Photodiode with transistor

Discussion in 'The Projects Forum' started by nacoolp, Jan 2, 2013.

  1. nacoolp

    Thread Starter New Member

    Jan 2, 2013
    I am using a photodiode sfh206k to detect the laser light. I want either 0 or 2.5 V as an output notifying ON or OFF respectively with the help of a transistor. Is it possible? and if it is, can you suggest me the load resisors and capacitors required for the same...??

    My reference circuit diagram is attached with the post. In this, what should be the output at P1 when laser light is detected and when there is no light.

    Thanks in advance
    Last edited: Jan 2, 2013
  2. Sensacell

    Well-Known Member

    Jun 19, 2012
    Note that the circuit is AC coupled via the two capacitors, only the AC component of the light signal will pass through this circuit.

    I believe you are looking for a digital ON-OFF 0-2.5 Volt output, this circuit will not work for this purpose.

    This circuit would need a detector circuit after it to demodulate the incoming AC light signal, it will not respond to un-modulated light signals at all.
  3. DickCappels


    Aug 21, 2008
    This one works with the photodiode being the only exotic component. The pot is the sensitivity adjustment. The 1N916 provides a temperature compensated threshold voltage. The 10k resistor from the collector to ground reduces the output voltage when the transistor in in the dark to 2.5 volts. Note: this really only works when you are driving a high resistance load.

    It is not optimized because we don't know much about your application, but it should get you started.

    You can also do with with just a photo transistor and two resistors.

    absf likes this.
  4. nacoolp

    Thread Starter New Member

    Jan 2, 2013
    Thank You Sensacell and DickCappels for your support. Your help did give me directions to move further with my circuit.:)
  5. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    Here is another method that makes a Darlington pair from the receiver.


    The gain should be high enough that you get an on/off output, but you could always add a comparator to the output of any of the above circuits to get a logic output, rather than a linear out.
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    The 5k resistor is way too high in value. The total LED current is about 40mA. 2N3904 is spec'ed for Vce(sat) with Ic/Ib=10, so the base current needs to be about 4mA. The resistor should therefore be 750Ω.
    With 5k, the base current will only be about 860uA. The saturated beta would therefore have to be about 50. Some transistors might work, some might not.
  7. nacoolp

    Thread Starter New Member

    Jan 2, 2013
    Thanks thatoneguy and Ron H for the reply but I want to strictly follow the photodiode only as I already it in stock.