Download - Examples Zener Diode
Zener Diode
Problems
MENJANA MINDA KREATIF DAN INOVATIF
Zener Diodes
• The analysis of Zener diodes can be divided into 3 categories:– Fixed Vi and RL
– Fixed Vi, variable RL– Fixed Vi, variable RL
– Variable Vi, fixed RL
• To make the analysis simple, the analysis will be explain directly from the examples
Example 1a (Fixed Vi and RL)
• Determine VL, VR and IZ
• To check whether VZ is in the “on” or “off” mode, the value of VL must be determine first
• To do that, take out the Zener diode from the diode• The circuit become:
Example 1a (Fixed Vi and RL)
• The circuit become:
Example 1a (Fixed Vi and RL)
• By doing a nodal analysis for the node VL
V 73.82.11
16
=∴
=−
L
LL
Vk
V
k
V
• As we can see, the value of VL is smaller than VZ, so the Zener diode is in the “off” mode
• Which will result in:
• And:
V 73.8L
A 0=ZI
V 27.773.816 =−=∴+=
R
LRi
V
VVV
Example 1b (Fixed Vi and RL)
• Repeat Example 1a with RL = 3kΩ
3 kΩ
• The same analysis is repeated from Example 2.26a where the Zener diode is taken out to examine the value of VL
• The circuit becomes:
Example 1b (Fixed Vi and RL)
• By doing a nodal analysis for the node VL
V 1231
16
=∴
=−
L
LL
Vk
V
k
V
Example 1b (Fixed Vi and RL)
• As we can see, the value of VL is larger than VZ, so the Zener diode is in the “on” mode
• When the Zener diode is in the “on” mode, it will maintain the voltage of 10V. Because of that VL becomes:
• And VR becomes:
V 01== ZL VV
V 61016 =−=RV
Example 1b (Fixed Vi and RL)
• Using current divider theory:
−=
−=VV
III
LR
LiZ
mA 67.23
10
1
6
=
−=
−=
kk
RR L
Example 2 (Fixed Vi , Variable RL)
• Determine the range of RL and IL that will result in VL being maintained at 10 V
Example 2 (Fixed Vi , Variable RL)
• To maintain VL at 10 V, the Zener diode must be in the “on” mode
• For IZM = 32 mA, the current at load:
1050
(min)
−−= III ZMRL
• The load would be:
mA 8
321
1050
=
−−= mk
Ω=== k 25.18
10(max)
mI
VR
L
LL
Example 2 (Fixed Vi , Variable RL)
• For IZ(min), the Zener diode are assume “off” but the voltage VZare maintained at 10 V
• The load current would be:
1050 −= II RL
• The load would be:
mA 401
1050
=
−=k
Ω=== 25040
10
mI
VR
L
LL
Example 2 (Fixed Vi , Variable RL)
Use given minimum value of zener voltage, VZ as a load voltage, VL
The load would be:
( ) Ω=== 250)1(10
(min)kRV
R LL
The load current would be:
Ω=−
=−
= 2501050
(min)VV
RLi
LL
mA 40250
10
(min)(max)
=
=
=L
LL R
VI
Example 2 (Fixed Vi, Variable RL)
• Retrieve back all the IL and RL value:
IL (min) RL (max)
Ω==Ω==
250mA 40
k 25.1mA 8
LL
LL
RI
RI
IL (min)
IL (max) RL (min)
RL (max)
Example 3 (Variable Vi, Fixed RL)
• Determine the range of Vi that will maintain the Zener diode in the “on” mode
Example 3 (Variable Vi, Fixed RL)
• To maintain Zener diode in “on” mode, VZ must equal to VL:
• Taking the maximum current of the Zener diode, input current
V 20== LZ VV
becomes:
• The input voltage will become:
mA 67.762.1
2060
=
+=
+=
km
III LZMR
V 87.36(max)
)220)(67.76(20(max)
=∴==−
i
Ri
V
mRIV
Example 3 (Variable Vi, Fixed RL)
• For IZ(min), the Zener diode are assume “off” but the voltage VZare maintained at 20 V
• Using nodal analysis at node VL:
2.1
20
220
20 =−i
k
V
• Retrieve back all the value of Vi:
V 67.23(min)2.1220
=∴
=
iVk
V 67.23V 87.36 == ii VV
Vi (max) Vi (min)