ELECTRONICS 1 Lecturer: Yaroub Ghazi 1 Tikrit University Electrical and Electronic Engineering Department Second Year
ELECTRONICS 1
Lecturer:
Yaroub Ghazi
1
Tikrit University
Electrical and Electronic Engineering Department
Second Year
LECTURE 6: ZENER DIODE
ZENER REGION
WHEN THE APPLIED REVERSE POTENTIAL BECOMES MORE AND MORE NEGATIVE,
A FEW FREE MINORITY CARRIERS HAVE DEVELOPED SUFFICIENT VELOCITY TO
LIBERATE ADDITIONAL CARRIERS THROUGH IONIZATION.
WHEN VZ DECREASES TO VERY LOW LEVEL, THIS MECHANISM, CALLED ZENER
BREAKDOWN, WILL CONTRIBUTE TO THE SHARP CHANGE IN THE CHARACTERISTIC.
IT OCCURS BECAUSE THERE IS A STRONG ELECTRIC FILED IN THE REGION OF THE
JUNCTION THAT CAN DESTROY THE BONDING FORCES WITHIN THE ATOM AND
βGENERATEβ CARRIER. DIODE EMPLOYING THIS UNIQUE PORTION OF THE
CHARACTERISTIC OF A P-N JUNCTION ARE CALLED ZENER DIODES.
ZENER DIODE
β’ THE ZENER DIODE IS A DEVICE THAT IS DESIGNED TO MAKE FULL USE OF THIS
ZENER REGION. ZENER REGION OCCURS AT A REVERSE BIAS POTENTIAL OF VZ.
ZENER DIODE APPLICATIONS
1- AC VOLTAGE REGULATORS (LIMITERS OR CLIPPER)
TWO BACK-TO-BACK ZENER CAN BE USED AS AN AC REGULATOR OR A
SIMPLE SQUARE WAVE GENERATOR AS SHOWN
A. SINUSOIDAL AC REGULAROR
ZENER DIODE APPLICATIONS
B. SIMPLE SQUARE WAVE GENERATOR
ZENER DIODE APPLICATIONS
2- DC VOLTAGE REFERENCE
TWO OR MORE LEVELS CAN BE ESTABLISHED BY PLACING ZENER DIODE IN A
SERIES AS SHOWN IN FIGURE AS LONG AS VI (E) IS GREATER THAN THE SUM OF
VZ1 AND VZ2, BOTH DIODE WILL BE IN BREAKDOWN STATE (ON STATE) AND THE
THREE REFERENCE VOLTAGE WILL BE AVAILABLE.
ZENER DIODE APPLICATIONS
3- DC VOLTAGE REGULATORS
A. FIXED VI, VARIABLE RL
πΌπ =ππ β πππ π
(πΆπππ π‘πππ‘)
πΌπΏ(MIN) = πΌπ β πΌππ
π πΏ(πππ₯) =ππ
πΌπΏ(πππ)
πΌπΏ(πππ₯) = πΌπ β πΌππΎ
π πΏ(πππ) =ππ
πΌπΏ(πππ)
Example:
(a) For the network of Figure, determine the range of RL and IL that will result in VRL
being maintained at 10 V.
(b) Determine the maximum wattage rating of the diode.
Solution:
(a)
π πΏ(πππ) =π πππππ β ππ
=(1 πΞ©)(10)
50 β 10=
10 π 103
40= 250 Ξ©
ππ π = ππ β ππ = 50 β 10 = 40 π
πΌπ π =ππ π π π
=40
1 πΞ©= 40 ππ΄
πΌπΏ(πππ) = πΌπ π β πΌππ = 40 β 32 = 8 π
π πΏ(πππ₯) =ππ
πΌπΏ(πππ)=
10
8 ππ΄= 1.25 πΞ©
(b)
ππππ₯ = πππΌππ = 10 32 ππ΄ = 320 ππ
B. FIXED RL, VARIABLE VI
β’ FOR FIXED VALUE OF RL IN FIGURE THE VOLTAGE VI MUST BE SUFFICIENTLY TO TURN THE ZENER
DIODE ON. THE TURN ON VOLTAGE IS DETERMINED BY:
πΌπΏ =πππ πΏ
(πΆπππ π‘πππ‘)
πΌπ(πππ) = πΌππΎ + πΌπΏ
ππ(πππ) = πΌπ(πππ)π π + ππ
πΌπ(πππ₯) = πΌππ + πΌπΏ
ππ πππ₯ = πΌπ(πππ₯)π π + ππ
Example:
Determine the range of values of Vi that will maintain the Zener diode of
Figure in the βonβ state.
Solution:
ππ(πππ) =(π πΏ + π π)ππ
π πΏ=
1200 + 220 (20)
1200= 23.67 π
πΌπΏ =ππΏπ πΏ
=πππ πΏ
=20
1.2 πΞ©= 16.67 ππ΄
πΌπ (πππ₯) = πΌππ + πΌπΏ = 60 + 16.67 = 76.67 ππ΄
ππ(πππ₯) = πΌπ (πππ₯)π π + ππ = 76.67 ππ΄ 0.22 πΞ© + 20
= 16.87 + 20 = 36.87 π