AD8220 Noise calculations

Dear Matt,

I've been doing a very critical desing using AD8220. But i'm afraid that the IC might fail the design requirement at the very last moment as i found difficulties. Hence need your worthy help.

Fist of all i need to explain my design requirement. I'm doing Neural Interfacing. The Neuronal spikes are very very small (about 50uV extracellular). We need to record those signal (about 1kHz). The problem is you can't draw much current from the cell. Hence we preffer 1pA. But anything less than 10pA is safe... So, 2 main desing requirements: ultra small signal (<50uV) and ultra low bias current (<10uA). I tried with AD8236. But it failled. The conventional system uses INA116 (Texus Instruments). But i don't like it becuase of it's bulky size. I want to use AD8220. I'm using the B version. so the current is okay for me (10pA). But i missed out the noise. Input noise seems okay. But output noise density is too high (90n//V). I just connected AD8220 without any gain resistor. Hence the gain should be 1. Without any input signal i found the baseline noise about 20uV. My question is: Is it correct? If so, then how i calculate the baseline noise? Or how i know by reading the datasheet? I need anyting less than 5uV baseline noise so that i can record 20uV spikes with good fedality. Any suggestion?

Alam,

Applied Neuroscience lab,

Hong Kong Polytechnic University.

Parents
    •  Analog Employees 
    on Aug 13, 2011 2:46 AM

    Hi Alam,

    It sounds like you've been doing a very good job testing your circuit. High frequency noise is often a problem in instrumentation circuits and it is very good practice to filter this out. I would recommend placing this circuit at your inputs anyway because RF interference can often be rectified in the In Amp and seen as a DC offset at the output if you don't filter it out first. Take care to put the capacitor, called Cd in the datasheet, between the + and - inputs of the amp, because otherwise component mismatch in the LPFs can reduce your CMRR. I would also like to call your attention to how this changes the differential corner frequency. Hence, if you took the advice of the datasheet and made your Cd 1uF, then you would want to put a approximately 50 ohm resistor for your LPFs to have about the same corner frequency. Of course, you probably want smaller caps and a larger resistor than this, but that's how the math would work out anyway.

    With respect to deriving the baseline noise for the circuit, what you have done makes sense. Once you have calculated this total noise level that you learned from Matt in the video, you still have two basic components of the baseline noise level. The first is 1/f noise, which is the 0.1Hz to 10Hz p-p value listed in the datasheet. That will probably be close to 0.8uV at a gain of 495, but for safety we can estimate 2uV p-p. The rest is the broadband noise. To find this, you multiply your total noise level by the square root of the bandwidth of interest, which is any signal that is not filtered out. So if you have 14nV/rtHz, then you multiply by sqrt(1590hz - 10hz) and you get an rms voltage of about 0.56uV. Now to combine these, we can convert the first number from peak to peak voltage to rms voltage by dividing by 6.6 and then adding the same way we have been: sqrt( (2uV/6.6)^2 + (.56uV)^2) = 0.64uV rms. Which is 4.2uV peak to peak if we convert back. This means that at the output, with a gain of 495, you should expect to see about a 2mV noise level, and a 20uV input spike would be visible above your noise level because it would be gained to 9.9mV.

    Sorry to write so much on the subject, but I hope this helps!

    Scott

Reply
    •  Analog Employees 
    on Aug 13, 2011 2:46 AM

    Hi Alam,

    It sounds like you've been doing a very good job testing your circuit. High frequency noise is often a problem in instrumentation circuits and it is very good practice to filter this out. I would recommend placing this circuit at your inputs anyway because RF interference can often be rectified in the In Amp and seen as a DC offset at the output if you don't filter it out first. Take care to put the capacitor, called Cd in the datasheet, between the + and - inputs of the amp, because otherwise component mismatch in the LPFs can reduce your CMRR. I would also like to call your attention to how this changes the differential corner frequency. Hence, if you took the advice of the datasheet and made your Cd 1uF, then you would want to put a approximately 50 ohm resistor for your LPFs to have about the same corner frequency. Of course, you probably want smaller caps and a larger resistor than this, but that's how the math would work out anyway.

    With respect to deriving the baseline noise for the circuit, what you have done makes sense. Once you have calculated this total noise level that you learned from Matt in the video, you still have two basic components of the baseline noise level. The first is 1/f noise, which is the 0.1Hz to 10Hz p-p value listed in the datasheet. That will probably be close to 0.8uV at a gain of 495, but for safety we can estimate 2uV p-p. The rest is the broadband noise. To find this, you multiply your total noise level by the square root of the bandwidth of interest, which is any signal that is not filtered out. So if you have 14nV/rtHz, then you multiply by sqrt(1590hz - 10hz) and you get an rms voltage of about 0.56uV. Now to combine these, we can convert the first number from peak to peak voltage to rms voltage by dividing by 6.6 and then adding the same way we have been: sqrt( (2uV/6.6)^2 + (.56uV)^2) = 0.64uV rms. Which is 4.2uV peak to peak if we convert back. This means that at the output, with a gain of 495, you should expect to see about a 2mV noise level, and a 20uV input spike would be visible above your noise level because it would be gained to 9.9mV.

    Sorry to write so much on the subject, but I hope this helps!

    Scott

Children
No Data