Diffusion Current in Semiconductors: Coulomb's Law?

[tonyjk]

Hello,

The diffusion current in a semiconductor is caused by difference in concentration of holes and electrons. My question is it related to Coulomb's law (electric force due to electric field) ?

Thank you

 

[zoki85]

Yes it is:
http://en.wikipedia.org/wiki/Diffusion_current

 

[tonyjk]

I have already read it but my question is because of what if there is difference of concentration of charged carriers there is diffusion current? is it related to Coulomb's law about electric force between 2 charged particules?

 

[zoki85]

I have already read it but my question is because of what if there is difference of concentration of charged carriers there is diffusion current? is it related to Coulomb's law about electric force between 2 charged particules?

What do you think what kind of force generates the electric field?

 

[tonyjk]

I mean that the electric charges will generate electric field thus attraction of 2 opposite charges like what is stated in Coulombs law

 

[zoki85]

Aha, if you want to distinguish between cause and effect, read first about Fick's diffusion law.
See also on p.15 about diffusion current densities in semiconductors in this pdf.
Hope, it helps.

 

[tonyjk]

haha yes but why here coulombs law does not apply here

 

[zoki85]

It does apply after equillibrium state in semiconductor is established and the potential barrier formed.

 

[tonyjk]

No I am talking about diffusion current why it is not explained with Coulombs law

 

[zoki85]

No I am talking about diffusion current why it is not explained with Coulombs law

Becouse classical phenomenon of diffusion is not caused by electrostatic repulsion among particles, but concentration changes.
However, it is important to remember diffusion current is essential for explanation of properties of semiconductors P-N junctions (diodes, tranzistors etc.) And there is a famous Einstein's relation between diffusion constant D and electrical mobility of carriers μ_q:
D/μ_q=k⋅T/q

 

[tonyjk]

ah ok thanks a lot

 

[tonyjk]

Hello again,

In a diode, there is a resistance also in the p and n region, and electric field is required to move the charges in forward bias conditions. My question is when there is diffusion current is the electric field helping the diffusion current because of the resistance? thus we can say that there is also drift current?

Thank you

 

[zoki85]

Hello again,

In a diode, there is a resistance also in the p and n region, and electric field is required to move the charges in forward bias conditions. My question is when there is diffusion current is the electric field helping the diffusion current because of the resistance? thus we can say that there is also drift current?
Thank you

It is not good to think of a semiconductor resistances as key point here. In p-n junction, n-type region has a majority electrons carrier concentration, and p-type region has a majority holes carrier concentration. Electrons diffuse from n side to p side, and the holes diffuse from p side to n side. If the electrons and holes were not charged, diffusion would continue until concentrations of holes and electrons on both sides were the same. Since they are charged, charge separation in material creates E-field. This E-field opposes diffusion and at some point there is balance where diffusion process stops. This results in a built in potential due to E-field generated and so called depletion layer. Now, in a forward biased state the net effect is such that when external voltage is applied, it reduces the "barrier voltage": when it exceeds the barrier voltage the current flow through the p-n junction quickly increases. Example of this depedence is Shockley's ideal diode equation:

Of course, there will be also a drift current. But remember drift and diffusion components are just helpful tool in the analysis, to distinguish between causes and effects. In reality there is just one current through the diode. For example, another indirect effect of the external voltage to diffusion current can be as follows. The rate at which diffusion occurs depends on the average velocity of carriers and distance between scattering events in the material. Raising the temperature increases the termal velocity of carriers, and diffusion occurs faster at higher temperatures in semiconductors. Higher the external voltage to forward biased diode, higher the current through it and diode acquires higher operating temperature. Therefore, diffusion current also increases.

 

[tonyjk]

Great. But why when we calculate the loss of power due to resistance of the diode we take into consideration the total current in the diode? why can't we just take the drift current caused by the applied electric field?

 

[zoki85]

Becouse, in reality it is the "total" current that flows and counts (so loss is P_D=V_D⋅I). I made that clear:

But remember drift and diffusion components are just helpful tool in the analysis, to distinguish between causes and effects. In reality there is just one current through the diode.

regards

 

[tonyjk]

Just to make things clear. Here in the photo they represented a real diode by 4 resistances and an ideal diode. I know it is a representation but when we represent the p and n region by 2 resistors so there is a drift current. But inside this 2 regions there is drift and diffusion current. So you mean by same current is that the overall current consisting of drift and diffusion current is all the same inside the diode right? (because it is a serie circuit)

 

[zoki85]

Looks like you didn't read pdf in post #6... Yes, overall (total) current consists of drift and diffusion component. When forward biased diode conducts, diffusion component is dominant and only small increase of applied voltage is needed for diffusion and overall current to increase dramatically (Like Shockley's equation shows).

 

[tonyjk]

haha yes. Really that is my last question, the electric field due to resistance is acting on all the current thus on the diffusion right?

 

[zoki85]

No. Voltage drives the current, and resistance regulates current magnitude. That's Ohm's Law.
Resistance of diode is highly nonlinear due to properties of materials involved.

 

[tonyjk]

Yes voltage drives the current due to electric field.So electric electric field is driving also the diffusion current to overcome the resistance right?

 

[zoki85]

Distribution of electric fields in a diode is very nonuniform due to some of the effects already mentioned. In forward biased diodes there are zones of very low field, and thin zones of very high field. Applied external voltage (or if you want external field) disturbes preexisting diode's fields and makes it to conduct more or less. So you may just say there is some average field across diode that drives overall current.

 

[tonyjk]

so diffusion current that does not required external electric field is being drived somehow by it

 

[zoki85]

so diffusion current that does not required external electric field is being drived somehow by it

Not the happiest description/choice of words, but if you like it, yes sort of. That "somehow" is well described by diode equation.

 

[tonyjk]

Sorry maybe this is a dumb question but : when we say resistance of a diode (linear or non-linear) it is opposing the flow of current but does that mean that there's an electric field like in a resistor "helping the current against the resistance of the diode", or just an internal resistance like inside an EMF ?

 

[zoki85]

Resistance says almost nothing about electrical fields. It says how much current will flow given the applied voltage.

 

[tonyjk]

When I was talking about external electric field and its action on the diffusion current I was talking about Resistance's one but it appears it is an internal resistance no electric field is present like a normal resistor. But reading Fundamentals of Microelectronics by Behzad Razavi and in its introduction of semiconductor, P and N Semiconductors( Or region) acts like a normal resistor (External electric field is applied to make a drift current inside it). But when mentioning about the diode, he only talks about drift current due to internal electric field of the depletion region of the diode. That's why I was asking about the resistance of the P and N region and I am still confused lol...

 

[zoki85]

I don't know what did you conclude from that textbook, but participation of drift current in forward biased conducting mode of a diode is minimal. The field in the depletion region is difference between applied field (due to external voltage) and the preexisting (built in) field of the junction. As the field is reduced, reduced is the barrier to the diffusion of carriers from one side of the junction to the other. This increases the diffusion current. The drift current changes very little becouse the depletion region is reduced only slightly and the number of minority carriers swept across the junction changes insignificantly.

 

[tonyjk]

ok thank you now it is clear. Can you tell me what kind of resistance exist in a diode other then the resistance in the P and N Region. I was confused because the P and N region acts like a resistor that's why so it requires an electric field like a normal resistor to conduct the current(it is mentioned in the book also see the pic). But other than the resistance of the P and N Junction what else do we have to make the total resistance nonlinear?

 

[zoki85]

In a classical p-n junction designed diode almost all of the applied voltage in a forward biased conduction mode drops across the depletion region. That's becouse the conductivity of the depletion region is much,much lower than in the rest of a diode due to limited number of carriers in the depletion region. Therefore, most of diode's resistance is the resistance of the depletion region: forward voltage drop around 0.7 V is practicaly drop across the junction. Power diodes have a higher voltage drop (1-1.5 V) becouse they are engineered for high currents and part of the voltage drop occurs across the rest of the silicon body, contact resistance silicon-metal etc.

 

[tonyjk]

In a classical p-n junction designed diode almost all of the applied voltage in a forward biased conduction mode drops across the depletion region. That's becouse the conductivity of the depletion region is much,much lower than in the rest of a diode due to limited number of carriers in the depletion region. Therefore, most of diode's resistance is the resistance of the depletion region: forward voltage drop around 0.7 V is practicaly drop across the junction. Power diodes have a higher voltage drop (1-1.5 V) becouse they are engineered for high currents and part of the voltage drop occurs across the rest of the silicon body, contact resistance silicon-metal etc.

Hello again,

The resistance of the depletion region is due to the electric field? or it act like a resistor by dissipating energy as heat?