Electrolinux
New Member
Hi, everyone !!
The circuit in the attached file is a transimpedance amplifier or a current to voltage converter. The tricky thing here is that the current is extremely small, some microamperes only, because it is generated by a photodiode, as you can see in the scheme.
When the photodiode is biased (imagine there's no op-amp, so the anode is directly connected to the ground), the photodiode generates a current value which depends on the received light (just the way it should be !!). This current value is very small, as I said above. For example, if the lab has the lights on, the current equals 1.25 :mu: A. If the photodiode lens is covered by a black paper, this current value is only 25 nA. I've realized that the higher the resistor value is, also the higher the current value is. With a 1 Khm: resistor, there's no any current, so I guess using a 10 Ghm: resistor would be better, but those high value resistors aren't readily available to me.
However, if the converter is added, the photodiode isn't working anymore. The voltage drop in the photodiode is always 0.28 V, which is also the value measured when the photodiode isn't biased. I'd say there are two possible solutions: using an op-amp with a better signal to noise ratio (although the TL071, from T.I., is a low-noise JFET-input op-amp) and using resistors of Ghm: But I don't really know what's going in this circuit whatsoever
I would appreciate any kind of help. Thank you.
The circuit in the attached file is a transimpedance amplifier or a current to voltage converter. The tricky thing here is that the current is extremely small, some microamperes only, because it is generated by a photodiode, as you can see in the scheme.
When the photodiode is biased (imagine there's no op-amp, so the anode is directly connected to the ground), the photodiode generates a current value which depends on the received light (just the way it should be !!). This current value is very small, as I said above. For example, if the lab has the lights on, the current equals 1.25 :mu: A. If the photodiode lens is covered by a black paper, this current value is only 25 nA. I've realized that the higher the resistor value is, also the higher the current value is. With a 1 Khm: resistor, there's no any current, so I guess using a 10 Ghm: resistor would be better, but those high value resistors aren't readily available to me.
However, if the converter is added, the photodiode isn't working anymore. The voltage drop in the photodiode is always 0.28 V, which is also the value measured when the photodiode isn't biased. I'd say there are two possible solutions: using an op-amp with a better signal to noise ratio (although the TL071, from T.I., is a low-noise JFET-input op-amp) and using resistors of Ghm: But I don't really know what's going in this circuit whatsoever
I would appreciate any kind of help. Thank you.
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