Load Line Method with Transistors (19-Transistors)
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- Опубликовано: 12 сен 2024
- If the emitter is grounded, the transistor has a nonlinear transfer function. How can you determine the gain? Let's use the load line method to find the operating point and then show what happens for both DC and AC. With example problems.
Aaron Danner is a professor in the Department of Electrical and Computer Engineering at the National University of Singapore.
danner.group
Video filmed and edited with help of CIT, NUS.
Thank you, your lectures takes me back 50 years or so… thanks for the memories and also showing me I really need to relearn the basics. So often, as you grow older you forget all the basics…. Grrrrrr, getting old totally sucks. I mean, I did work as a design-engineer for 30+ years, but watching some elementary design videos really brings you back to how much you have forgotten in those years…
I love your explanation... Thanks!
Videos in the past were more interesting when circuits were built and tested, compared to recent videos in which just theory and algebra are presented.
Thank you professor ❤
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The last example raise a technical problem that I never solved in my life: for the "static" load line you obtained the line equation using Ohm's and Kirchhoff laws in the circuit: Ic = (Vcc-Vce)/Rc but how can the line equation be obtained for the "dynamic" load line (considering the effect of RL)? For the dynamic circuit the AC part of the supply is zero and in the axes of the graph we don't represent the AC values but only the instantaneous values. In other words I'm not so sure that the dynamic load line concept is correct. What do you think? Thank you for your very appreciated work.
Excellent videos. I am watching all of them on this channel.
Question:
In the first example Vbe = 0.9V was larger then Vce = 0.8V. So the transistor must be in saturation mode. Because Vcb = -0.1 V. Then why is the dot of the oparation point in the right graph in de active region? Or am I missing something.
Transistor is more voltage control device than current control, it is IB =IC/B not IB = IC x B .
When IB = 20 mA, IC is about 620 mA, so Beta shall be about 31 not 12! :)