Electronics - Diodes and Applications - Discussion

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).

Ideally, a diode would act like a short circuit (0V across it) if it was conducting current. When a diode is conducting current it's forward biased (electronics jargon for "on"). The current-voltage relationship of an ideal diode. Any negative voltage produces zero current " an open circuit.

THE DIODE IS IDEAL AND HENCE HAS NO VOLTAGE DROP, except for cases where it is framed as a silicon diode with voltage drop of 0.7 so, in this case, using Kirchoff's voltage law,

VT=V1+V2 VT=V2 where v1=0 from V=IR I=V/R.
I=12/12000 I=0.001.
I=1mA.

As they haven't mentioned about diodes condition whether it is ideal or not.
so,assuming;

1) diode as ideal = drop will be 0v,
Hence answer = 1mA.

2) diode as practical(non-ideal) = drop will be 0.7 for silicon diode.
Hence answer = 0.942mA.

In series current across every resistor in a branch is remains same in an ideal case. The diode is forward biased acts like conductor so current across resistor equal to the current across the diode.

Current accross resister is 12/12k=1mA.

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.

Ideal means perfect.
So 0.7v is disregard.
We only account 0.7v in dealing 2nd approximation. The first approximation is the ideal approximation so I=V/R.

Here in the circuit, the diode in the circuit is an ideal diode so No voltage drop in the diode.

So,
Current I = V/R = 12/12k(12000) = 0.001 A or 1 mA.

We will assume diode as ideal because there is no specify that it is Si or Ge.

So I = v/r, given v=12v & r=12K.

I = 12v/12k = 1mA.

Generally si or ge diode has consider 0.7v barrier. But in case of ideal it should be consider 0v, that's why answer is 1mA.