EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: elki on April 19, 2024, 10:47:56 am
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Would you consider this amplifier to be of the voltage-feedback type or the transimpedance one? If I am not mistaken, this configuration should work as a transimpedance amplifier with the gain I(R2)xR1?
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Hello elki,
your circuit is a ac coupled standard inverting amplifier with some input filtering. A transimpedance amplifier (https://en.wikipedia.org/wiki/Transimpedance_amplifier) translates an input current into a voltage at the output, so you need a current source (e. g. a photo diode) for that.
Guido
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function may also be:
bandpass with gain
or differentiator with upper+lower frequency limit
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With the non-inverting (+) input tied to the negative ground rail I doubt your circuit will work at all?
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@ CaptDon
With the non-inverting (+) input tied to the negative ground rail I doubt your circuit will work at all?
yes, you're right. - I missed that ...
Guido
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In the circuit shown the gain at the opamp is R1 divided by R2, in your case a voltage gain of 10X, however you are feeding an input circuit with loss (the series 1K resistor) and the loss increases due to the capacitors to ground on the input circuit. I suppose a rail to rail opamp can work with the non-inverting input connected as you have shown. I think your drawing would actually require a bi-polar power supply to work as shown if you connect that non-inverting input to the common ground reference. Guess it doesn't matter what type of amplifier it is if it won't work as drawn?
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With the non-inverting (+) input tied to the negative ground rail I doubt your circuit will work at all?
It will. Just will cut out all negative output part and add distortions to positive. Such an 'unideal rectifier' :bullshit:
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Thank you for your replies. I am actually using this circuit connecting it to a photodiode and sourcing its current from anode as input. That’s why I thought this amplifier can be considered as a trans impedance amplifier, since it converts current to voltage, but I guess I am wrong.
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Thank you for your replies. I am actually using this circuit connecting it to a photodiode and sourcing its current from anode as input. That’s why I thought this amplifier can be considered as a trans impedance amplifier, since it converts current to voltage, but I guess I am wrong.
The input is ac-coupled so it can't be the only thing connected to the photodiode... Photodiode amps usually have the diode direct on the inverting input and the frequency response set by the feedback network.
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The circuit won't work, at least not properly.
If it works in simulation, then it's because the op-amp power supply isn't being modelled correctly.
If it appears to work in real life, then the op-amp must be going into phase inversion or something weird.
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Even if you get the dc bias sorted out the ac voltage gain is less than unity. At frequencies where the 10pF can be ignored it is a classic virtual ground opamp circuit with gain equal to the ratio of feedback resistor to input resistor. The effective input resistor is 1000 + 100. So 1000/1100 = 0.9091.
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Thanks again for your feedback.