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Lesson 2402-8
Topic 26 – The Zener Diode as a Voltage Regulator (Part II)
Diagram
15 V
470Ω
1.5kΩPower
Supply
+
_
24 V
Equations for circuit analysis
• 𝐼𝑆 =𝑉𝑆−𝑉𝑍
𝑅𝑆 (𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)
• 𝑉𝑇𝐻 =𝑅𝐿
𝑅𝑆+𝑅𝐿𝑉𝑆 (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
• 𝑉𝑇𝐻 =𝑅𝐿
𝑅𝑆+𝑅𝐿𝑉𝑆 =
1.5kΩ
470Ω+1.5kΩ24V =
1.5k
1.97k24 =
18.274V
Second example
• Calculate all three currents.
Second example work
• 𝐼𝑆 =𝑉𝑆−𝑉𝑍
𝑅𝑆=
24V−15V
470Ω=
9
470=
19.149mA
• 𝑉𝐿 = 𝑉𝑍 = 15V
• 𝐼𝐿 =𝑉𝐿
𝑅𝐿=
15V
1.5kΩ= 10mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 19.149mA − 10mA =9.149mA
Third example
• Assuming a tolerance of ± 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=
24V−15V
446.5Ω=
9
446.5= 20.157mA
• 𝐼𝐿 =𝑉𝐿
𝑅𝐿=
15V
1.575Ω= 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=
25
470=
53.191mA
• 𝑉𝐿 = 𝑉𝑍 = 15V
• 𝐼𝐿 =𝑉𝐿
𝑅𝐿=
15V
1.5kΩ= 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=
24V−12V
470Ω=
12
470=
25.532mA
• 𝐼𝐿 =𝑉𝐿
𝑅𝐿=
12V
1.5kΩ= 8mA
• 𝐼𝑍 = 𝐼𝑆 − 𝐼𝐿 = 25.532mA − 8mA =17.532mA
Process for solving
Analyzing a Loaded Zener Regulator
Process Comment
Step 1 Calculate the series current (Eq. 1) Apply Ohm’s Law to RS
Step 2 Calculate the load voltage (Eq. 3) Load voltage equals diode voltage
Step 3 Calculate the load current (Eq. 4) Apply Ohm’s Law to RL
Step 4 Calculate the zener current (Eq. 5) Apply the current law to the diode
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