INA849: if the output noise of INA849 is good for me

Hi Nan,

Nan Li said:

My requirement is output noise less than 10pA/sqrt(Hz) in DC frequency band.

I'm a bit confused. Output noise is usually given in Volts not Ampères. And what do you mean by "DC frequency band"? What is your signal frequency range?

Nan Li said:

I only find that the RTI voltage noise is about 1000nV/sqrt(Hz), when G = 1, in this band from INA849 datasheet.

The INA849 is optimized for very high gains. And for these very high gains the noise of INA849 is ultra low.

Nan Li said:

If I need use one resistor which is larger than 100kOHM to meet my requirement?

What do you mean by this? That your source impedance is larger than 100k? What is your load resistance?

Nan Li said:

Or shall I choose one more accurate INA chip

This depends on your load resistance and the gain. What gain do you need?

Do you need an instrumentation amplifier at all? Can the job also be done with a standard OPAmp circuit?

Kai

Hi Nan,

I only find that the RTI voltage noise is about 1000nV/sqrt(Hz), when G = 1, in this band from INA849 datasheet.

Could you share the design requirements or application with us?

An instrumentation amplifier (IA) is designed for large gains, low input differential signals that operate under a noisy environment. As Kai pointed out, does your application require IA at G = 1? 

In addition, the voltage spectral noise density at 1uV/sqrt(Hz) at approx.100mHz (single point) does not say much about the overall noise contributions, which is a part of 1/f noise contribution. If your application's BW is over 10kHz or more, then the 1/f noise is contributed very little or insignificantly to overall noise integration.  We have many zero drift IAs and/or prevision op amps that can eliminate the most of 1/f noise for a low bandwidth application. 

Please watch the topics of the noise video series in the link below, if you are interested (there are total 9 videos in the noise subject). 

https://training.ti.com/ti-precision-labs-op-amps-noise-1f-flicker-noise?context=1139747-1139745-14685-1138803-13846

If you have additional questions, please let us know. 

Best,

Raymond

Hi Kai,

1) I want to output one DC bias current, so "DC frequency band" means that the frequency of output from INA849 is DC.

2) My sample resistor is larger than 100KOHM I think, not the load or source impendence. Below R1.

3) I want to use one good instrumentation amplifier, because I need the output is low noise level, smaller than 10pA/sqrt(Hz) ,,, by the way, I think "10pA/sqrt(Hz)" is equal to "1000nV/sqrt(Hz)" when R1 is 100KOHM based on Ohm's law.

Best Regards,

Nan

Hi Raymond,

I want to design one high precision DC current source with low output noise, and uA level.

The RTO noise is equal to G*RTI. So if the RTI noise is so large at DC point, the output noise is large too.

My design like below.

I want to use INA849 to generate good output with low noise, below 10pA/sqrt(Hz) as I said.

Raymond Zhang1 said:

If your application's BW is over 10kHz or more

I only want to have a good DC output. 

Best Regards,

Nan

Hi Nan,

now I understand

Kai

Hi Nan,

noise estimation becomes more simple when you short-circuit the output of your circuit to signal ground. This is allowed with a current source. Then the noise output voltage of INA849 is transformed into a noise current by dividing the noise output voltage by 100kOhm.

First of all, how much noise you will see in your application mainly depends on your bandwidth. Without any low pass filtering, which you can install at the load, e.g., you will see the total wideband noise of INA849 which will be quite a lot. If you provide a bandwidth limiting at 100kHz, on the other hand, your noise output voltage across the 100k resistor will be about

45nV/SQRT(Hz) x SQRT(100kHz) = 14.2µVrms

assumed a flat noise voltage spectral density. But because the noise is increasing at the low frequencies it has to be corrected to about 14.8µVrms, according to the TINA-TI noise simulation:

nan_ina849_1.TSC

14.8µVrms is about 14.8µVrms x 6.6 = 98µVpp, resulting in a noise current of 980pApp.

If your signal bandwidth is only 1mHz...10Hz, then your output voltage will be about 8µVpp resulting in a noise current of 80pApp. (For a bandwidth of 0.1Hz...10Hz the noise output voltage would be 5µVpp according to the datasheet.)

This is a huge difference, demonstrating how important it is to know the bandwidth of application.

By the way, the 100k resistor also adds to the noise because every resistor shows thermal noise. For a bandwidth of 100kHz the noise would be:

SQRT (4 x 1.38^-23 x 300 x 100k x 100k) x V = 12.9µVrms = 85µVpp

resulting in a noise current of 850pApp.

Adding the two noise sources will result in

SQRT (980pApp^2 + 850pApp^2) = 1.3nApp

And for a bandwidth of 1mHz....10Hz, on the other hand, the resistor noise would be 129nVrms = 850nVpp resulting in noise current of 8.5pApp.

Adding the two noise sources will result in

SQRT (80pApp^2 + 8.5pApp^2) = 80pApp

So you see that there's a huge difference depending on the bandwidth: In a wide bandwidth application the noise contribution of INA849 is way smaller than in a DC (very low frequency) application.

Kai

Hi Nan,

I only want to have a good DC output. 

Thanks for Kai's noise analysis and estimation!

I normally would not recommend to build one's own instrumentation or difference amplifiers for an application. The reasons are that the resistors inside of these op amp topology have much superior matching than the discrete components, and often it is lower in overall cost as well when dealing with 100ppm tolerance resistors. 

I simulated INA849's noise scenarios and I do not see that the it is difficult to lower Vnoise than the circuit below. OPA2388, OPA2205 or others in discrete op amps will work out better in noise than INA849 configuration in this application. 

In INA849's Howland current pump design, you have to consider 1/f, resistor noise and Vos, BW, temperature drift parameters etc.. It is easier to control the list of variables in a discrete constant current source in this case. We do have other precision op amps with even lower voltage noise spectral density figures (down to 1.1nV/sqrt(Hz)), see the attached link below. I think that zero drift op amps may work better in the application. 

OPA388 Howland I pump + OPA388 06182022.TSC

https://www.ti.com/amplifier-circuit/op-amps/precision/products.html#p7typ=1.1;3&sort=p7typ;asc

If you have additional questions, please let us know. 

Best,

Raymond

Hi Nan,

Raymond's idea with the Howland current pump built by discrete OPAmps has the advantage that you can add low pass filtering and by this heavily decrease the wideband noise. Also, I do not see any reason to use an instrumenation amplifier like the INA849 at the input of your circuit, because there's no differential signalling at the input and no common mode input voltage to reject. A standard OPAmp should do in this circuit.

But to find a good circuit we first need to know what your load is. If it's a simple resistive load, it will be easy to make the current pump stable. But if it is a complex, capacitive and/or inductive load, then some phase lead compensation may be necessary.

Also, a chooper OPAmp with its extremely low LF noise should only be used, if the load in your circuit (seen by the chopper OPAmp as source impedance !) is low ohmic, because chopper OPAmps do not like high ohmic source impedances. One trick to make them work with high ohmic source impedances is to make them see exactly the same impedance from both inputs to signal ground. But this would only work with a constant load impedance in your circuit and in this case you would not need a current pump anyway. So it can absolutely make sense to not take a chopper OPAmp here.

By the way, for lowest noise current the current sense resistor (R1 = 100k in your schematic) should be as high as possible because the noise voltage is only proportional to SQRT(R) but the noise current is proportional to 1/R, or by other words, the noise current follows the relation

SQRT(R) / R

Increasing the current sense resistor by a factor of 100, e.g., results in a decrease of noise current by a factor of 10.

I have seen current pumps with a high supply voltage to be able to generate a constant current through a very high current sense resistor, only to keep the noise current ultra low.

Another trick is to use a high voltage and a high current sense resistor -or better call it current limiting resistor- to generate a quasi constant current through the load without any regulation. Because, if the load is very small compared to the current limiting resistor, even changes of load resistance will not relevantly alter the current through the load:

nan_ina849_2.TSC

This method is super simple and does not need any current pump circuit and can be very precise at the same time. It will work even with the most sophisticated complex loads without needing any stability control or current regulation. Just a simple constant voltage is enough.

Another idea is to regulate the current by the help of the microcontroller which also feeds the DAC. This should work very well in a DC constant current application: You measure the voltage drop across the load and because you know the output voltage of your DAC you can calculate the current through the load by

I_LOAD = (U_DAC - U_LOAD) / 100k.

If the current through the load is too small, increase the DAC's output voltage and vice versa.

This circuit also has the advantage that it will work with sophisticated, complex loads making it difficult otherwise to phase compensate the (Howland) current pump.

Kai