diode v-i characteristics curve, diode in series connection
TRANSCRIPT
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
V
ηVD 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 = 30mA
I = 35mA
R =R =R =100kΩ
-V = -V - V
V = V - V
V VI + =I +
R RV R
V = + (I -I )2 2
5kV 100kV = + (35Χ10 -30Χ10 ) = 2750Volts
2 2V = 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 = 30mA
I = 35mA
VV = V = = 2.5kV
2V V
I + =I +R R
V RR =
V -R (I -I )
R =100kΩ
2.5kVΧ100kΩR =
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