Hey,

The image attached is what I am currently using to sense the output from a relay (28V). These OptoIsolators are connected to GPIO pins so when the Relay is closed, the GPIO goes High providing me with positive feedback.

The problem I am having is that the Current Limit resistor plus the diode in the isolator are going to draw more power than I want. Plus the cost for the parts is much higher than I would like because the resistor is rated for 2 watts.

Is there some other way to sense this voltage and provide 5vdc to my GPIO when it is present that will cost less than ~10 dollars? The trouble is that I need to use this circuit several times which makes the price go up rapidly.

 

...because the resistor is rated for 2 watts

If it hurts when you do that, don't do that! You don't need a 2W resistor for an LED.

And are you saying that the maybe 5-10mA drawn by the LEDs is too much?

 

Currently the LED draws 2.5mA (if the input voltage is 5V). The 1k5 resistor dissipates 9mW.
The LTV costs about 15 to 30 cents. The resistor even less.

 

Bah I did my calculations wrong. But because I am trying to keep my cost down the OptoIsolator that I am using requires 20mA which means that in my system which has 112 of the optoisolators, my power consumption is ~60W. I was hoping there was something besides the optoisolator that I could use.

 

Currently the LED draws 2.5mA (if the input voltage is 5V). The 1k5 resistor dissipates 9mW.
The LTV costs about 15 to 30 cents. The resistor even less.

So the optoisolator's input is 28V and its output is 5V. So as I understand it, the LED is drawing closer to 20mA meaning the resistor is dissipating ~.6W (actually a little less because the proper resistor value for the current limit is closer to 1.35k)

 

You could take all those resistors up to 4.7K or maybe higher, taking the heat dissipation down to 1/8w, allowing you to use 1/4W or 1/2W resistors. The LED will work down to 2.5mA according to the datasheet, so aiming for 4-5mA is realistic.

 

No idea what the rest of the schematic is but how about using a double pole relay instead of a single pole relay (I'm assuming you have a single pole). Then you can simply apply 5V to the common terminal and then connect the NO terminal to your micro pin.. It will then go high when the relay is energized.. No need for a resistor or opto or anything.

 

You could take all those resistors up to 4.7K or maybe higher, taking the heat dissipation down to 1/8w, allowing you to use 1/4W or 1/2W resistors. The LED will work down to 2.5mA according to the datasheet, so aiming for 4-5mA is realistic.

How can you tell that the LED will work down to 2.5mA from the datasheet? I have been looking over it but a lot of the graphs mean so little to me.

No idea what the rest of the schematic is but how about using a double pole relay instead of a single pole relay (I'm assuming you have a single pole). Then you can simply apply 5V to the common terminal and then connect the NO terminal to your micro pin.. It will then go high when the relay is energized.. No need for a resistor or opto or anything.

Due to space requirements and contact current requirements, I haven't been able to find a DPDT or SPDT Relay that I can use. If you know of any that have a footprint equal to or smaller than .2" x .8", 12V coil and 3A+ Contact Current let me know

 

So the optoisolator's input is 28V and its output is 5V. So as I understand it, the LED is drawing closer to 20mA meaning the resistor is dissipating ~.6W (actually a little less because the proper resistor value for the current limit is closer to 1.35k)

Ok, I was assuming 5V optocoupler input.

works above 1.6mA but is single channel. http://www.futureelectronics.com/en...s-optocouplers/Pages/6390355-H11L1M.aspx?IM=0

or something like this but it's not high speed.
http://www.clare.com/home/pdfs.nsf/www/CPC1301.pdf/$file/CPC1301.pdf

 

How can you tell that the LED will work down to 2.5mA from the datasheet? I have been looking over it but a lot of the graphs mean so little to me.

From the LiteOn datasheet

 

Do you need the isolation? If not it seems like a voltage divider would work.

 

Ic in your application is 0.5mA (Vcc 5v and 10k pulldown) plus port input current.
from graph 1 about 1.5mA LED current will cause a suitable transition, go for the full 2.5mA as a safety factor..
As suggested above though, if isolation is not required use a couple of resistors for a voltage divider.

 

Do you need the isolation? If not it seems like a voltage divider would work.

I don't necessarily need isolation but I thought it would help avoid any voltage spikes that may occur from damaging my GPIO chips. I could probably use a LPF to help with that though