Review of my "Optocoupler as Relay Board" circuit

Other Hardware General Electronics
21

dlloyd:
Replace 220R with 1K. No resistor needed from 12V to opto's collector.

Like that?

22

Looks better, but shouldn't 12V IN be renamed to LOAD or RELAY OUT or RLY OUT and 12V OUT be renamed to 12V IN or 12V PWR?

23

dlloyd:
Looks better, but shouldn't 12V IN be renamed to RELAY OUT or RLY OUT and 12V OUT be renamed to 12V IN?

12V in is the main 12V input, and 12V out is the load. I'm not going to use any relay.

24

Check your 12V IN connections ... no GND.
Check your subject and circuit title. OK, instead of RELAY OUT maybe LOAD OUT, SWITCHED OUT, OPTO OUT, etc.

25

dlloyd:
Check your 12V IN connections ... no GND.

I just don't understand...

Check your subject title. OK, instead of RELAY OUT maybe LOAD OUT, SWITCHED OUT, OPTO OUT, etc.

I agreed. I'm going to change that. Thanks :slight_smile:

26

Oh my, I didn't mean to change the title.

Just think of your 12V IN and 12V OUT connectors ... do they work as described?

A user might want to connect 12V power (12V and GND) to 12V IN. (there's no GND here)

Now, where would the user connect the LOAD? Where's the switched output?

27

I think i know what you mean now. The two outputs is marked wrong.

Here is an update.

28

Looks good. +1

29

dlloyd:
Looks good. +1

Thanks :slight_smile:
But should i add an diode for polarity protection in the circuit?

30

Sounds like a great idea.

31

While everything DLoyd said about the optocoupler current is true, it is for all practical purposes irrelevant in your case because the load placed on that optocoupler is a mosfet gate which is a voltage controlled gate and therefore current limited to the extent that it is limited to the current inrush of the gate capacitance charging. I don't honestly know what that is this case but I suspect it is << less than the opto CTR limited output current. (the gate shunt resistor draws about 1.2mA)

32

I meassured the current through optocoupler to be 11mA.

Thanks for clarifying :slight_smile:

33

The PC817 Vf = 1.2V and the led resistor was 1k in the last schematic you posted but if that were the case the
led current would be (5V-1.2V)/1000=0.0038A (3.8mA). If the CTR is 50% you should see half that through the collector but 11mA would be more typical of a led resistor of 150 ohms : (5V-1.2V)/150=0.0253A,
50% of 0.0253=0.0126A,
The 10k gate resistor draws 12V/10000=0.0012A.
I'm surprised the collector is passing 11mA with only 3.8mA going through the led with a 1k led resistor.
Did you replace the 220 ohm resistor or is it still there ?

34

You are right. I meassured this yesterday when the 150 ohm resistor was connected :slight_smile:

35

You are right. I meassured this yesterday when the 150 ohm resistor was connected

Hey , how about that math , eh ?
What I would be asking is "Is that a coincidence or would the current increase through the collector if the led resistor were smaller (we know it would decrease if the led resistor were larger) so the question is 'How much
current does the mosfet gate draw if it is a voltage controlled device and not a current controlled device ?
I'm guessing it would stay about the same if the led resistor were reduced to 100 ohms.

36

raschemmel:
Hey , how about that math , eh ?
What I would be asking is "Is that a coincidence or would the current increase through the collector if the led resistor were smaller (we know it would decrease if the led resistor were larger) so the question is 'How much
current does the mosfet gate draw if it is a voltage controlled device and not a current controlled device ?
I'm guessing it would stay about the same if the led resistor were reduced to 100 ohms.

hehe, Good work on the math mate :slight_smile:
So if i understand you right...
the current between collector and emitter is approximately the current between the anode and cathode X 1.5?

37

NEGATIVE
The CTR is the Current Transfer Ratio
Led current: Collector current is 50% so
if the Vf (forward voltage) of the led is 1.2V, that means the voltage drop across the led is 1.2V
which makes the led resistor voltage drop Vcc-led voltage = 5V - 1.2V = 3.8V and the led current = current through the 150 ohm resistor = 3.8V/150 = 0.0253A (25.3mA) and the collector current = CTR x led current = 0.5 x 25.3mA = 12.6mA, of which 1.2mA is the gate shunt resistor current =(in schematic in post #20) 12V/10,000 ohms = 0.0012A = 1.2mA.

39

Continuing the discussion from Review of my "Optocoupler as Relay Board" circuit:

In the schematic from the above post, we see gnd on the IC side and gnd2 on the relay side. In most arduino user situation those ground are electrically connected. Should we use a decoupling capacitor between them to better isolate the IC? For example if there is noisy device on the load side like a pump?

thank you