I am a novice in the field of electronics, and I would like to get the community's recommendation regarding the design of a breakout board for a PCB solid state relay.
I got this PCB solid state relay as it suits the requirement(control input: 5VDC , load: 5VDC <5A) for my arduino project.
I am assuming it is a "plug&play" SSR similar to a panel-mount unit. Is this assumption correct? Or should I consider adding components(fuse, diodes) to make it safe and reliable just like a panel-mount solid state relay?
Since it will be soldered on a stripboard, I am planning to add 2 indicator LEDs, one on the input side and the other on the load side. What are the factors to be considered when choosing the LED?
Lastly, should the LED be on the positive terminal or negative terminal? Below are 2 diagrams to illustrate my question.
Im can't recommend a resistor because I know nothing about your circuit, e.g. voltage, what LED you are going to use. Even then, there are LED calculators online to do the math for you.
@missdrew alright, I will have a look at LED calculators online. But I will be using 5VDC for input signal and the load will be 5VDC as well. I guess I will have to use 5V LEDs.
How do you suggest I should connect the LED for the output side? The load I am connecting is approximately 1A.
The second diagram on that page shows how to connect it - the input already has a resistor which is why it claims to operate from 3 to 32V directly on the input in that diagram - for more details try searching "FCGT-P5DD60 datasheet" in your favorite search engine.
The reason it won't work for the output is because an SSR is an electronic switch is SERIES with the HOT (L) (line) lead, so the way you have the diode would be the same as putting it across a closed switch (direct short)
In order to add an indicator across the output it would need to be across L & N.
Also, the resistor value would be a function of the line voltage (peak Voltage, not RMS).
ie:
Let V(AC) = 230VAC
Let I(led) = 20mA (led current rating @V(forward)
V(peak) = 230SQRT(2) = 2301.4142 = 325V.
You would need to know the forward voltage of the led to calculate the resistor value but
if we assume V(forward) = 2.2V,
.'. R(current limiting) = (Vpeak-Vforward)/I(led)
= (325.6-2.2)/0.020A
= 323.4/0.020
= 16171 Ohms = 16.1k Ohms
P(resistor) = 323.4*0.02 (P = V * I ) (power = voltage drop times current)
= 6.48W
Next standard value would probably 10W
Led current limiting resistor for 230vac = 16.1k/10W
@MarkT I have tried searching for the datasheet online, but there is none. I even contacted the retailer for a datasheet with internal schematic, even they do not have one.
Post a vendor link for the SSR.
I can't find the datasheet.
Can't tell what the specs are or if it is an AC or DC SSR
WHY do you need an SSR for a 5Vdc load ?
Solid state relays are used for AC loads, although DC versions are available but nobody uses
solid state relays for a 5vdc load. If the load is 5V@5A, you would use a MOSFET,
like FQP30N06L
Since they're motors and not solenoids or relays, you probably don't need the diodes.
Yes, you are right I can just use a MOSFET or I guess an octocoupler with some addons if I want to isolate the two circuits. It is very hard to source specific IC components(mosfet, octocoupler, etc) locally, hence it was much easier for me to grab a dc-dc solid state relay.
@raschemmel I reworked the schematic following the guideline you offered in post #8, to add the indicator across positive and negative terminal for the output side.
