P-N Junction Forward Bias PHYSICS help

Discussion in 'Physics' started by Digin8918, Sep 5, 2009.

1. Digin8918 Thread Starter New Member

Sep 5, 2009
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0
Hello, I am trying to understand the physics of a PN junction in the forward bias mode( I feel like I need to understand that one before I can move on to understanding reverse-bias). I have read several resources( including my electronics book ), and it is difficult for me to understand. I feel like I understand most of it, but what I do not understand is how there is a total net current in ONE direction. From what I THINK I know, there is a voltage that propels holes from the P-side toward the N-side, and electrons from the N-side towards the P-side. The external voltage that is applied to the junction overcomes the barrier voltage that is produced by diffusion so that the charge carries can move into their opposite sides, but so again, how is there a net current in one direction if the holes and electrons recombine and cancel eachother out? Thank you for your time in reading this. I will appreciate any input regarding the topic you may have.

2. Ratch New Member

Mar 20, 2007
1,068
3
Digin8918,

It appears to me you have a good idea of what is going on. Let's take the N side first. Some of the N side electrons cancel the holes coming over from the P side, and other N side electrons diffuse over into the P side. This lowers the concentration of N side electrons The negative voltage causes some more electrons to enter the N side to reestablish the concentration equilibrium, thereby creating an electron current that enters the N side.

Some of the P side holes cancel the electrons coming from the N side and other holes diffuse over to the N side. This lowers the concentration of P side holes. The positive voltage causes some more holes to enter the P side to reestablish the concentration equilibrium, thereby creating a hole current that enters the P side.

A hole going one way is the same as a electron going in the opposite direction, so the holes entering the P side are equal to electrons leaving the P side. Therefore, electrons enter the N side and leave the P side. Charge movement is current, and that is how a forward biased diode supports a charge flow, also known as current.

Ratch

Last edited: Sep 5, 2009
3. Digin8918 Thread Starter New Member

Sep 5, 2009
6
0
So, let me get this straight...the net current is moving from N side to P side across the diode because the electrons move from N to P, and the holes move from P to N, but everytime a hole moves from P to N, another electron is moving from N to P as well....so it's like a larger ratio of electrons moving to one side than it is holes moving to the other side? So, in an actual circuit with a diode, and for the purposes of analyzing the circuit currents, the diode when it is forward-biased has a current moving from the N side to P side, and so the pointed end of the diode represents the P side? If this " <| " were a diode, then when this diode is forward-biased, the current will be moving from right to left? I may have completely botched that, but if I'm wrong, please try one more time to set me straight. Thanks.

4. Ratch New Member

Mar 20, 2007
1,068
3
Digin8918,

First of all, describing net current as moving is wrong. Charge moves, current exists. Current already implies charge movement. It does not move twice.

Every time a hole in the P side annililates a electron from the N side, a electron leaves the P side via the lead attracted by the positive voltage and creates a hole in the P type material. Every time a hole diffuses into the N side, a electron leaves the P side via the lead attracted by the positive voltage and creates a hole in the P material. Vice versa for the N side. Either activity causes a charge flow. A larger forward voltage causes this activity to increase.

As long as the diffusion and recombination are equal, the holes and electrons will balance each other.

The arrow of the diode is the P type material and the bar is N type material. The parts manufacturers use conventional charge flow which is a mathematical concept that assumes current consists of moving positive charges. If the charges are negative, they are equivalent to a positive charge moving the opposite direction.

Ratch

5. bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
Here's how I look at a PN junction: the P side is doped to make it deficient in electrons and the N side is doped to make it fat with electrons. Stick the two pices together: at the boundary where they join the excess electrons will jump across and fill up some holes in the close area making what is called a "depletion region" which is more or less neutral. To "forward bias" the junction and get current to flow, you need to increase the voltage to get a strong enough E field to force electrons to cross the depletion region and create current flow.

6. Ratch New Member

Mar 20, 2007
1,068
3
bountyhunter,

So far so good. Deficient in electrons means fat with holes on the P side.

The electrons from the N side and the holes from the P side will cross the junction because of diffusion. The electrons from the N side ionize the P atoms making them negative, and the holes from the P side ionize the N atoms making them positive. This sets up a junction barrier voltage that opposite to the diffusion voltage, and stops further diffusion from taking place. Forward biasing lowers the barrier voltage and allows the diffusion to take place again. So the mechanism for charge movement is different for a PN junction than that for a wire. The depletion region is caused by the barrier voltage and is anything but neutral. It is full of ionized atoms. If fact, it is often represented by a battery connected in opposite polarity to the diffusion voltage.

For a PN junction, it is the diffusion voltage that moves the electrons. The forward bias lowers the barrier voltage so it counters the diffusion voltage less. By the way, that creates a charge flow where current exists, not current flow.

Ratch