i am going to make electronic spirometer... i will use led and photodiode... in between them there is a fan which rotates when we blow in that chamber... when the flaps of fan will cut the light which falls on the photodiode, the output will zero... and when it not cut the light it will give some voltage of 0.5v.. pls give me idea about this logic... i am not sure it will be work or not .... . . :-(
This is an op-amp circuit that I copied from the "Worksheets" section on the top of the "Projects" page. The formula for gain is: input voltage times 1+R1/R2. You only need to give the amplifier some voltage to work with and decide how much amplification you want. Tell us more if you want more help.
My recommendation is to operate the diode in the photocurrent mode because it will be more linear. If you operate it in the photovoltaic mode (open circuit), you will find that its output starts getting compressed as it approaches 0.5 volts very early on. If you aren't using a long cable, you can dispense with the 22 uf capacitor and get a frequency response mostly limited by the capacitance of the photo diode, the characteristics of the opamp and the feedback resistor. You can actually take this basic circuit up to the hundreds of MHz, but that would be a whole different discussion.