Revered members,
Kindly see my attachment.
From the options given, i think the answer is ON switch, because when the input is low, that is when the applied voltage is below the breakdown voltage(0.7 V for Silicon) the current is low and when the applied voltage is equal or above the breakdown voltage, there is saturation.
Am I correct?

 

Hello,

Yes, you are right:

From the eBook Page: http://www.allaboutcircuits.com/vol_3/chpt_3/1.html

• A diode is an electrical component acting as a one-way valve for current.

When forward biased the "switch" is closed.
When reverse biased the "switch" is opened.

Bertus

 

Note however that the term "breakdown voltage" is also commonly used to denote the reverse breakdown voltage at which a diode begins to conduct, i.e. the limit of its reverse blocking capacity. http://en.wikipedia.org/wiki/Breakdown_voltage

I believe that technical English (at least in UK), normally refers to the "forward voltage" or "cut-in voltage" for conduction in the preferred direction.

 

Adjuster,
I stand corrected. It is forward voltage ( 0.7V for Si and 0.3V for Germanium). Thanks for clarifying.