a) Distinguish between metals, insulators and semiconductors on the basis of their energy bands. b)Why are photodiodes used preferably in reverse basis condition? A Photodiode is fabricated from a semiconductor with band gap of 2.8 ev. Can it detect a wavelength using 6000nm? Justify.
Metals - Band Gap - either there is no energy gap b/w conduction and valence band or they both overlap each other
Insulators - Valence band is completely filled - conduction band is completely empty and energy gap is quite large that small energy from other source cannot overcome it - Eg> 3eV
Semiconductors - Valence band totally filled - conduction band empty but energy gap is very small - Eg< 3eV
When photo diode is illuminated with light due to breaking of covalent bonds, equal umber of additional electrons and holes come into existence whereas fractional change in minority carrier is much higher than fractional change in majority carrier charge . Since fraction change in minority carrier current is measurable significantly in reverse bias that that of forward bias . Therefore photo diode are connected in reverse bias
Long answer shot - it increases sensitivity ! : P minority carrier is easily measure in reverse bias than forward one !
Energy Gap = 2.8 eV
wavelength = 6000nm = 6000 x 10^-9 m
Energy of Signal is given by E = hc/wavelength
h - plank's constant - 6.626 x 10^-34 Js
c = 3 x 10^8
E = 6.626 x 10-34 x 3 x 108
----------------------------- = 3.313 x 10-20 J
6000 x 10-9 .
But 1.6 x 10^-19 J = 1ev
Therefore 3.313 x 10-20J
------------------- = 0.207eV
1.6 x 10-19 .
Energy of signal is 0.207 eV which is less that 2.8eV energy band gap of photo diode , hence photo diode cannot detect it !
Insulators - Valence band is completely filled - conduction band is completely empty and energy gap is quite large that small energy from other source cannot overcome it - Eg> 3eV
Semiconductors - Valence band totally filled - conduction band empty but energy gap is very small - Eg< 3eV
When photo diode is illuminated with light due to breaking of covalent bonds, equal umber of additional electrons and holes come into existence whereas fractional change in minority carrier is much higher than fractional change in majority carrier charge . Since fraction change in minority carrier current is measurable significantly in reverse bias that that of forward bias . Therefore photo diode are connected in reverse bias
Long answer shot - it increases sensitivity ! : P minority carrier is easily measure in reverse bias than forward one !
Energy Gap = 2.8 eV
wavelength = 6000nm = 6000 x 10^-9 m
Energy of Signal is given by E = hc/wavelength
h - plank's constant - 6.626 x 10^-34 Js
c = 3 x 10^8
E = 6.626 x 10-34 x 3 x 108
----------------------------- = 3.313 x 10-20 J
6000 x 10-9 .
But 1.6 x 10^-19 J = 1ev
Therefore 3.313 x 10-20J
------------------- = 0.207eV
1.6 x 10-19 .
Energy of signal is 0.207 eV which is less that 2.8eV energy band gap of photo diode , hence photo diode cannot detect it !