Transimpedance Amplifier - Transimpedance gain Vs GBW product - Page 1
Forum section: electronics / projects

author: arivalagan13
author_rank: Contributor
timestamp: July 14, 2019, 10:51:23 am
content: Hi allI am designing a precision transimpedance amplifier. I've got a question regarding the design of the TIA.What is the relationship between the transimpedance gain and the gain-bandwidth product of the operational amplifier used in the design?Example: Let's choose a non-inverting amplifier.If I choose an opamp with a GBW product of say, 500 MHz, the product of the closed loop gain and the bandwidth can never exceed 500 MHz.Loop Gain x Bandwidth = 500MHzIf I choose to design a transimpedance amplifier in the place of the non-inverting amplifier the above relationship still holds?i.e. transimpedance gain x Bandwidth = 500 MHzAm I clear in my question?Relevant answers and documentation will be of much use for my project.RegardsM ArivalaganResearch ScholarAnna UniversityChennaiIndia

author: magic
author_rank: Super Contributor
timestamp: July 14, 2019, 11:47:00 am
content: Closed loop gain is a concept in current to current or voltage to voltage amplifiers. I'm not quite sure how to define it in a current to voltage amplifier. I am however quite sure thatopenloop gain is GBW divided by frequency. Although not always and not at every frequency - check your opamp's datasheet for exact OLG vs freq plot. Note that OLG may vary a few dB from sample to sample.OLG is defined as output swing divided by differential input swing. So it determines precision - higher OLG means that the inverting input is held closer to ground for the same output swing. You do the math.There is no direct connection between OLG and TI except for one: for the same input current, higher TI produces higher output swing and therefore higher input swing. However, the same effect can be caused by higher input current, regardless of TI gain.Note that when the inverting input moves away from ground, voltage across the feedback resistor changes and TI gain is no longer what you expect.

author: awallin
author_rank: Frequent Contributor
timestamp: July 14, 2019, 12:01:57 pm
content: usually the photodiode capacitance (together with the op-amp input capacitance) together with the transimpedance-resistor will limit the bandwidth.the op-amp just has to be fast enough not to limit the overall bandwidth.much more bandwidth in the op-amp won't help much - I think it just gives you more noise at high frequencies where you don't have transimpedance gain.the mentioned OPA656 has a 1MHz design on the datasheet front page, with 1 MOhm transimpedance. Your other posts talked about 300MHz bandwidth, and now you suggest a 1 GOhm transmipedance - so things don't quite add up...

author: pwlps
author_rank: Frequent Contributor
timestamp: July 14, 2019, 12:12:08 pm
content: As magic pointed out you can't define a "closed loop voltage gain" for a TIA.Quote from: arivalagan13 on July 14, 2019, 10:51:23 amit can't be like this: "transimpedance gain" is voltage/current, the units don't match!However you can easily do the calculation yourself starting from first principles. For a simple TIA with just a resistor R between V- and Vo the equations are:Vo = A(V+ - V-) = -A V-IR = V- - Vowhere A is the open loop gain that can be approximated by A=GBP/fThe solution isVo = -IR A/(1+A)therefore, in contrast to a voltage amplifier, the transimpedance "gain" Vo/I will stay high up to GBP frequency. However at high frequencies you have to include parasitic capacitances (also check for stability if you want to have high gain), you might look here for a more advanced calculation:http://www.ti.com/lit/an/sboa122/sboa122.pdf

author: magic
author_rank: Super Contributor
timestamp: July 14, 2019, 12:22:33 pm
content: Quote from: pwlps on July 14, 2019, 12:12:08 pmYes, you are right. That's actually a simple formula for "close loop transimpedance" and I checked the math, it's legit. I though about this problem just yesterday and ended up concluding that no such formula can exist because it depends on input current magnitude. Clearly I was wrong, probably by forgetting to calculate output errorrelativeto input current