1
\$\begingroup\$

i've just done this tutorial: http://www.arduino.cc/es_old/Tutoriales/Rele

it works great! i've done it in a breadboard (with two 5v circuits for test) and then soldered (for just does not have 220v on a breadboard) and it works well for 1 light.

now i need more lights and i thought to make a relè for switch on/off a multiple adapter.

the thing i want to know: the pin of the relay are thin (just all other 5v-breadboard-based component) and i thought that more volts you have more bigger have to be the wire.. how many normal 220v ac light bulb can i control with one relay?

\$\endgroup\$
  • 1
    \$\begingroup\$ Can you give us the relay's specifications, preferably a datasheet? \$\endgroup\$ – stevenvh Aug 25 '12 at 17:46
5
\$\begingroup\$

The wire/pin thickness is not relevant to voltage, but to current. But even a 16 A relay will have thin pins, so that shouldn't be a problem.

What is something to take into account is that incandescent lamps cause a high inrush current, due to the fact that their resistance is low when they're cold. That inrush current may be as high as 10 \$\times\$ the nominal value. For instance, a 60 W bulb will draw 60 W/220 V \$\approx\$ 250 mA. Inrush current will be 2.5 A, then a 10 A relay will only switch four(*) 60 W bulbs, instead of the forty one might expect.


(*) Strictly speaking even only three, if you take a 10 % tolerance on the mains voltage into account

\$\endgroup\$
  • \$\begingroup\$ ok thanks. what can happen fif i plug more? and so what is the best way to do for more? use more relay? \$\endgroup\$ – nkint Oct 2 '12 at 18:49
  • \$\begingroup\$ @nkint - If you overload the relay the contacts will weld together, and it won't switch off anymore. More relays is a solution, or a special "high inrush current" relay. We had one type with custom contact material at work, but since it was custom that's not commercially available. But for instance the Schrack RT1 Inrush Power relay is specified at 165 A inrush for 20 ms. \$\endgroup\$ – stevenvh Oct 2 '12 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.