• State
  • Locked Locked
  • Replies 6 replies
  • Subscribers 46 subscribers
  • Views 1307 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Current Suppression Techniques before Amplification of a Low Noise Signal

Fatin Haque
Prodigy 80 points
Other Parts Discussed in Thread: ISO120

Hi again,

This is an interesting problem. I wish to amplify my signal of interest that's in the few nA range that is riding on top of a few hundred mA DC signal. The bandwidth of the signal of interest is 100kHz (related to my other post).

Part A)

I need to amplify the signal of interest as much as possible (10^9 V/A gain if possible). But, to get this sort of gain from any I->V amplifier, I would think that I need to suppress the current as much as possible without introducing too much more noise since the signal is in the nA range.

What is my best option to suppress this current? The only thoughts that I have is either having a dedicated voltage source (adjustable by hand by some potentiometer?), invert the source signal, and then subtract the current before amplification. Or, something like use a second amplifier, and subtract the from the original signal.

Part B)

To get a gain of 10^9 V/A would require large feedback resistor value for an opamp. Is this a problem?

I'm not sure what to do here. Thanks!

  • 0
    Guru 21975 points

    Fatin,

    If you could provide more detail on your measurement problem we may be able to provide better advice.

    What is the frequency range of your signal of interest?  100kHz is the upper limit. What is the low end of the frequency range of interest?

    What is the voltage level at which you are measuring the current? Is this measured at some high voltage level? (Perhaps the reason for the ISO120 mentioned in a previous post... ISO120 100kHz BW boost component selection.

    If you could provide information on the source of the current and signal type, it could be very helpful.

    Regards, Bruce.

  • 0 in reply to Bruce Trump
    Prodigy 80 points

    The low end of the frequency of interest is DC.

     

    The voltage levels are due to the biasing voltages on a MOSFET (within a few volts for each...enough to cover IV curve. Maximum 20V).

     

    I just posted in my other thread some details (maybe I should'v combined the threads). Here is some more details:

    "What I am trying to do is measure RTS / 1/f noise in a MOSFET. This current signal is on the order of a few nA. But, I need to drive Ids to different values on the MOSFET IV curve (triode region, and weak inversion). This signal needs to go into some current preamplifier that can take this small signal and amplify it enough to be detected by the oscilloscope.

     

    The isolation amplifier would be placed between the amplifier and the oscilloscope since the power supplies biasing the MOSFET is on a different ground than the oscilloscope. So, the input of the ISO would be a few volts (1-5V) and the output would be of the same order (unity gain)."

  • 0 in reply to Fatin Haque
    Guru 21975 points

    Fatin,

    Okay, I understand much better now. I don't think that you need the isolation amp.

    I will come back with a possible circuit approach. Do you have clear access to the gate, source and drain of the MOSFET? Can I bias it in any convenient circuit to facilitate the noise measurement?

    Bruce

  • 0 in reply to Bruce Trump
    Prodigy 80 points

    Thanks! 

     

    I plan to place the MOSFET and amplification circuit onto a PCB. The PCB will be in a shielded box where there are connections to shielded twisted pair. The power supplies will send their power through the twisted pair. This will go onto the PCB where I will have voltage dividers and low pass filters on the PCB to clean up the signal. This cleaned up signal will have access to the MOSFET.

     

    Thanks!

  • 0 in reply to Fatin Haque
    Guru 21975 points

    Fatin,

    Here is a simplified circuit that may be a starting point. The two supplies on the MOSFET set up the operating point. Any noise current in the source (simulated by source IG1) is forced to flow into the summing junction of U1 through C1 and is converted to a voltage output by flowing through R2.

    There are many subtle points and tricky issues with this approach. I can discuss with you if you want to provide contact information.

    Regards, Bruce.

  • 0 in reply to Bruce Trump
    Prodigy 80 points

    Thanks. Sure we can discuss this:

    519-888-4567 x31259

    fhaque [at] uwaterloo.ca