Equivalent circuit for light

[lgbeno]

So I've been thinking about this over the last few days.  I'm doing a pulse oximetry project and have a constant current led driver sending light to a photodiode amplified with a Tia.  I used a third op amp to regulate the led current such that the common mode output of the Tia is always .5v.  The circuit works but I gets me to thinking.

I know in mechanics, voltage=force, current=velocity, resistance=friction, cap=1/spring & inductance=mass but what about light?

Specifically what would the equivalent circuit be for a led emitting light, a finger attenuating said light and a
Photodiode receiving the result?

I'm thinking the led is a current source, finger is a "resistor" and photodiode is a voltage sense of the drop across the finger.  Ground is ambient light.

[penfold]

Unfortunately, i think light is light in both domains, well, heat or some other kind of radiation!

If you're working purely in the domain of light, then linking light flux to electric current works quite well, except when you try and find an analoge of voltage or start to consider wavelength,

Not exactly a useful post, but thought I'd contribute anyway

[lgbeno]

Unfortunately, i think light is light in both domains, well, heat or some other kind of radiation!

If you're working purely in the domain of light, then linking light flux to electric current works quite well, except when you try and find an analoge of voltage or start to consider wavelength,

Not exactly a useful post, but thought I'd contribute anyway

I think it's useful, at least tells me that I'm not the only one confused!

[qno]

Warning:
Leds age. Are sensitive to temperature.

Phototransistors are even worse than LED's.

For any repeatable results you have to compensate for above problems.

[IanB]

Light is a flow of energy, in other words neither current nor voltage but power. We could say the intensity of the light (power per unit area) is analogous to voltage, and the quantity of light (surface area illuminated) is analogous to current. Power is then intensity times surface area.

If you shine a beam of light through a sheet of translucent paper (a resistance), you could say that the intensity (voltage) of the light is reduced, but the surface area illuminated (current) remains about the same.

[ejeffrey]

The analogy is not so simple.  The reason that the mechanical/electrical/hydraulic analogy is useful is because they are described by the same differential equation.  That is why you can reason about one system in terms of the other: they have the same equations of motions so they have mathematically the same solutions.

You need to include wave mechanics to understand the flow of power in an optical system.  So you can think of your optics as equivalent to sound waves.  Unfortunately that doesn't usually help as most people don't have good intuition about sound waves, either.  The LED and phototransiter are non-linear and harder yet to include.

[jmole]

You need to include wave mechanics to understand the flow of power in an optical system.  So you can think of your optics as equivalent to sound waves.  Unfortunately that doesn't usually help as most people don't have good intuition about sound waves, either.  The LED and phototransiter are non-linear and harder yet to include.

For this particular example, you likely don't need to worry about wave mechanics. In fiber optics, you might.

You can basically equate current to light input/output in the linear range of the LED/sensor. The non-linear part comes with the obstruction of the LED with your finger.