200 X-Frame-Options: SAMEORIGIN Content-Security-Policy: frame-ancestors 'self'; X-UA-Compatible: IE=Edge X-XSS-Protection: 1; mode=block X-Content-Type-Options: nosniff X-Telligent-Evolution: 12.x Pragma: no-cache Connection: close Accept-Ranges: bytes Cache-Control: no-store, no-cache, max-age=0 Date: Wed, 04 Dec 2024 16:06:14 GMT ETag: "0x8DD02F56CCD7D1F" Set-Cookie: .te.csrf=FdJ_lCM7uX4Xn49lRNJMr__K-lNDf7uMgko5vr5M6L4; path=/; secure; SameSite=Lax Set-Cookie: .te.auth=hPs8%2bJQBNWZ5hIIBGd9Ruv%2btWh2axSfvQuC3OV44wL5etz0eMpHOs9RBNP8Cs4B9r7P1SrDqZCA59V4tUVYT%2ff0yF2ktou75OOcg8r4QL92nzE0ndK0SgosYl9dSeahSsCWjfDLJt6ILx4W6%2fZoZJCPeJny2wjg7auCLNW2raX7%2fJM3n3PRrO6%2fDgIi0XGf0ld1lP60Y5aGjDhthEYVKyNcMfoESncewz5ZSzUPVP1kQo00inwx%2fp%2fgie15H8RNPxs6t4IIhdU72eQTlm9O0Cg%3d%3d; path=/; secure; HttpOnly; SameSite=Lax Set-Cookie: e14Guest=1; expires=Fri, 31-Dec-9999 23:59:59 GMT; path=/; secure; HttpOnly Set-Cookie: e14Country=NL; expires=Fri, 31-Dec-9999 23:59:59 GMT; path=/; secure; HttpOnly Set-Cookie: e14Language=en; expires=Fri, 31-Dec-9999 23:59:59 GMT; path=/; secure; HttpOnly Set-Cookie: e14TimeZone=Etc/UTC; expires=Fri, 31-Dec-9999 23:59:59 GMT; path=/; secure; HttpOnly Server-Timing: cdn-cache; desc=MISS Server-Timing: edge; dur=4 Server-Timing: origin; dur=19 Server-Timing: ak_p; desc="1733328374118_3090373299_175826992_2340_8733_6_10_-";dur=1 Strict-Transport-Security: max-age=86400 ; includeSubDomains ; preload #set ($UserId = $core_v2_user.Accessing.Id) #set ($Roles = $core_v2_utility.MakeList('Moderators', 'Almost Admins', 'Administrators')) #set ($IsAdminLike = $e14_utils.IsUserInRoleOrAdmin($UserId, $Roles)) #if($IsAdminLike == 'False') #end
Actions
Forum Thread Details
  • State Verified Answer
  • Locked Locked
  • Replies 18 replies
  • Answers 1 answer
  • Subscribers 14 subscribers
  • Views 4813 views
  • Users 0 members are here
Related
This discussion has been locked.
You can no longer post new replies to this discussion. If you have a question you can start a new discussion

Opto-isolotor module implementation suggestion(s)

colporteur

G'Day,

How can the following opto-isolator circuit output be used?

 

I purchased (on sale) a number of 4 channel opto-isolator modules from a Asian manufacture. I made the buy without due diligence on my part. The circuit works but there doesn't appear to be a way to use the existing connection as an output to do work.

 

image

 

There are four physical connection points on terminal strips. IN1 & G on the left are two physical connection points and V1 & G are the other two physical connections shown on the right. Assuming IN1 & G on left are input and V1&G on right are output, there doesn't appear to be a connection point on the output to do actual work.

 

In my haste to purchase such a great sales deal, I failed to realize there is no external connection on the south side of the 3K resistor. If V1 is supply and G is ground, then where is the output connection?

 

I do have a few electronic marbles rolling around in my head from post secondary education. That knowledge provides me no solution to how I get output from this circuit to do work without adding an additional physical connection. That is hard given the board uses surfaces mount components and very little space to make solder connections. I did make a connection to the south side of the resistor to test operation and it worked.

 

If someone has a suggestion on how to use the circuit without modification, I would appreciate it. Other wise the great sales deal was a bust.

 

Sean

 

image

Top Replies

  • 0

    what do you like to pay for my development support? image

     

    1. What does the opto output drive? Is this a logical or analog input of the following circuit?

    2. how high is the voltage?

    3. do you really need to use the emitter follower?

     

    "south side", "north side" .... is a funny language (of course it exist this slang).

     

    This "jumper cap" can be really a problem, if you like to nail that input directly to GND, one for the opto, one for the circuit.

    I think this is a "switch off" for the following circuit.

    But here, I think, is needing more than just nailing with a jumper.

     

    A bit more information would help thinking better with your idea.

    I have some solutions for you.

     

    Best Regards

    Gerald

    ---

  • 0

    How did you come up with that schematic?  It would be helpful if you labeled the pin numbers on the optocoupler..  Pictures of the board (both sides) would also be helpful.

  • 0 in reply to geralds

    G'Day,

    Much appreciate your feedback. Compensation for your initial response is in the high zero's. You can have as many zero's as you feel are necessary to compensate you for any future time you invest in development support.

     

    I am not looking at the opto-isolator as a driver but rather an output logic producer, whether that be high or low when the input is triggered. I didn't see the voltage as being important but if that garners a result then +5V be the value for input and output. I have no preference on how the output is generated (i.e. emitter follower).

     

    I elected to use South side and North side reference to minimize my effort and maybe appeal to a wider audience. The drawing is a cut&paste from the manufacture docs. Ideally it would have more detail. I updated the drawing in Inkscape to improve the notation. I made the navigation assumption most maps providing direction have a indicator of North facing the top.

     

    The "jumper cap" is not a capacitor but rather a jumper that enables use of an external ground (jumper removed) or use the same ground (jumper installed) provided from the input. You would assume the abbreviation cap means capacitor but not in this schematic.

     

    This is a commercial module still being sold. My knowledge suggests there is a design flaw (i.e. no output connection). I am challenging that notion by asking a wider audience. With a solution forthcoming I am hoping the sale item I took advantage of actually has some value.

     

    Sean

  • 0 in reply to ralphjy

    The schematic is manufacture provided. I have done some some drawing notation to clarify the resistor and the manufacture cap notation.

     

    I confess, it is not uncommon for me to provide an insufficient detail response to technical questions. I challenge that response in this case. I appreciate your feedback but in my opinion, the details you ask for contribute little to determining if there is a practical use. If all the pins were labelled, the 817 opto-isolator box fully drawn, the voltage references provided and a full printed circuit board layout provided, I don't see how that would arrive at a solution.

     

    I don't want to modify the board. I don't want to add components to the board. The circuit works.   WORKS, in the sense that the 817 behaves the way it should when a signal is applied. The box is an generic 817 opto-coupler. A signal on the input triggers a response on the output. The question remains how is the output provided obtained given the limited connections.

     

    Sean

  • 0 in reply to colporteur

    Hi Sean,

    image All of your drawings also your questions I understood.

    The point, please, we need a bit more information about the following circuit.

    Yes, saw that it is a logical schematic, and the emitter follower drives the "black box" of its output.

    Respective a description of the input signal, the active logic state (active low, or active high), is necessary to know.

    Has that input connector a pull-down resistor?

    The jumper is a bridge, not a capacitor (cap).

     

    The optocoupler PC817 is a easy to use device.

    https://www.farnell.com/datasheets/73758.pdf

     

    Gerald

    ---

  • 0 in reply to colporteur

    Hi Sean,

     

    "it works....", well it works, or it don't work.

    If you make a logic interface then it is important that you have sharp edges of the signal, without any distortions.

    An emitter follower with an opto coupler, it is formally a transistor amplifier driven via the photo-current of its transmitter_LED.

    So it drives as a current source, not as a voltage source which what is necessary for the logical input. Please refer e.g. TTL or CMOS behavior, at input or output as well.

    So you need some informations, working parameters, of that module.

    Then we can create an interface.

    I think it would be a two transistor interface, NPN(opto transistor) and a PNP transistor, if you need an active high for the module.

    The collectors will be the driver outputs.

     

    Gerald

    ---

  • 0

    The output is V1 - it should not be connected to a power supply although a power supply won't hurt it.

    It should be connected to an input that is pulled high externally. However, the current that the opto-coupler can sink is limited by the 3K resistor. If you want a logic output, the external pullup might be around 47K.

    Inside the optocoupler is a phototransistor that turns on when an input is applied.

  • 0 in reply to geralds

    I have updated the drawing with the 817 pinout. Your suggestion of an emitter follower output, made me realize the need for those details.

     

    I still challenge your need for the other details. The discussion doesn't need to extend into that level of detail. My focus was output only. I was aware of the emitter follower output but had hoped there was another approach.

     

    Sean

  • 0 in reply to dougw

    Can you help me better understand your response?

     

    If V1 is connected to an input that is pulled high, then the voltage divider created between the 3K resister and pull-up to high comes into play in determining what is high and low output.

     

    What ratio of divider to you see practical to create an output that transitions from high to low at the V1 point of this circuit?

     

    Sean

  • 0 in reply to colporteur

    You can try this circuit

    image

     

    Gerald

    ---

  • 0 in reply to colporteur

    Ok, I questioned the schematic because it seemed odd to connect a 3K resistor and LED in series with the input diode.  I guess just using a low current LED as an indicator.

     

    Anyway, depending on the input impedance of what you are trying to drive, you could probably just get away with putting a resistor across the "jumper cap" pins.  That way you get the bottom of the voltage divider at the output "G" pin.  Just size the resistor for the swing that you want and use that pin for the output.  Put the resistor external if you want an isolated ground.