Electronics - Diodes and Applications - Discussion

I=V/R
V1=Vcc-0.7
=12-.7
=11.3
I=11.3/12k
=0.942 mA

Usually we take LED drop voltage as 1.5 . Why here as 0.7 ?

Diode is ideal then why use 0.7v

If diode is idial then why taken voltage drop as 0. 7.

Forward Voltage Drop:

Electricity uses up a little energy pushing its way through the diode, rather like a person pushing through a door with a spring. This means that there is a small voltage across a conducting diode, it is called the forward voltage drop and is about 0.7V for all normal diodes which are made from silicon. The forward voltage drop of a diode is almost constant whatever the current passing through the diode so they have a very steep characteristic (current-voltage graph).

In case of ideal diode in forward bias voltage drop is 0.

Hence i= v/r
i=12/12k
=1mA

Answer is 1mA as they have mentioned it as a ideal diode.

Ans Should be 1mA.
Beacause An ideal diode offers zero resistance in foreword bias & infinite resistance in reverse bias.
So Drop across ideal diode is Zero.

Answer is A, b/c voltage drop across an ideal diode is 0v.

Actually the diode is in ideal condition so there is no cut in voltage (0v). Answer should be 1mA.