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I am new to Power Electronics, I am working on to AC line detection using H11AA1 IC. Schematic is as shown hereenter image description here. With the following calculation, I figured out the resistor value R1=8K R1= ( Vs - V_led)/(10mA) which is in my case R1=( 230 - 1.5)/(10mA) =22.8K; So I added R1 20 K/1Watt but I could not achieve the result; Has anyone work with H11AA1 IC for the similar purpose, Do I need to change R1 value?? DataSheet of H11AA1 here vishay.com/docs/83608/h11aa1.pdf

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  • \$\begingroup\$ What voltage is Vcc? What do you measure on pin 5 with an oscilloscope? What’s the purpose of C1? \$\endgroup\$ – winny Nov 27 '18 at 8:41
  • \$\begingroup\$ @winny here VCC is 5V, at pin 5 using voltameter I observed 1.7v., but when I provide AC supply at the input side of H11AA1 it should change, but it not showing any changes. \$\endgroup\$ – dss2309 Nov 27 '18 at 8:55
  • \$\begingroup\$ You are going to need an oscilloscope to see that waveform. Why is C1 there and what value does it have? \$\endgroup\$ – winny Nov 27 '18 at 10:57
  • \$\begingroup\$ I added that cap (of 4.7uF) to stabilize the output if any noise occurs. regarding oscilloscope, I don't have one. \$\endgroup\$ – dss2309 Nov 27 '18 at 12:15
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I did the same schematic with R1=330K 1/2w. and C1=10uF. The AC current is 0.737 mAAC.(got from amp meter). So, the power dissipation is 230 x 0.737 = 169.51 mW( approximately). That's ok with R1 1/2w.

If we don't have C1, this will be a zero crossing circuit. If we have C1, the voltage on pin5 is almost 0 when the AC line still alive and = Vcc when the AC line was gone.

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  • \$\begingroup\$ Thanks for the answer, Would you explain to me the R1 calculation? \$\endgroup\$ – dss2309 Mar 24 at 6:50
  • \$\begingroup\$ Hi, sorry for late reply. I was decided to try an error for the R1 with 3 values (100K, 220K, and 330K). Just need to minimize the AC current as much as possible that H11AA1 still working as normal. By using the amp meter without calculation. :-) \$\endgroup\$ – Pathman Chiewchitboon Apr 9 at 13:23
  • \$\begingroup\$ Okay, So you mean for AC circuitry like above one should work for minimal current but within the operating conditions! Thank you \$\endgroup\$ – dss2309 Apr 10 at 1:02
  • \$\begingroup\$ Definitely.! It's still in the operating condition. \$\endgroup\$ – Pathman Chiewchitboon Apr 10 at 14:14
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Update: OP's original schematic showed a single LED in the opto-coupler. It now shows either a reverse protection diode or LED.


You have no reverse polarity protection on the opto-coupler LED. These can usually withstand about 5 - 20 V reverse bias so it was probably destroyed on the first negative half-cycle.

Replace the H11AA1 and add in a reverse protection diode across the LED.

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  • \$\begingroup\$ vishay.com/docs/83608/h11aa1.pdf \$\endgroup\$ – replete Nov 27 '18 at 8:19
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    \$\begingroup\$ In the datasheet, it is given as AC or polarity insensitive input, with that assumption I did not add the diode \$\endgroup\$ – dss2309 Nov 27 '18 at 8:21

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