Using 4n33 optocoupler?

[codesuidae]

4n33 example

Greetings,

I am working on a project that will as part of its function control current to a 12v lightbulb that draws up to 2A (signal bulb from an American car).

I'm using an IFR540 to switch the bulb. I'm also using a 4n33 optocoupler to provide feedback to the system as to whether or not the bulb has actually turned on. To do this I want to place the 4n33 input in parallel with a resistor on the ground side of the light bulb. If the light bulb is burnt out, no current will flow and the 4n33 output side will not show current flow.

My problem is that I need to have at most 2mA flowing into the 4n33, and I'm not sure how to do that correctly. The diode in the 4n33 needs at least 2v to turn on, which means that if I wanted to run it directly off the sense resistor I'd have to have at least a 2v drop there, which would be much more power than i want (since its a 12-14v bulb I'd lose some brightness too). I think what I need is an NPN transistor with its base and emitter across the sense resistor to feed the diode.

I don't know if i need a resistor in the base lead of the transistor though. I think the voltage drop across the sense resistor is small enough that I can do without a base resistor (R1 in the diagram below).

Is this a reasonable way to use the 4n33 to confirm that the bulb is actually on?

         VCC 12v                         VCC 5v
          ^                               +
          |     ___                       |
          o----|___|----+                .-.
          |    R2 10k   |                | | R3
        ,---.           |                | | 10k
        | X | Lamp 2A   +----+           '-'
        '---'           |    |            |
          |    ___    |/     |            |
          o---|___|---|      |        +---o--->Sense
          |   R1???   |>     |        |   |
         .-.            |    |      |/    |
   0.005R| |            |    V 4N33 |     |
     1W  | |            |    - Opto |>    |
         '-'            |    |        | |/
          |             |    |        o-|
          o-------------+    |          |>
          |                  |            |
          |                  |            |
       ||-+                  |            |
       ||<-  IFR540          |            |
 In >--||-+                  |            |
          |                  |            |
          +------------------+            |
          |                               |
         ===                             ===
         GND                             GND

Thanks for any advice.

 

[DarkJedi]

4n33 examples

Well, the way i see it i think u have gone too far for what u want to do, a simple foto resistor would do the job, using a couple of opmas as window comparator to reach ttl levels, optocouplers are used when you need to isolate two circuits, for example a motor from it's digital control, noise can pass on to the digital part, so you use an optocoupler there to avoid the motor noise, but useing it as a light sensor, u think u better try a fotoresistor.

 

[Fish4Fun]

4n33 optocoupler

well, my skills with a text editor are just not up to the task at hand, but if you will consider the following spec sheet I think you might find an easier cicuit:

h**p://www.irf.com/product-info/datasheets/data/ircz24.pdf

These devices are designed with you purpose in mind.

 

[codesuidae]

12v optocoupler

Well, the way i see it i think u have gone too far for what u want to do, a simple foto resistor would do the job.

I presume you mean by using the photoresistor to check receive light from the bulb? I considered that, but since the bulb operates from complete darkness to full sunlight, I thought it might be difficult to set up any kind of photosensor to reliably detect the light from the bulb while in full sunlight.

The bulb is also housed inside a reflector, so I'd have to put the photosensor in front of the bulb, which blocks some of the light, and also exposes it to the environment, which would be pretty harsh (ice, snow, mud, etc).

optocouplers are used when you need to isolate two circuits, for example a motor from it's digital control.

That is also a concern here, and the reason I've shown two seperate power supplies. The 12v system is a car electrical system and is very noisy, not something I want to connect to my microcontrollers.

Thanks

 

[DarkJedi]

4n33 spec

Well, cosidering the specifications of the enviroment you mention your choice seems more logical now.

Sorry to have bothered you.