How to Design a Phototransistor Circuit for Specific Output Voltages?

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Discussion Overview

The discussion revolves around designing a phototransistor circuit to achieve specific output voltages based on varying irradiance levels. Participants explore the application of load lines and the interpretation of a log-log graph from the phototransistor's specification sheet to determine circuit components and configurations.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using Fig. 5 from the phototransistor's spec sheet to trace desired voltages onto the irradiance curve to find matching collector currents.
  • Another participant introduces the concept of load lines and recommends marking points on the graph to determine the resistor size and supply voltage.
  • A participant questions the application of slope from the log-log graph, expressing uncertainty about its implications for resistance calculation.
  • Further clarification is provided regarding the need to convert the log-log graph to a linear format for easier interpretation of the intercept points.
  • One participant shares their approach of calculating the series resistor size using Ohm's Law based on voltage and current changes at specified output voltages.

Areas of Agreement / Disagreement

Participants generally agree on the utility of load lines and the importance of understanding the graph, but there is uncertainty regarding the interpretation of the log-log plot and its application to resistance calculations. The discussion remains unresolved on the best approach to utilize the graph for circuit design.

Contextual Notes

Participants express uncertainty about the scales on the axes of the graph and the difficulty in extracting actual current values from it. There are also varying levels of familiarity with concepts like load lines and Ohm's Law, which may affect their approaches.

ihggin
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Homework Statement



Use the following phototransistor (http://www.vishay.com/docs/81532/bpw96.pdf) to design a circuit that gives an output voltage of 8 V when irradiance is 0.05 mW/cm^2 and 1 V when irradiance is 1 mW/cm^2 (both at wavelength 950 nm).

Homework Equations





The Attempt at a Solution



I'm assuming that Fig. 5 on page 421 of the spec sheet is useful? My best guess would be to trace the desired voltage onto the appropriate irradiance curve and then find the matching collector current? I haven't had any background with transistors and am pretty uncertain as to what kind of elements the circuit should even contain (op amps?, resistors?). A conceptual explanation of what is going on would be much appreciated. (Btw I searched the web for explanations of phototransistors, etc. but nothing made too much sense to me.)
 
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Have you studied load lines? If so, could you apply that to figure 5 ?

You would need to mark both points on the graph and then draw a line through them to the axes.

The slope of this line gives the size of the resistor you would put in series with the phototransistor, and the supply voltage is given by the intercept on the horizontal axis.
 
Thanks for the help. I haven't studies load lines before, but I just read the wiki, and it mostly makes sense now.

Just a clarifying question though: in the Fig. 5 plot, the slope that gives resistance shouldn't be of the line on the log-log graph, right? (I'm pretty sure this is the case, since the units wouldn't even work out, but I'm just wondering a bit why they give a log-log plot.)
 
Yes, I think you would have to move the graph to some linear graph paper.

You only need the two intercept points, so you don't need to move everything across.

You could also just calculate the size of the series resistor needed and work out the supply voltage from there.

It is not easy to get the actual currents from that graph, either, because it is hard to work out the meaning of the scales on the axes. I took it to mean 1, 2, 4, 6, 8, 10.

I tried it and got the slope of the line between the points and then applied this slope to the small triangle on the bottom RHS of the graph. This gave a small extra voltage which I had to add to the 8 volts to get the supply voltage.

Working it out with Ohms Law was easier, though. Calculate the size of the resistor from the changes in voltage and current at the 8 Volt and 1 Volt points, then work out the voltage across that resistor with 0.2 mA and 4 mA (or whatever you read from the graph as the currents).
Add this to 8 volts and 1 volt respectively to get the supply voltage. This should be the same in each case.
 
Okay, thanks again for the help! (I used Ohm's law and it worked out fine enough.)
 

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