Amplifier type - Page 1
Forum section: electronics / beginners

author: elki
author_rank: Regular Contributor
timestamp: April 19, 2024, 10:47:56 am
content: 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?

author: r6502
author_rank: Regular Contributor
timestamp: April 19, 2024, 12:37:43 pm
content: Hello elki,your circuit is a ac coupled standard inverting amplifier with some input filtering. Atransimpedance amplifiertranslates an inputcurrentinto a voltage at the output, so you need a current source (e. g. a photo diode) for that.Guido

author: Andree Henkel
author_rank: Regular Contributor
timestamp: April 19, 2024, 12:51:36 pm
content: function may also be:bandpass with gainor differentiator with upper+lower frequency limit

author: CaptDon
author_rank: Super Contributor
timestamp: April 19, 2024, 12:56:35 pm
content: With the non-inverting (+) input tied to the negative ground rail I doubt your circuit will work at all?

author: r6502
author_rank: Regular Contributor
timestamp: April 19, 2024, 01:01:30 pm
content: @ CaptDonQuote from: CaptDon on April 19, 2024, 12:56:35 pmyes, you're right. - I missed that ...Guido

author: CaptDon
author_rank: Super Contributor
timestamp: April 19, 2024, 01:07:12 pm
content: 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?

author: xvr
author_rank: Frequent Contributor
timestamp: April 19, 2024, 01:42:49 pm
content: Quote from: CaptDon on April 19, 2024, 12:56:35 pmIt will. Just will cut out all negative output part and add distortions to positive. Such an 'unideal rectifier'

author: elki
author_rank: Regular Contributor
timestamp: April 19, 2024, 03:48:48 pm
content: 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.

author: MarkT
author_rank: Frequent Contributor
timestamp: April 19, 2024, 05:51:25 pm
content: Quote from: elki on April 19, 2024, 03:48:48 pmThe 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.

author: Zero999
author_rank: Super Contributor
timestamp: April 19, 2024, 06:20:52 pm
content: 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.

author: wasedadoc
author_rank: Super Contributor
timestamp: April 19, 2024, 07:00:53 pm
content: 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.

author: elki
author_rank: Regular Contributor
timestamp: April 19, 2024, 07:08:51 pm
content: Thanks again for your feedback.