Could someone please help me comprehend this exercise and guide me towards its solution??

 

Do you know how much diode leakage current and forward voltage drop changes with temperature?
If not, then you need to research it.
You need that information to solve the problem.

 

At 20°C form the data we know that Vd2 = VR1 = 520mV therefore


I = VR1/R1 = 520mV/520kΩ = 1μA (we know now the D1 diode reverse leakage current)

At 40 degrees of Celsius D2 diode forward voltage will drop by witch number? Do you know this "magic number"?
Also what about reverse leakage current? Will it go up or down with the increase in temperature?

 

View attachment 138288Could someone please help me comprehend this exercise and guide me towards its solution??

Hi,

Questions about diode circuits really have to have a model associated with the diode(s) being used. If the model is not given, then some background information about the course work would help to determine what is allowed and what is not allowed. This is because there are so many different ways of thinking about diodes.

For example, there is a linear approximation for a diode under a change of temperature. If that is an allowed approximation then you could use that idea to help solve the problem. That's because the 520k resistor acts almost like a constant current source for small changes in diode voltage. In real life the current will change a little too, but as an approximation it may not change enough to alter the simpler approximation by a significant amount. Note that if the 520k resistor was really just 5 Ohms for example then the linear approximation involving voltage only would be much less valid.

I would say to start you could think of this in terms of this linear approximation as i see Jody was suggesting. Your chore is to figure out what happens with a change of plus and minus 20 degrees C.