0
\$\begingroup\$

I'm using phototransistors to sense low speed data (about 7 bits a second). With one phototransistor and a 300K resistor, I found my sweet spot and can transmit correct data with lazer.

I find if the resistor value is too high, I won't be able to sense data, and if the value is too low, then the sensitivity is too restrictive.

The phototransistor I use is wired in a common-emitter fashion (with collector to +5V through 300K resistor). The model I'm using is PT334-6c. I'm not entirely sure on the value of the capacitance in that model but I know one exists somewhere since the manual indicates the rise and fall time is 15uS if the resistor is 1K.

For the remainder of my question, let's give this capacitance a value of X.

Now if I connected two phototransistors in parallel to each other (collector to collector and emitter to emitter) and have one resistor as the load, will the capacitance value be 2 times X and will I have to make the resistor half the value?

And what about if of those two phototransistors, one is completely covered so no light enters it, and the other one has the valid lazer pulses going to it? would the capacitance still be 2 times X? or just X because only one phototransistor is active?

How would the number of phototransistors that detect the pulses out of the total number of phototransistors in parallel affect the total capacitance of the phototransistor network?

\$\endgroup\$
  • \$\begingroup\$ Is this for some kind of papertape reader? Just curious about the application, right now. \$\endgroup\$ – jonk Feb 2 at 4:12
  • \$\begingroup\$ LASER is an abbreviation of 5 words \$\endgroup\$ – Sunnyskyguy EE75 Feb 2 at 4:20
  • \$\begingroup\$ 15uS and 1Kohm indicates 15,000 pF equivalent capacitance. Might this be Miller Capacitance multiplication? where most of the BASE charge is consumed in a wasteful charging of the base-collector capacitance? Thus a cascade circuit should help/ \$\endgroup\$ – analogsystemsrf Feb 2 at 6:10
  • \$\begingroup\$ This is the datasheet to my phototransistor everlight.com/file/ProductFile/PT334-6C.pdf and it states the rise and fall time are 15uS each if Vce = 5V, Ic = 1mA, and RL=1000 ohms. I can't understand how 1mA is achieved with 5V and 1000 ohms when ohms law says that with 5V and 1000 ohms that you get 5mA current, not 1mA. \$\endgroup\$ – Mike Feb 3 at 2:19
  • \$\begingroup\$ Oh, and the base of the transistor is not available as a pin on this phototransistor. Only the collector and emitter. \$\endgroup\$ – Mike Feb 3 at 2:21
1
\$\begingroup\$

You can get far more gain-bandwidth with a stable photo-diode, PD and OpAmp with feedback R than any phototransistor.

Using tandem PT’s can double I and C so the dV/dt=I/C is the same for same R . But using a TIA can achieve 1e6 gain at 1Hz, which would be excessive so R feedback reduces the voltage gain using a TIA configured Op Amp.

Choice of narrow beam emitter and narrow beam PD increases gain by 2x for each reduction of beam angle /2 on the detector. A black daylight blocking filter is integrated into most lenses for IR transmission is highly recommended.

\$\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.