# 2402-8 Zener Diode Regulation Part II Lecture 4-26-12

```Lesson 2402-8
Topic 26 – The Zener Diode as a
Voltage Regulator (Part II)
Diagram
+
Power
Supply
24 V
_
470Ω
15 V
1.5kΩ
Equations for circuit analysis
• πΌπ =
ππ −ππ
ππ
• πππ» =
(πΈππ’ππ‘πππ 1)
ππΏ
ππ
ππ +ππΏ
(πΈππ’ππ‘πππ 2)
• ππΏ = ππ (πΈππ’ππ‘πππ 3)
• πΌπΏ =
ππΏ
ππΏ
(πΈππ’ππ‘πππ 4)
• πΌπ = πΌπ − πΌπΏ (πΈππ’ππ‘πππ 5)
First example
• If the zener diode is disconnected from the
source, what is the load voltage?
First example work
•
ππΏ
πππ» =
ππ
ππ +ππΏ
1.5kΩ
24V
470Ω+1.5kΩ
18.274V
=
=
1.5k
24
1.97k
=
Second example
• Calculate all three currents.
Second example work
• πΌπ =
ππ −ππ
ππ
=
24V−15V
470Ω
=
19.149mA
• ππΏ = ππ = 15V
• πΌπΏ =
ππΏ
ππΏ
=
15V
1.5kΩ
9
470
=
= 10mA
• πΌπ = πΌπ − πΌπΏ = 19.149mA − 10mA =
9.149mA
Third example
• Assuming a tolerance of &plusmn; 5 percent in both
resistors, what is the maximum zener current?
– To maximize the current, we need to find the
lower tolerance for RS and the upper tolerance for
RL.
Third example work
• ππ − 5% = 446.5Ω
• ππΏ + 5% = 1.575k
• πΌπ =
ππ −ππ
ππ
• πΌπΏ =
ππΏ
ππΏ
=
=
24V−15V
446.5Ω
15V
1.575Ω
=
9
446.5
= 20.157mA
= 9.524mA
• πΌπ = πΌπ − πΌπΏ = 20.157mA − 9.524mA =
10.633mA
Fourth example
• Suppose the supply voltage can vary from 24
to 40 V. What is the maximum zener current?
– Calculate the values at the maximum supply
voltage.
Fourth example work
• πΌπ =
ππ −ππ
ππ
=
40V−15V
470Ω
=
53.191mA
• ππΏ = ππ = 15V
• πΌπΏ =
ππΏ
ππΏ
=
15V
1.5kΩ
25
470
=
= 10mA
• πΌπ = πΌπ − πΌπΏ = 53.191mA − 10mA =
43.191mA
Fifth example
• The zener diode is replaced with a 1N963B
diode. What are the load voltage and the
zener current?
Data Sheet information
Fifth example work
• ππΏ = ππ = 12V
• πΌπ =
ππ −ππ
ππ
25.532mA
• πΌπΏ =
ππΏ
ππΏ
=
=
24V−12V
470Ω
12V
1.5kΩ
=
12
470
=
= 8mA
• πΌπ = πΌπ − πΌπΏ = 25.532mA − 8mA =
17.532mA
Process for solving
Step 1
Step 2
Step 3
Step 4
Analyzing a Loaded Zener Regulator
Process
Comment
Calculate the series current (Eq. 1) Apply Ohm’s Law to RS
Calculate the load voltage (Eq. 3) Load voltage equals diode voltage
Calculate the load current (Eq. 4) Apply Ohm’s Law to RL
Calculate the zener current (Eq. 5) Apply the current law to the diode
Any questions?