Consider a silicon p-n junction at room temperature having the following parameters-Doping on the n-side - 1 - 1017 cm-3Depletion width on the n-side - 0-1 - mDepletion width on the p-side - 1-0 - mIntrinsic carrier concentration - 1-4 - 1010 cm-3 Thermal voltage - 26 mVPermittivity of free space - 8-85- 10-14F-cm-1Dielectric constant of silicon - 12The peak electric field in the device is

Ans : (b) Electric field, E= qN_oϵ(x w) At the centre x=0 So, E_max = qN_oϵ ( W_n) = 1.6× 10^ 19× 10^17×10^ 512× 8.85× 10^ 14 = 0.15 MV/cm ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

PN Junction in Equillibrium

Consider a si...

Question

Consider a silicon p-n junction at room temperature having the following parameters:
Doping on the n-side = $1 \times 10^{17} c m^{- 3}$
Depletion width on the n-side = $0.1 μ m$
Depletion width on the p-side = $1.0 μ m$
Intrinsic carrier concentration = $1.4 \times 10^{10} c m^{- 3}$ Thermal voltage = $26 m V$
Permittivity of free space = $8.85 \times 10^{- 14} F - c m^{- 1}$
Dielectric constant of silicon = 12
The peak electric field in the device is

1.80 M V . c m^{- 1}

, directed from p-region to n-region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.15 M V . c m^{- 1}

, directed from p-region to n-region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.15 M V . c m^{- 1}

, directed from n-region to p-region

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.80 M V . c m^{- 1}

, directed from n-region to p-region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.15 M V . c m^{- 1}$ , directed from n-region to p-region
Ans : (b)

Electric field,
$E = \frac{q N_{o}}{ϵ} (x - w)$
At the centre $x = 0$
So, $E_{m a x} = \frac{q N_{o}}{ϵ} (- W_{n})$
$= \frac{- 1.6 \times 10^{- 19} \times 10^{17} \times 10^{- 5}}{12 \times 8.85 \times 10^{- 14}}$
$= - 0.15 M V / c m$

Suggest Corrections

Similar questions

Q. Consider a silicon p-n junction at room temperature having the following parameters:
Doping on the n-side =

1 \times 10^{17} c m^{- 3}

Depletion width on the n-side =

0.1 μ m

Depletion width on the p-side =

1.0 μ m

Intrinsic carrier concentration =

1.4 \times 10^{10} c m^{- 3}

Thermal voltage =

26 m V

Permittivity of free space =

8.85 \times 10^{- 14} F - c m^{- 1}

Dielectric constant of silicon = 12
The built-in potential of the junction

Q. Consider a silicon p-n junction with a uniform acceptor doping concentration of

10^{17} c m^{- 3}

on the p-side and a uniform donor doping concentration of

10^{16} c m^{- 3}

on the n-side. No external voltage is applied to the diode.
Given:

k T / q = 26 m V, n_{i} = 1.5 \times 10^{10} c m^{- 3}, ϵ_{S i} = 12 ε_{0}, ε_{0} = 8.85 \times 10^{- 14} F / c m

, and

q = 1.6 \times 10^{- 19} C

.
The charge per unit junction area (

n C c m^{- 2}

in the depletion region on the p-side is

Q. The doping concentrations on the p-side and n-side of a silicon diode are

1 \times 10^{16} c m^{- 3}

and

1 \times 10^{17} c m^{- 3}

, respectively. A forward bias of

0.3 V

is applied tothe diode. At

T = 300 K

, the intrincic carrier concentration of silicon

n_{i} = 1.5 \times 10^{10} c m^{- 3}

and

\frac{k T}{q} = 26 m V

. The electron concentration at the edge of the deplection region on the p-side is

Q. When a silicon diode having a doping concentration of

N_{A} = 9 \times 10^{16} c m^{- 3}

on p-side and

N_{D} = 1 \times 10^{16} c m^{- 3}

on n-side is reverse biased, the total depletion width is found to be

3 μ m

. Given that the permittivity of silion is

1.04 \times 10^{- 12} F / c m

, the depletion width on the p-side and the maximum electric field in the depletion region, respectively are

Q. Consider an abrupt PN junction (at

T = 300 K

) shown in the figure.The depletion region width

x_{n}

on the N-side of the junction is

0.2 μ m

and the permittivity of silicon

(ϵ_{S i})

1.044 \times 10^{- 12} F / c m

. At the junction, the approximate value of the peak electric field (in kV/cm) is

Join BYJU'S Learning Program

The correct option is C 0.15MV.cm−1, directed from n-region to p-regionAns : (b) Electric field, E=qNoϵ(x−w) At the centre x=0 So, Emax=qNoϵ(−Wn) =−1.6×10−19×1017×10−512×8.85×10−14 =−0.15 MV/cm