I have a controller with a digital input and digital common measuring 3vDC. When shorted, it will activate the controller.
Question: I want to create a short by using an opti transistor so that an LED can activate it. Do I need a resistor in the circuit? Is there a general opti transistor that would work? Thanks!

 

I have a controller with a digital input and digital common measuring 3vDC. When shorted, it will activate the controller.
Question: I want to create a short by using an opti transistor so that an LED can activate it. Do I need a resistor in the circuit? Is there a general opti transistor that would work? Thanks!

Draw out what you want to do. As long as you don't have more than 5 mA through your phototransistor. And, remember that a fully illuminated phototransistor will have 0.6 volts across Emitter to Collector.

 

I was confused as digital common measuring 3vDC, what is that means?
Your question should be easy to solve if you attach the circuit.

 

Draw out what you want to do. As long as you don't have more than 5 mA through your phototransistor. And, remember that a fully illuminated phototransistor will have 0.6 volts across Emitter to Collector.

Why?
If it is saturated Vce should be no more that a tenth of a volt or so. It will be less than Vbe.

 

Here is the circuit showing the emitter and collector that is to short my digital input of 3vDC. Do I need a resistor on the emitter or collector? Is there a standard type photo transistor I should use? Thanks!

 

Your description has some problem that you can't shorted the 3Vdc to Ground.
I think you should draw a block diagram and label the input/output and voltage to show all you want.

 

Here is a diagram on how I think it will work. I am hoping I don't need a resistor.

 

What do you mean by "When shorted, it will activate the controller"?
What signal does the controller need?
Show the complete circuit.

 

You can decide the V1 and adjust the values of R1, also the R2(4.7 K) can be higher.

 

Here is a diagram on how I think it will work. I am hoping I don't need a resistor.View attachment 85125

If you'd told what you're trying to control, it would probably help, but I assume it's a simple contact that you need to ground - if you try shunting the contacet to ground via a diode (cathode to ground), you'll see if the device will accept a semiconductor actuation (it probably will, I have used both regular and photo transistors for contact actuation at numerous occasions).

You can use almost any regular photo transistor, while a photo Darlington transistor may have a too high drop and don't even try with a photo diode, it won't be sensitive enough on its own.

Please post more info on what you're trying to control and why you've taken this route.

 

This is the controller I am using: http://www.eddywireless.com/yahoo_site_admin/assets/docs/ML8032_GSM_RTU_User_Manual.90103454.pdf
On pg 3 & 4, it shows a digital input that you can short pin 7 & 8 with a switch, etc. I want to replace the switch with a phototransistor. It measures 3vDC across pin 7 & 8.
Thanks!

 

They don't say much about the DI (Digital Input) other than it is a Dry Contact Input. Pin 7 is labeled Digital input common port leading me to believe it may be circuit common and pin 8 is simply labeled Digital input. They want a dry contact input. What that likely does is place pin 8 at a logic low. Pin 8 might be being held high using an internal pullup? Using a DMM you could measure across pins 7 & 8 and see what you have. Worst case a small photo transistor could drive a small relay and provide the dry contacts.

Ron

 

When I used a DMM, it measures 3vDC on pin 7, with pin 8 the ground. When I short 7 & 8 with a switch, it activates the controller relay. I want to use a phototransistor across 7 & 8 to make the short and activate the relay. Is there a general phototransistor I can use & can I use it without a resistor?

 

Did you see what I posted in #9, all you want is there.

 

When I used a DMM, it measures 3vDC on pin 7, with pin 8 the ground. When I short 7 & 8 with a switch, it activates the controller relay. I want to use a phototransistor across 7 & 8 to make the short and activate the relay. Is there a general phototransistor I can use & can I use it without a resistor?

Yeah, see what Scott posted in #9. Scott uses an opto coupler but if you only want a phototransistor you could try one. For example a NPN Photo Transistor with emitter to common and the collector to the 3V pin. Since the 3 volts gets to common via a switch there is an internal pullup. So you can use a phototransistor or photodiode and provide the light source or as Scott suggested use an opto coupler.

Ron

 

This is the controller I am using: http://www.eddywireless.com/yahoo_site_admin/assets/docs/ML8032_GSM_RTU_User_Manual.90103454.pdf
On pg 3 & 4, it shows a digital input that you can short pin 7 & 8 with a switch, etc. I want to replace the switch with a phototransistor. It measures 3vDC across pin 7 & 8.
Thanks!

Knowing the input impedance of your multimeter and the supply voltage (or the Vcc after an eventual voltage regulator), you can calculate the value of the pull-up resistor on the board.

If your voltmeter has got an input impedance of 10MOhm, the current would be 3V/10MOhm = 300nA. If the supply was 5V, that would mean a (5V-3V)/300nA = 6M7 pull-up.

That doesn't sound likely though, so your voltmeter probably have a much lower impedance and I'm guessing the voltage as well.
Either way, a pull-up is very unlikely to pass enough current to hurt a photo transistor (not a photo diode), so just put it on, emitter to pin 7 and collector to pin 8.

If you get one of the 2-pin photo transistors that looks like an LED and you're in doubt of which pin is which, just plug it in and if it doesn't work, replug it the other way around.