In PN junction diode, P side is doped with acceptor concentration of 2 × 1016 cm-3, N side is doped donor concentration of 5 × 1017 cm-3. The contact potential of diode is: Assume thermal equivalent voltage equal to 26 mV and intrinsic carrier concentration equal to 1.45 × 1010 cm-3.

This question belongs to Electronics And Communications Engineering Electronic Devices And Circuits

Electronic Devices and Circuits

In PN junction diode, P side is doped with acceptor concentration of 2 × 10¹⁶ cm^-3, N side is doped donor concentration of 5 × 10¹⁷ cm^-3. The contact potential of diode is:
Assume thermal equivalent voltage equal to 26 mV and intrinsic carrier concentration equal to 1.45 × 10¹⁰ cm^-3.

A) 0.82 V B) 0.2 V C) 1.2 V D) 0 V

Answer & Solution

Correct Answer: Option A

Let's discuss the solution. Join the discussion

Question posted by Examveda

Community

Join the Discussion

1 Comment

Join the discussion

Jubair Khan 1 year ago

Formula:
Vbi = Vt ln (NA.ND/ni^2)
where
Vt = KTq = 26 mv = 0.026 v
NA = 2 * 10^16
ND = 5 * 10^17
ni = 1.45 * 10^10
Vbi =( 0.026 ln 1 * 10^34) / (2.1025 * 10^20)
= 0.82 v

Keep Learning

In PN junction diode P side is doped with acceptor concentration...

In PN junction diode, P side is doped with acceptor concentration of 2 × 1016 cm-3, N side is doped donor concentration of 5 × 1017 cm-3. The contact potential of diode is: Assume thermal equivalent voltage equal to 26 mV and intrinsic carrier concentration equal to 1.45 × 1010 cm-3.

Join the Discussion

Related Questions on Electronic Devices and Circuits

In PN junction diode, P side is doped with acceptor concentration of 2 × 10¹⁶ cm^-3, N side is doped donor concentration of 5 × 10¹⁷ cm^-3. The contact potential of diode is:
Assume thermal equivalent voltage equal to 26 mV and intrinsic carrier concentration equal to 1.45 × 10¹⁰ cm^-3.