0
\$\begingroup\$

I have two phototransistors. If I measure the impedance (what I understand to be essentially the resistance of anything that's not a resistor) across just one of the phototransistors, then I measure around 235K. I've checked that they both measure the same.

Two phototransistors stuck in a breadboard.

But if I then put the phototransistors in parallel, then I measure 36k.

Two phototransistors in parallel in my breadboard.

My understanding is that if I have two of the same impedances in parallel then they should measure as half the impedance, which should be around 112K not 36K.

The light conditions are not changing in the room. I am measuring with the red lead on the flat side (emitter) and the black on the collector. Can this really be happening?

\$\endgroup\$
  • 1
    \$\begingroup\$ The light in the room might not have changed, but probably when you rearranged the devices, you got one pointing more directly at the light source. \$\endgroup\$ – The Photon Feb 8 '17 at 23:56
3
\$\begingroup\$

Since phototransistors are not linear devices, and you are not giving these any power to begin with, they are not acting linearly. Sometimes the approximation is valid, for example when biased in a near-linear region, but just floating like that they are not.

I suspect that if you biased both phototransistors (using like a 1-10k resistor and a 5V source) that their impedance would respond more closely to how you are describing.

\$\endgroup\$
  • \$\begingroup\$ I did as you suggested, and put 5 volts through it with a 10k resistor in series. Things started to then behave as I would expect. Thanks for your explanation. \$\endgroup\$ – Chase Roberts Feb 9 '17 at 0:17
3
\$\begingroup\$

They are phototransistors, not photo-resistors. They will not obey ohms law.

They will probably act more like constant current sinks with the current being somewhat constant even of the voltage varies. At the low voltage you get out of a meter on the resistance range to may be more complicated.

For example this is a plot of the current vs voltage for a phototransistor.

An additional issue may be that the polarity of the voltage may not be what you expect. Analog meters tend to have the positive output on the black/negative connection because it makes for easy arrangement of the switches in the meter. Digital meters tend to have the red probe be positive but it is not guaranteed.

enter image description here

\$\endgroup\$

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.