electrical conductivity of a semiconductor
Answer (1)
7th Oct, 2020
Hello dear student
Let's derive the expression the expression for conductivity of semiconductor.
In a Semi-Conductor of length (L) and area of Cross-Section A.
Let n e and n h be the no. of electrons and holes with drift velocities V e and V h respectively.
So, the total current in the semi-conductor will be the sum of current due to electrons as well as holes,
i.e. I = I e + I h (i)
As we know that
I = n e A V d
\I e = n e e A V e
And I h = n h e A V h
I = n e e A V e + n h e A V h
I = eA[n e V e + n h v h ]
I/A = e[n e V e + n h v h ] (2)
As we know that
E = V/l (in magnitude)
Also, R = l/A
Or = RA/l
Where is resistivity and R is resistance
Or E/ = (V/l)/
Or E/ = (V/l)l/ RA
E/ = (V/RA)
E/ = I/A (3)
Put (3) in (2) we get,
E/ = e[n e V e + n h v h ]
I/ = e / E[n e V e + n h v h ]
= e [n e V e /E + n h v h /E] ( = Conductivity)
Here, mobility (m) = Drift vel./ Electric field
= e [n e m e + n h m h ]
This is the derivation and expression for the conductivity of a semiconductor.
Hope that's helpful
Best of luck
Let's derive the expression the expression for conductivity of semiconductor.
In a Semi-Conductor of length (L) and area of Cross-Section A.
Let n e and n h be the no. of electrons and holes with drift velocities V e and V h respectively.
So, the total current in the semi-conductor will be the sum of current due to electrons as well as holes,
i.e. I = I e + I h (i)
As we know that
I = n e A V d
\I e = n e e A V e
And I h = n h e A V h
I = n e e A V e + n h e A V h
I = eA[n e V e + n h v h ]
I/A = e[n e V e + n h v h ] (2)
As we know that
E = V/l (in magnitude)
Also, R = l/A
Or = RA/l
Where is resistivity and R is resistance
Or E/ = (V/l)/
Or E/ = (V/l)l/ RA
E/ = (V/RA)
E/ = I/A (3)
Put (3) in (2) we get,
E/ = e[n e V e + n h v h ]
I/ = e / E[n e V e + n h v h ]
= e [n e V e /E + n h v h /E] ( = Conductivity)
Here, mobility (m) = Drift vel./ Electric field
= e [n e m e + n h m h ]
This is the derivation and expression for the conductivity of a semiconductor.
Hope that's helpful
Best of luck
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