connect IR receiver
I have a remote control kit.
How do I identify the pins (1,2,3) on the IR receiver device?
How do I connect the IR device to PCB (a,b,c)
Using the diode function on a DMM.
The following connections show mV while passing ~1mA
legs . . . . mV+ . . mV-
1->2 . . . open . . 650
2->3 . . . 540 . . . open
1->3 . . . 740 . . . open
If the device is NPN then leg 2 is base and -ve current flows leg1 to leg2 and also leg3 to leg2. Legs 1 & 3 being collector and emitter.
If the device is PNP then leg3 is base and +ve current flows leg1 to leg3 and also leg2 to leg3. Legs 1 & 2 being collector and emitter. Does the highest mV reading of 740 indicate that these are collector and emitter with light/heat holding the junction as apparently open?
How do I tell from the voltage drops which is collector and which is emitter?
How do I tell if it's PNP or NPN?
how do I connect 1,2,3 to a,b,c ?
R1 & R2 are both 22k.
The supply is 5Vdc
Those things are usually a self-contained receiver with a tuned circuit for a particular frequency. You supply power and ground and the decoded signal appears at the other pin, active low IIRC. The transmit signal is pulsed at 30~50kHz depending on model, but you don't see the modulation on the receiver output. I identified mine from the datasheet + I scoped the output with a soundcard. 38k is a common frequency, I just blasted it with a few remotes I had handy.
Isn't there any more info in the package?
You check the data sheet. It isn't an NPN or PNP. It's a receiver with ground, power and signal output at TTL levels. They typically run around a mA to operate and as Wakibaki said, the carrier is already removed so if you put a scope on it you'll see the envelope of the data which makes the micros job much easier.
Didn't the kit come with a PCB? I would expect a silk screen layer showing part positions.
Yes the PCB has the atmel control chip on it and three pads labeled a,b,c as shown in my diagram.
The IR receiver is mounted on or in the front facia. A 3 wire connection with three pin 0.1" pitch sockets can be used to connect them together.
The ground comment ticks a box.
Pin C is connected to PCB ground.
Pin 2 is connected to the metal enclosure around the IR receiver. That fits with the wiki description that the base is not connected and the light is the base input signal.
Maybe these two connect to each other.
Then that just leaves pins 1 & 3 to pins a & b
But why am I getting transistor type readings across pins 1, 2 & 3. Pin 2 is apparently connected to the IR junction.
I must assume that the IR receiver is a IR module, not simply a IR phototransistor. Nobody would use a phototransistor for remote control, when an IR module does so much of the hard work for you. If what you have is indeed a phototransistor, then that begs the question: is that what your kit needs, or does it need an IR module? (i'd bet it is the latter).
An IR module has power, ground and output pins. If you bought it then you know the part number and should look up the datasheet. If you salvaged it from an old VCR or something, then you should trace the original PCB to find the power and ground connections.
The output of IR modules is almost always an open-collector style digital output, which is active low (it idles at Vcc and is pulled to ground when an IR signal is present). They usually have a built-in pull-up resistor, but not always. I assume that R1 or R2 is a pull-up. Usually power is supplied via a small resistor (100 ohm) followed by a decoupling cap. This forms a filter to prevent noise from the supply from affecting the module.
If what you have (and what you indeed need) is a phototransistor, then you can use a multimeter to find the pinout. What you want to do is measure resistance from collector to emitter, red on collector for NPN. You will know you have the two leads on the emitter/collector when exposing the thing to dark/light changes the measured resistance significantly (light acts the same way as base current, causing conduction from collector to emitter). You can determine NPN/PNP just as you would for any transistor (try not to change light level while doing this; wrap it in electrical tape if necessary).
The two RC kits all came together as a new unused unassembled kit on 4PCBs, each PCB was pre-assembled. rotary encoder which I don't intend to fit and a receiver.
Is DIYaudio still damaged.?
Upload pic is reporting
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here's the pic that the broken DIYaudio would not let me upload.
From the photo, it could equally be a phototransistor or a IR receiver module. But your original circuit diagram would not support a module. In fact, the circuit would nicely support a dual phototransistor, which would likely have two collectors and a common emitter. There need not be any base connection on a phototransistor, though a base is sometimes provided. With no more information than you have given, I assume that it is a dual NPN, and would try pins 1 and 2 to a and b and pin 3 to c. Try measuring resistance (not voltage drop) from 1 to 3 and 2 to 3, with 3 being the black lead. Vary the light level and see if the resistance changes.
Hi Andrew, these IR receivers are pretty common and output a TTL level signal.
Three connections, one is ground, one is 5 volts, and the output (comparator output here) is usually an open collector PNP stage requiring a resistor of 1 to 5k to ground. Depends on the input C of the following stage really.
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they are all much the same.
As they draw so little current I find that using a 470 ohm limiter in the 5 volt supply allows for unidentified devices to be tested on a trial and error basis. Use a 10 k or so on the output to ground and scope to confirm signal. Get the connections wrong and no damage done.
As mentioned, the remote handsets don't just transmit the 0 and 1's as an on/off of the IR transmitter, they modulate a carrier of between 30 and 40 khz. 36 and 38k are common frequencies.
Also you'll find when viewing the output on a scope that pressing the same button on the handset (such as Philips RC5) appears to give one of two codes each time it is pressed. This is due to "parity and toggle bits" being added that give a 0 and then a 1 at the start of the code.
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