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I would like to measure accurately (as possible as) using an IR LED and an IR phototransistor. I am using the following parts

SFH4550 - IR LED
BPV11 - IR Phototransistor

Obviously, the transistor gives a current output proportional to light. That current has to be converted to a proportional voltage.

What are the pros and cons of active vs passive current to voltage converters?

Passive:
Something like this where the transistor is the detector which is a current source: enter image description here

versus

Active:
Something like this where current i is from the phototransistor: enter image description here

Also how can I improve each method to give the best signal...reduce noise, flicker, etc?

Thanks, any help is appreciated!

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  • \$\begingroup\$ Not directly in answer to your question, but if you want to do good analog measurements with an LED/phototransistor pair in a setting where you can't provide full optical shielding, consider using an oscillator to drive the LED and a bandpass filter or lock-in amplifier on the receiver to separate the signal from noise. \$\endgroup\$ – Chris Stratton Dec 20 '11 at 17:13
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The main difference is that the passive circuit has an output impedance roughly equal to the resistor value. The active circuit has very low output impedance. That could mean you can get much more gain from the active circuit before the response time gets to be limited by the capacitance of whatever load is receiving the circuit output.

Also, in the passive circuit the photodetector bias will vary as its output current changes. The bias on the photodetector in the active circuit will be constant. This could be important if you care about the capacitance of the detector changing.

Both of these effects might mean that you can use a higher resistance value in the active circuit, which will reduce the Johnson noise contribution from that resistor, which gets to your question about accuracy. In either case you'll want to add a capacitor in parallel with the resistor in your circuit to limit the noise bandwidth of your receiver. Op-amp vendors have app notes describing how to optimize the noise in transimpedance amplifier circuits (here's one from National)

If you go with the active circuit, you'll also want to pick an amplifier with low thermal drift of the offset current and low flicker noise (assuming your main goal is to measure the dc light power).

I'm not much familiar with phototransistors, but I noticed the datasheet you linked shows nearly 2-to-1 change in collector current for temperature varying from 0 to 100 C. This is probably going to be the limiting factor in your sensor accuracy. Using a photodiode instead of the phototransistor, and making up the gain using the transimpedance amplifier circuit, and carefully selecting the op-amp and resistor you use, you ought to be able to achieve much better accuracy (I'm thinking on the order of a few % over 0 - 100 C, but I may be forgetting some critical issue).

Edit - One more thing

One minor difference (that might be important depending how you use the circuit) is that because its an inverting amplifier the active circuit will require both positive and negative power supplies (or some possibly accuracy-limiting hoop-jumping to avoid them), whereas the passive circuit may be easier to work in a single-supply environment. But even with the passive circuit you'll have to be pretty careful with the next amplifier circuit down the line to keep its accurate when the current is small.

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