Doping with gallium and arsenic

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In summary, Gallium doping of a silicon integrated circuit created a resistor with a concentration of 5x10^22 m^3.
  • #1
Conductor
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Homework Statement


The figure shows how a resistor in an integrated circuit made of silicon has been created by doping with gallium with the concentration ##5\cdot10^{22} m^{-3}## in an area that since before contained arsenic with the concentration ##1\cdot10^{22} m^{-3}##.
doped.png

The following questions concern the grey area in the figure.
a) Is the material n-type or p-type?
b) Is it electrons or holes that are minority charge carriers?
c) Calculate the minority- and majority charge carrier concentrations at ##300 K##.
d) Calculate the resistance if the dimensions of the doped are are: length ##100 µm##, width ##1 µm## and thickness ##100 nm##.
e) Calculate the distance between the valence band edge ##E_V## and the Fermi level ##E_F##. Sketch in a figure where ##E_F## is in comparison with ##E_V##, ##E_C## and ##Ei##.
(##E_i## is the intrinsic fermi energy level)
(concentration = density)

Homework Equations


Fermi level:
##E_F=\frac{E_V+E_C}{2} + \frac{kT}{2} \cdot ln \frac{n}{p}##
where ##E_F## is the Fermi energy level, ##E_V## is the energy level of the valence band edge, ##E_C## is the energy level of the conduction band edge, ##n## is the electron concentration and ##p## is the hole concentration.

Electron concentration ##n## in the conduction band:
##n=N_C\cdot e^{\frac{E_F-E_C}{k\cdot T}}##
where ##N_C## is the "effective state density" of the conduction band

Hole concentration ##p## in the valence band:
##n=N_V\cdot e^{\frac{E_V-E_F}{k\cdot T}}##
where ##N_V## is the "effective state density" of the valence band

Resistivity

##\rho = \frac{1}{e\cdot(\mu_n \cdot n + \mu_p \cdot p)}##
where ##\mu_n## = electron mobility, ##\mu_p## = hole mobility

The Attempt at a Solution


a) Is the material n-type or p-type?
P-type, since the concentration of gallium is higher than the concentration of arsenic.

b) Is it electrons or holes that are minority charge carriers?
Electrons, since it's p-type.

c) Calculate the minority- and majority charge carrier concentrations at ##300 K##.
Minority charge carriers (electrons):
##n=N_C\cdot e^{\frac{E_F-E_C}{k\cdot T}}##
##k=0.0259 eV## (Boltzmann's constant)
##T=300 K##

I need help with this question. How are the minority and majority charge carrier concentrations related to the concentrations of gallium and arsenic? (They are given as ##5\cdot10^{22} m^{-3}## and ##1\cdot10^{22} m^{-3}##)

I'm guessing one of the equations I posted can be used somehow, but it seems we lack information, since we don't know either ##E_F##, ##E_C## or ##E_V##?

d) Calculate the resistance if the dimensions of the doped are are: length ##100 µm##, width ##1 µm## and thickness ##100 nm##.
I'm not sure how to calculate this. I wrote an equation for resistivity above, but that requires knowledge of mobilites of electrons and holes. That equation also doesn't include any geometric variabels which are given in the question.

What is the relation (equation) between geometry (length, width and thickness) and resistivity?

e) Calculate the distance between the valence band edge ##E_V## and the Fermi level ##E_F##. Sketch in a figure where ##E_F## is in comparison with ##E_V##, ##E_C## and ##Ei##.
I wrote some equations above that relates these energy levels to each other, but again like in question c) it seems we need more data. I guess I've missed some relation (equation) that can be used in order to answer it. I appreciate if someone can help me find out these relations.

The questions I've answered so far, are they answered correctly? I also appreciate assistance with the questions I couldn't answer. Please see each question for specifics.
 
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  • #2
(a) and (b) look right.

(c) Majority charge carriers are, too a good approximation, just those you injected (minus those needed to cancel the other doping). There is a formula for the product of majority and minority charge carriers (the formulas you have here are sufficient to find it). You can look up the bandgap of silicon.
Conductor said:
but that requires knowledge of mobilites of electrons and holes.
I think that is also something you have to look up.
Conductor said:
That equation also doesn't include any geometric variabels which are given in the question.
It gives you a material constant. The resistance depends on this constant and the geometry - dimensional analysis and some basic knowledge about electric current should be sufficient to figure out how.

(e) should be possible with (c).
 
  • #3
mfb said:
(a) and (b) look right.

(c) Majority charge carriers are, too a good approximation, just those you injected (minus those needed to cancel the other doping). There is a formula for the product of majority and minority charge carriers (the formulas you have here are sufficient to find it). You can look up the bandgap of silicon.
I think that is also something you have to look up.It gives you a material constant. The resistance depends on this constant and the geometry - dimensional analysis and some basic knowledge about electric current should be sufficient to figure out how.

(e) should be possible with (c).
So in this case, the majority charge carriers are ##5 \cdot 10^{22} - 1 \cdot 10^{22} = 4 \cdot 10^{22}## ? And minority charge carriers will be 0 (zero), since all of them will be used up to cancel the other doping?

Are those the values the question is asking for? Then why do we need to find the product of them? By product, do you mean the formula for mass action law ##n\cdot p = n_{i}^2##? That's the only formula I found that contains the product of them. You said the formulas I have here are sufficient to find it, but what confuses me about those formulas is that there are so many unknowns. We don't know ##N_C## and ##N_V## which are "effective densities of states" for example. And even if we can replace ##\frac{E_V+E_C}{2}## with ##\frac{E_g}{2}##, that doesn't give individual values for ##E_V## and ##E_C## which I think are needed for question (e).

I looked up the bandgap of silicon, that gives a value for the bandgap ##E_g##. I also found a bandgap value for GalliumArsenide, which I would have thought should be used since it's doped. But you're saying it's the value for silicon that should be used in this case?

About the mobilities of electrons and holes, I found some values for silicon. Is it again the values for silicon that should be used, not for any of the doped material? Anyway using this value together with the formula I posted (with ##n=0##) and using the geometric values I got a resistance value of roughly ##1 M\Omega##. Is that a reasonable value? I think it seems very high.
 
Last edited:
  • #4
Conductor said:
And minority charge carriers will be 0 (zero), since all of them will be used up to cancel the other doping?
Not zero, there are still some left - but I expect the concentration to be much smaller than the doping concentration. And the product allows to determine their number.
Conductor said:
That's the only formula I found that contains the product of them.
That is what I mean.

Conductor said:
I also found a bandgap value for GalliumArsenide
That would be a crystal made out of GaAs, that is not what you have here. The bandgap and the electron and hole mobilities are dominated by the crystal material, which is silicon.
 

Related to Doping with gallium and arsenic

1. What is doping with gallium and arsenic?

Doping with gallium and arsenic is a process in which small amounts of these elements are intentionally added to a semiconductor material, such as silicon, in order to alter its electrical properties.

2. How does doping with gallium and arsenic affect the electrical properties of a semiconductor?

Gallium and arsenic are both considered "dopants" because they have either one more or one less electron than the atoms in the semiconductor material. This creates an imbalance in the material's electron structure, allowing it to conduct electricity more efficiently.

3. What are the benefits of doping with gallium and arsenic?

Doping with gallium and arsenic can increase the conductivity and improve the performance of semiconductor devices, such as transistors and diodes. It can also allow for the creation of more complex electronic circuits.

4. Are there any potential risks associated with doping with gallium and arsenic?

While doping with gallium and arsenic is generally considered safe, there are some concerns about the toxicity of these elements. However, the amounts used in semiconductor fabrication are very small and are tightly controlled to minimize any potential risks.

5. How is doping with gallium and arsenic controlled during the semiconductor fabrication process?

Doping with gallium and arsenic is a highly controlled process in which the elements are precisely added to the semiconductor material using various techniques, such as diffusion or ion implantation. This ensures that the desired electrical properties are achieved and that the final product is safe for use.

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