HA17431HLP HITACHI [Hitachi Semiconductor], HA17431HLP Datasheet - Page 11

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HA17431HLP

Manufacturer Part Number
HA17431HLP
Description
Shunt Regulator
Manufacturer
HITACHI [Hitachi Semiconductor]
Datasheet

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Practical Example
Consider the example of a photocoupler, with an internal light emitting diode V
power supply output voltage V
the shunt regulator V
Next, assume that R
following values are found.
G
f
f
R
R
1
2
1
2
2
= 1 / (2 × π × 0.022 µF × 316 × 10 kΩ) = 2.3 (Hz)
= 1 / (2 × π × 0.022 µF × 3.3 kΩ) = 2.2 (kHz)
= 3.3 kΩ / 10 kΩ = 0.33 times (–10 dB)
=
=
0.5mA
5V – 1.05V – 3V
1.05V
2.5mA + 0.5mA
3
K
= R
= 2.1(kΩ) (2.2kΩ from E24 series)
= 3 V, the following values are found.
4
= 10 kΩ. This gives a 5 V output. If R
2
= 316(Ω) (330Ω from E24 series)
= 5 V, and bias resistance R
2
current of approximately 1/5 I
5
= 3.3 kΩ and C
Rev.0, Sep. 2001, page 11 of 15
F
= 1.05 V and I
HA17431H Series
1
= 0.022 µF, the
F
at 0.5 mA. If
F
= 2.5 mA,

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