# Transimpedance Amplifier for (TIA) for nanoAmps

Discussion in 'Analog & Mixed-Signal Design' started by Leotixidas, Jan 17, 2016.

1. ### Leotixidas Thread Starter New Member

Jan 17, 2016
5
0
Hello all,

I am trying to design a circuit to convert current to voltage.

The problem is that I have a 1kHz AC signal of 200 nA of amplitude that is carried on top of a DC current offset of 2mA . I would likemake the conversionl from current to voltage and then clear the the DC offset so I can get just the AC signal.

Do you think that a Transimpedance Amplifier circuit can help me doing that?

I am trying to understand the limitations (if any) of the TIA regarding the bandwidth of the signal I am trying to isolate (1KHz)

I am also worried about the fact I try to detect a small signal on top of a much larger DC offset.

Finally how what id the best way to deal with the noise of the system?

Thanks in advance for any and all help!

2. ### sailorjoe Member

Jun 4, 2013
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3. ### Leotixidas Thread Starter New Member

Jan 17, 2016
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i will spent some time on the article though just to make sure that the math is on my side on this circuit. (I am not sure about that at the moment)

Do you think TIA is the only way for a conversion like that?

4. ### Bordodynov Active Member

May 20, 2015
641
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See

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5. ### crutschow Expert

Mar 14, 2008
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A TIA is not the only way, but it is likely the best way.

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6. ### SWer New Member

Dec 20, 2015
16
4

Let's start with the transimpedance amp and the current source: a 1 kHz 200 nA signal riding on top of a 2 mA offset. A TIA is a good choice for the "current to voltage" converter so I suggest sticking with that approach. However, as shown on your schematic, the 1M feedback resistor is MUCH too large. Reduce it to something in the range of 3k9. This will then isolate the rest of your signal processing circuitry from the input current source - 3k9 x 2mA will result in a 780 uV signal riding on a 7.8V offset. You should use a low noise, both current and voltage, FET input opamp for the TIA amp.
Next you need to get rid of this very large DC offset, as you suggested in your original post. One simple way to do this is to use a simple RC high pass filter into the non-inverting input of a voltage follower, say 10uF followed by a 10K resistor to ground. Consider using an opamp with very low input bias current to avoid introducing another offset into your output signal. The Fc of a 10uF/10K single pole HPF is ~1.6 Hz. This is a very simplistic approach and needs careful attention to avoid adding DC offsets to your recovered signal all throughout the rest of your processing/filtering chain. The actual signal is still of very low amplitude.
Now with the large offset voltage eliminated you can concentrate on amplifying your signal of interest to usable levels. Noise is going to be an ongoing problem that you will need to address when you are amplifying the very low amplitude 780 uV signal to usable signal levels but I think you already know that.
Best regards!

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7. ### Leotixidas Thread Starter New Member

Jan 17, 2016
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To begin with thank you all guys for the quick reply and your suggestions. I really appreciate it!

I have 3 questions

Question 1
About the noise (which is definitely an issue and SWer mentioned it correctly ). How do we usually deal with it (i mean the standard procedure)?

Are the filters what usually used when we have to deal with noise? I was thinking a band pass filter that with a bandwith 100Hz to 10 KHz (my signal is at 1kHz so I was thinking 10 times lower for the low fc and 10 times higher the high fc..). That bandpass will also allows us to get rid off the DC offset as SWer suggested

what does sound to you as an approach?

Question 2
I started studying a bit the theory behind the opamps and I see that open loop gain ( AOL) is typically very low. (a few Hz)

Does that affect the function of the TIA?.....is it possible that the drop of the AOL at 1KHz (the frequency that I care about) will cause the opamp NOT to function correctly and NOT to convert current to voltage correctly?..

In other words the non-inverting and the inverting inputs will not both be at (almost) the same potential any more (their difference is not zero as the output gain is not infinite anymore...).

To state it differently is there any danger that the bandwidth of the desired conversion does not include the 1kHz which really interests me. Is that a problem with the signal I try to convert? (Crutchow that is the reason why I asked if the TIA is the best approach )

forgive me if my question is simplistic but I am just starting to understand circuit design

Question 3

the schematic is not mine but a suggestion from Borbodynov (Thank you for that Borbodynov)

Borbodynov I quite follow you with the transimpedance amplifier , the two nothch filters (...why a notch filter at 100Hz?) I dont quite undersatnd the last two blocks. What is the purpose of the 1kHz opamps? are those filters?

Thank you all again guys!

Best Regards

8. ### sailorjoe Member

Jun 4, 2013
361
63
On Q1, in basic terms, there are two kinds of noise, narrow band and broadband. An example of narrowband noise is 50 or 60 Hz hum from power lines, and 100 or 120 Hz noise from fluorescent lights.
These can be eliminated with notch filters.
Broadband noise is noise (random signal) that covers a broad spectrum. You can't filter it with a notch filter. But since it's energy is spread out over a large spectrum, you can reduce it a lot with a band pass filter that is as narrow as possible. This is precisely what we do by tuning a radio. The filter should be no wider than what is needed to pass the good signal, yet block everything else.
All of this is exactly what bornodynov did for you in his circuit.

Last edited: Jan 18, 2016
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Dec 20, 2015
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10. ### SWer New Member

Dec 20, 2015
16
4
I'll start with Q2. Without going into a detailed explantion about open-loop gain and closed loop gain with an opamp let me just answer your basic question about whether the transimphedance amplifer will accurately convert the current at 1 kHz and the answer is a very definite yes. Your bandwidth needs (1 kHz) are very modest so an opamp like a LF356 will easily satisfy that need, however noise is going to be your greatest problem.
More study will reveal the differences between open-loop gain and closed-loop gain and resulting bandwidth of an opamp.

Before I go on about noise, I'd like to ask a few questions that could lead to a much better solution when recovering small signals from massive amounts of noise

What is the source of the 1kHz signal? Can you externally control whatever it is that is generating that signal? Is your current source ouputing that signal as the result of some physical phenomenon that you are sensing or measuring with with a transducer of some kind? Will the signal be changing in amplitude and/or frequency? Will the 2mA offset remain steady or will it possibly change as well?

I'm not asking you to divulge any proprietary information, just some basic info.

Regards!

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11. ### Leotixidas Thread Starter New Member

Jan 17, 2016
5
0
Dear SWer

No the signal will nor be changing in amplitude or frequency. the 2mA may change a bit but lets assume it steady for the shake of the argument and lets say that I can not control the external signal .
I hope that gave you a more specific idea. If you need more information we can talk more about it if you like.

Does that changes the analysis somehow? How does that affect the noise analysis? (you are right noise is a big issue)

I have anotheer question. What about connecting a resistor just after the output of the current source? (i know that is not optimal as the resistor will interfere somehow with the current but my question is how "unacceptable" is this interference?)In other words how imperative is the use of the TIA?

Best regards

12. ### SWer New Member

Dec 20, 2015
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13. ### SWer New Member

Dec 20, 2015
16
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Leotixidas,
I'll first answer your question about terminating the output from the detector with a resistor. I have seen some working designs that do exactly that, usually a 50 ohm resistor followed by a opamp connected as a standard inverting voltage amplifier. I didn't ever try this configuration except with SPICE simulations mostly to explore the impact on bandwidth. I don't remember very much about those simulations. My instances of implementing TIAs was for photomultiplier signal currents in the 100uA-300uA amplitude range and having1MHz to 30MHz bandwidth requirements.

I still don't quite understand exactly what you need but I will offer some opinions that you might consider.

You are trying to amplify and filter a very small signal that is accompanied by a steady state signal (the offset current) that is 10,000 times larger than the signal. That in itself is something of an engineering challenge. Then you need to convert the current from your detector to a voltage for further processing-filtering. To get a useful ouput from this circuitry (I'm guessing about the actual value) you will need to amplify the 200nA current (in voltage form) by approximately a factor of 5E06 to get 1 volt at the output of your circuit.
Your detector will no doubt have some noise in the current output so that noise will also get amplified - that is referred to as noise in signal. Your final 1kHz output signal is going to be swamped with noise.
Each stage of amplification is going to add noise to your signal as well as DC offsets that will need to be dealt with. You are undertaking a major engineering challenge that will include low noise circuit design and developing some very stable offset voltage adjustments!

Is this an academic exercise or do you have a real need for such a circuit ?

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14. ### GopherT AAC Fanatic!

Nov 23, 2012
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SWer, use the reply button and then type below the (slash "QUOTE) tag. You don't have to make two posts each time.

15. ### Leotixidas Thread Starter New Member

Jan 17, 2016
5
0

SWer,

Yes the way you said it is correct. Except we I don't really need to get up to 1 volt V at the output of the circuit.(mV or even μV could do the job).

It is not an academic exercise I am really trying to built the circuit.

Do you think it can be done? Yes the signal is swamped in noise that is why I was asking a bout clearing the output from anything else than my signal.

About terminating the output with a resistor. i know that it is not optimal but it was just an idea for a fast conversion. I haven't seen many circuits using this approach myself though and I was wondering that there is may be a reason for that. (Every time I search I to V conversion I found myself studying a TIA circuit).Do you have any idea why it is not used so often? I know that adding a resistor alters the very current you are trying to detect but is that fact so critical?

Regards

16. ### Bordodynov Active Member

May 20, 2015
641
188
I offer the alternative variant of the unset of the big parasite currents (2 mA) by resonance scheme.
His(its) necessary to adjust on frequency 1 кГц.
Here is see to my scheme and results calculation:

17. ### SWer New Member

Dec 20, 2015
16
4
First to GopherT: Thank you for explaining about replying. I apologize to you and the other members of the forum for my earlier double postings. The process is not very intuitive.

Leotixidas and Bordodynov,

Leotixidas: The reason why a resistor to ground followed by an amp is not used is because (among many reasons) the technique requires an opamp with a very large gain-bandwidth product for any significant amount of gain whereas a simple TIA configuration does not. That, in a nutshell, is why you only find TIA amps for this type of application. A TIA might not be the only way to convert current to voltage but it is the only practical way.
Now that you have explained more about what you are trying to accomplish and the levels of the signal that would suffice for your application, Bordodynov seems to have provided a simple solution.

Best wishes to you both!

18. ### GopherT AAC Fanatic!

Nov 23, 2012
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What is the goal? Do you need a very accurate measurement of the 1kHz signal strength or do you just need to know that it exists (variable level of amplification is allowed?). Do you just need to prove that you can attenuate the noise in the circuit, what is it that you are trying to do.

Depending on your goals, there are much easier ways than the 6-op amp design proposed, and with a much lower noise result (if you share your goals). The source of the signal would also be helpful to know.

19. ### Bordodynov Active Member

May 20, 2015
641
188
I researched the possibility to improve my circuit. It turned out that you can throw away the night filters, and use a couple of filters at 1 kHz. At this frequency will be well filtered from 1 Hz to 500 Hz. Those. only one operational amplifier for a transimpedance amplifier. As well as two dual operational amplifiers to filter 1 kHz.

20. ### GopherT AAC Fanatic!

Nov 23, 2012
6,060
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UF noise is a concern, common mode input should be used for the DC and take advantage of the common mode rejection that a differential amplifier can provide. No 50Hz filtering needed.

Simply divide the whole signal. Make sure resistance is not so high that the constant current supply saturates (I x R voltage on the differential legs cannot exceed max supply on path to ground). In this example, the 2ma is divided into two equal paths to ground at 1mA each. Multiply by 10,044 ohms so you will need 10.044 volts of compliant constant current.

The two identical paths to ground will cancel all noise (normally). The twin T notch filter on the inverting input takes the 100nA AC signal and attenuated by 40db. Hence, you now have a differential input (AC signal is still at the non-inverting input).

The filter is centered at 1k and extends back to 200hz or so and up to 8k Hz or so. An active notch filter may be narrower or deeper but this is a start.

Main take-home, take advantage of the common mode rejection so you don't need to filter so many frequencies of interference.

The following first stage should yield about 1-4 mV p-p signal and 1-4V out of second stage.