View Full Version : Looking for details on UIRT2 IR transmission function

April 19th, 2004, 02:29 AM
I have a UIRT2 that I'm trying to connect directly to the IR input of an amplifier. The IR input on the amp consists of a ground and a signal connector. The amp and computer share the same ground through the audio cables - i.e. the IR input ground on the amp = computer ground. The amp requires a 5V signal - "active high". It works fine when I disconnect the audio cables - amp ground does not equal computer ground.

I don't understand how the IR output works on the UIRT2.
Tip (always) = +5V with respect to ground
Ring (not transmitting) = ?, 0V with respect to Tip, 0V with respect to ground, maybe not connected.
Ring (transmitting) = ?, 5V signal with respect to Tip, 0V with respect to ground.
Sleeve (always) = Ground

Somehow I need to get a 5V signal with respect to ground out of the UIRT2. Does anyone have any ideas?

April 19th, 2004, 08:45 AM

You could look at the IR linc from smarthome. It will optically connect the 2.

April 19th, 2004, 06:05 PM
Yeah, I saw that. It costs $20 though - I could use an IR blaster which would be much cheaper. Maybe I'm being crazy, but I want to use the IR input connection on the back of the amp.

After looking at the UIRT2 schematic and doing some more reading, I understand how it works now. The ring (signal) is not connected most of the time. When the base of the darlington transistor receives a signal it turns on and connects the ring to ground.

Now, somehow I need to convert this to a 5V signal. I could place a resistor between the tip and ring and connect a digital inverter to the tip. Or I could use an optoisolator as suggested. Or, maybe, I could modify the UIRT2 so that it bypasses the darlington and drives the IR input on the amp directly.

So here are my new questions: How much current can an output pin on the PIC supply? IR input circuits are probably transistor based and don't need much current, right? Is it safe to simply short the darlington to bypass it?