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I want to detect logic of the mains supply (230VAC, 50Hz) and feed the output of HCPL-3700 into ATmega16L. I read through the datasheet but could not understand how to calculate the resistor values. Does anyone have an idea of how to go about doing this?

EDIT 1

I am referring to application note on HPCL-3700, page 5 under AC Operation With No Filtering

Assuming 60% criterion, 230 x sqrt(2) x 60% is 195V. choosing closest whole value of 200V instead of 195V.

Rx = (V+ - VTH+) / ITH+ = (200V - 5.5V) / 2.5mA = 38.9 Kohms.

choosing round figure of 40 Kohms for Rx. I can use two 22 Kohm resistors.

Calculating power dissipation in 22 Kohm resistors
P = (I*I) * R
= (4.4mA * 4.4mA) * 22K (where 4.4mA is Iin)
= 426 mW

So, choosing 22 Kohm resistors with 500mW power rating.

How does this answer look? I think I'm wrong because there are few things that still don't make sense.

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  • \$\begingroup\$ What do you mean by "logic" of mains supply? \$\endgroup\$ – miceuz Jun 25 '12 at 6:25
  • \$\begingroup\$ It's not clear to me why you suddenly want to use a 4 times more expensive optocoupler, after all the trouble we took to calculate for the cheaper SFH620A. \$\endgroup\$ – stevenvh Jun 25 '12 at 10:01
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    \$\begingroup\$ I don't know where you got the 4.4 mA, but the power calculation is wrong. You have 230 V across 44k, that's 230\$^2\$/44000 = 1.2 W, so each of the 22k should be 0.6 W. I would take at least 0.75 W. \$\endgroup\$ – stevenvh Jun 25 '12 at 10:07
  • \$\begingroup\$ i know man, but my supervisor wants me to take a look at this optocoupler instead of the previous one. He says its 'better' because of it has a bridge rectifier. I am pretty pissed too \$\endgroup\$ – David Norman Jun 25 '12 at 10:08
  • \$\begingroup\$ Uh-oh, supervisors :-(. You could tell him that the other one doesn't need a bridge rectifier because it handles both halves of the phase by the two LEDs in antiparallel. Does the same thing for 2.50 dollar less. \$\endgroup\$ – stevenvh Jun 25 '12 at 14:08
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The datasheet should have been a lot more clear on this. The only relevant information on the output pull-up I could find was on page 3:

Logic low output voltage = 0.4 V maximum, at 4.2 mA
Logic high output current = 100 \$\mu\$A maximum

That last value must be the off-leakage current, but it's very high. It would mean that a 1 kΩ pull-up resistor would drop 100 mV, a 10 kΩ would give an unacceptable 1 V drop. I would stick to 1 kΩ.

For the input resistors you need figure 8 of the datasheet.

enter image description here

This shows the threshold values for the input voltage hysteresis as a function of series resistance. For instance, at 40 kΩ the levels are 50 V and 100 V. That means the output goes low if 100 V is exceeded, and high again if the input voltage goes below 50 V. At 230 V AC the output will be low between 18° and 171° of a half cycle, that's a 85 % duty cycle.

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