An IR receiver is tuned to a modulated IR signal. It doesn't do anything with a steady IR beam.
An IR receiver is tuned to a modulated IR signal. It doesn't do anything with a steady IR beam.
Uncle $crooge
oh ok thanks. See I thought it wouldn't work but couldn't tell you why, I guess because it seemed to simple haha.
ok now that I've got the push button part working I was going to work on the relay part because I won't get to go get more IR stuff till thuresday so I'm stuck for now on that but I have the relay stuff and wanted to make sure that this relay will work. The ratings on it are:
Coil voltage: 9VDC
Coil Resistance: 500 Ohms
Nominal Current: 18mA
Contact Rating: 2A at 125VAC
I''m not sure this will work in fact the reason I'm asking is because I think the leads on it look to small to use on ac so I have a slight feeling that this would just melt but I wanted to make sure just in the off chance it would work.
I bought another relay before this one but I don't think it can operate on just 9vdc. Its specs are
continuous coil voltage: 132VAC
rated voltage: 110/120VAC
Coil resistance: 4.430 ohms
pull in voltage: 9.6VAC max
dropout voltage: 36VAC min
nominal coil current: 8.4/9.2mA
contact rating 10A at 110VAC or 24VDC
maximum operating voltage: 250VAC/125VDC
minimum load: 100mA 5VDC
I know the second one is way more robust and I believe it could take it but its got a clear case on it and just out of curiosity I hooked a 9VDC up to the coils and nothing happened(the switches didn't move) so I believe it takes more than 9v to throw the switch, am I thinking right?
Basically I was going to put a relay on notQ so that when the relay is off the led is on to show that its in standby then when the led is off the relay would be thrown.
But the contact rating is 2A @ 125 VAC so it should be OK unless you want to switch more current than 2A or a higher voltage.Originally Posted by psecody
This does not make sense. The operate voltage (they call it "pull in") is less than the release (dropout) voltage.Originally Posted by psecody
The operate voltage of a relay is always greater than the release voltage.
It is designed for AC so it is not suitable for your application.
It is designed for AC so it is not suitable for your application.Originally Posted by psecody
This sounds right. But the correct terminology for relays (I spent many years designing complex relay circuits) is - a relay is either operated or released.Originally Posted by psecody
Len
Your circuit needs a transistor to drive the 9V relay. The relay coil needs 18ma which can't be supplied by a CD4013 IC.
Uncle $crooge
ok but ya'll think the first relay would be ok to just turn my lights (ac) on and off with?
The CD4013 doesn't have 18mA for a relay. The CD4013 can drive a transistor that can drive a relay.
Uncle $crooge
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.Originally Posted by psecody
For example, a 60 Watt light at 120 Volt would be a current of 60/120 = 0.5 amp.
Len
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