I have seen this rather simple circuit for end position detection of a DC motor drive mechanism. An infrared LED (V402) and a phototransistor (V401) are used as a light barrier.Infrared Position Detector

Can someone give me a hint concerning the circuit topology name and function of the additional BC856A (V403) PNP transistor in this circuit? (Is it a current amplifier?)

Isn't this circuit design very susceptible to ambient light and also temperature variation?

Edit1: Here is the simulation result of this circuit. It seems to have a current to voltage gain of 20V/mA.

enter image description here

  • 1
    \$\begingroup\$ Have you looked up cascode or common base? Also, note that 20V/mA just happens to be what a 20k Ohm resistor achieves. \$\endgroup\$ – jonk Jul 5 at 7:15
  • \$\begingroup\$ It seems to me that T1 is always operated in the saturation region and thus is constantly switched on. So it has basically no function and the 20V/mA output is caused by the 20k Ohm resistor R1, right? \$\endgroup\$ – Stefan Wyss Jul 5 at 7:53
  • 1
    \$\begingroup\$ It helps to keep the Vce of the detector BJT constant -- eliminating early effect issues. And no, it's not saturated until VM1 equals VF1, at which point it moves gradually into saturation. \$\endgroup\$ – jonk Jul 5 at 7:58
  • \$\begingroup\$ @jonk Ok, I think you got it right. The purpose of the BJT (V403/T1) is to keep the voltage across the photo transistor (V401) constant. This reduces the early-effect, i.e. it decouples the photo current from the collector-emitter (Uce) voltage. (Would you write this as an answer so I can give you credit for?) \$\endgroup\$ – Stefan Wyss Jul 5 at 13:57
  • \$\begingroup\$ I could. But Dave has done a yeoman's job already and spent more time thinking about the circuit than I did. I could provide some additional examples to help clarify a few things. But if you are happy with Dave's answer I'm fine with that. \$\endgroup\$ – jonk Jul 5 at 16:45

You made a good start by setting up a simulation. Unfortunately, your simulation is misleading you. There are two issues:

  • The phototransistor is not an ideal current source; instead, you should be modeling it as a current limiter. Putting a diode in parallel with it is one way to model this.

  • The circuit was undoubtedly designed for a supply voltage higher than 3V. 5V is much more likely.

Jonk's comment comes close to the answer — the transistor is functioning as an emitter-follower, with the emitter terminal held at roughly 2/3 the supply voltage plus a diode drop. This means that the voltage across the phototransistor is limited to slightly less than 1/3 the supply voltage.

While this will help with some dynamic effects, it's much more likely that it was done in order to meet a voltage rating limitation on the phototransistor.

  • \$\begingroup\$ Thanks for your insightful answer. Which way around would I have to place the diode? I can confirm that V_ON is always 3V (I got that from another part of the schematic). The photo transistor (V401) has a collector-emitter breakdown voltage of min. 5V. So the voltage rating limitation is off the table. \$\endgroup\$ – Stefan Wyss Jul 5 at 12:19
  • \$\begingroup\$ Think about it -- which way does the diode need to point in order to all "excess" current from the ideal source to flow through it and bypass the rest of the circuit? \$\endgroup\$ – Dave Tweed Jul 5 at 12:30
  • \$\begingroup\$ I do not get the point of this current limiter modeling of the phototransistor. Did you propose that for keeping the graph for VM1/VF1 right for currents in excess of 100uA? Don't you think we would rather need to add some internal resistance to the phototransistor for a better model? \$\endgroup\$ – Stefan Wyss Jul 5 at 12:40
  • 1
    \$\begingroup\$ The point is, a phototransistor won't force current through an external circuit, while an ideal current source will -- and this will mess up your interpretation of the simulation results. \$\endgroup\$ – Dave Tweed Jul 5 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.