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- 3.1 ZENER DIODES
- VOLTAGE REGULATION
A voltage regulator circuit automatically
maintains the output voltage of a power
supply constant, regardless of
- a change in the load
- a change in the source voltage
- ZENER DIODES
The simplest of all voltage regulators
is the zener diode voltage regulator.
A zener diode is a special diode that
p
is optimized for operation in the
g
breakdown region.
- ZENER DIODE CIRCUIT
The zener diode is typically connected
reverse biased, in parallel with the load.
Resistor Rs limits current to zener.
zener
- ZENER DIODE CHARACTERISTICS
In the forward
region, the zener
diode acts like a
regular silicon
diode, with a 0.7
volt drop when it
conducts.
- ZENER DIODE CHARACTERISTICS
In the reverse bias
region,
region a reverse
leakage current
flows until the
breakdown voltage
is reached.
At this point, the
reverse current,
called zener
current Iz,
increases sharply.
sharply
- ZENER DIODE CHARACTERISTICS
Voltage after
breakdown is also
b kd i l
called zener
voltage Vz.
Vz
Vz remains nearly
a constant, even
constant
though current Iz
varies
considerably.
- ZENER DIODE RATINGS
A zener spec sheet typically provides
- the maximum power rating Pz
- the nominal zener voltage Vz at test
current IzT
- the maximum DC zener current IzM
Example:
- 1N752 has a power rating of 500mW, a
h ti f 500 W
nominal zener voltage of 5.6V at a test
current of 20 mA, a maximum zener
mA
current of 80 mA.
- ZENER DIODE CIRCUIT EXAMPLE
In a zener diode voltage regulator circuit,
VDC(IN) = 20V Vz = 9V, Rs = 82Ω, RL = 100
20V, V 9V R 82 100Ω
Q: Calculate IL.
A: IL = VL/RL = VZ/RL = 9V/100 Ω = 90 mA
82Ω
20V 100Ω
9V
- ZENER DIODE EXAMPLE (CONT’D)
Q: Calculate IT.
A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs
= [20 - 9]/82 Ω
= 134 mA
82Ω
20V 9V 100Ω
- ZENER DIODE EXAMPLE (CONT’D)
Q: Calculate Iz
A: Iz = IT - IL
= 134mA - 90mA
= 44mA
82Ω
20V 9V 100Ω
- ZENER DIODE EXAMPLE (CONT’D)
Q: If zener Pz = 500mW, is Iz safe ?
A: IzM = Pz / Vz
= 500 mW / 9V
= 55 mA
Since circuit Iz
= 44 mA (< 55mA),
zener is safe
- ZENER DIODE EXAMPLE (CONT’D)
Q: Suppose load changes from 100 Ω to 200 Ω.
Q pp g
Find new IL, Iz, IT. Is Iz safe ?
A: IL = VL/RL = VZ/RL = 9V/200 Ω = 45 mA
20V 9V 200Ω
- ZENER DIODE EXAMPLE (CONT’D)
A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs
= [20 - 9]/82 Ω = 134 mA (no change)
Iz = IT - IL = 134mA - 45mA = 89mA
Since IzM = 55mA, zener is not safe.
20V 9V 200Ω
- ZENER DIODE EXAMPLE (
(CONT’D)
)
Q: Suppose load changes from 100 Ω to 80 Ω.
Find new IL, Iz, IT. Is Iz safe ?
A: IL = VL/RL = VZ/RL = 9V/80 Ω = 112 5 mA
112.5
20V 9V 80Ω
- ZENER DIODE EXAMPLE (CONT D)
(CONT’D)
A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs
= [20 - 9]/82 Ω = 134 mA (no change)
Iz = IT - IL = 134mA - 112 5mA = 21.5 mA
112.5mA 21 5
Since IzM = 55 mA, zener is safe.
20V 9V 80Ω
- ZENER DIODE EXAMPLE (CONT’D)
Q: Suppose zener regulator circuit has no load
connected, is Iz safe ?
A: IT = VRs/Rs = [VDC(IN) - Vz]/Rs
= [20 - 9]/82 Ω = 134 mA (no change)
Iz = IT = 134 mA this is very Unsafe
20V 9V
- ZENER DIODE EXAMPLE (
(CONT’D)
)
Q: In the previous example, suppose VDC(IN)
changes from 20V to 30V. Is Iz safe ?
A: IL = 9V/100 Ω = 90 mA
IT = [30-9]/82 Ω = 256 mA
Iz = IT - IL = 256 mA - 90 mA = 166 mA
Very Unsafe
82Ω
30V 9V 100Ω
- ZENER DIODE EXAMPLE (
(CONT’D)
)
Q
Q: Suppose VDC(IN) changes from 20V to 18V.
pp ( ) g
Is Iz safe ?
A: IL = 9V/100 Ω = 90 mA
IT = [18-9]/82 Ω = 110 mA
Iz = IT - IL = 110 mA - 90 mA = 20 mA
Safe
82Ω
18V 9V 100Ω
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