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We are trying to hook up phototransistors in parallel so we can sum the output currents and put it over a resistor to get an output voltage.

When we do this, the first phototransistor (closest to the voltage source) seems to be greatly affecting the output voltage, and the others are negligible

We are sending 3.1v to the phototransistors in parallel.

What can we change so that each phototransistor will equally affect the output voltage?

enter image description here

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  • \$\begingroup\$ you can add low value resistors to either the collectors of each photo-transistor, or to the emitters of each photo-transistor. This will help balance the transistors. Also, are the photo-transistors equally illuminated? \$\endgroup\$ Commented Mar 17, 2021 at 0:14
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    \$\begingroup\$ Please show a schematic - the phototransistors should add equally if it is wired up correctly. \$\endgroup\$ Commented Mar 17, 2021 at 0:15
  • \$\begingroup\$ Added schematic. And they are all getting equal light. However, when we cover the first one, the voltage significantly drops, if we cover the 2nd one, the voltage drops less, and so forth. \$\endgroup\$
    – EEIsCool
    Commented Mar 17, 2021 at 0:35
  • \$\begingroup\$ 500 kOhms is likely too big. Get the operating current up to at least 1 mA per transistor. Try a 1 kOhm or smaller for you output sense resistor. \$\endgroup\$ Commented Mar 17, 2021 at 0:54
  • \$\begingroup\$ @EEIsCool - if only illuminate only one at a time do they give the same voltage? \$\endgroup\$ Commented Mar 17, 2021 at 1:59

2 Answers 2

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The problem is that the first transistor to be illuminated brings the output voltage too close to the supply, it is not possible for the output voltage to go higher than that so the second and subsequent transistors cannot raise the voltage any higher.

If you reduce the value of the 500k resistor so that a single transistor produces a small voltage much less than the supply (say 500mV) illumination on additional transistors will increase the voltage by the same amount.

It could be done more accurately if the combined signal from the photo-transistors is fed into a trans-impedance amplifier instead of using the output directly. The voltage across the photo-transistors would not depend upon the illumination.

The output will depend upon the strength of illumination. If you want to avoid that one way would be to separate the four transistors to provide a digital signal then process the signals digitally, maybe using a microcontroller or by adding the digital signals in ana analog fashon.

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When photo-transistors are connected in parallel, the way that is shown in your schematic, the \$V_{ce}\$ of each transistor will be equal, (assuming negligible voltage drop in the wires between the photo-transistors).

As a result, the photo-transistor that is most conductive will dominate the circuit. If you are familiar with how open-collector output pins may be "wire-or"ed together, you will understand this aspect of your circuit.

If this is not behavior that you want, then your circuit must be altered.

If you want to measure the average light intensity, perhaps the best way would be to using a summing amplifier, with each photo-transistor contributing to an input.

Edit:

Kevin White's answer states:

The problem is that the first transistor to be illuminated brings the output voltage to close to the supply, it is not possible for the output voltage to go higher than that so the second and subsequent transistors cannot raise the voltage any higher.

If you reduce the value of the 500k resistor so that a single transistor produces a small voltage much less than the supply (say 500mV) illumination on additional transistors will increase the voltage by the same amount.

I think this is clearer and more accurate than my own answer. Please change the accepted answer to Kevin White's (that is, if you agree).

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