How to calculate the current in a 400 nm wire of graphene

[grahas]

Hi, I need to calculate the current in a 400nm piece of graphene. Can some one help me please? I found online that graphene has electron mobility of 200,000 cm^2/( V s). My best attempt is as follows. Using the equation vd = uE where E is the magnitude of the electric field applied to a material, vd is the magnitude of the electron drift velocity (in other words, the electron drift speed) caused by the electric field, and µ is the electron mobility (wikipedia). I found that vd= 200,00 cm^2/s. From here I am not sure of I should divide by 400nm to get a speed and then use that to calculate total electrons passing per second to find coulombs. Please help.

 

[mfb]

I moved the thread to our homework section.

The drift speed is independent of the 400 nm, but the current is not. You'll need another equation to relate the two.

 

[grahas]

This is not homework. I trying to get the attention of a research group and need to have some ballpark estimates for what I am going to need their help to build. I don't have the background in these physics to really do this my self, I don't have any teachers or anyone to get help from, also I don't have any textbook that I could get help from. Really I only have what I can remember from my AP physics E&M course I took almost 2 years ago.

 

[mfb]

It is still homework-like, even if it is not an actual homework question.

Your vd has wrong units.

You'll need the electron density. That plus the width of the strip and the drift velocity are sufficient to calculate the current. There is just one way to combine them to get an electrical current out of them.

 

[grahas]

Sorry for long wait, I am still trying to figure this out.

 

[grahas]

Ok I am stuck. Really sorry I don't have a formal education, I had to read wikipedia for answers. When I researched (googled, youtubes, and wikipedia) all I found is that electron density is the probability of finding an electron in a region in an atom. Original I thought it was the amount of electrons in an area so I figured electron density would be 1 cm / 70 pm = 142,857,143 carbon atoms. If every carbon atom has one free electron then 142,857,143 carbon atoms ^2 is 2x10^16 electrons /cm^2. Is that much barking up the right tree?

 

[mfb]

Electron density is the number of electrons per volume (that contribute to electric currents), and some material constant to look up. This publication quotes 10¹² /cm².

 

[grahas]

I came up with an answer. PLEASE don't just gloss over it. The math needs to be done correctly. I came up with 1.28A. Using the following variables. Voltage is equal to 1V, electron mobility is equal to 200,000cm² /(s*V), Electron density is 10¹²/cm² I had a question about that number; since the I couldn't make the units work I multipyed that number by the square root of itself to make it a cm³; is that ok to do? How am I supposed to get around that?, any way assuming that is acceptable to do I then multiplyed Electron density by drift velocity and the width of the strip to get 8x10¹⁸electrons/s. When I divide this number by 6.25x10¹⁸electrons I get 1.28A. In my mind I think that that is a lot of current for such a small wire but then when I think in my head the carbon is really acting as a guide for the electrons and the electrons can flow freely because their is nothing to run into.

 

[mfb]

since the I couldn't make the units work I multipyed that number by the square root of itself to make it a cm³; is that ok to do?

You can't do that.
Graphene is a 2-dimensional layer, per square centimeter is the right thing.

Speed * surface density * width has units of 1/second: $\frac m s \frac{1}{m^2} m = \frac{1}{s}$, multiply by the elementary charge to get a current.

 

[grahas]

I got 0.00000128A. Is that the same as what you got?

 

[mfb]

Is your strip of graphene really 1 meter long? Otherwise check your calculation of E.

 

[grahas]

Where did the 1 meter come from? I am going to cut the graphene from a sheet into a really long skinny wire so it will be a lot longer than 1 meter. Also is E just the voltage being applied to the wire? I just set E to equal 1 volt to make calculations easy.

 

[mfb]

E is the electric field strength. You should have gotten mismatching units earlier if you plugged in the voltage.

 

[grahas]

I think that I got that the voltage to cancel out whe calculating the drift velocity. 200,000 cm^2/( V s) * 1V . Another question, how much voltage could this strip take before it broke? And just checking, electric field strength is measured in voltage right? Also how would the length impact the amound to current that the wire could carry? The device I am going to propose has an extreamly long wire.

 

[mfb]

cm^2/s is not a velocity.

Electric field strength is voltage per length, e. g. V/cm.

Mechanical breaking? I don’t know. I guess it can overheat, and we can calculate the power once the resistance is known, but the temperature will depend on the environment and I have no idea how heat-tolerant graphene is.

 

[grahas]

The strip will be about 3000 meter long, 300000 cm. So E = 50 V / 300000 cm = 1.666x10^-4 V/cm . Then v_d = 200,000 cm²/( V s) x 1.666x10^-4 V/cm = 33.33 cm / s

v_d = 33.33 cm / s
Electron / surface density = 10¹² /cm²
width = 4x10^-5
elementary charge = 1.60217662 × 10^-19 coulombs

Current = elementary charge x width x Electron (surface) density x drift velocity

Current = q x m x 1/m² x m/s

Current = 1.60217662 × 10^-19 C x 4x10^-5 cm x 10¹² /cm² x 33.33 cm / s = 2.13602 x10^-10A

Is that the amount of current that flows through a 400 nm graphene strip?

 

[mfb]

50 V over 3000 m is a tiny electric field.

I can confirm your result.

 

[grahas]

Good to know. I am going to try and write a python program to find the variables that maximize force and are feasible to manufacture. For the community I will write up the conclusion of the post into a little nugget that might be useful.

 

[grahas]

Current in a strip of graphene can be calculated with following equation.
w is the width of the strip
V is the voltage on the two ends of the strip
l is the length of the strip

$$A=1.60217662\times10^{-19}C\times w\times \frac{10^{12} }{cm^{2}}\times \frac{V}{l}\times \frac{200000cm^{2}}{Vs} $$

example calculation
w = 1 cm
V = 120 V
l = 100 cm

$$A=1.60217662\times10^{-19}C\times 1 cm\times \frac{10^{12} }{cm^{2}}\times \frac{120V}{100cm}\times \frac{200000cm^{2}}{Vs} =0.03845A $$

If you wanted 10 A you would need 260 layers. This would be 18,200 picometers thick or $1.82\times10^{-6}cm$