np=ni^2 in doped semiconductors

[george2525]

Hello

I have seen the relation np=ni^2 a lot when analysing doped (and intrinsic) semiconductors

where:
n= free electrons in C band
p=free holes in V band
i = subscript ''intrinsic''

I wonder if anyone can explain how this relation holds true for a doped semiconductor using minimal mathematics.

Lets take Silicon doped with Arsenic (Group 4 with 5)

For Si intrinsic - n=p so np=ni^2

thats fine. makes sense

But lets say we add an arsenic atom with 5 valence electrons...

this atom will add one extra conduction band electron to the matereal right....

So ''n'' increases. Ok thats fine

But why would ''p'' decrease?

I understand that an extra electron increases the probability of recombination but then if the extra electron from the arsenic recombines then we have also lost a free carrier so why does the relationship hold?

anyone understand this in a basic way?

 

[Arouse1973]

I believe it's because it is a "concentration" P=(ni^2)/n so if n goes up p goes down and the material is n-type because n=ND-NA. And on the flip side n=(ni^2)/p is a p-type material because P=NA-ND. NA and ND are acceptor and donor atoms respectively.
Thanks
Adam

 

[Harald Kapp]

This is called the "law of mass action" and is described here. A special reference to semiconductor physics is a bit down the page uner "In semiconductor physics".

 

[george2525]

yes I know it can be proved mathematically via probability equations etc but ive yet to see anyone describe it in words.

The formula and its derivation are often quoted but never explained.

Im looking for something that says ''the reason the relation is true is because when one carrier does this, another does that and therefore the term ni^2 is always true''

Tha maths is fine but boring and not very insightful to me

 

[Ratch]

Hello

I have seen the relation np=ni^2 a lot when analysing doped (and intrinsic) semiconductors

where:
n= free electrons in C band
p=free holes in V band
i = subscript ''intrinsic''

I wonder if anyone can explain how this relation holds true for a doped semiconductor using minimal mathematics.

Lets take Silicon doped with Arsenic (Group 4 with 5)

For Si intrinsic - n=p so np=ni^2

thats fine. makes sense

But lets say we add an arsenic atom with 5 valence electrons...

this atom will add one extra conduction band electron to the matereal right....

So ''n'' increases. Ok thats fine

But why would ''p'' decrease?

I understand that an extra electron increases the probability of recombination but then if the extra electron from the arsenic recombines then we have also lost a free carrier so why does the relationship hold?

anyone understand this in a basic way?

When an atom of intrinsic (pure crystalline) silicon ionizes, an electron-hole pair is released. When arsenic is substituted for silicon within the structure, the ionization only releases an electron, because the remaining 4 valance electrons of arsenic fit nicely into the neighboring silicon atoms. The more silicon atoms that are replaced by arsenic atoms, the fewer silicon atoms are available to ionize and release an electron-hole pair. If it were possible for every silicon atom to be replaced by an arsenic atom, no electron-hole pairs would be released. This is the reason why doping with arsenic causes an increased concentration of free electrons, and automatically decreases the concentration of holes. Without silicon ionizing, there are no holes. If you want to know why the relationship corresponds to the formula you gave, then you have to assimilate the math.

Ratch

 

[Arouse1973]

It also may help to understand that even though intrinsic semiconductor material doesn't have any dopants, it still has active electron hole pairs generated by thermal excitation this is the ni and is independent on any dopant levels. The np product is constant because the recombination rate equals the thermal generation rate.
Thanks
Adam

 

[george2525]

Thanks

so are you saying that an arsenic atom that donates one electron cannot release any more electrons once bound to silicon? IE by thermal energy?

I didnt know that!

so - adding one atom of arsenic adds one C band electron (100% probability) and also removes any prospect of an EHP being generated at that exact location because the bond is so strong.

Is that correct?

so 1 arsenic atom = +1 C band electron (100%) , -1EHP (possible)

it kind of makes sense

also does the relation only work under equilibrium conditions?

I mean at 0 kelvin p=0 right (because EHP=0) so n=Nd only

so is it only at one particular temperature when the rates match up?

Thanks

 

[Arouse1973]

The arsenic forms a very strong covalent bond with the silicon atom when the semiconductor is manufactured at very high temperatures. It forms part of the semiconductor crystal when everything cools down and is locked in.

Thermal equilibrium is when the semiconductor has no other influence upon it like, temperature change, electric field, magnetic field etc.

Adam

 

[Ratch]

Thanks

so are you saying that an arsenic atom that donates one electron cannot release any more electrons once bound to silicon? IE by thermal energy?

I didnt know that!

so - adding one atom of arsenic adds one C band electron (100% probability) and also removes any prospect of an EHP being generated at that exact location because the bond is so strong.

Is that correct?

so 1 arsenic atom = +1 C band electron (100%) , -1EHP (possible)

it kind of makes sense

also does the relation only work under equilibrium conditions?

I mean at 0 kelvin p=0 right (because EHP=0) so n=Nd only

so is it only at one particular temperature when the rates match up?

Thanks

What is EHP? You should define acronyms at least once before you use them. What is a C band electron? You wanted to know why the hole concentration goes down when the electron concentration goes up and vice versa. Did you understand my explanation? Now you want to know about zero K? What rates? To whom are you directing your question?

Ratch

 

[george2525]

C = Conduction
EHP = electron hole pair

i was speaking to all on the thread who know the answer

the rates are those mentioned by ''adam''

 

[Ratch]

C = Conduction
EHP = electron hole pair

i was speaking to all on the thread who know the answer

the rates are those mentioned by ''adam''

Good, now I know what you are talking about.

so are you saying that an arsenic atom that donates one electron cannot release any more electrons once bound to silicon? IE by thermal energy?

You are asking if the arsenic atom can be ionized twice? Perhaps, but more likely if there is that much energy around, it will ionize another silicon atom and produce an electron-hole pair. Even if the arsenic atom ionized twice, the second time will produce a electron-hole pair like any silicon atom and reduce the hole concentration.

so - adding one atom of arsenic adds one C band electron (100% probability) and also removes any prospect of an EHP being generated at that exact location because the bond is so strong.

At room temperature, only so much thermal energy is available, and only so many silicon bonds can be broken. The fifth electron on arsenic is easily separated and is available as a free electron. If the arsenic atom ionizes again, it will release an electron-hole pair, and 1 less silicon atom will ionize.

also does the relation only work under equilibrium conditions?

I mean at 0 kelvin p=0 right (because EHP=0) so n=Nd only

so is it only at one particular temperature when the rates match up?

What are you defining as being in equilibrium? At 0 K, nothing is ionized, and the semiconductor is inert, frozen out, and inoperable.

Ratch

 

[george2525]

ok by equilibrium I mean whatever temperature that may be. I dont know what that is. I guess its defined as something but im not sure.

I thought that at 0K there will be 0 EHP's but the arsenic free electrons would still be available for conduction right? I was under the impresion that there would still be some minimal conduction at 0K due to the extra electrons having nowhere to go other than the conduction band.

my point was this - if at 0K, there will be no holes at all so p=0.
But if there are still a few free electrons (from arsenic) n=some small value
and therefore np=0

an intrinsic semiconductor at 0K will have p=0 AND n=0, so np=0

so in fact, the relation np=ni^2 does hold. But is it correct that at ANY temperature the added electrons are always in the conduction band, even if they are effectively useless at low temperature?

Here is my thought process (if you have the time to read it)

1. take tiny piece of silicon made up of 4 si atoms only and cool it to 0 Kelvin

2. n=0 and p=0

3. increase temperature, and np=ni^2 (here we assume that the ni value is increasing to a value corresponding to the current temperature) so the relationship holds.

- lets assume that 2 atoms ionize and so we have 2 free electrons and 2 holes so ni^2 = 4 = np

4. remove one si atom and replace with arsenic. We now have 3 silicon atoms and one arsenic.

5. Cool to 0 Kelvin.

- n=1 and p=0 and therefore np=0 - ok good.

6.increase temperature to the same point where 2 electron hole pairs were previously created in the intrinsic silicon...(where we had n=2, p=2 and ni^2=4)

7. Now we have the arsenic electron plus the 2 silicon electrons (from EHP formation) and n=3 and p=2 and np=6 which is wrong!

- so im aware that in order to equal 4 p must be 4/3 !

Hence where my problems are arising from

I do see from what you said that every arsenic atom reduces the probability of an intrinsic EHP forming.

So is it the case that this probability is reduced by just the right amount that when analysed over time the average n=3 and p=4/3 ? that would almost make sense for me

sorry im aware that was quite convoluted!

 

[Arouse1973]

Hello

Any covalent bond can be broken but in most cases it is a very strong bond. And as Ratch said the energy will be used up liberating the silicon covalent bonds and arsenic free electrons. It takes approx. 1.1 eV to free a covalent bonded electron from silicon. And about 50 meV to free a donor electron from the covalent bonded silicon and arsenic. Normally full ionization takes place at room temperature
.
If I have time tonight I'll give you my answers to your questions.

Thanks
Adam

 

[Ratch]

georger2525

ok by equilibrium I mean whatever temperature that may be. I dont know what that is. I guess its defined as something but im not sure.

I thought that at 0K there will be 0 EHP's but the arsenic free electrons would still be available for conduction right? I was under the impresion that there would still be some minimal conduction at 0K due to the extra electrons having nowhere to go other than the conduction band.

my point was this - if at 0K, there will be no holes at all so p=0.
But if there are still a few free electrons (from arsenic) n=some small value
and therefore np=0

If you take a slab of N-type or P-type silicon. and measure its resistance, it will be very high. That is because the concentration of arsenic atoms in the N-type silicon is quite small, so it does not affect the conductivity of the silicon very much. Unlike a metal, which has scads of electrons itching to hop off one atom and onto its neighboring atom, the free electrons in N-type silicon are very, very sparse. A PN diode does not work by propelling charge carriers by electrostatic fields like a resistor does. It works by controlled diffusion. Anyway, lowering the temperature will not increase the already low concentration of electrons available for conduction at even room temperature. That formula you keep quoting does not work below a certain temperature.

Ratch

 

[Arouse1973]

ok by equilibrium I mean whatever temperature that may be. I dont know what that is. I guess its defined as something but im not sure.

I thought that at 0K there will be 0 EHP's but the arsenic free electrons would still be available for conduction right? I was under the impresion that there would still be some minimal conduction at 0K due to the extra electrons having nowhere to go other than the conduction band.

my point was this - if at 0K, there will be no holes at all so p=0.
But if there are still a few free electrons (from arsenic) n=some small value
and therefore np=0

an intrinsic semiconductor at 0K will have p=0 AND n=0, so np=0

so in fact, the relation np=ni^2 does hold. But is it correct that at ANY temperature the added electrons are always in the conduction band, even if they are effectively useless at low temperature?

Here is my thought process (if you have the time to read it)

1. take tiny piece of silicon made up of 4 si atoms only and cool it to 0 Kelvin

2. n=0 and p=0

3. increase temperature, and np=ni^2 (here we assume that the ni value is increasing to a value corresponding to the current temperature) so the relationship holds.

- lets assume that 2 atoms ionize and so we have 2 free electrons and 2 holes so ni^2 = 4 = np

4. remove one si atom and replace with arsenic. We now have 3 silicon atoms and one arsenic.

5. Cool to 0 Kelvin.

- n=1 and p=0 and therefore np=0 - ok good.

6.increase temperature to the same point where 2 electron hole pairs were previously created in the intrinsic silicon...(where we had n=2, p=2 and ni^2=4)

7. Now we have the arsenic electron plus the 2 silicon electrons (from EHP formation) and n=3 and p=2 and np=6 which is wrong!

- so im aware that in order to equal 4 p must be 4/3 !

Hence where my problems are arising from

I do see from what you said that every arsenic atom reduces the probability of an intrinsic EHP forming.

So is it the case that this probability is reduced by just the right amount that when analysed over time the average n=3 and p=4/3 ? that would almost make sense for me

sorry im aware that was quite convoluted!

This is my take on what you have said. I think you got everything pretty much correct, apart from the ni bit. Remember the ni constant is for intrinsic semiconductor material, this is what the (I) is. It is independent of doping levels. So you can't use it for anything else.

Without getting too much into it, you maybe better to think about band-gap energy levels. So for an electron to reach the conduction band it has to have enough energy. The energy needed depends on the type of semiconductor material and the strength of the bond the electron has with the atom. Remember that although adding arsenic adds one free electron to the mix, it will only break free if it has enough energy. Room temperature generally does this.

Think about getting a ping pong ball and a straw and place the ball on a tilted table. Stretch a piece of tape across the table and label the bottom half of the table Valence and the top half conduction. Blow through the straw and try and move the ball into the conduction band area. If takes a certain amount of blow (energy). Now go and stand in the freezer for an hour and try again. You won't have enough blow (energy) to move the ball very far. This is the best way I can explain it.

Basically it's the contraction and expansion of the crystal lattice at low and high temperatures that cause the band gap energy to go up and down, making it harder or easier (more or less energy) for the electrons to move away.

Ratch will correct me if I have somthing not quite right.
Adam

 

[Ratch]

george2525,

This link starting at page 65 gives a good explanation of temperature dependence of carrier concentration. It shows that at about 100 K, the semiconductor starts to freeze out, and at about 450 K, the semiconductor becomes intrinsic no matter how much it is doped. Read that, and it should start make sense. Your formula is valid in the extrinsic region.

Ratch

http://www.pdfboi.com/books/eee/Semiconductor Device Fundament by Robert F. Pierret --pdfboi.com.pdf

 

[EdBurke]

I have not seen, via many sources, an answer that makes sense to me. I'll try suggesting one. Let's say silicone has 10^23 atoms/cm^3. ni=pi= 10^10 at 70 degF. np=10^20/cm^6. We add 10^15/cm^3 arsenic. If all of the silicon holes are filled because of the abundance of donor electrons then np=0. This is not experimentally found. I suggest, and hope some mathematical expert will comment, that statistically just about all of the silicon holes are filled by donor electrons, but not all just because of chance and thermal activity. For normal temperatures of people's interest and usual doping concentrations, is there a simplified equation that matches experimental data?

 

[Ratch]

I have not seen, via many sources, an answer that makes sense to me. I'll try suggesting one. Let's say silicone has 10^23 atoms/cm^3. ni=pi= 10^10 at 70 degF. np=10^20/cm^6. We add 10^15/cm^3 arsenic. If all of the silicon holes are filled because of the abundance of donor electrons then np=0. This is not experimentally found. I suggest, and hope some mathematical expert will comment, that statistically just about all of the silicon holes are filled by donor electrons, but not all just because of chance and thermal activity. For normal temperatures of people's interest and usual doping concentrations, is there a simplified equation that matches experimental data?

I believe I answered your question in post #5 of this thread. What do you not understand about that explanation?

Ratch

 

[Harald Kapp]

Resurrection of an old thread -> closed.