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why would something like that happen? all the relay has to do is to close its contacts, what difference would the load do?

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  • \$\begingroup\$ What powers the load? \$\endgroup\$ – jonk Dec 24 '16 at 0:32
  • \$\begingroup\$ @jonk AC mains 220v ac \$\endgroup\$ – iMohaned Dec 24 '16 at 0:46
  • \$\begingroup\$ You might want to define "doesn't work" with something more descriptive. Assuming there is no galvanic connection, I'd also be curious about what your AC powered load actually is. And also curious about the current required by your 12 V relay (though I don't expect that to be more than \$180\:\textrm{mA}\$.) \$\endgroup\$ – jonk Dec 24 '16 at 1:00
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    \$\begingroup\$ Consider redrawing your schematic diagram. The section driving the relay is especially problematic. It is difficult to read, with the battery at the bottom. A first glance says "the diode is upside down" followed by "what's driving the relay" followed by " oh, geeze, its all connected right but what mess." \$\endgroup\$ – JRE Dec 24 '16 at 10:15
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In theory it shouldn't, but perhaps your relay drive circuit is operating at the margins and is being thrown off a bit by interference of some sort when the load is present.

Looking at things, you have a PIC micro, which if the IO pin is putting 5V into R3, delivers 500 microamps into the base of Q2. The DC gain of a 2N2222, let's optimistically call it 100, would yield 50mA current through the collector. LED2, powered by the 12V battery, would want nearly 50mA of current by itself ((12V Battery - 1.8V LED voltage drop) / 220 ohms = ~48mA). Perhaps the relay is only barely getting enough juice in this setup?

Try change R3 to something lower, like 1K or something, and increase R5 to at least 1K if not more. Or even take out LED2 entirely to see what happens. And check the specs on the relay coil - how much current does it need to operate? You could just test it with the battery and a meter, too.

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  • \$\begingroup\$ I agree about \$R_3\$. I'd anticipate that the total collector load on the BJT is around \$200\:\textrm{mA}\$ from what I see there and guessing about the relay. That's a lot to expect from \$400\:\mu\textrm{A}\$ base drive current. \$\endgroup\$ – jonk Dec 24 '16 at 1:03

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