Hello all

I have a project and going to replace a Phototransistor. The previous circuit is using a sensor has photo Darlington and the replace sensor just a common phototransistor

Previous sensor:


The replace sensor:


The previous sensor has a test circuit made as:

sensor A (Anode) as pin#1, K (Cathode) & E (Emitter) as pin #2, C (Collector) as pin #3.

Connect a 12Vdc PSU return (-) to pin #3;
Connect 12Vdc PSU (+) to one end of 1k resistor and other end to pin #1;
Connect a LED (+) to +5Vdc and other end to pin #2.
Will power on and sensor is not blocked, the LED should turn on. When sensor is blocked the led should be off.

I think if the circuit is correct, this should work on both sensors. But I am not seeing LED on and Off. Any idea of what is happening here?

 

Hello,

Do you have more details on the used sensors?
The darlington type will likely be more sensitive.

Bertus

 

Hello,

Do you have more details on the used sensors?
The darlington type will likely be more sensitive.

Bertus

I am trying to replace EE-sx298 with EE-sx198. I am not sure which part I did wrong. Even with EE-sx298, I am not seeing the LED on and off

 

Hi h,
Post a diagram showing how the Opto and LED are connected, also the power supply.
E

 

Hello,

So these are the devices:
https://eu.mouser.com/ProductDetail/Omron-Electronics/EE-SX298?qs=SMvRj3nRgHV2Sd%2BIsiGLrg==
https://eu.mouser.com/ProductDetail/Omron-Electronics/EE-SX198?qs=SMvRj3nRgHUjmBCeulmcNQ==

Can you tell us how you are limiting the current for the led emitter?
The leds are tested with 20 mA and have an absolute max current of 50 mA.

Bertus

 

Hi h,
Post a diagram showing how the Opto and LED are connected, also the power supply.
E

I do not have a diagram, but this is what I understand

 

Hello,

So these are the devices:
https://eu.mouser.com/ProductDetail/Omron-Electronics/EE-SX298?qs=SMvRj3nRgHV2Sd%2BIsiGLrg==
https://eu.mouser.com/ProductDetail/Omron-Electronics/EE-SX198?qs=SMvRj3nRgHUjmBCeulmcNQ==

Can you tell us how you are limiting the current for the led emitter?
The leds are tested with 20 mA and have an absolute max current of 50 mA.

Bertus

Well, I use an adjustable dc PSU. select in voltage model and set to 12V and 5 V. in this case the PSU should auto adjust based on the load circuit, is it?

 

Hello,

No, the led will need a current limiting resistor.
When you connect the led directly to the powersupply, the led will already be burned.

Bertus

 

Hi h,
That wiring totally wrong.
I will draw a diagram, what are the battery voltages?

E

 

Hi h,
This is an example circuit.
Assuming 5V supplies
E

 

But I am not seeing LED on and Off

Is it an infrared LED?

 

Hi h,
This is an example circuit.
Assuming 5V supplies
E
View attachment 328450

Wow, very cool thing.

I had similar circuit as you show here. I test it, and it works fine. However, I am trying to understand how the previous test circuit works with 12V dc on the LED/diode side, and it is not connect to ground but return only.

By the way, may I ask what software do you have here? Is it free tool online?

 

Is it an infrared LED?

No, it is not. The LED on/ off, I refer to the additional LED I connected on 5Vdc PSU, not the LED inside of the sensor.

 

By the way, may I ask what software do you have here? Is it free tool online?

hi h,
It is LTSpice, it is free, but can be hard to learn. Many members use LTSpice and will offer help.

On my circuit diagram, the right side of the opto is only connected to 0v, to show it working. it is not required in an actual circuit.
E

Link.
https://www.analog.com/en/resources/design-tools-and-calculators/ltspice-simulator.html

 

hi h,
It is LTSpice, it is free, but can be hard to learn. Many members use LTSpice and will offer help.

On my circuit diagram, the right side of the opto is only connected to 0v, to show it working. it is not required in an actual circuit.
E

Link.
https://www.analog.com/en/resources/design-tools-and-calculators/ltspice-simulator.html

Thanks. I had same circuit as you show and it is work correctly. The part I do not understand is I had a previous testing circuit. On left side, it is connect a 12V DC PSU. On right side bottom is connect 12V dv return. I do not understand how that works. Do you have any idea of this?

 

Thanks. I had same circuit as you show and it is work correctly. The part I do not understand is I had a previous testing circuit. On left side, it is connect a 12V DC PSU. On right side bottom is connect 12V dv return. I do not understand how that works. Do you have any idea of this?
View attachment 328669

The 5v circuit has to use the 5v gnd (return) via the 5v power supply ground or it won't work.
The 12v circuit will work as shown, but the 5v side won't unless the ground terminal of both 12v and 5v supplies are connected together to form a common ground.

 

The 5v circuit has to use the 5v gnd (return) via the 5v power supply ground or it won't work.
The 12v circuit will work as shown, but the 5v side won't unless the ground terminal of both 12v and 5v supplies are connected together to form a common ground.

Thanks. I verified the test circuit with someone else. And they miss that point.

 

Hi hs,
An opto coupler is designed so that the Input and output do not have to share the same 0V/common.

It is primarily used as an Opto-isolator.

Why are you using an opto-coupler in your project?

E

 

Thanks. I verified the test circuit with someone else. And they miss that point.
View attachment 328683

Be aware, however, that they way you have connected the opto is unusual, in that opto's are generally used to provide isolation of the input circuit from the output circuit. Since ground is "common" at both input and output, there is no isolation.
Removing the ground connection between the Optos' LED cathode and transistor emitter will re-establish isolation, but you'll need to ensure that the appropriate ground returns are used.