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I Am designing a circuit to detect the presence of AC line voltage.It is basically to detect the switch On/OFF for a load. Here is the circuit. enter image description here

Here Input resistor ladder to step down the voltage. Varistor added to get it away from surge (Fuse included). zener to protect the opto. Opto is darlington with transistor to reduce the opto input current and therefor the input resistor losses.

Since this going to commercial product i would like to gather some info on reliability and surge protection.

1) Is this circuit good enough to handle potential surges/Back emf from inductive loads. 2) Is there any better/cost effective alternatives. 4) I am using 8 in my board to detect 8 channels all are powered from same AC source. Is there a way to consolidate these and reduce overall cost?

Thanks in advance!

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  • \$\begingroup\$ In my opinion it's too much complicated. I would eliminate varistors, caps, zener , R29 and R26. Instead of zener I would add a diode. Or maybe better if you use an optocoupler with two internal LEDs. \$\endgroup\$ – Marko Buršič Sep 20 at 11:41
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This is simplified version of your schematics, that should work equal.

schematic

simulate this circuit – Schematic created using CircuitLab

For example SFH628A-4 has a CTR ratio of 250 at 0.5mA IF current. The absolute maximum allowed forward current is 50mA. The insulation resistance is 5.3kV. So at 5.3kV spike, the current trough diode would be 5300/440= 12mA, which is far from absoulte rating.

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  • \$\begingroup\$ It will work. How about the high voltage/Surge? \$\endgroup\$ – Electron Sep 20 at 12:01
  • \$\begingroup\$ And resistor is rated to 200V. So is it not a good idea to connect resistors in series to get more bandwidth like energy meter? \$\endgroup\$ – Electron Sep 20 at 12:02
  • \$\begingroup\$ Of course, you can make a string of resistors if you want. The surge won't burn your led easily, you can calculate Vsurge/Rin to get the current. 5kV/440kohm= 11.mA that current is easily handled by LED. If the current of 0.5mA through LED is enough to detect on your MCU, that's another story. \$\endgroup\$ – Marko Buršič Sep 20 at 12:08
  • \$\begingroup\$ Yes I am using this circuit and its working fine. Since the Darlington coupled transistor gives out put atleast 2/3 of total waveform \$\endgroup\$ – Electron Sep 20 at 12:31
  • \$\begingroup\$ The panasonic ERJUP8 SMD resistor 1206 accepts max 1kV spike. \$\endgroup\$ – Marko Buršič Sep 20 at 12:37

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