I was told in another poll that it was easy to build one on your own.· Is it?· If it is, can you show me a·VERY DETAILED·guide on how to do it?· I would really be happy if you could do that.·········-SciTech02.·
It did?· Well, I wanted to ask how to make it so it could you only three pins, but I see you provided it.· Can this work through a servo extender cabble? (The one that came with the mounting bracket for the ping senser) That type of cabble.· Also, I'm mediocre at schematics.· What is the thing behind the resistor?· It says·1n5231 on it.· That's about it.···· -SciTech02.
Post Edited (SciTech02) : 3/19/2006 10:25:45 PM GMT
If you go to Parallax.com, find 350-00014, which is the infrared receiver component. While you are there download the zip file. Within the zip file their is a pdf called "Stamp Weekend Application Kit". If you read that document, I think it should answer all your questions.
1. Yes, it should work through a servo extender cable, as long as it's not too long. 3 feet might be pushing it.
2. The IR Detector is an 'active' component. It expects the IR output of an IR led blinking at a 40 Khz rate (or 36 Khz, or 38 Khz). When it detects this signal, it pulls its output LOW for as long as it detects this signal.
The element you were asking about is a "diode" -- which will be reverse-biased (ie not-conducting) when the IR-Detector pulls the signal (on the "Data" pin) low.
You use "PULSIN" to detect how long the signal is LOW.
3. The IR-LED is that device that has the arrows coming out of it. You drive this signal with a 'FREQOUT DataPin, 40000' to give the LED a 40 Khz blink rate, so it can 'talk' to some other IR-Detector -- or even its own to use the unit as a reflected-IR distance detector.
Really, the "Weekend Application Kit" explains much of this in more detail.
LED's make rather poor diodes. Their forward biased voltage drop is 1.4 to 2.8 volts (where a diode is 0.7 volts).
But otherwise, I think your theory is correct, the LED itself would keep current from flowing backwards. The second diode is probably for faster response -- which you need when you're measuring pulse-widths and 'echo' signals.
Oh, and the reason there isn't a 'short' is because the active device only pulls low -- I believe it lets the signal 'float' high.· Now, why there isn't an additional resistor in place to insure it 'floats' high, I don't know. ·
Comments
Post Edited (SciTech02) : 3/19/2006 10:25:45 PM GMT
Ray
2. The IR Detector is an 'active' component. It expects the IR output of an IR led blinking at a 40 Khz rate (or 36 Khz, or 38 Khz). When it detects this signal, it pulls its output LOW for as long as it detects this signal.
The element you were asking about is a "diode" -- which will be reverse-biased (ie not-conducting) when the IR-Detector pulls the signal (on the "Data" pin) low.
You use "PULSIN" to detect how long the signal is LOW.
3. The IR-LED is that device that has the arrows coming out of it. You drive this signal with a 'FREQOUT DataPin, 40000' to give the LED a 40 Khz blink rate, so it can 'talk' to some other IR-Detector -- or even its own to use the unit as a reflected-IR distance detector.
Really, the "Weekend Application Kit" explains much of this in more detail.
1. wouldn't there be a short between the pin and the sensor's output when you send the FREQOUT command to the pin?
2. why is there a second diode on the LED side of the circuit, wouldn't the LED itself keep current from flowing backwards?
3. wouldn't the LED be on constantly unless the output of the sensor was internally current limited?
did some datasheet reading on the sensor.
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Who says you have to have knowledge to use it?
I've killed a fly with my bare mind.
Post Edited (CJ) : 3/19/2006 11:53:10 PM GMT
But otherwise, I think your theory is correct, the LED itself would keep current from flowing backwards. The second diode is probably for faster response -- which you need when you're measuring pulse-widths and 'echo' signals.
Oh, and the reason there isn't a 'short' is because the active device only pulls low -- I believe it lets the signal 'float' high.· Now, why there isn't an additional resistor in place to insure it 'floats' high, I don't know.
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