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)