0
\$\begingroup\$

Excuse the crude ms paint drawing.enter image description here

The above is based on an actual relay whose coil has a resistance of 320 ohms. The common to normally closed contact has a resistance of near zero. When the switch is open, no current goes through either of the parallel paths. However, when the switch is closed (this is where i get confused (the rest is my guessing)), first current would flow through the coil. This would thus close the contacts. However when this happens wouldn't the current (by Kirchhoff rule) go only through the contact and bypass the coil. This would thus open the internal contact and current would be forced to flow through the coil again. Then the current by kirchoff would go through the contact. Thus the process would repeat.

What actually happens? Does it flip on and off extremely fast or does something else happen? In real life it does not flip on and off and rather current flows normally.

\$\endgroup\$
3
  • \$\begingroup\$ If the intended function is to have the relay switch on/off repeatedly then maybe you haven't supplied enough voltage to "trigger" the relay. BTW your circuit diagram is awful and makes no sense to me. Is it meant to be a reed relay? \$\endgroup\$
    – Andy aka
    Commented May 27, 2013 at 10:31
  • 1
    \$\begingroup\$ @user1950278 We have a schematic editor built into the site. Please consider replacing the MS-Paint drawing with a schematic. \$\endgroup\$ Commented May 27, 2013 at 15:25
  • \$\begingroup\$ Please draw the relay coil and contacts in such a way that we can see how you have actually connected them, and the switch and battery. \$\endgroup\$ Commented Sep 30, 2020 at 15:56

1 Answer 1

4
\$\begingroup\$

The idea you describe can work, but it is very crude to remove the power from the coil by shorting the battery. At the very least you should insert a resistor to limit the current somewhat, as is you rely on the internal resistance of the battery, which can be OK for a 9V radio battery, but disastrous for 12V car battery.

The more common way to achieve roughly the same is to use a 'relay' with a Normally Closed (NC) contact, so you can open the circuit instead of shorting the battery. This is in fact how the common electro-mechanical doorbell works.

\$\endgroup\$
8
  • \$\begingroup\$ I am using a 12V power supply not a battery, so there is very little resistance. My question is what happens to the relay? Does it flip on and off continuously? \$\endgroup\$ Commented May 27, 2013 at 16:18
  • \$\begingroup\$ That depends to a large part on your supply. You are shorting the supply. Some supplies respond to that by cutting down the power for a certain time, or maybe until they are reset. Try it with a 100 Ohm resistor in series with the supply and the relay will probably vibrate (untill it breaks down, because it was not designed for this type of operation). \$\endgroup\$ Commented May 27, 2013 at 16:31
  • \$\begingroup\$ So what i should do is have a 320 ohm resistor in parallel with the coil of the relay. \$\endgroup\$ Commented May 27, 2013 at 17:32
  • \$\begingroup\$ I said 'in series with the power supply' and you interpret that as 'in parallel with the coil'???? \$\endgroup\$ Commented May 27, 2013 at 17:41
  • \$\begingroup\$ i was responding to the comment because you said putting a resistor in series with the supply would break down the relay. what i was thinking was: If i put a resistor in series with the contact, but parallel to the coil, then both resistance would be the same and the power supply wont be shorted. Would this solve the issue of the "vibrating relay" and the shorting of the power supply? \$\endgroup\$ Commented May 27, 2013 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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