PN -Junction Diode | PN-Junction Diode Characteristics

PN -Junction Diode | PN-Junction Diode Characteristics


1.    When P and N extrinsic semiconductors are formed as a PN-junction, the diffusion of holes from P-side to N-side and diffusion of electrons from N-side to P-side occurs


2.    The region of P and N materials near the junction are depleted of charges.  So, immobile ions are left over.

3.    I0 is the reverse saturation current in reverse bias condition.
4.    I0 is doubles for every 100 rise for germanium and for every 60 rise in silicon.

5.    Reverse saturation current is a micro-amphers for  germanium  &  nano -ampheres in silicon diodes
6.    Practically Barrier Potential Is 0.5 To 1 Micron (106).

7.    Barrier potential width is depends is depends on the temperature .when temperature increases potential barrier width is decreased.

8.    Barrier potential is decreased approximately 2.5 milli volts per degree rise in temperature.

9.    Generally potential barrier potential voltage is 0.3v for germanium 0.7v for silicon
Ø Saturation Current    I= I0(E‑V/Nvt  - 1 ).

Ø Channel length modulation does not exist in pn-junction diode. This is for only in MOSFETS.

 Barrier Potential Is Depends On The Following

o   Type Of Semi-Conductor
o   Donor Impurity Added
o   Acceptor Impurity Added
o   Temperature

 Mainly PN-Junction Diode Having Two Types Of Resistance

o   Static Resistance (Dc Condition )
o   Dynamic Resistance (Ac Condition)

Static Resistence of PN-Junction Diode:

Static Resistance Is Measured In Dc Condition. In Dc Condition Resistance is Zero Means It Is In Short Circuit

Dynamic Resistance of PN-Junction Diode:

  It Is Defined As The Reciprocal Of Slope Of Forward Characteristics. Typically It’s Value Is 1-25ohms

Ø In Reverse Bias Condition Resistance Of Diode Is Infinity Because This Is Open Ckt Condition.

Ø Reverse Break-Down Voltage Of Silicon Is Higher Than The Germanium Diode

Important Formulas In Pn-Junction Diode

Vt  =  K * T
K=8.62*10-5ev/0k
T=Temp   (At Room T= 27 + 273= 3000k)

   When They Give One Saturation Current And Two Temperature Values We Can Find Another Saturation Current Through This Formulae i.e
I02 = I01 (2t2-T1/ 10)