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?
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.
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.
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:
