hi 123,
I use the 4 and 2 relay version of that module.
The relays are operated via a opto coupler and transistor.
The opto coupler circuit requires a LOW [0v] level in order to operate the relay. [refer the circuit diagram on your link]
A 15millAmp thru 20milliAmp current enables the opto coupler.
If you were for example using a MCU/PIC, powered by 5v, the opto driving pin would have to 'sink' this 15/20mA current [ from a 5v external source] in order to operate the relay.
Sinking this current would not be a problem for most MCU/PICs powered with 5v.
However other users powering their MCU's from a 3v supply find the opto does not get sufficient current to enable the opto.
What is the application for the relay module.?
E
Good question. In most applications, it's probably useless. Relays and controlling processor will be typically supplied by the same 5V source. esp1 already pointed out that a lower driver voltage than 5V isn't suitable. In any case driver input and 5V relays supply are routed through adjacent pins of the input header, without utilizing the possible isolation strength or noise suppression capabilities of the opto coupler.What is the purpose of this?
24VDC @10Amps reads 10 x 100,000 0r 1,000,000
hi Gary,
Checking the d/s data plots, I would agree with your maths, 10^6 ops 24Vdc @10A ~ 2years.
Eric
I was able to find a schematic for this relay device (see below). I see it has a feature for the input control pin to be optically isolated from the relay control input.
What is the purpose of this? How could the input pin get damaged if the relay contacts themselves are already electrically isolated?
View attachment 137893
this is double isolation...
What really going on here is, the control is done with a low voltage, in this case it looks like a 5 volt signal from Vcc to IN0. The voltage for the relay coil could be higher (like 12V or 24V). This switches the higher voltage on the relay contacts.
you have two LED drops, and depending on your opto coupler (U1) the current could be much lower than 20 ma, the value 20 ma was probably from the external LED indicator. typically people ran LEDs at 20 ma, The modern ones, I use in test fixtures, I have to run a 5 ma, or people complain they are too bright.
Also with the above circuit, I would move the relay coil (with diode) to the emitter side of the transistor and the other end to ground (emitter follower), this way you tie the two collectors (U1-4 and Q1) to the to the JDD-Vcc (not sure why it's called that) and lose R2, connect U1-3 directly to the base of the transistor.
When they talk of voltage hysteresis:
It is not desirable to operate this driver circuit with analog voltages, use logic levels. A small resistance (couple of 100 ohms) in series with the base is a good habit. You can also use a capacitor from the base to the ground to prevent chatter.
You may use a Schmitt trigger comparator in the input side if you want some hysteresis.
A relay has some hysteresis build in. It takes more current to energies the relay then it does to hold it close.
yes a base resistor is a good idea, but with a emitter follower, the load will limit base current.
It is not desirable to operate this driver circuit with analog voltages, use logic levels. A small resistance (couple of 100 ohms) in series with the base is a good habit. You can also use a capacitor from the base to the ground to prevent chatter.
You may use a Schmitt trigger comparator in the input side if you want some hysteresis.
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