Draw the current-voltage characteristic of a diode?

Question

Hi, Unfortunately, I'm stuck on this task. Can someone please help me. I would be grateful for any tips! Thank you in advance

easylife2 · Answer

From Figure 1, you can read the required data to calculate the internal resistance of the diode. The breakdown voltage of the diode is also easy to read.

lutz28213 · Answer

Even if the diocesan line has been simplified by a (shifted) straight line, it is a non-linear characteristic (because it does not go through 0/0).

In such a case, a graphic solution is available, but it consists of two parts:

(1) For Uq positive (only D1 effective) and 2) For Uq negative (only D2 effective).

However, both parts are calculated according to the same principle (break voltages do not need at all). For the left diode:

For this purpose, the characteristic curve of the resistor R must be drawn into the given characteristic curve of the diode Id=f(Ud), but of course expressed by Id (which also flows through R due to the series connection) and Ud.
Like this: Ur=Uq-Ud and therefore Ud=Uq-Id*R. This characteristic is easy to record with two points. To this end (a)Ud=0 set for a point on the Id axis and (b) Id=0 set for the point on the Ud axis.
Then you have 2 characteristic curves in the picture and the intersection is the required terminal voltage (point b)
For point a), simply add both partial voltages (Ud and Uq) point by point for the same current.

Similar procedure in principle for the other (right) diode – and finally both sub-images merge.