CP123 optocoupler

Discussion in 'General Electronics Chat' started by Jotto, Apr 1, 2011.

  1. Jotto

    Thread Starter Member

    Apr 1, 2011
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    Used the test circuit that BMorse posted here for a 5vdc optocoupler. LED turned on at about 3.5vdc, brighter at 5vdc.

    My question is, what parameters are you using to determine value of resistors that are used? Can the same 5vdc circuit be used on all photocouplers and optocouplers? The one I tested is a photocoupler.

    Great site you have here, glad I found it. I am a bench tech at a casino. Like being able to test components.

    Thanks Jim
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Hello Jotto,
    You didn't provide a link to BMorse's test circuit, so it's hard to say.

    It's best to look at the optocoupler/photocoupler's datasheet prior to testing to determine what the actual Vf and current rating is, or you might burn the thing up by accident. Lots of optocouplers have an IR emitter Vf of 1.2v and 10mA current; higher if given in short pulses - but you really should read the datasheet beforehand, and test to the specs in the datasheet.
     
  3. Jotto

    Thread Starter Member

    Apr 1, 2011
    159
    17
    http://forum.allaboutcircuits.com/attachment.php?attachmentid=13097&d=1256662637

    5/.01=500, so I could actually use a 500 ohm resistor? or 5/.00001=500k ohm if using 10ua?

    This is the circuit I used. I also gave the wrong P/N its PC123. This is 1.2 Vf and 10ua (this one is listed in micro amps, might be a misprint). What I would like to do, is have the ability to test these when suspected bad. I have never seen one go bad at this time. I have built a small test board that has a variable power supply, an the circuit listed above. I would like to add more as required.

    Thanks Jim
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I received your E-mail. Sometimes it can take awhile to get a reply. I am very likely going to lose Internet access soon, for what may be awhile.

    The circuit:
    [​IMG]

    You calculate the current limiting resistor R1 as:
    Rlimit >= (V+ - Vf_LED) / Desired_Current
    So, if V+ = 5v and Vf_LED = 1.2v, and your Desired_Current is 10mA, then:

    Rlimit >= (5v - 1.2v) / 10mA = 3.8v/0.01a = 380 Ohms.
    380 Ohms is not a standard E24 value of resistance.
    A table of standard resistance values can be found here:
    http://www.logwell.com/tech/components/resistor_values.html
    Bookmark that page.
    Looking at the green E24 columns, you'll see that 360 and 390 are the closest values.
    However, to prevent excessive current, you'd use 390 Ohms instead of 360.

    On the output side, the transistor is wired as an emitter follower, so you'll lose ~0.7v across the base-emitter junction. Let's just say that the LED on the output has a Vf of 2.0v at 10mA.
    So, to calculate R2:
    R2 >= (V+ - (Vbe + Vf_LED))/Desired_Current
    R2 >= (5v - (0.7v+2v)) / 10mA
    R2 >= (5 - 2.7)/0.01
    R2 >= 2.3/0.01
    R2 >= 230 Ohms
    As you can see on the table, 230 Ohms is not a standard E24 value either, but 240 Ohms is.

    I found a datasheet for the PC123 here:
    http://www.datasheetcatalog.com/datasheets_pdf/P/C/1/2/PC123.shtml
    For the emitter side, typical Vf =1.2v for If=20mA.
     
    Last edited: Apr 8, 2011
  5. Jotto

    Thread Starter Member

    Apr 1, 2011
    159
    17
    Thanks for the information. I am going to try one next week an post example here for you to critique if you don't mind.

    Have a good weekend.

    Jim
     
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