HA17431 Series
Practical Example
Consider the example of a photocoupler, with an internal light emitting diode VF = 1.05 V and IF = 2.5 mA,
power supply output voltage V2 = 5 V, and bias resistance R2 current of approximately 1/5 IF at 0.5 mA. If
the shunt regulator VK = 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.
G2 = 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.1, Sep. 2002, page 19 of 24