How is the current through the diode calculated?
Hello, unfortunately I have absolutely no idea how to solve the problem.
Can someone explain to me what I have to do here?
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Not at all… there is no voltage indication without which the current through the diode cannot be calculated.
That’s wrong. It’s all there is.
Then tell me the total voltage, which one is on.
what do you want to tell me? The task is solved below. Already seen?
Or are you just rolling? Slowly I think it can only be that…
I = U/R ….
She didn’t ask? Why do you need these? The question is already answered…
There is
All this is clearly defined (even the total voltage, which is not in demand here)
This can be solved exactly graphically: The diode sees as a source
and
Now you cut the corresponding work line
with the characteristic curve. The intersection is about 0.62V.
Do I see that you have accepted that on your account the voltage at R2 is determined by the total current (and not by a partial current) at the upper branch? There’s something wrong with that!
Sorry, I made a stupid mistake. You’re right.
If you neglect the differential resistance of the diode, you can determine the substream of a branch and thus the diode current via the current divider.
@AMG38 can you explain to me what is meant by differential resistance?
Don’t be irritated – the differential resistance (start of the characteristic) doesn’t matter!
You can see at the voltage-current curve of the diode that the current is dependent on the voltage. This dependency is not proportional but exponential.
For example, if you look at 0.5V, a current of 20 mA flows. The resistance of the diode at this point would then be at 0.5V/20mA=25Ohm. This, however, cannot be taken as a generally or permanently valid value for all voltages, for you can see, for example at 0.6V, there flows a current of 40mA. The associated resistance is then at 0.6V/40mA 15Ohm.
As you can see, the resistance has decreased. It thus changes as a function of the applied voltage. Thus, the resistance is not constant but differential.
It is also possible to solve exactly.