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Relay Driver Current

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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
 
esp1 said:
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
Click to expand...

Hope the original poster doesnt mind me joining the conversation.
Cheers, @danner123.

I also have these relays: Relay Board Module Optocoupler LED for Arduino PiC ARM AVR
Relay Data Sheet: **broken link removed**
Note: SRD-05VDC-SL-C relay ends with "C" for Form C.
Making these "Ebay Board Module" relays Contact Ratings actually: 7 Amp Resistive or 3 Amps Inductive at 28V DC.

We have been discussing this relay elsewhere:

Quote: The minimum "Electrical Life Expectancy" is specified at 10^5 operations at "nominal coil voltage" (sic). I presume that's an error for nominal contact current, but perhaps they're implying that the drive (and thus the switching speed) must be as specified. However, much the same lifetime seems to be indicated in the adjacent graphs. If the relay operates once per minute, then 10^5 corresponds to just under 3 months of operation.

PERFORMANCE (at initial value) what that is???
contact resistance 100 milliohms

Keen to here how you interpreter the Data Sheet specifications.

Cheers,
Gary
 

hi Gary,
The way I read it, the Vertical Log Ops scale, each division is 10^5 Ops , at the contact loading along the Horizontal axis.

Eric
 

Hi Eric,

24VDC @10Amps reads 10 x 100,000 0r 1,000,000
24x60=1440 minutes a day.
694 Days at 1 operation per minute.

Not sure what the other fellow I was chatting with meant with 3 months operation?
Have I missed anything.

Thanks,
Gary
 

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?

Sainsmart Relay.png
 
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What is the purpose of this?
Click to expand...
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.

But as a module buyer, just take it as is.
 
esp1,

The application for the relay module is to switch in AC voltages from values of 105VAC-215VAC. The max load is 3A. As FvM mentioned, we don't really think the opto coupler has any use in this. Do you know what is the main reason for it?

Thanks!
 

Constructer said:
24VDC @10Amps reads 10 x 100,000 0r 1,000,000
Click to expand...

How is that? Have I forgotten how to read a graph?

Are you talking about the graphs in Relay Data Sheet: **broken link removed**

10% charged!!!!
 

hi danner,
Checking the 0v copper track on a 2 & 4 relay version of this module, you can see that Gnd of the JD-Vcc, Vcc select pins is directly connected to the Gnd pin of IN1 & IN2 select pins.
This means the Gnd's are connected whatever JD-Vcc or Vcc you select, which means the Signal Input circuitry cannot be isolated from the Relay drive transistor Emitter. [Refer your posted circuit]

The Relay contact set is isolated from the relay driver circuit.

E
 

esp1 said:
hi Gary,
Checking the d/s data plots, I would agree with your maths, 10^6 ops 24Vdc @10A ~ 2years.

Eric
Click to expand...

It would seem I am being singled out about this data, perhaps that is what I missed.
 

danner123 said:
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
Click to expand...

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.
 
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RobertFalbo said:
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.
Click to expand...

View attachment 137893[/QUOTE]

Recently reintroduced my self to Basic with purchase of Picaxe IC.
Everything I wanted and could not afford in the eighties, and a lot more.

I purchased a few of the 12 channel relays and other sizes to cover any module size requirements.
As a general switch, it would be advantageous to change connections for a different voltage.

If I understand correctly, the relay is ok as long as operating voltage is within specification.
The base needs direct connection to drive relay in this way.

When they talk of voltage hysteresis:
I understand from a delay coming into equilibrium and fluctuation, but this relates to other science.
Is there any issue with the electrical phenomenon in this application?
 

Constructer said:
When they talk of voltage hysteresis:
Click to expand...

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.
 

c_mitra said:
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.
Click to expand...

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.
 

RobertFalbo said:
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.
Click to expand...

Not here, but hysteresis can be the problem and solution.
As in the Johnson Controls thermostat on the Cool Room.
Tuning off several degrees lower then on.

I read that hysteresis can be a problem with emitter follower, it could be switching off where I remember a problem circuit!
It seems logical that the relay might only take what it needed, or are we protecting the Transistor?

c_mitra said:
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.
Click to expand...

I am not interested in protecting from hysteresis, more just a question.
35 years ago I designed irrigation control systems to be used on farm,
the electronic work was just a requirement to an ends.
Using Weatstone bridges then AC to measure resistance between probes,
later embedded in gypsum to buffer external salts.
A lot of Hysteresis before settling.
Cheers on use of resistor
 
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hi,
A cautionary note.
There is 3Vrelay coil version of this relay module.
Users have found that the module still requires a 5V voltage input for the resistor/opto-emitter and signal indicator LED, a 3V supply will not work.
I understand the sales literature has been modified to warn buyers/users.

A quick fix to enable these 3V relay modules, is to tack a lead out wire from the junction of the Base resistor and the Opto emitter output.
Pulling this lead out wire upto +3V will switch On the relay drive transistor.

As the opto coupler and common 0V copper track on the PCB is already not isolated, switching the transistor in this way will not further degrade isolation.

E
 
Re: 5V 1/2/4/6/8 Channel Relay Board Module Optocoupler LED for Arduino PiC ARM AVR

I am planing to use Arduino Micro running from 5v Regulator to operate the optocoupled relay, only one at a time!

I am using a 5v Reference IC for Arduino Aref.
The reference IC needs 8v minimum! (Using 9v)

How can I rewire the optocouple relay board so relays coils run from 9v.
 
Last edited:

Makes only sense if you replace the relays by 9V type. Otherwise you need to drop down the voltage with series resistor, same overall power consumption as if supplying the relays by 5V linear regulator. Or use 9 to 5V switched mode regulator.
 

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