The reason I am writing is I was curious about hooking up a higher external power supply, Can I hook up a 120v to the 4channel 5v relay module? or does it have to be 5v. I am trying to only use my arduino for i/0 not power. I am also looking at powering high wattage devices, over 1000w.
Electronic – Relay power supply
relay
Related Solutions
Assuming your supply isn't collapsing somehow, to me it's most likely that the CTR of the opto isn't sufficient to saturate Q1 and meet the pull-in voltage requirement of the relay:
- The typical forward drop of an optoisolator photodiode is on the order of 1.5V
- There's another LED in series with the path, dropping at least another 0.7V
The photodiode current is going to be on the order of:
\$I_D = \dfrac{5V - 1.5V - 0.7V}{1k\Omega} = 2.8mA\$
The vast majority of optoisolators that I know of only guarantee their CTR at 5mA current or higher. Even a high gain opto (100-300% CTR) will underperform at this current level. It's quite possible that many optos in this application will tend towards higher CTR and work without circuit modification. Also, some Q1s may have much higher \$H_{FE}\$ and handle the weak drive.
I would consider soldering another 1k in parallel with the existing R1 and see if the relay performance improves. Most optoisolators can handle up to 50mA photodiode current; that being said, the diode current should be set to the lowest possible current that allows your application to operate robustly, since optoisolators do age (the CTR degrades over time: faster as the photodiode current increases).
tl;dr 9V or 5V adapter, Arduino powers a 5V latching relay
If the load isn't going to be switched very frequently, you should go with a latching relay. Actually, the only proper use for a non latching relay is in a device that switches to one state (on or off) occasionally, for short periods of time. If the device isn't switched more frequently than once in a couple of seconds, go with a latching relay.
If you use a single-winding latching relay, you can connect its coil to 2 digital outputs and set them to input-high-impedance when the relay is in a stable state. When you want the relay to switch over to one position, you pulse one of its coil pins with high, and the other with low for several milliseconds (look at the datasheet). If you want the relay to switch the other way, you do the opposite - pulse the other one high, and the first one low.
The actuation voltage has nothing to do with the voltage or power that goes through a relay. There are no advantages to using a 12V relay, and you'd need an external driver for it (or at least a transistor / darlington pair). The solution I proposed above, with a 5V latching relay, is optimal. Just put two diodes in reverse parallel across the coil to prevent collapsing-magnetic-field-induced voltage spikes from damaging your Arduino.
Best Answer
There are two sides on a relay, the coil side for activating the relay, and the contact side, for switching.
Your relay is a 5 V relay, that's the activating side. If you look at the (messy) datasheet you'll see that it has a resistance of 55 Ω, for a 450 mW power. Some of that power is used to create the magnetic field needed to close the contact, but most is dissipated in the coil's resistance as heat. Power can be calculated as
\$ P = \dfrac{V^2}{R} \$
and at 5 V that's indeed 450 mW. If you would power it with 120 V then
\$ P = \dfrac{(120 V)^2}{55 \Omega} = 260 W\$
will most likely will make the relay explode. The datasheet says the maximum voltage is 110 % of the rated voltage, so that's 5.5 V maximum.
That activation power will switch any voltage and current as long as they're below the rated values. You don't need a higher activation voltage to switch 10 A than for 1 A, or for 120 V instead of 12 V.
The contacts are rated at 10 A, for DC a maximum voltage of 30 V is given for the form A contact (SPST). This is a form C contact (SPDT) however, and this doesn't specify a DC rating for a resistive load. For AC it's 240 V. (JeeShen's answer is not correct here.) But power is maximum 240 W, so at 120 V AC you can only switch a resistive load of 2 A.