PN Junction Q&A: Reverse Bias Voltage at Cathode

[CKaiL]

Homework Statement

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[h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2]

It has becomes a P type semi conductor due to the fact that there is more acceptor than donor. From definition , forward bias is the direction of easy current flow while reverse bias is in the direction of little to no current flow. I check online and it say that when it is in reverse bias , voltage at the cathode (P type junction) is higher but why?

 

[CWatters]

I check online and it say that when it is in reverse bias , voltage at the cathode (P type junction) is higher but why?

That's down to how diodes work. It's the polarity of the applied voltage that determines if the diode is conducting or not. That should be in your textbooks. If not try..

http://www.learnabout-electronics.org/diodes_01.php

Diodes are made from two differently doped layers of semiconductor material that form a "PN junction". The P type material has a surplus of positive charge carriers (holes) and the N type, a surplus of electrons. Between these layers, where the P type and N type materials meet, holes and electrons combine, with excees electrons combining with excess holes to cancel each other out, so a thin layer is created that has neither positive nor negative charge carriers present.

Since there are no charge carriers in this Depletion Layer no current can flow across it. In effect a small natural potential is set up within the semiconductor material that has an opposite polarity to the P and N type layers, and because of this narrow band of reversed potential, no current can flow through the diode. When a voltage is applied across the junction however, so that the P type anode is made positive and the N type cathode negative, provided that the applied voltage is greater than the natural junction potential of the depletion layer, the positive holes are attracted across the depletion layer towards the negative cathode, also the negative electrons are attracted towards the positive anode and current flows.

When the diode is reverse biased (the anode connected to negative and the cathode to the positive voltage), the positive holes are attracted towards the negative voltage and away from the junction. Likewise the negative electrons are attracted away from the junction towards the positive voltage applied to the cathode. This action leaves a greater area at the junction without any charge carriers (either positive or negative) left. This causes the depletion layer to widen. It is depleted of charge carriers and so is an insulator. As higher voltages are applied in reverse polarity to the diode, the depletion layer becomes wider still, since the applied voltage is attracting more charge carriers away from it. The diode will not conduct with a reverse voltage (a reverse bias) applied. Once the voltage is applied in the forward direction (positive to anode and negative to cathode) again, current will flow; in this case as the voltage is increased more current flows. The increase in current does not follow a straight-line relationship, as it would do if the voltage was being increased across a resistor. To begin with no current flows until the applied voltage reaches the "junction potential". Once this is overcome (at about 0.15V for germanium diodes and about 0.6V for silicon), current rises sharply as the diode conducts.

 

[CKaiL]

Hi
Could I check with you when it is reverse bias , does the positive side of the battery point to the A and negative point to the B in this question? or another way round?

 

[TSny]

Hi
Could I check with you when it is reverse bias , does the positive side of the battery point to the A and negative point to the B in this question? or another way round?

This might help: http://www.circuitstoday.com/pn-junction-diode-characteristics

 

[rude man]

Hi
Could I check with you when it is reverse bias , does the positive side of the battery point to the A and negative point to the B in this question? or another way round?

Right.
A reverswe-biased junction has the p side negative and the n side positive.

 

[CKaiL]

One more question from me.
I understand that under reverse bias condition , there is an increase in barrier voltage. But i would like know which terminal has a higher voltage with respect to another ? The one closer to the P junction (point B) or the one closer to N junction (point A)

 

[TSny]

Point A will be at a higher potential compared to point B for reverse bias.