I have performed a small experiment to obtain the VI characteristics of a semiconductor diode using ANALOG DISCOVERY circuit design kit. I have connected the arbitrary waveform generator (function generator) to a series combination of a 1N4007 diode and a 1 kilo ohm resistor. To obtain the VI characteristics, I connected the terminals of oscilloscope channel 1 across the diode (to obtain the diode voltage as a function of time) and channel 2 terminals across the resistor. The voltage across the resistor is proportional to the current in the circuit according to ohm's law. (since I have chosen a 1 kΩ resistor, the numerical value of voltage (in volts) across R is the same as current (in milliamps)... In XY mode, I obtained a plot channel 1 (V) vs channel 2 (I). Then I changed the current scale of the VI plot by varying volts/division for channel 2. It appears to me that the cut in voltage of the diode is varying with the scale chosen. for 5V/div resistor voltage (5mA/div current), the knee voltage is about 6.5 V but as i decrease the scale, the knee shifts toward the left... For 20mV/div (20μA/div) the knee voltage is less than 0.4V. My inference to this observation is that the cut in voltage depends on the current range in which the diode is being operated in the circuit. Does this mean that for a silicon diode, the knee voltage need not be 0.6 to 0.7V? Or is that defined only for a particular current range? If i am using the diode as a rectifier, I might be dealing with currents as high as 2 to 3 amps. If i plot VI characteristics for that range, i believe the knee voltage will be more than 1V... Why is it said that the knee voltage or cut in voltage of diode is 0.6 to 0.7 V? Any detailed explanation about knee voltage of diode please reply! Thank you!