So I've decided to make a circuit based on this: enter image description here

I have a few questions. First of all, when the photo-transistor switches off and Q1 switches off, as per the article, what is the voltage going through Q1? Can anyone explain R3 and Q1. I know how it functions generally but as the photo-transistor forms essentially an open circuit with R4, wouldn't there be negative voltage going through R3? If anyone can explain that It'd be much appreciated.

Also, how would I go about choosing resistor values (mainly R1, R3, R4)?

  • \$\begingroup\$ Do you have a datasheet of the photodarlington? \$\endgroup\$
    – stevenvh
    Aug 13, 2012 at 16:59
  • \$\begingroup\$ Ah, no not at the moment. I'll sort out specific components I'm going to use later on, I have no problem with that I just needed the theory. Also, If I understand the operation of photodarlington's then theoretically could I just use a normal photo-transistor with appropriate lighting/resistances? Thanks for the explanation, it was great. \$\endgroup\$ Aug 13, 2012 at 17:44
  • \$\begingroup\$ Thanks. You can use a common phototransistor. That already amplifies the current from a photodiode, but of course the current will be much less than the Darlington's. The other components' values will also change, and for Q1 you may need a transistor with a higher hFE. All depends on the phototransistor's specs. \$\endgroup\$
    – stevenvh
    Aug 13, 2012 at 17:48

2 Answers 2


Voltages don't go "through", that's current. The photodarlington is a current source, and small currents will go through R3 until about 150 µA is reached. Then the voltage drop across R3 is 0.7 V and Q1's base-emitter voltage will fix R3's voltage to that level, and therefore R3 won't get more than the 150 µA. If the photodarlington sources more that will go into Q1's base.

So if Q3 gets enough light Q1 will conduct and pull the base of Q2 to ground. Q2 won't conduct anymore and the LED will extinguish. If there's less light than needed for the 150 µA then Q1 won't conduct, and Q2's base will get current through R1, and the LED will light up.

R3 is there to make sure that Q1 won't get current if it's dark. A photodarlington has a dark current of probably 1 µA or so, and that would be enough to get Q1 conducting. R3 takes care of the first 150 µA, and this way creates a threshold for Q1.

R4 will limit Q1's base current in case the photodarlington will want to source too much of it. With the current 330 Ω the base current will be limited to 25 mA, which will prevent damage to the transistor.

The resistor values will depend on the photodarlington's characteristics, and the light level you want the LED to go on/off.

Kudos to dextorb for this Falstad simulation.

  • 5
    \$\begingroup\$ To supplement this great answer, I have drawn up the schematic for OP in a simulator so you can visualize the current flow when Q3 on/off: goo.gl/5ELTv \$\endgroup\$
    – dext0rb
    Aug 13, 2012 at 17:17

stevenvh had explained all things very good, so no more explanation is required.

However for the resistances values, some values of resistances are not properly selected.

R1 should be 4.7Kohm

R3 is OK

R4 should be a variable Resistor of 1K to 5K, use only one end point with middle one movable point and set resistance value according to your requirement

  • \$\begingroup\$ Why a potmeter for R4? What's wrong with the 25 mA limit the current value gives? \$\endgroup\$
    – stevenvh
    Aug 13, 2012 at 17:21
  • \$\begingroup\$ POT on R4 will set when to Light LED ON and OFF, it will help to chose correct value of resistance for required darkness \$\endgroup\$ Aug 13, 2012 at 17:29
  • 4
    \$\begingroup\$ No, that's what R3 is for. R4 only limits Q1's current long after the LED has gone on, it's not relevant to the threshold level. \$\endgroup\$
    – stevenvh
    Aug 13, 2012 at 17:41

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