Give that Vx is 3v.Diode cuttoff V=0.7V.It is said that that the diode is in off state for a specific value of R1 and R2.What is the criteria for R1 and R2 such that the diode remains off?
Lets draw the forward current thru the diode on a logarithmic scale of current:
Notice the diode is never off; instead the current drops by 10:1, e.g. from 1mA to 100uA, for each 0.058 volts reduction in the forward diode voltage.
That factor of 0.058 depends on steepness of the internal doped junction, and upon the junction temperature.
It is common for a diode to be considered to have an 'on'/'off' threshold at 0.7V in early school work. This is an oversimplification, but useful prior to understanding the junction characteristics more fully.
There is an almost infinite number of R1/R2 values that meet the simplistic criterion. What you really want to know is the minimum ratio of R1, R2 that results in the voltage across R1 being less than 0.7V.
Let's do a back of the napkin …..set the current through the R1-R2 network at 1mA, this give a total resistance of 3 / 0.001 --> 3000 Ohms.
Since the threshold for diode 'on' is considered 0.7V, then the value of R1 must be 0.7 / 0.001 --> 700 Ohms. This means R2 is 1300 Ohms and gives a ratio of 7:13. Or simplistically R1 = 0.538 * R2
Notice at this point the diode would be considered 'on', but in reality no current (in the simplistic model) flows through it.
How can a forward biased diode be off?
- That's an oxymoron.
There must be a misunderstanding. 0.7V is the switched "ON" voltage. Perhaps it is just an arbitrary threshold for 1 Amp for some diode or some other reason and the question doesn't want you to think too much about the knee of the threshold.
=> 0.7V is the normal saturated ON condition for a Silicon diode.
0.6V is a low current threshold so less than 0.6V may be considered "off" for some
or < 0.5V for others with high values of R resulting in << 1uA.
So what R ratio gives < 0.6V drop from 3V or 2.4V across R2?