You can save a lot of math, by understanding that the load line draws the line between two extreme situations, a short across diode when v=0 and an open circuit, when i=0. Just take the equation expressed at 3.21sec of the video, set i=0, you'll get v=5. Then take the same equation and set v=0, to get i=1/25. Then draw the line between the points on the graph through the diode characterists to find the ideal operational point, Q. Its good to know math but its only a tool.
Ive been thinking, why you have to go through all that process when just substituting 0.7 volts for Vd will give you the current right away. And then what happened to the 20ma Q point?
your i-v characteristic graph is not plotted well, there can't be such rise in current as the voltage rises, therefore you got totally incorrect readings from the graph. unless you are operating at very high temperatures. another case could be that the diode is LED, which has Vd in between 2V and 4V. that's why you got two different current readings. current can't be 34 mA and 20 mA at the same time through the diode. on the graph we see that it's probably LED that you're dealing with, and when you try to confirm this by calculation, you totally disregard that fact and say "yo, it's just a regular silicon diode, with forward voltage of 0.75V. try being consistent next time and use an appropriate graph
you should use load line method when there is only one diode in the circuit. Otherwise you should use other dc analysis methods like piecewise linear circuit model.
Find the Thavenin equivalent circuit without a diode....that will give you one voltage source and one resistor. Then you add the diode and Work out the load line in normal way....
his i-v characteristic graph is not plotted well, there can't be such rise in current as the voltage rises, therefore he got totally incorrect readings from the graph. unless he's operating at very high temperatures. another case could be that the diode is LED, which has Vd in between 2V and 4V. that's why he got two different current readings. current can't be 34 mA and 20 mA at the same time through the diode. on the graph we see that it's probably LED that he's dealing with, and when he tries to confirm this by calculation, he totally disregards that fact and says "yo, it's just a regular silicon diode, with forward voltage of 0.75V, I don't care about those 2.5V from earlier. he took wrong Vd in the end of the video, totally not caring about the graph. that's why
You can save a lot of math, by understanding that the load line draws the line between two extreme situations, a short across diode when v=0 and an open circuit, when i=0. Just take the equation expressed at 3.21sec of the video, set i=0, you'll get v=5. Then take the same equation and set v=0, to get i=1/25. Then draw the line between the points on the graph through the diode characterists to find the ideal operational point, Q. Its good to know math but its only a tool.
Thank u for a very simple detailed explanation!
Ive been thinking, why you have to go through all that process when just substituting 0.7 volts for Vd will give you the current right away. And then what happened to the 20ma Q point?
your i-v characteristic graph is not plotted well, there can't be such rise in current as the voltage rises, therefore you got totally incorrect readings from the graph. unless you are operating at very high temperatures. another case could be that the diode is LED, which has Vd in between 2V and 4V. that's why you got two different current readings. current can't be 34 mA and 20 mA at the same time through the diode. on the graph we see that it's probably LED that you're dealing with, and when you try to confirm this by calculation, you totally disregard that fact and say "yo, it's just a regular silicon diode, with forward voltage of 0.75V. try being consistent next time and use an appropriate graph
What if you also have a resistor parallel to the diode? How do you find the load line in that case?
one year old question, no one answered it. I really wish I knew it too
you should use load line method when there is only one diode in the circuit. Otherwise you should use other dc analysis methods like piecewise linear circuit model.
Find the Thavenin equivalent circuit without a diode....that will give you one voltage source and one resistor. Then you add the diode and Work out the load line in normal way....
how did you assume the 0.75 voltage of the diode. the graph says it lies in 0.8
why choose supply voltage 5V? why not say 4V or 3V? in that case Q point will change. how to deal with this?
How can the current in the diode be 20 mA and 34 mA?
Didn't understand that as well!
his i-v characteristic graph is not plotted well, there can't be such rise in current as the voltage rises, therefore he got totally incorrect readings from the graph. unless he's operating at very high temperatures. another case could be that the diode is LED, which has Vd in between 2V and 4V. that's why he got two different current readings. current can't be 34 mA and 20 mA at the same time through the diode. on the graph we see that it's probably LED that he's dealing with, and when he tries to confirm this by calculation, he totally disregards that fact and says "yo, it's just a regular silicon diode, with forward voltage of 0.75V, I don't care about those 2.5V from earlier. he took wrong Vd in the end of the video, totally not caring about the graph. that's why
very good explanation
thanx a lot.u made it toooo easy :)