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I ran into a problem with my switch

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The CD4013 doesn't have 18mA for a relay. The CD4013 can drive a transistor that can drive a relay.
 
yes I know how to wire it with the transistor but I'm worried about the relay because it looks awful small and was making sure that it could handle the 120AC coming out of the wall thats powering my lights. I didn't want to mess up the whole circuit because of that one part. It says its rated for 2Amps but isn't the power in your house more like 6+ amps? I was just making sure this relay would work and could handle the AC load that it would have before I test it and blow something up haha. Thanks
 
psecody said:
yes I know how to wire it with the transistor but I'm worried about the relay because it looks awful small and was making sure that it could handle the 120AC coming out of the wall thats powering my lights. I didn't want to mess up the whole circuit because of that one part. It says its rated for 2Amps but isn't the power in your house more like 6+ amps? I was just making sure this relay would work and could handle the AC load that it would have before I test it and blow something up haha. Thanks
It depends on what you are switching.

For example, a 60 Watt light at 120 Volt would be a current of 60/120 = 0.5 amp.
 
hi pescody,

the question about 'available' current from a power source, driving a device that requires less current than the 'available' current is a very old one.

If the device is correctly rated say, 1amp to operate at say 120vac 50/60hZ, then you can it connect to any 120vac 50/60hZ power source, providing the power source can 'deliver' more than 1amp [as per the example]

Regarding driving incandescent lamps, remember the 'cold' resistance of
lamp will be quiet low. So the 'inrush' current during lamp wamp up time, which is usually quite short, can be 10 times the running current.

These inrush spikes can cause interference in your low level circuits, so
use good filtering.

Regards
EricG
 
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