Hello, i've got a newbie question, i hope some of you more experienced guys can give me an answer. I've recreated a simple circuit, that closes relay contacts using optocoupler and transistor, just for fun and i've found out that it's not working, after some time i've succeded to get the circuit working as it should, but i still don't understand why it didn't worked the first time. I posted 2 pictures of my circuit (please forgive me for the ugly looking circuit, but i hope you will get it). The circuit in the first picture, named "This circuit doesn't work" is my non working circuit, and the other schemtic is my working circuit. I want to know more about electronics, and i am just confused, why the circuit in the first schematic doesn't work, and in the other one it is working just fine? Can somebody please tell me what am i doing wrong and point me to the topic i need to research to get more knowledge in this area. If i've made some spelling mistakes, forgive me i am not a native speaker. Thanks, have a great day

 

Does the opto supply share ground with the output circuit in the one that doesn't work?

 

Does the opto share ground with the output circuit in the one that doesn't work?

No, the GND of the optocoupler is independant of the other side of the circuit.

 

Well there is your answer, basic electronic theory...

Current can only flow in a complete circuit.

 

The current for the relay coil is VERY important but is not shown. Also the relay coil must have a diode parallel with it to absorb the flyback high voltage produced when the transistor turns off.

The circuit with the 10k resistor produces a base current of about 0.41mA in the C1815 transistor.
The circuit with the 40k resistor produces a base current of about only 0.28mA which might not be enough (the datasheet for most little transistors say that the base current should be 1/10th the collector current when a transistor is used as a switch.

The datasheet for the PC817x optocoupler shows that without a letter (the x I show) example PC817x, then the sensitivity of the coupling has a wide range. so your coupler might be very sensitive or has a poor sensitivity.

 

The relay is already a coupler so why use an optocoupler plus two power supply voltages??

 

Hello,

Optocouplers are frequently used to decouple grounds, so it possible and normal that the ground at the input stage and the output stage are different. In this kind of circuits, they are used to avoid that the 5Volt power supply is damaged by the relay currents. For that reason, it is also important to install the freewheeling diode in parallel with the relay.

 

Well there is your answer, basic electronic theory...

Current can only flow in a complete circuit.

Well there is your answer, basic electronic theory...

Current can only flow in a complete circuit.

So the base of the transistor Q2 and the emiter need to be from delivered from one supply. Can i make the voltage divider and close a circuit with a resistor to a V1 power supply.

 

The current for the relay coil is VERY important but is not shown. Also the relay coil must have a diode parallel with it to absorb the flyback high voltage produced when the transistor turns off.

The circuit with the 10k resistor produces a base current of about 0.41mA in the C1815 transistor.
The circuit with the 40k resistor produces a base current of about only 0.28mA which might not be enough (the datasheet for most little transistors say that the base current should be 1/10th the collector current when a transistor is used as a switch.

The datasheet for the PC817x optocoupler shows that without a letter (the x I show) example PC817x, then the sensitivity of the coupling has a wide range. so your coupler might be very sensitive or has a poor sensitivity.

I attached the flyback diode but forgot to put it in this schematic, i know i can use only a relay but i i recreated this circuit just for fun, thank you for all the answer. I will research this more.

 

The circuit with the 10k resistor produces a base current of about 0.41mA in the C1815 transistor.

Wrong...the circuit with the 10k resistor produces zero mA.

 

Hello,

Optocouplers are frequently used to decouple grounds, so it possible and normal that the ground at the input stage and the output stage are different. In this kind of circuits, they are used to avoid that the 5Volt power supply is damaged by the relay currents. For that reason, it is also important to install the freewheeling diode in parallel with the relay.

Thanks for the answer, but why didn't it work for the first time (when the GNDs are different)

 

Can i make the voltage divider and close a circuit with a resistor to a V1 power supply.

Why would you need to do that, just use the circuit that works.

 

Why would you need to do that, just use the circuit that works.

I know, but i just want to know why doesn't it work, i want to understand the concept. Btw, thanks for all the answers !

 

I already gave you the answer...

Current cannot flow from the positive terminal of the 5volt supply thru the transistor to the negative terminal because there is no return path.

It's the most basic electronic theory, it's the first thing you learn when learning electronics.

 

I already gave you the answer...

Current cannot flow from the positive terminal of the 5volt supply thru the transistor to the negative terminal because there is no return path.

It's the most basic electronic theory, it's the first thing you learn when learning electronics.

I am an idiot, i forgot that transistor closes with the current flow, not with the voltage being present, took me long to realise that haha. Thanks for everything. Best regards!