Mega2560 on less than 7V, shared with SSR relays

Hello,

I see 7V is the recommended minimum for powering these boards through the small barrel plug. But I would also like to power ssr’s using the same power supply if possible. These are the standard Opto-22 thin ssr’s as I think they are known as, or specifically 1781-OB5S modules from western reserve controls. The logic input on these is 5V, and I have been able to dig and find the maximum is 7V. I dont want to overpower the relays too much, so does powering both the boards and the relays using a shared 6V power supply seem reasonable? Is there really an issue dropping below the 7V recommended for the Mega? The board will be running about 16 ssrs and thats it. Or the alternate is to just use two power supplies at 5V and 7.5V for the ssrs and mega.

Thanks,
Scott

The minimum voltage into the barrel connector to get 5V out of the regulator is 6.8V

The outputs of the Arduino are 5V (or less if you don’t provide enough voltage).
The input of your SSR is 5V. The SSR can switch down to 3V (and up to 60V)at 3A.

I don’t see what the problem is. What do you mean “power the SSR’s”? They don’t use a power supply. They use a 5V logic signal to switch 3-60V on an off.

If you've got clean 5VDC then you can skip the barrel jack and the regulator and just feed that directly to the Ardunio through the 5V pin.

WRC4 Series Discrete I/O Modules

From what I can see, the nominal input resistance is 240Ω, so the nominal current would be 5V - 1.2V (IRLED) / 240 = 15.8mA. With 16 SSRs, that's about 250mA when all SSRs are turned on. Yep, it's best to use just one 5V power supply for the Mega and skip the barrel jack as suggested by Delta_G.

You can get a step down regulator like these which will bring anything from 5.1V to 36V down to 5V. As previously stated, you can use the output of this regulator to power the Arduino directly and bypass the on board regulator. If power consumption is important to you, they should also be more efficient than the Arduino regulator.