Design phototransistor circuit

[ihggin]

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.)

 

[vk6kro]

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.

 

[ihggin]

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.)

 

[vk6kro]

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.

 

[ihggin]

Okay, thanks again for the help! (I used Ohm's law and it worked out fine enough.)