Just one question: Speaking about a "transimpedance amplifier" - are you referring to the AD844 ? If yes - this is a very special type and NOT a typical TI. Otherwise, I do not understand your question because "classical" transimpedance amps have, of course, also a feedback capacitor.

It depends on the frequency of the signal.  For low frequencies, the capacitor has little effect and the TI behaves like a current-sensitive amplifier.

Usually the feedback capacitance is added to a TI to compensate for the capacitance of photo-diode (or other semi-conductor-based light detector) - without the appropriate feedback capacitance, TIs exhibit a response peak instead of the flat frequency response that is usually desired, because of the capacitance of the photo-diode (or other stray capacitance) seen at the amplifier input.

Hi Vangenheim,

Sorry, I am not talking aboout a specific Op-Amp. I am talking about the Design.  If you can, please draw an integrator with feedback resistor and TIA with feed back capacitor. What is the difference you see there?

You can create the desired function over a broad frequency response, but not including DC for the integrator. you can improve the integrator by splitting the feedback resistor and adding a capacitor to ground, forming a bridge T network in feedback.

A transimpedance amplifier (TI) can be used in a similar way a classical  opamp is used - with one important exception: There must be always a minimum ohmc resistor in the feedback path with a value as specified from the manufacturer (typically, some hundreds of ohms). Therefore, using a TI as an inverting intergator, you can use the same combination as for an opamp (feedback capacitor C and input resistor R1 to the inverting input) - and, in addition, a resistor R2 between the common node of R1 and C and the inv. input. The value of R2 must be as recommended in the data sheet for the TI. 

CFA (TI) "does not like" a capacity between inverting input and a common node. Very simple description you can see on pgs 69 to 72 in

https://www.researchgate.net/publication/281480600_Theory_of_electronic_circuits

Book Theory of electronic circuits

More information you can find in: Franco, S., “Current Feedback Amplifiers Benefit High-Speed Designs,” EDN, January 5, 1989,pgs 161 to 172

or see below

You did not give any schematics.

Anyway, I'm quite sure that in both cases you haven't got an 'ideal' integrator but some low-pass to get rid of unwanted high-frequency signal contents.

Josef, CFA has nothing to do with this circuit... The question is about the circuit solution made by the conventional voltage op-amp.

Hi Cyril - as far as I know the term "transimpedance amplifier" (TI) is frequently used instead of "current-feedback amplifier" (CFA). Both have an inverting low-resistive and a non-inv. high resistive input.

(The fact that a voltage opamp can be wired and used as a TI is another story).

Yes Lutz, but TI also has nothing to do with this circuit...  and "the fact that a voltage opamp can be wired and used as a TI" is exactly this story... Let me tell another version of this "tale"...

Both circuits exploit the same idea. They consist of a passive element (resistor or capacitor) that serves as a current-to-voltage converter or a current-to-voltage integrator. A current source passes an input current through the element, and the voltage drop across it should serve as an output quantity. The problem is when the input current source is not perfect - then the voltage drop across the element introduces an error (decreases the current).

So the remedy is to neutralize this voltage drop and then to use it as an output quantity. This neutralization can be implemented in two ways - without and with using a negative feedback. The former can be realized by connecting a negative resistor (capacitor) in series to the element. Thus the resulting voltage and impedance is zero and the input source is shorted.

In the latter (more often used), an op-amp is forced (by the mechanism of a voltage-type negative feedback) to add an equivalent voltage to the voltage drop across the element. The result is the same but the error is less.

This real op-amp implementation needs these additional elements in parallel to the main element - a resistor (in the case of a charge amplifier) and a capacitor (in the case of a transimpedance amplifier).

But the fact that a main element from the one circuit is used as a secondary element to the other one, has given a rise to confusion in Gyanendra...

In this combination of two elements in parallel, depending on the situation (frequency), the one element dominates over the other... and changing the proportion between their values, we will see how the circuit fades in/out from the one kind to the other... and how the (name of the) circuit gradually changes from TIA to CA or v.v.... as if the circuit incarnates from one body to another:) 

https://www.researchgate.net/post/Is_there_any_connection_between_the_humble_resistor_and_the_transimpedance_amplifier_What_does_the_op-amp_really_do_in_this_electronic_circuit

https://www.researchgate.net/post/Can_we_virtually_increase_the_capacitance_up_to_infinity_How_do_we_make_such_virtual_capacitors_Have_they_the_same_properties_as_real_capacitors

Cyril - after carefully reading again the question, I must admit that you are right. The OP is asking for two different applications of the classical voltage opamp. From the beginning, this was not clear to me - my fault.

Therefore - here is my updated answer:

Gyanendra Bhattarai, you are right - if we (a) use a (small) capacitor across the feedback resistor of the TIA  and (b) a (large) resistor across the feedback capacitor of an intergator circuit, both feedback blocks look similar. But watch the values!.

For the TIA, the capacitor  (if it is necessary) is very small, but it does not cause any integration procees. Instead, it creates a high frequency pole, thereby improving the phase margin somewhat. But realize that in most cases, such a capacitor is not necessary.

For the integrator, the feedback capacitor has a major (integrating) influence and the resistor should be made as large as possible. But note that there are cases where such a resistor is not needed (filter and oscillator circuits). This is the case when the integrator is part of  a system with overall feedback.  

Lutz, I agree... Also, the capacitor can be (if possible) periodically discharged...

The current integrator is a particular case of CCVS  transimpedance amplifier (current compare and voltage sampling amplifier), having as feedback 1/sC network. 

By the way, in my diploma thesis that I developed in 1981, I had to design such an AC charge amplifier...

As David Ashby said above, it's a matter of frequency response and gain. In the integrator circuit, the resistor across the integrator capacitor adds a zero in the gain at some low frequency (1/(2*pi*R*C) based on the time constant of the RC feedback network, so the gain does not continue to rise at 20dB/decade as frequency moves towards DC, as in the ideal integrator. The maximum low freq gain is due to the resistor, and at frequencies below the RC zero the circuit looks identical to a transimpedance (current to voltage converter) amplifier. This concept applies to the transimpedance amplifier as well, where the capacitor added across the feedback resistor creates a pole (1/(2*pi*R*C) in the gain such that the gain rolls off at -20dB/decade beyond this cut-off frequency.