OPA380: How to measure current using Trans-impedance amplifier

Prodigy 120 points

Replies: 8

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Part Number: OPA380

Hello,

I am trying to build a current sensing circuit using a trans impedance amplifier(TIA). I wanted to implement the I to V conversion stage(as shown in attached diagram) using a trans-impedance amplifier and also be able to GND the line at the same time. I am unsure how to implement this and needed help. Please let me know if

1) This is possible using TIA.

2) How would I go about implementing the I to V stage using TIA.

Thanks a lot,
Sufyan Khan

8 Replies

  • Morning Sufyan, 

    This transimpedance thing seems to be coming up a lot, an area I have published quite a lot on -going back to the mid-90's actually. Where to start? Many different permutations here, 

    1. key thing is the source, you show voltage source with R? That sounds more like a voltage gain stage. - normally, you have a cap source of a detector diode or something. 

    2. pretty much all ADC's are diff input now, for which we have FDA's to drive those,  

    As part of the OPA838 product launch, I wrote this detailed two stage (noise analysis mainly) article - just an example, click on these EDN figures to expand in new window, 

    https://www.edn.com/design/analog/4458766/Controlling-spot-and-integrated-noise-in-a-two-stage-transimpedance-design

    And, more recently, this general decomp discussion (most I/V end up using decomp)

    pdf here, as planet analog broke their figure expand links in august 2019, 

    3364.Applying High Speed DeCompensated VFAs July1_2019.pdf

    Michael Steffes

  • In reply to Michael Steffes:

    Hello,

    Thank you for your prompt reply.

    The circuit that I have attached in my previous post is a simplified representation of my actual application circuit. I want to be able to measure the current flowing through the load in the range of 10nA-100uA. If TI has any reference design or if you can suggest a suitable IC for this purpose, it would be helpful.

    Please guide.

  • In reply to Sufyan Khan:

    So that sounds like a high or low side current sense app - the sense resistor is usually very small, followed by an INA - there are parts that do that, just look on the TI web site for current sense or INA

    Michael Steffes

  • In reply to Michael Steffes:

    Most of the integrated solutions like INA250, include a sense resistor that is very small(mOhm), so if I am trying to sense currents in nA,the output of the chip would have a low voltage amplitude even in the highest gain variant of the chip(A4), since the drop across the sense resistor is extremely small. It is my understanding that these parts are designed for Ampere level currents. Are there IC's for precison uA/nA sensing from TI?

    Also the sense resistor method of sensing is one way to do it. I was actually trying to build a circuit such that it could measure current without using a sense resistor. Since my current amplitudes are low, the noise will also come into the picture when I am trying to use the drop across the sense register as the sole basis for calculating current. If there was a direct method such that the trans-impedance amplifier could be connected in series with the load, and give an amplified voltage output proportional to the current through the line, it would definitely solve the issue.

    Please let me know what you think.

    Thanks and regards,

    Sufyan Khan

  • In reply to Sufyan Khan:

    You are correct in those INA Ampere type sense solutions, 

    Perhaps a list of specs would prove useful, 

    1. signal current range

    2. Desired gain

    3 Desired frequency  span, 

    Those articles I linked earlier would perhaps help. And yes the SNR comes into this eventually, 

    Michael Steffes

  • In reply to Michael Steffes:

    Thanks for your reply Michael,

    1. Signal current range: 10nA to 100uA

    2. Desired Gain: For 10nA, output should be 0.5mV, and for 100uA, output should be 5V. Gain can be set accordingly.

    3. Maximum frequency of input signal: 1Mhz

    Please let me know if you need more information from me.

    Thanks and regards,

    Sufyan Khan

  • In reply to Sufyan Khan:

    Thanks, that is good info. 

    To create this "50kohm" gain, it is still perplexing how do deal with your voltage source through an R source. Not really a current source, is the voltage unknown or the R unknown, and that is what you are trying measure? 

    A classic transimpedance design sees a high DC impedance in shunt with a C source - not the case here, so maybe a voltage amplifier instead. In any case, look at the LMP7718 for a low input Ibias high GBP solution - of some sort.

    Michael Steffes

  • In reply to Michael Steffes:

    The V source is known. What I'm trying to achieve is measure very high resistances using this circuit.

    For example, if V is 10V, I want to measure resistances from 1Gohm to 100kohm, i.e currents ranging from 10nA to 100uA. So, in such a case, I know that V=10V, I will feed the output of my amplifier to the ADC and then I will be able to calculate the resistance present in the line. Thus, V is known, I is measured and hence R is calculated.

    Thank you for your patience, I hope we can find a solution for my query.

    Regards,

    Sufyan Khan