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Transimpedance amplifier

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houly

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hello
I would want to know what is the function of theis particular feedback circuit ... ?

and how can I calculate it ? I can consider the photodiode as a current generator ... because we are in linear mode and V+=0v, V- is equal to 0V too. so does this equivalent schematic is good ? I'm not sure that I can consider that because the current generator should also be shorted to the ground ?!

here is the original schematic

**broken link removed**​

here is my equivalent 1 and 2 schematics to calculate the ouput with all elements.

whith virtual ground of the op amp, we should have :
**broken link removed**​

without considering the virtual ground, we should have :
**broken link removed**​

I hope that someone could help me to understand the aim of the feedback and how can I calculate it.
 

I think it increases the effective transimpedance by a factor of 10 (so 15meg) at high frequencies but I would have to check properly when I have more time.

Keith

Actually, I think that is incorrect. I will check later.
 

Thanks keith, and do you know what equivalent schematic is good to calculate the feedback ?
 

OK, I have had a chance to look properly and the feedback with RA, RB and CF is a method of providing a small feedback capacitor across RF. It would actually seem a little unnecessary with 36pF - you could remove RA and RB and simply put 3.6pF across RF instead. The method makes more sense if you want a capacitor of say 0.1pF across RF. Then you could use the tapped feedback of RA/RB so you can use 1pF for CF or change the resistor ratios to use an even bigger CF while still having the effect of 0.1pF across RF. However, stray capacitances could still make that approach fail.

Keith.
 

albbg said:
This is not a generally desirable reference since it uses a Tee feedback network which is usually not recommended since this depreciates the signal/noise ratio. See the application note: Photodiode monitoring with op amps (originally from Burr-Brown, now part of Texas Instruments , with reference number AB-075 and sboa035.pdf). In the case of the paper quoted above the use of the Tee feedback configuration is acceptable since there is a high gain preamplifier in the form of a photomultiplier so the noise performance of the transimpedance stage is largely irrelevant.
Click to expand...
 

The circuit linked by albbg does the opposite of the original circuit, dividing the R feedback branch instead of the C branch, so it doesn't explain anything.

A reason for the divider may be, that the OP isn't unity gain stable, but the effect depends on the diode capacitance and OP GBW, without this information, everything is just a guess. I should add, the assumption, that the original circuit designer did something resaonable isn't more than a guess.

As Keith mentioned, it can be used as a means to set small C feedback values. But 3.6, or more exactly 3.3 pF is still a convenient capacitor size.
 

Sorry, you are definitely right.
I didn't read carefully the paper and at first glance the proposed circuit seemed to me the same as the discussed in the article.
 

FvM said:
The circuit linked by albbg does the opposite of the original circuit, dividing the R feedback branch instead of the C branch, so it doesn't explain anything.

A reason for the divider may be, that the OP isn't unity gain stable, but the effect depends on the diode capacitance and OP GBW, without this information, everything is just a guess. I should add, the assumption, that the original circuit designer did something resaonable isn't more than a guess.

As Keith mentioned, it can be used as a means to set small C feedback values. But 3.6, or more exactly 3.3 pF is still a convenient capacitor size.
Click to expand...

The observations made by Keith1200rs and FvM that the resistive divider is used to reduce the effective value of CF are correct, but this configuration does raise some questions. The divider resistance needs to be kept low so the effective source resistance of the divider is low and hence that the impedance of the feedback of this leg is primarily capacitive. But this means that in the present case for say a typical output swing of 10V the opamp must supply a current of ~20mA.
The other question is why is it necessary to have CF at all. It is required in order to achieve dynamic stability of the system. There is a component missing from the original circuit schematic i.e. the capacity of the photodiode (say CD), which in this zero bias configuration can be substantial. Introduction of the appropriate CF in effect neutralizes the RFxCD time constant and hence makes the loop phase shift much less than the ~360degree it would otherwise tend to and hence be on the verge of instability with most undesirable transient response. It may seem an unnecessary complex arrangement to obtain a low value effective capacity but with much higher transimpedance gains (I use as high as 100Tohm) the required capacity becomes very low and the CF leakage resistance must be >>100Tohm. For a fuller treatment of the transimpedance amplifier see the TI application note referred to in the previous post (sboa035 at TI.com) or the book Hamilton (2007):An analog electronics companion (p488-504) ISBN 9780521687805 and the many references there. CF and CD also have a significant effect on the noise performance and this should be considered as well.
 

Besides diode capacitance and OP GBW, also the circuit application respectively signal characteristic are unknown. So you can't decide, if Cf is only providing frequency compensation, which is surely required already for moderate OP bandwidths, or if a charge amplifier characteristic is intended for the frequency range of interest.

I just realized, that the original post is a few weeks old, however one question hasn't been answered yet: Clearly, the last equivalent circuit is correct for the feedback factor calculation, but you have to add the diode (and also OP input) capacitance.
 

FvM said:
Besides diode capacitance and OP GBW, also the circuit application respectively signal characteristic are unknown. So you can't decide, if Cf is only providing frequency compensation, which is surely required already for moderate OP bandwidths, or if a charge amplifier characteristic is intended for the frequency range of interest.

I just realized, that the original post is a few weeks old, however one question hasn't been answered yet: Clearly, the last equivalent circuit is correct for the feedback factor calculation, but you have to add the diode (and also OP input) capacitance.
Click to expand...

Prompted by the post of FvM, it is not clear what the original post by houly was looking for. The equivalent circuits suggested are missing both the diode capacity and the op amp. Both of these are essential if the circuit is to be simulated to find the frequency and transient response. Additional to the previous reference sboa035 from Texas instruments, here are two more that should be of considerable use:
1) AN1803 from National Semiconductor: Design Considerations for a Transimpedance Amplifier by Maithil Pachchigar, available at www.national.com
2) Understand and apply the transimpedance amplifier by D Westerman, available at
(and this has a link to part 1)
 
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