intro to nanoscience exercise help

[dragisto]

hi

I started reading a text book called introduction to nanoscience by s m Lindsay and looking at the exercises and I am unsure what equation he was using to get the answer.

the question was "what is the size of the current fluctuations (shot noise) in a current of 100PA[ over a counting period of 1 second? use e=1.6x10^-19."

the working in the back says "given the fact that 1A=1C/s.

(100pC/1s)*(1c/10^12 pC)*1=(1e-/1.6x10^-19c)=6.25x10^8e-/s"

any help would be apprecieated.

[studiot]

The standard equation is

The mean square shot noise current, i, in a circuit carrying a current i_A is given by 

i² = 2ei_AB

where e is the charge on the electron and B is the bandwidth of the measuring equipment.

[dragisto]

 

10 hours ago, studiot said:

The standard equation is

The mean square shot noise current, i, in a circuit carrying a current i_A is given by 

i² = 2ei_AB

where e is the charge on the electron and B is the bandwidth of the measuring equipment.

ok right so the 100pC/1s=i_aand e=e- I'm guessing then does the 1C/10¹² pc *1/1.6x10^-19c=2B

at which point where does the 10¹² come from and the 1.6x10^-19 not cancel the e-?

[studiot]

OK so here is my working.

 

The question asks for a current change.

So the units of this are amps.

 

So looking at my equation this is the square root of the left hand side.

On the right hand side we have a current times a charge times a bandwidth.

The units of bandwidth are seconds^-1

The units of charge are coulombs.

So the right hand side is in amps *coulombs per second 

But coulombs per second is amps.

So we have (amps)²

We are given e = 1.6x10^-19 and i_A = 100*10^-12 amps and B = 1/1 second

Substituting we have

i2 = 2 * 1.6x10^-19 * 100*10^-12

=3.2x 10^-29 = 32 x 10^-28

i = √(32 x 10^-28)

=√(32) x 10^-14

=0.06 x 10^-12 amps

= 0.06 picoamps

[dragisto]

16 minutes ago, studiot said:

OK so here is my working.

 

The question asks for a current change.

So the units of this are amps.

 

So looking at my equation this is the square root of the left hand side.

On the right hand side we have a current times a charge times a bandwidth.

The units of bandwidth are seconds^-1

The units of charge are coulombs.

So the right hand side is in amps *coulombs per second 

But coulombs per second is amps.

So we have (amps)²

We are given e = 1.6x10^-19 and i_A = 100*10^-12 amps and B = 1/1 second

Substituting we have

i2 = 2 * 1.6x10^-19 * 100*10^-12

=3.2x 10^-29 = 32 x 10^-28

i = √(32 x 10^-28)

=√(32) x 10^-14

=0.06 x 10^-12 amps

= 0.06 picoamps

ok that makes sense but the answer in the back is different to the one you've got there.

"(100pC/1s)*(1c/10^12 pC)*1=(1e-/1.6x10^-19c)=6.25x10^8e-/s

The shot noise is given by N^1/2 i.e. 2.5x10^4 e/s= 4x10^-15 A"

this is what the book gives as the answer in the back.

to get the answer in the book using your equation i'd need something to cancel the 2 while keeping the rest of the numbers the same.

Edited June 28, 2018 by dragisto

[studiot]

Sorry, you will have to ask you teacher about this, I don't know much about the use in nanotechnology.

My equation comes from its use in electron ballistic devices (valves mainly).

 

The only thing I can think of would be that it is balanced or two sided, whereas your app might be one sided ie all the variation is extra counts.

This would account for a 2.

 

I would be interested when you find out, though.

Edited June 28, 2018 by studiot

[dragisto]

it's fine, no worries 

I was just trying to figure out why it's written the way it is, perhaps it is balanced or two sided I just don't see it in the question itself.

thanks for the assistance anyway though.