PN-Junction Diode Characteristics

Forward Bias --- External battery makes the Anode more positive than the Cathode --- Current flows in the direction of the arrow in the symbol.

Reverse Bias --- External battery makes the Cathode more positive than the Anode --- A tiny current flows opposite to the arrow in the symbol.

Graphical PN-Junction Diode V-I Characteristic

Reverse breakdown

Forward Bias RegionReverse Bias Region

Mathematical Approximation

D

T

VηV

D sI = I (e -1)

Ideal PN Junction Diode V-I Characteristic

Forward Bias – Short Circuit

Reverse Bias – Open Circuit

Diode Reverse Recovery Time

ta is the time to remove the charge stored in the depletion region of the junction

tb is the time to remove the charge stored in the bulk semiconductor material

Reverse Recovery CharacteristicsSoft Recovery

Reverse recovery time = trr = ta+tb

Peak Reverse Current = IRR = ta(di/dt)

Reverse Recovery CharacteristicsAbrupt Recovery

Reverse recovery time = trr = ta+tb

Peak Reverse Current = IRR = ta(di/dt)

Series-Connected Diodes

• Use 2 diodes in series to withstand higher reverse breakdown voltage.

• Both diodes conduct the same reverse saturation current, Is.

Diode Characteristics

• Due to differences between devices, each diode has a different voltage across it.

• Would like to “Equalize” the voltages.

Series-Connected Diodes with Voltage Sharing Resistors

Series-Connected Diodes with Voltage Sharing Resistors

Series-Connected Diodes with Voltage Sharing Resistors

• Is = Is1+IR1 = Is2+IR2

• IR1 = VD1/R1

• IR2 = VD2/R2 = VD1/R2

• Is1+VD1/R1 = IS2+VD1/R2

• Let R = R1 = R2

• Is1 + VD1/R = Is2 +VD2/R

• VD1 + VD2 = Vs

Example 2.3

• Is1 = 30mA, Is2 = 35mA

• VD = 5kV

• (a) – R1=R2=R=100kΩ, find VD1 and VD2

• (b) – Find R1 and R2 for VD1=VD2=VD/2

Example 2.3 (a)

s1

s2

1 2

D D1 D2

D2 D D1

D1 D2s1 s2

DD1 S2 S1

-3 -3D1

D2 D D1

I = 30mAI = 35mAR =R =R =100kΩ-V = -V - VV = V - V

V VI + =I +R RV RV = + (I -I )2 2

5kV 100kV = + (35Χ10 -30Χ10 ) = 2750Volts2 2

V = V - V = 5kV - 2750 = 2250Volts

Example 2.3 (a) simulation

D1DIODE_VIRTUAL*

D2DIODE_VIRTUAL**

R1100kOhm

R2100kOhm

U1DC 1MOhm -2.727k V

+

-

U2DC 1MOhm -2.273k V

+

-

V15000 V

Example 2.3 (b)s1

s2

DD1 D2

D1 D2s1 s2

1 2

D2 12

D1 1 s2 s1

1

2 -3 -3

2

I = 30mAI = 35mA

VV = V = = 2.5kV2

V VI + =I +R R

V RR =V -R (I -I )

R =100kΩ2.5kVΧ100kΩR =

2.5kV -100kΩΧ(35Χ10 -30Χ10 )R =125kΩ

Example 2.3 (b) simulation

D1DIODE_VIRTUAL*

D2DIODE_VIRTUAL**

R1100kOhm

R2125kOhm

U1DC 1MOhm -2.500k V

+

-

U2DC 1MOhm -2.500k V

+

-

V15000 V