The discussion revolves around the resistance of a diode when subjected to direct current (D.C.) versus alternating current (A.C.), focusing on the interpretation of I-V characteristics and the implications of slope on resistance values.
There is an ongoing exploration of the relationship between D.C. and A.C. resistance, with participants providing insights based on graphical interpretations. Some participants express uncertainty and seek clarification, while others assert their understanding of the resistance behavior.
Participants are considering the effects of reverse bias and breakdown voltage on resistance, which introduces additional complexity to the discussion. There is also a mention of the definition of linear devices in relation to the problem context.
Can you show us on the I-V curve? I think I know what the answer is, but I'm not sure.Pushoam said:
Yes, correct. With that graph you can show that the tangent to the graph always has a greater slope than the line from the origin to the point on the graph (at least in the 1st quadrant).Pushoam said:
Before the breakdown voltage, the DC resistance is less than AC resistance.berkeman said:Quiz Question -- is it still true in the 3rd quadrant for reverse bias? Include low-level reverse bias and as the diode starts to go into reverse breakdown...
Can you show on the I-V curve why you think this is true?Pushoam said:
berkeman said:Can you show on the I-V curve why you think this is true?
Yep! Good job. Looking at the slope really helps on this type of question. You can also show it mathematically by taking the first derivative of the Diode Equation, but this one is pretty easy to see graphically.Pushoam said:
