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I'm doing an amplifier for a photodiode. In my search I see the transimpedance amplifier is the most used. I will use a ADC just after. The value of my feedback resistor will be 100 MOhm.

I've done some more research and I found some information about the integrator. I know how it works but what are the mainly advantages/disadvantages between the integrator and the transimpedance configuration (bandwidth, precision, level of current to measure,...)

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  • \$\begingroup\$ Have you seen an application that uses an integrator instead of a TIA or is this just some kind of guess? \$\endgroup\$ – Andy aka May 16 '14 at 14:41
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    \$\begingroup\$ Do you really need 1x10^8 V/A? Like others mentioned diode and PCB leakage currents and parasitics will be a challenge. What kind of bandwidth do you need? Even though a TIA theoretically has the full gaind-bandwidth of the op-amp, a 100 MOhm resistor along with parasitic capacitance could be an issue. Also a resistor that large has a lot of thermal noise. \$\endgroup\$ – John D May 16 '14 at 15:35
  • \$\begingroup\$ Yes I know all application use a TIA. My thesis is about photodiode amplifier with high gain. I know the PCB leakage current and my circuit works. The question was just to know what a integrator is not use. \$\endgroup\$ – laurent May 16 '14 at 21:29
  • \$\begingroup\$ As others said using a 100MOhm resistor is calling for trouble. Why not use a more sane value like 1Meg and follow the TIA with a second 'gain of 100' amplifier stage? \$\endgroup\$ – Nils Pipenbrinck Jun 15 '15 at 15:17
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Please be aware that you are asking for trouble. A feedback resistance of 100MOhm will cause you all sorts of problems. You are talking such low currents that leakage will be a big problem, and if you are making a pc board, scrupulous cleanliness and removal of ALL traces of solder flux will be critical.

That said, do not use an integrator. Any problem a transimpedance amp has with leakage applies to integrators.

Furthermore, you have not described your system well enough to determine why your light levels are so low, and this opens up speculation about other potential problems. If your light levels are low because the source is far away, then you will have major problems with background and stray light. It's true you can deal with this (more optical effort, source modulation/demodulation, narrowband filters, etc), but you haven't given any indication of where to start.

I would suggest that you give serious thought to some form of optical enhancement, using a lens to focus incoming light on your detector and increase optical flux.

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  • \$\begingroup\$ My thesis is about high gain photodiode amplifier. I'm electronician. The opticians say they must measure this level of light. The goal of the thesis is also to deal with all these problems (noise, leakage,..). So the main problem of integrator is the same of the TIA. It's the leakage currents ? \$\endgroup\$ – laurent May 16 '14 at 21:34
  • \$\begingroup\$ @Laurent - I don't know where to start. This is your thesis? And you have no clue about the differences? Oh my. Well, let's start at the beginning. What wavelengths are you working at? What optical power levels? What sort of time duration (that is, are you looking at pulses, AC, or DC? If pulses, is it single shot or repetitive, and in either case what pulse width and what peak intensity? \$\endgroup\$ – WhatRoughBeast May 16 '14 at 23:48
  • \$\begingroup\$ MY thesis is almost finish. I've a transimpedance amplifier (and an ADC, MCU and everything else) that works. I had to do a photodiode amplifier with high gain (the photodiode used is the S2592 (Hamamatsu, hamamatsu.com/eu/en/product/category/3100/4001/4103/S2592-03/…). The bandwidth of the amplifier must be 100Hz. I know the gain must be 100*10^6 (could be less or more). My question is just to know why a TIA and not an integrator. Maybe someone will ask me this question and I don't know what to answer. \$\endgroup\$ – laurent May 17 '14 at 10:51
  • \$\begingroup\$ @Laurent - Your OP asked "what are the mainly advantages/disadvantages between the integrator and the transimpedance configuration". But it's not an either/or choice. You always use a TIA to convert light to voltage. The question is whether or not to use an "integrator" to post-process the TIA output. And by "integrator" I think you mean "box-car integrator". And you use a BCI to increase signal to noise for a pulsed light source. If the light is steady DC, no BCI. See ecse.rpi.edu/~schubert/Course-Teaching-modules/… \$\endgroup\$ – WhatRoughBeast May 17 '14 at 11:08
  • \$\begingroup\$ @laurent "MY thesis is almost finish." Wait a minute. You consider it "finished" when you don't even have a working first stage?! What signals did you use to test the rest of the design? I smell trouble... \$\endgroup\$ – Kuba Ober Jun 15 '15 at 18:57
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I've been working on building instruments with a photodiode and a high gain TIA. I started out with a 100M feedback resistor, which actually worked pretty well. I have a heater serpentine trace on an inner layer under the photodiode and analog stage, with a thermistor placed near the detector. I have a PID temp control loop keeping that part of the PCB at 40C, with a PWM'd FET from my microcontroller putting power into the heater trace. There is a groundplane between the heater trace and the analog parts. The high gain parts and the photodiode traces are all on the top side, no vias. Temp controlling all the parts is pretty important, and even then, I need to use resistors with <100ppm tempco. Everything is in a metal light tight box. I've made almost 100 of these, and at 100M feedback the performance is good. They do have to be calibrated using a precision blackbody source, but after that they are extremely accurate. I am only interested in <1Hz signal though, so not sure how this would work out for higher bandwidths.

For other system level reasons, the light flux has been cut to 1/4 of what it was, so now I'm up to 400M feedback resistors, and having some trouble with measurement stability. So I'm thinking of moving to an integrator to see how that works. Another benefit of the integrator is that you can change the gain by changing the integration time. Right now I use an SMT reed relay to switch in a lower value parallel feedback resistor to change gain, which has some annoyances and is not fast.

If you do it, make sure to use a very low leakage switch to reset the integrator and a very good integration capacitor like NP0 ceramic or a silicon capacitor. I'll write again later to tell you how my integrator solution worked out.

BTW, all in one digital light sensors like those from Hamamatsu and TAOS use an integrator internally.

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    \$\begingroup\$ Phil Hobbs disclosed an interesting setup where he used a bootstrapped telcom relay pair to eliminate most of the capacitive effects switching fb resistors up into the G ohms. \$\endgroup\$ – Spehro Pefhany Jul 10 '14 at 14:59
  • \$\begingroup\$ Such high resistance systems are absurdly hard to mass manufacture. Anything that comes off a PC-board assembly line has to be, pretty much, manually cleaned to get rid of contamination and resulting leakage. Given the current noise of op-amps, I don't quite see the point of such high single-stage gains, unless I'm missing something obvious. A 2 stage design would be presumably much easier to manufacture? \$\endgroup\$ – Kuba Ober Jun 15 '15 at 18:56

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