# 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
• 𝐼𝑆 =
𝑉𝑆 −𝑉𝑍
𝑅𝑆
• 𝑉𝑇𝐻 =
(𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)
𝑅𝐿
𝑉𝑆
𝑅𝑆 +𝑅𝐿
(𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
• 𝑉𝐿 = 𝑉𝑍 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3)
• 𝐼𝐿 =
𝑉𝐿
𝑅𝐿
(𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4)
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 5)
First example
• If the zener diode is disconnected from the
source, what is the load voltage?
First example work
•
𝑅𝐿
𝑉𝑇𝐻 =
𝑉𝑆
𝑅𝑆 +𝑅𝐿
1.5kΩ
24V
470Ω+1.5kΩ
18.274V
=
=
1.5k
24
1.97k
=
Second example
• Calculate all three currents.
Second example work
• 𝐼𝑆 =
𝑉𝑆 −𝑉𝑍
𝑅𝑆
=
24V−15V
470Ω
=
19.149mA
• 𝑉𝐿 = 𝑉𝑍 = 15V
• 𝐼𝐿 =
𝑉𝐿
𝑅𝐿
=
15V
1.5kΩ
9
470
=
= 10mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 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
• 𝑅𝑆 − 5% = 446.5Ω
• 𝑅𝐿 + 5% = 1.575k
• 𝐼𝑆 =
𝑉𝑆 −𝑉𝑍
𝑅𝑆
• 𝐼𝐿 =
𝑉𝐿
𝑅𝐿
=
=
24V−15V
446.5Ω
15V
1.575Ω
=
9
446.5
= 20.157mA
= 9.524mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 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
• 𝐼𝑆 =
𝑉𝑆 −𝑉𝑍
𝑅𝑆
=
40V−15V
470Ω
=
53.191mA
• 𝑉𝐿 = 𝑉𝑍 = 15V
• 𝐼𝐿 =
𝑉𝐿
𝑅𝐿
=
15V
1.5kΩ
25
470
=
= 10mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 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
• 𝑉𝐿 = 𝑉𝑍 = 12V
• 𝐼𝑆 =
𝑉𝑆 −𝑉𝑍
𝑅𝑆
25.532mA
• 𝐼𝐿 =
𝑉𝐿
𝑅𝐿
=
=
24V−12V
470Ω
12V
1.5kΩ
=
12
470
=
= 8mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 25.532mA − 8mA =
17.532mA
Process for solving
Step 1
Step 2
Step 3
Step 4
Process
Comment
Calculate the series current (Eq. 1) Apply Ohm’s Law to RS