> From: Mike Singer[SMTP:m_singer@POLUOSTROV.NET] > Sent: Monday, January 05, 2004 2:44 AM > To: PICLIST@MITVMA.MIT.EDU > Subject: Re: [EE]: How to measure light? Phototransistor? > John N. Power wrote: >> Phototransistors convert photons to electrons, which means >> current output. > I'd rather say - Photons form electron-hole pairs in the base which > creates a base current. >> A good opamp solution can be made with the LM3900. It works from >> 5 volts and has current sinking inputs. Connect the phototransistor >> to the plus input of one of the amplifiers (the chip has 4), and a >> feedback resistor from the output to the inverting input. Additional >> current into the inverting input will subtract an offset from the >> output, >> and choosing the value of the feedback resistor will allow setting the >> gain. > By the way, "plus input" current is about 0.02 ma for LM2900. Ever > seen a phototransistor with the 0.02 ma output current? This particular amplifier doesn't work that way. Each input appears as a diode to ground. A current into the positive input is subtracted from the current into the negative input by means of a current mirror. The current difference drives the output. Because of this current sinking property, this amplifier is good for sources which produce current directly, such as phototransistors. Try to make a conventional virtual ground input using a normal opamp and a single 5 volt power supply. You might try to bias the positive input with a divider, but remember that this limits the voltage on the phototransistor. Also, the output is inverted which causes problems if the amp output can't go within a volt of ground. The 3900 can sink current without inverting. The maximum current into each input is 5mA max at 25C, and reduces to 3.8 mA at 70C. Within this range, the inputs can accomodate a perfectly reasonable amount of current. > Not to say, two transistors without negative feedback (phototransistor > and "plus input" transistor) are real killers for the converter > linearity > and even stability. Temperature signal shift will be much more than > working signal range. Consider a current source like a phototransistor working into a diode load to ground. The current in the diode controls the output voltage. This is quite linear. > Best Regards, > Mike. > P.S. On a non-related wave: For me Olin's stile of writing is like > president Reagan's style of communication, John N. Power's > is like Bush II's, Russell's The Great Pacifier (or Pacifist - which > is better) is like Pope's :-) > Hope nobody is offended, if so a thousand apologies, guys. Bush is a Yale graduate, so I find that flattering. Which "stile"[sic] is yours? John Power -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body <> -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body