In a practical diode, there is a voltage drop across a diode and its starts conducting heavily after its cut in voltage Vγ is exceeded by the forward biased voltage applied.Similarly it is not a perfect conductor with zero resistance in forward biased condition.It has some finite resistance in forward bias condition.

So voltage drop across forward biased diode is made up of

The drop equal to cut-in voltage which is to be maintained across the diode to keep it forward biased.


Can anyone explain me the red marked sentences clearly ??

When frward biased voltage is > cut in voltage the diode conducts right?? then why the sentence is opposite.

Why do we have to maintain the drop that is Vf across the diode to keep it forward biased ??

Please clear my confusion

 

There are quantum effects. The Vf is a threashold, where you have exceeded the barrier gap in the PN junction. Once the diode starts conducting there is resistance, but it is not linear. As you increase the voltage the internal resisance apparently increases.

In some diodes, such as tunnel diodes, this characteristic is cultivated to the point the curve becomes that of negitive resistance, which makes it ideal for certain types of oscillators.

 

Once the diode starts conducting there is resistance, but it is not linear. As you increase the voltage the internal resisance apparently increases.

Bill,

As the voltage increases the resistance decreases - small incremental increases in voltage beyond the threshold voltage result in increasingly larger increases in current and consequently the resistance is decreasing. Naturally the diode will eventually destroy itself when the current gets to high.

Dave

 

I stand corrected, I knew this, I just mis-spoke.

 

Quantum effects? That sounds a bit heavy, especially for a beginner.

mujju433,

It's all to do with the graph of current v voltage graph for a PN junction. Did your teacher show you one?

 

basically it says that you need to apply voltage higher than the "cut-in" voltage of the diode to keep it conducting (ON)

so if cut in of diode is 0.7v, you need to apply at least 0.8v to keep it on (from a battery or other external power source). that is not considering the internal resistance and other stuff, 1v is a more practical number. if it goes lower than that the diode will stop conducting (turn OFF)

 

Assume an ideal diode but which needs 0.7 Volts to conduct.

When the voltage across the diode is less than 0.7V no current flows.

When the voltage across the diode is of the right polarity and greater than 0.7V the diode conducts and current flows. To maintain this current flow the diode has to remain in the conduction state and this happens when the voltage across the diode is greateror equal than 0.7V. Thus the voltage drop across the diode will be 0.7V which is the minimum voltage it conducts.

In a real diode, the voltage drop across it varies with the current through the diode according to the diode characteristic.

 

I want to ask about the breakdown voltages in reverse bias

 

I want to ask about the breakdown voltages in reverse bias

Start a new thread and ask away.

Dave