ZENER DIODES AND REGULATORS

Textbook section 4.4

INEL 4201 Electronics I

Oxford  University  Publishing  Microelectronic  Circuits  by  Adel  S.  Sedra  and  Kenneth  C.  Smith  (0195323033)

4.2.  Terminal  Characteristics

of  Junction  Diodes

▪ Most  common  implementation  of  a  diode  utilizes  pn  junction.  

▪ I-­‐V  curve  consists  of  three  characteristic  regions  ▪ forward  bias:  v  >  0  ▪ reverse  bias:  v  <  0  ▪ breakdown:  v  <<  0  

discontinuity  caused  by  differences  in  scale

Oxford  University  Publishing  Microelectronic  Circuits  by  Adel  S.  Sedra  and  Kenneth  C.  Smith  (0195323033)

= − Si I<< − Si I −= / Tv VSi I e

V  =  -­‐V

ZK

V  =  -­‐V

T

V  =  10V T

= −/( 1)Tv VSi I e

-

vD

+

iD

VZ

Zener diodes - reverse-bias operation - keeps an approx. constant voltage independent of diode’s current

- Operation in this regime can destroy a regular diode

Voltage regulatorVS R

Let:

• VZ

: Zener diode’s nominal voltage.

• iZ,min

: minimum current through diode required for proper operation.

• iL,max

, iL,min

: maximum and minimum load’s current.

• VS,max

, VS,max

: maximum and minimum source voltage levels

For proper operation, select R so that

R VS,min

� VZ

iZ,min

+ iL,max

Maximum power dissipated in the resistor:

PR,max

=

(VS,max

� VZ

)

2

R

Maximum power dissipated in the diode:

PD,max

= VZ

⇥✓VS,max

� VZ

R� i

L,min

◆

D and R must have a power rating greater than PD,max

and PR,max

, respec-

tively.

• Minimum zener current ≥ iZ, min when load current = iL, max

• VZ is specified at a specific current; use piece-wise linear model to estimate the diode’s voltage at a different current

Voltage regulatorVS RLet:

• VZ

: Zener diode’s nominal voltage.

• iZ,min

: minimum current through diode required for proper operation.

• iL,max

, iL,min

: maximum and minimum load’s current.

• VS,max

, VS,max

: maximum and minimum source voltage levels

For proper operation, select R so that

R VS,min

� VZ

iZ,min

+ iL,max

Maximum power dissipated in the resistor:

PR,max

=

(VS,max

� VZ

)

2

R

Maximum power dissipated in the diode:

PD,max

= VZ

⇥✓VS,max

� VZ

R� i

L,min

◆

D and R must have a power rating greater than PD,max

and PR,max

, respec-

tively.

=>

rZ

VZ

+−VS

R1

D1 RLvL

+

-

+−VS

R1

D1RL vL

+

-

rZ

VZ

=>

Piece-wise linear approx.iD

vD

iD vD

+

-

VZ

slope = 1/rZ

Analysis of Zener diode circuit using piecewise linear approximation:

1. Transform voltage source to current source then back to voltage source:

V 0S =

VS

R1(R1||RL) =

VS

R1

R1 ⇥RL

R1 +RL=

RL

R1 +RLVS

in series with a resistance R01 = R1||RL.

2. Find diode’s current:

iD =

V 0S � VZ

R1||RL + rZ

3. calculate VL by finding voltage across diode:

VL = VZ + rZ ⇥ iD

0

0

0

PROB. 4.61• Design a regulator with output voltage VL = 7.5V

• VZ = 7.5V at IZ = 12mA, knee (minimum)

current is 0.5mA, incremental resistance rz = 30Ω

• Power Supply: VPS = 10V

• Load: RL = 1.2kΩ

PROB. 4.61 DESIGN

IL =7.5V

1.2k�= 6.25mA

IR = IL + IZ = 12mA + 6.25mA = 18.25mA

R =VPS � VL

IR=

10V � 7.5V

18.25mA� 137�

10V

139.51

RL12001 10V

139.51

RL12001

301

7.14V

IZIL

IR

WARNING: Diagram should show R=137 instead of 139.5

What’s the output voltage if the supply is 10% high?

VZ = IZ � 30� + VZ0 � VZ0 = 7.5V � 12mA � 30� = 7.14VVPS � VL

R=

VL � 7.14V

30�+

VL

RL

VL = (R�30��RL)

�VPS

R+

7.14V

30�

�

For R = 137�, VPS = 10V and RL = 1200�,

VL = (24.116�)

�10V

137�+

7.14V

30�

�= 7.5V

as expected. For VPS = 11V ,

VL = (24.116�)

�11V

137�+

7.14V

30�

�= 7.676V

What’s the output voltage if the supply is 10% low?

VL = (24.116�)

�9V

137�+

7.14V

30�

�= 7.32V

What’s the output voltage if the load voltage is 10% high and the load isremoved?

VL = (137��30�)

�11V

137�+

7.14V

30�

�= 7.83V

What’s the smallest load resistor that will allow the zener to operate atIZK = 0.5mA if VPS = 10V ?

VZ = 7.14V + 0.5mA � 30� = 7.155V

IR =10V � 7.155V

137�= 20.8mA

IL = 20.8mA � 0.5mA = 20.3mA � RL,min =7.155V

20.3mA� 352�