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1 ECE 255: Zener Diodes (Sedra and Smith, 4.4) Mark Lundstrom School of ECE Purdue University West Lafayette, IN USA Lundstrom: Fall 2019 ECE 255: Fall 2019 Purdue University
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ECE 255:
Zener Diodes (Sedra and Smith, 4.4)
Mark Lundstrom School of ECE
Purdue University West Lafayette, IN USA
Lundstrom: Fall 2019
ECE 255: Fall 2019 Purdue University
Goal for this lecture
Lundstrom: Fall 2019 2
VDD >VBR
−VD
+
I
R
Analyze circuits like this.
ID
ID = IS eVD VT −1( )
Outline
Lundstrom: Fall 2019 3
1) Zener diodes and applications 2) Zener diode model 3) Zener model parameters 4) Zener diode circuits
Diode IV: reverse breakdown
4
VD
ID mA( )
Lundstrom: Fall 2019
“breakdown”
-Zener (negative temp coeff) -avalanche (positive temp coeff.)
Diode IV: reverse breakdown
5
VD
ID mA( )
−IZK“knee current”
1/ rzslope
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−VZK
+VZ−
IZ
Zener diode
Recall: ideal voltage source
6
V
I
V0V0
I
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Recall: real voltage source
7
V
I
V0V0
I
+V−
V =V0 + IR0
R0
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“battery regulator”
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+VL
−
60 ±1 V
IL 1kΩ
+50 V−
RL VL
IL
50 V
Zener diodes
9
VD
− VD +
ID mA( )
0.6 − 0.7 V
ID = IS eVD VT −1( )
Lundstrom: Fall 2019
need a model here
Modeling the Zener diode
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+VZ−
IZ
IZ
+
Vz
−
VZ 0
rz
if VZ >VZKIZ > IZK
Model parameters:
Lundstrom: Fall 2019
Outline
Lundstrom: Fall 2019 11
1) Zener diodes and applications 2) Zener diode model 3) Zener model parameters 4) Zener diode circuits
Finding the model parameters
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IZ
+
Vz
−
VZ 0
rz
if VZ >VZKIZ > IZK
VZ =VZ 0 + IZrz
IZ = 10 mA VZ = 6.9 V
IZ = 20 mA VZ = 7.0 V
VZ 0 = ? rz = ?
Outline
Lundstrom: Fall 2019 13
1) Zener diodes and applications 2) Zener diode model 3) Zener model parameters 4) Zener diode circuits
“Zener shunt regulator”
Lundstrom: Fall 2019 14
+VL
−
60 ±1 V
IL 1kΩ
RL
IZ
At the nominal input voltage, what is VL when IL = 0?
rz = 20 Ω
IZK = 0.3 mA
VZ 0 = 49 V
Zener shunt regulator
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+
VL
−
60 ±1 V
IL = 0 1kΩ
At the nominal input voltage, what is VL when IL = 0?
49 V
20Ω
IZ = 60− 491 k + 0.02 k
= 10.8 mA
VL = 49 +10.8 × 0.02= 49.2
IZKVL
“Line regulation”
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+
VL
−
60 ±1 V
IL = 0 1kΩ
What is the change in output voltage for a +/- 1 V change in input voltage?
49 V
20Ω
ΔVL = ΔV
rz
rz + R
ΔVL
ΔV=
rz
rz + R
ΔVL
ΔV= 0.02
1.02= 0.02
line regulation = 20 mV/V
KVL
“Load regulation”
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+VL
−
60 V
IL = 2 mA
1kΩ
RL
What is the change in output voltage if the load draws 2 mA of current?
49 V
20Ω
KVL
KVL 60− IR ×1− 49− IZ × 0.02 = 0
IR = IZ + IL
IR
IR = 10.75 mA
IZ = 8.75 mA
IZ RL = ∞( ) = 10.8 mA
ΔIZ = −2.05 mA
IZ
“Load regulation” (the easy way)
Lundstrom: Fall 2019 18
+VL
−
60 V
IL = 2 mA 1kΩ
RL
What is the change in output voltage if the load draws 2 mA of current?
49 V
20Ω
ΔVL ≈ rzΔIz = 0.020× −2( ) = −0.040
ΔVL
ΔIL
= −0.0402
VmA
= −20 mVmA
IZ ΔIZ ≈ −2 mA
Maximum load current
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+VL
− 60 ±1 V
IL max
= ?
1kΩ
RL
What maximum load current can be drawn if the regulator is to operate properly?
49 V
20Ω
IZK = 0.3 mA IR
VL ≈ 49 V
IR ≈60 − 491k
= 11mA
IZ
IL max= IR − IZK = 10.7 mA
Summary
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A good Zener diode in breakdown is like a good battery.
Lundstrom: Fall 2019
The Zener diode model consists of an ideal battery in series with a resistor.
To insure that we are operating in the breakdown region, must be sure that IZ > IZK.
Zener diodes
Lundstrom: Fall 2019 21
1) Zener diodes and applications 2) Zener diode model 3) Zener model parameters 4) Zener diode circuits
