HA17432VUP 查看數據表（PDF） - Hitachi -> Renesas Electronics

零件编号

产品描述 (功能)

比赛名单

HA17432VUP

Shunt Regulator

Hitachi -> Renesas Electronics 

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

Share Link: