i am begginer in electronics and this is my first circuit its about a small motion sensor 220v AC switch i want to make myself but i want to use Triac as a solid state relay insteade of normal relays i searched alot about Triac circuits and i coudn't understand some of them. can i trigger the Triac gate using output power from motion sensor module HC-SR501 without damaging sensor module as shown in circuit diagram?? is it necessary to use optoisolator or can the circuit done using triac only?

Note: i am trying to make the circuit as simple as possible cause i want the final PCB the same size as the HC-SR501 motion sensor module PCB

Switch circuit schematic

Edit: using your tips and with some googling around i made that modified schematic, will this work. i don't want to test until i am sure i will not fry anything.

i used this opto-triac isolator in this modified schematic.

Modified Switch circuit schematic

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    \$\begingroup\$ A dangerous first project. Ignoring the many other issues, your first question seems to be whether you can use the sensor output to drive the BC547. You should know the base current required by the rest of your design, so read the datasheet for the sensor module. It shows that the BISS0001 IC is used by the module. Read the BISS0001 datasheet which tells you that the absolute maximum current the output pin can drive is 10mA. There is the answer to your first question. It is generally not a great idea to make a PCB layout if such questions are unanswered at the schematic level. \$\endgroup\$ – replete Dec 11 '17 at 1:54
  • \$\begingroup\$ i made the circuit using a normal 5v relay on a breadboard and it worked, this means sensor module output can drive BC547, to improve the size i wanted to replace the 5v relay with Triac but i don't have enought knowledge to adjust the circuit to it. \$\endgroup\$ – Tarek Helmy Dec 11 '17 at 2:13
  • \$\begingroup\$ This will end in smoke and flames if you power it up. If you use an optotriac optoisolator (MOCxxx) or a mechanical relay to drive the triac you can make it work. The triac may require a heatsink. \$\endgroup\$ – Spehro Pefhany Dec 11 '17 at 4:24
  • \$\begingroup\$ @SpehroPefhany will this be able to drive the triac lcsc.com/product-detail/SMD-Optocouplers_JC3H7C_C55287.html \$\endgroup\$ – Tarek Helmy Dec 12 '17 at 7:30
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    \$\begingroup\$ Not gonna put up with you thumbing your nose at us with such sloppiness. -1 for the disrespect, and closing as unclear since I stopped reading on the first line, so I don't know what is being asked. Go away until you learn some respect for the volunteers here you are asking for a favor. \$\endgroup\$ – Olin Lathrop Dec 12 '17 at 11:53

It's not the best idea to drive a triac this way. The “ground” potential in your schematic is never too far from the earth potential, while the potential of the hot terminal can swing all the way from −310 V to +310 V. What happens when it swings negative? Well, the negative potential is applied to the collector of your BC547 through the triac, while the base and emitter are sitting near earth/ground. Therefore, the base and emitter become positively biased relative to the collector. Remember, bipolar junction transistors can work even with their collector and emitter reversed (albeit with greatly degraded performance), so, best case scenario, the transistor will turn on and conduct from the emitter to the collector; worst case, the applied voltage will fry the IC.

  • \$\begingroup\$ What OP refers as earth seems to be neutral and ground symbol is his common negative. Still, it won't work. \$\endgroup\$ – winny Dec 11 '17 at 7:02

You may find the following schematic helpful. This is what a temperature-controlled CT2091 soldering iron looks like on the inside. Notice how terminal A1 of triac Q1 is connected to the ground of U2. This way dangerous potential differences can be avoided. Also notice how the gate drive to Q1 is provided through C2. This way the polarity of gate current always matches the polarity of terminal A2 relative to A1. This achieves two things: square waves at the output of U2 are transformed to short pulses, which is all a triac needs anyway, and the triac operates in the 1st and 3rd quadrants, where the gate is more sensitive.

enter image description here

  • \$\begingroup\$ thanks for your reply, but actually this is to complicated for me to understand, i didn't study anything in electronics my knowledege comes from searching online i only love to make my stuff myself at home \$\endgroup\$ – Tarek Helmy Dec 12 '17 at 7:28

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