Hi all
Given the 5V relay in the schematic I think I can use a 3.3v GPIO pin to control it. Right?
What it the minimum voltage to give it the HIGH state having Vcc=5V?
The schematic shows 2 leds and a 1k res in series.
It's a curiosity.
Thanks
https://www.elegoo.com/products/elegoo-8-channel-relay-module-kit

 

hi,
What is the purpose of using opto isolators, when the resistor and transistor would be suitable.?

E

 

hi,
What is the purpose of using opto isolators, when the resistor and transistor would be suitable.?

E

you should ask to Elegoo...
Btw there on that board there is a jumper that allows to completely separate the circuit of the input from the circuit of the relay.

 

hi LL,
I am asking you Why, not Elegoo

E

 

I have bought this standard relay module and now I am trying to use it with a 3.3V board.
It's buildt like this. Why do you ask me why it's built like this?
An octocoupler is meant to separate input from output. There is a jumper on the board to allow this possibility.
I could just use a power mosfet for the load and a mosfet to drive it.
Maybe there is a reason if billions of these modules were sold though...
I cannot understand what is the meaning of your questions and why you don't just address the original question...

 

hi,
This is a common way of driving a transistor/relay.
E

 

One reason to use an Opto is to allow an active low output to drive the relay.
And if the board had both ends of the opto input available, then it is easy to swap to an active high.
Another advantage is to keep relay switching noise out of the micro as the relay supply can be fully isolated from the micro.
Like, a 24V power supply for the relays and have the relays switching mains.
Often, in industrial control, the 24V supply can have many devices on it so full isolation is a very good idea.

And on the transistor driver shown above, it is always wise to add a B-E pull down resistor, maybe a 10K just to make sure the transistor is off.

 

Hi,
I only use an opto in that type of driver if ground isolation is required.
E

 

And on the transistor driver shown above, it is always wise to add a B-E pull down resistor, maybe a 10K just to make sure the transistor is off.

In the module mentioned above it's not possible though...
Is IRL510 ok for that purpose, right? Can I drive it directly with the GPIO pin?
btw I still have not received an answer to my original question...

 

Given the 5V relay in the schematic I think I can use a 3.3v GPIO pin to control it. Right?
Yes

 

btw I still have not received an answer to my original question...

You will probably have to short out the indicator LED as 3.3V will nor be enough voltage to drive 2 LEDs in series.

 

Using that relay module could I think of a solution like this one using a 2n2222a?
Thanks

 

No, not with the relay module input connected to +5V always, as in your circuit.

Just connect the output pin to the relay board input, it will probably work.

Bob

 

Hi all
Given the 5V relay in the schematic I think I can use a 3.3v GPIO pin to control it. Right?
What it the minimum voltage to give it the HIGH state having Vcc=5V?
The schematic shows 2 leds and a 1k res in series.
It's a curiosity.
Thanks
https://www.elegoo.com/products/elegoo-8-channel-relay-module-kit

It should work. As one person (dendad) said, short out the indicator LED, as voltage drop may be too large across 2 LEDs for 3.3V to trigger the opto.
VCC connects to your 3.3V VCC. JD-VCC should be connected to a 5V supply (if possible). Remove any VCC to JD-VCC jumper to do this. This assumes the relay needs that full 5V to trigger.

 

It should work. As one person (dendad) said, short out the indicator LED, as voltage drop may be too large across 2 LEDs for 3.3V to trigger the opto.
VCC connects to your 3.3V VCC. JD-VCC should be connected to a 5V supply (if possible). Remove any VCC to JD-VCC jumper to do this. This assumes the relay needs that full 5V to trigger.

Why do I have to separate the Vcc if I have same ground in my circuit?

 

Because one side is 3.3V and the other is 5V?

Bob

 

It should work. As one person (dendad) said, short out the indicator LED, as voltage drop may be too large across 2 LEDs for 3.3V to trigger the opto.
VCC connects to your 3.3V VCC. JD-VCC should be connected to a 5V supply (if possible). Remove any VCC to JD-VCC jumper to do this. This assumes the relay needs that full 5V to trigger.

I would like to know if the relay is active: that LED would be appreciated...
btw today I have tried the board and 3.3V GPIO pin can activate the relay and the LED.
photocoupler
relay

 

A 5VDC relay you mention says minimum pull in voltage is 75% of rated voltage. So, a 5V relay would require 3.75V to ensure pull in. Anything less may work, but maybe for a while. Then it may fail to work.
Don't mess with specifications. If the relay requires 5V, use 5V. If you need a LED indicator, maybe look at other options for indicators.
Otherwise, use a different circuit, one where a 3,3V level can trigger a relay directly via a driver like a transistor. You could also use a transistor driver to trigger that relay board's opto circuit if everything on the board is wired to +5V.
Final option is to use a level converter for 3.3V to 5V logic levels. There are several on places like Ebay or Amazon. The circuit is basically a MOSFET wired up to convert the voltage levels, and is very compact. They are often called I2C level converters, but are basically just logic level converters (but may be limited in current capacity, may not have enough current sinking to trigger opto):
https://www.ebay.com/itm/134086962144?hash=item1f3834abe0:g:3yYAAOSw5CddN5bm

 

Otherwise, use a different circuit, one where a 3,3V level can trigger a relay directly via a driver like a transistor. You could also use a transistor driver to trigger that relay board's opto circuit if everything on the board is wired to +5V.
Final option is to use a level converter for 3.3V to 5V logic levels. There are several on places like Ebay or Amazon. The circuit is basically a MOSFET wired up to convert the voltage levels, and is very compact. They are often called I2C level converters, but are basically just logic level converters (but may be limited in current capacity, may not have enough current sinking to trigger opto):
https://www.ebay.com/itm/134086962144?hash=item1f3834abe0:g:3yYAAOSw5CddN5bm

level_shifter
0.22A right? If that's the case it's enough.

 

You only need 10mA or less to trigger the opto on that relay board.