# How to use IV characteristics of a diode in calculations

In the following question they have asked to calculate the negative and postitive peaks of the output voltage across load resistor and to determine the peak load current and the diode power dissipation

The peak values can be calculated as 49.3V and 0V

Hence the peak load current and power dissipation can be calculated

But they have also provided the IV characteristic of the diode also

How does that helps me in this question Or is there any errors in my method because i haven't used the characteristic curve • Looking at the graph - for any given current through the load you can determine the forward voltage drop across the diode - for about 100mA this is about 0.9V (not 0.7V) giving 49.1V. In reverse bias a current of about 1uA flows so there will be some negative voltage (-0.5mV - not quite 0) across the load resistor. – JIm Dearden Oct 23 '16 at 15:33

The graph give you a more accurate answer that assuming 0.7 which is only an approximation. The voltage and resistor are already in thevenin equivalent form. on the graph this is a line as the voltage goes from 0 to 50. It would be easy to plot the line if the graph when out to 50V because then out just use the end points 0.1A and 50V. Since this ins impractical on this graph, use the equation I=-V/500+0.1. If you do this at V=1.2 I=0.0976A. Plot this point and draw the line to 0V,0.1A. Where this line crosses the diode curve is the actual voltage for the diode when you have 50V and 500Ohms.

When a silicon diode reaches 0.6 V we call it "saturated" and then the slope becomes the bulk series resistance ( or as I like to call it, the incremental effective series resistance , ESR.

The ESR. is the almost linear slope after 0.6 V but current ratings are often given for Vf=0.7V or Vf @ xx amps, as the transition zone for lower than rated currents e.g.10% If max tends to be in this range from 0.6 to 0.7 drop ( almost like a 1.5V battery voltage ranges from 1.3 to 1.6V )

If Vo = 49.3V max then for any given diode we can compute the current that causes a 0.7V drop or choose a diode datasheet and get these parameters.

Here is the common 1N4148 small signal diode curves with more details from the datasheet.

See if you can find the current approximately (as there will be a tolerance for all parts ) for the forward conducting voltage Vf=700 mV .

Using only Fig 4. as this small is rated only for 10mA DC for heat reasons. 