HA17431H Datasheet, PDF (11/15 Pages) Hitachi Semiconductor – Shunt Regulator

HA17431H Series
Practical Example
Consider the example of a photocoupler, with an internal light emitting diode V_F = 1.05 V and I_F = 2.5 mA,
power supply output voltage V₂ = 5 V, and bias resistance R₂ current of approximately 1/5 I_F at 0.5 mA. If
the shunt regulator V_K = 3 V, the following values are found.
R₁ =
5V – 1.05V – 3V
2.5mA + 0.5mA
= 316(Ω) (330Ω from E24 series)
R₂ =
1.05V
0.5mA
= 2.1(kΩ) (2.2kΩ from E24 series)
Next, assume that R₃ = R₄ = 10 kΩ. This gives a 5 V output. If R₅ = 3.3 kΩ and C₁ = 0.022 µF, the
following values are found.
G
2
=
3.3
kΩ
/
10
kΩ
=
0.33
times
(–10
dB)
f₁ = 1 / (2 × π × 0.022 µF × 316 × 10 kΩ) = 2.3 (Hz)
f₂ = 1 / (2 × π × 0.022 µF × 3.3 kΩ) = 2.2 (kHz)
Rev.0, Sep. 2001, page 11 of 15