show steps
What is this book name
Microelectronic circuits
OMG AN ACTAL ACCENT
big up da mandem
Connelly Parkway
Jarrett Grove
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:
Lol
ok
Caleb Prairie
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.
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.
Asia Glen
Negative and positive 😊
Ledner Ramp
Kallie Court
Gonzalez Ronald Gonzalez Sandra Harris Donald
Turcotte Rue
TQ OIIIIII !!!🤣
Tad Oval