IR Led, Phototransistor Motion Detector Circuit Not Working.

Discussion in 'Homework Help' started by KLoong, Mar 4, 2011.

  1. KLoong

    Thread Starter New Member

    Mar 4, 2011

    Hi. May I know why is this circuit to detect motion not working?
    When the part in the red box is not connected, the point connecting to the base measures 2.2V when the phototransistor is not blocked and 4V when it is blocked. Even though I was expecting 0V and 5V respectively, I figure that could work since an increment of 1.8V is enough to turn on the BJT (BC537).
    So I thought I could just take the voltage at the collector of the phototransistor and connect it to the base of the BJT so that the BJT could be turned on when the beam from IR Led to the phototransistor is blocked. Unfortunately, after connecting the parts in the red box, the voltage measures 540mV when the phototransistor is unblocked and 580mV when blocked.
    I've tried to solve this problem by adding resistors to the phototransistor branch and the emitter branch of the BJT but the LED still doesn't light up when the beam is broken. I've ran out of ideas on how to solve this so I hope the AAC community can lend me a hand here. Any help would be much appreciated.

  2. mjhilger


    Feb 28, 2011
    Since the emitter of the BC537 is grounded, do you realize what the max voltage at the base of that transistor could become? Aside from that, 1M on the collector of the phototransistor will not supply much drive to your BC537 - what is the hfe. You need to determine the current (transistor is a current amplification device) needed to light the LED then work backwards to determine the collector resistor for the photodiode to supply that current. Either way first consider the maximum possible (without destorying the BC537) base voltage on your switch transistor.
  3. Adjuster

    Well-Known Member

    Dec 26, 2010
    Your measurements with the BC537 disconnected show that the phototransistor current is only changing by about (4V-2.2V)/1MΩ=1.8μA.

    This is not enough to drive an LED via one ordinary bipolar transistor: even if the LED current was only a few mA, the transistor would need to have a current gain in the thousands. In this situation, if you reduced the collector resistor to give enough current for the second transistor to light the LED, the small current from the phototransistor could not turn it off.

    You therefore need to do one of two things. If possible, increase the amount of light shining on the phototransistor so that it is able to pass more current. You may then be able to reduce its collector load to perhaps 100kΩ, while still getting the voltage to drop down to less than half a volt when the phototransistor is getting light.

    If you cannot get more light to shine on the phototransistor, you need to provide more current amplification. A Darlington pair of transistors might just about do it, but you would also need to increase the phototransistor load from 1MΩ to perhaps 3.3MΩ to ensure that the LED would turn off.
  4. mjhilger


    Feb 28, 2011
    The darlington is a great suggestion if you need to use BJT. You might also consider changing the BC537 to a 2N7000 FET. It requires almost no current for drive and would work well, you might need to decrease the collector load resistor from 1M to 330K or so. Just a different direction you might examine.