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I'm trying to sense IR wave with a phototransistor circuit. The circuit operates well under DC operations (I mean 3.3V through inputs). However, if a square wave is supplied from UART input, after 2kHz output voltage waveform become noisy DC about 200mV. What should I consider to increase bandwidth to 100kHz (maximum)? There is 1cm space between IR & phototransistor in vertical axis. LED draws 20mA current.

Here are the datasheets:

bc848b: http://www.nxp.com/documents/data_sheet/BC848_SER.pdf

bc858b: http://www.nxp.com/documents/data_sheet/BC856_BC857_BC858.pdf

phototransistor: http://www.megasan.com/service/pdfhandler.ashx?fileid=3565

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What's R4 for? Why not connect the 2nd transistor right to the photo-transistor (PT) emitter? What's the voltage at the PT emitter.. (maybe you have it in backwards?) \$\endgroup\$
    – George Herold
    Commented Dec 5, 2014 at 15:54
  • \$\begingroup\$ i don't remember the exact value; but, Vbe of BC847b was about 1.3V. Actually, everything seems normal in the presence of IR wave; however, the circuit continues to operate when IR light goes off (I even tried by putting circuit to a box in a dark room). Why does PT continue to conduct if there is no light source near to it? @GeorgeHerold \$\endgroup\$
    – ythey
    Commented Dec 7, 2014 at 21:03
  • \$\begingroup\$ Is that the whole circuit? No caps anywhere? could be leakage/ RF pickup/ voltage ripple.. Do you have a 'scope? \$\endgroup\$
    – George Herold
    Commented Dec 7, 2014 at 21:50
  • \$\begingroup\$ It is just a prototype design to use for further design process. IR LED is connected to NPN AND gate, and it just works fine(I've already tested it about more than 20 times.) The purpose is to transmit, and receive information through IR&PT pair. \$\endgroup\$
    – ythey
    Commented Dec 7, 2014 at 21:58

2 Answers 2

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First, you do not identify what phototransistor you're using, so there is no way anyone can answer your question about collector identification. "flat side" almost certainly identifies the photosensistive face, not collector or emitter.

It doesn't help that you haven't provided a number for the ouput voltage, either. "Low" could mean 1 volt, or .5 volt, or .1 volt.

That said, you need to do the following: put your circuit (without the LED) in a completely light-tight enclosure and see what the output is. Most likely you're picking up stray IR.

If you're still having problems, go back to the data sheet for your phototransistor and look at the dark current spec. If you're using a bc847b, your DC gain could be as high as 450. Assuming "low" means 0.5 volts, that's a collector current of about .5 mA, and a dark current of 1 uA will provide enough base current to do that.

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  • \$\begingroup\$ Actually, I don't know what the phototransistor is, which why I couldn't have provide the datasheet of it. It was a bit urgent test of a project, so I've used whatever I've found(bad engineering approach I know). Secondly, I've already tried the circuit in an opaque object, nothing changed. The output will be applied to UART of a uP, so anything below 0.5V is enough, I think. The problem is that PT continues to conduct even if there is no light near to it. \$\endgroup\$
    – ythey
    Commented Dec 7, 2014 at 21:11
  • \$\begingroup\$ Try this: Connect one end of the PT to ground, and the other end through a 1k resistor to +5. Measure the voltage across the PT with no light. Then reverse the PT and measure again. The orientation with the higher voltage is correct. Now find a resistor value that gives 4 volts across the PT when dark. This will tell you the PT dark current. Now you can set R2 and R3 so as to give 0.1 volt at their junction, with no transistor connected. Illuminate the PT with the LED, and if the R2/R3 voltage is 2 volts or more, connect the output transistor. \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 7, 2014 at 22:00
  • \$\begingroup\$ Let me correct my previous comment, which I now see as the result of a brain fart. With the LED shining full on the phototransistor, and a 1k to +5, choose the orientation which gives the lower voltage across the PT. The difference should be obvious. If it's not, I suspect you have a bad transistor. \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 8, 2014 at 2:18
  • \$\begingroup\$ Connected 4k resistor to the collector of PT. 1.1V across PT junction. I found dark current to be about 1mA. Isn't that a bit high? \$\endgroup\$
    – ythey
    Commented Dec 8, 2014 at 6:40
  • \$\begingroup\$ I finally reach the datasheet-> megasan.com/service/pdfhandler.ashx?fileid=3565 \$\endgroup\$
    – ythey
    Commented Dec 8, 2014 at 7:01
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With a rise/fall time of 10/15 uS, you'll have a difficult time reaching 100Khz. At best you might hope for 40Khz. A similar circuit is used for serial communications on utility meters, and isn't used over 19.2k baud.

enter image description here

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  • \$\begingroup\$ The rise/fall time of which component are you talking about? And why is it difficult to reach 100Khz with 10/15 microseconds of rise/fall time? As you give the 19.2k baud example, what are acceptable / maximum rise fall times for 19.2k baud and why? \$\endgroup\$
    – Pro Backup
    Commented Feb 26, 2016 at 10:03
  • \$\begingroup\$ The phototransistor. 100Khz is not a practical frequency with these rise/fall times, as a full bit includes a rise, fall, and hopefully some dwell time for the uart to register the bit properly. \$\endgroup\$
    – user76064
    Commented Mar 3, 2016 at 12:36

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