I am currently writing my masters thesis and have this one question I can't answer myself. I have a PSD sensor with a common cathode which is biased with +5V (positive reverse bias).
I found this article:
"The application of the VBIAS voltage (reverse bias) causes the photodiode’s junction capacitance to decrease. As light impinging on the photodiode becomes brighter, the photodiode current (IPD) flows from the cathode to the anode. As the luminance increases on the photodiode, the IPD current also increases. With this occurrence, the amplifier output terminal decreases in voltage."
Why is the amplifier voltage output DECREASING when the photodiode current is INCREASING with more light? I don't get it...
I always thought like this: increasing photocurrent = increasing amplifier output voltage. Thanks for the enlightenment.
That is an inverting amplifier (or current to voltage converter). The voltage at the output has be be reduced to drive more current through the feedback resistor.
Memorize the basic feedback op amp rule: the output does whatever it has to do to make the voltage difference between the input pins equal to zero.
A good overview on various photodiode amplifiers and bias configurations is here (mostly assuming op amps with bipolar power supplies).
But if I look at the attached files, it is shown, that more photodiode current = more amplifier voltage output. What's the difference between my circuit and this one?
Look at the polarity of the bias voltage in each circuit - they are different. The second example uses ground, which is more negative than Vref. The first uses +5V which is positive with respect to Vref.
Ah, I missed the diode polarity. The answer is, the first circuit uses photoconductive mode, and the second uses photovoltaic mode. Generally, PC mode is faster but noisier, and PV mode is more light sensitive.
I realized later that the 0.1V bias (offset) in circuit #2 may have been intended to avoid problems associated with using an op amp that is not rail to rail on input.
The OPA322 is, however, rail to rail on both input and output.
The second circuit forward biases the diode slightly, by about 0.1V.
I don't know why anyone would do that.
I suspect its actually reverse bias and the model is showing photocurrent.
@hpetersen1993
as has been said, the photocurrent is flowing from the +rail IN towards the op amp - input.
neg feedback maintains that -input at 2.5V - the SAME as the + input
by adjusting Vout so that all the photocurrent also flows through the feedback resistor (15k)
To do so Vout MUST go less positive as the photocurrent increases.
So the circuit is working fine, just reverse the readings in software.
Thanks everyone for the replies, it helped me a lot.
The PSD sensor has one cathode and two anodes for the both current outputs. For everyone not familiar with this kind of sensor, check the picture. If I want to calculate the position of the laser spot I need both currents Ix1 and Ix2, so the anodes have to be connected to the inverting inputs of the op amp. Am I right?
Is there any other way to connect the PSD sensor to the amplifier, other then the way I connected it in my first post?
(Both anodes to the inverting input of the op amp).
What bandwidth are you wanting? - the opamps used need to be fast enough, its not enough
just to use the correct circuit topology. And if speed isn't an issue you might get away with
simpler circuitry (although reverse bias is the most linear mode, which might be important
regardless of speed).