Using HCPL-3700 from Agilent Technology.
For the mode of Input Clamp Voltage, we have
Vihc2 = 6.7 v
|Iin| = 10 mA
we have
Vr = Vpk - Vihc2 = 240*1.414 - 6.7 = 332.7 v
So
Rx = Vr / |Iin| = 332.7 / 10 = 33.3 K
Rx/2 = 16.7 K
Use Rx/2 = 18.7 K, 1%, so
Px/2 = (Vr/2)2 / Rx/2 = 1.5w
Choose
Rx/2 = 18.7 K,1%, 1.5 w
Question:
Div2 is a flip-flop to divide 120 Hz for 60 Hz
RL is not required ???
Using HCNR-0201 from Agilent Technology.
Data sheet gives IF = 1 ~ 20 mA, choose IF = 2 mA, we have
R3 = (VCC - VF)/IF = 5/2 = 2.5 K
Data sheet gives
IPD1 = K1 * IF = 0.5 * 2 = 1 mA
so
R1 = VIN / IPD1 = 5/1 = 5 K
as VIN is an analog signal out of a 5-v DAC (Digital Analog Converter)
Data sheet gives
IPD2 = K3 * IPD1 = 1 * 1 = 1 mA
so
R2 = VCC / IPD2 = 5/1 = 5 K
C1 is for C1R1 to meet Fmax
C2 is for C2R2 to meet Fmax