INA333: Need help in Hardware design guidelines for Microvolt (µV) signals

Hi Jerry,

thanks a lot for your response.

Gerasimos Madalvanos said:

What is setting the common mode on the input of the INA333? Once the common-mode voltage is defined,

Please help me to define this. what parameters we need to identify?

Gerasimos Madalvanos said:

I see Rf and Rg are not connected. If you plan on amplifying this signal you will need to attach a gain setting resistor. The transfer function of the device is as follows.

Output of INA333 will be the input of TLV3603. Can you please help me to calculate the value of gain setting resistor?

Gerasimos Madalvanos said:

Please see section 8.2.2.5 in the datasheet on setting a bias current return path. You will need to have some path for the bias current of this device, and there are multiple options provided in the datasheet for how to provide this.

I will add 47k resistor.

Thanks,

Gaurav

Hello Guarav,

You may need to tie the 47kOhm resistors to a voltage potential that is not GND, so that the input is within a valid common-mode range. The common mode voltage is the sum of the DC voltage at IN+ and at IN-, divided by 2. If the magnetic sensor is not providing a DC biasing voltage, you will need to provide one yourself and tie that to the 47kOhm resistors.

To decide the gain setting resistor value, what is your input voltage value and your desired output voltage value. This gives you the required transfer function you need for your instrumentation amp. From there, the output will be set by the following equation

Vout = (Vin * (1+ 100k/Rg)) + Vref

Then you must pick an appropriate size Rg to realize the appropriate gain. For a gain of 101, you will select a 1k resistor.

When you say the signal generated is microvolts, do you mean it is single digit microvolts, or is it hundreds of microvolts?

Answering these questions will help you design your system.

Best,
Jerry

Hi Jerry,

Thanks for your response.

Gerasimos Madalvanos said:

the magnetic sensor

It is not a sensor. It is a wheel with magnets.

Gerasimos Madalvanos said:

Vout = (Vin * (1+ 100k/Rg)) + Vref

what is Vref?

Gerasimos Madalvanos said:

do you mean it is single digit microvolts,

Yes

Gerasimos Madalvanos said:

. If the magnetic sensor is not providing a DC biasing voltage, you will need to provide one yourself

Can you provide me as per your experience? it will be very helpful for me and a great learning also.

Thanks,

Gaurav

Guarav,

The inductor that is generating a microvolt signal in the presence of the wheel with magnets can be seen as a sensor. It is sensing the wheel with magnets, and in response is generating a microvolt value.

GAURAV UPADHYAY said:

what is Vref

Vref is an input into the instrumentation amplifier:

You may have trouble with noise if your input signal is single digit microvolts. With no input resistors, the low frequency noise on the INA333 is 1uVpp. This may introduce a significant error.

GAURAV UPADHYAY said:

Can you provide me as per your experience? it will be very helpful for me and a great learning also.

Typically the most common bias voltage value  is midsupply ((V+)+(V-))/2. This can be generated from a buffered resistor divider. This ensures your first stage amplifiers have the maximum swing range.

Best,
Jerry

Hi Jerry,

Gerasimos Madalvanos said:

You may need to tie the 47kOhm resistors to a voltage potential that is not GND, so that the input is within a valid common-mode range. The common mode voltage is the sum of the DC voltage at IN+ and at IN-, divided by 2. If the magnetic sensor is not providing a DC biasing voltage, you will need to provide one yourself and tie that to the 47kOhm resistors.

Sorry, still i did not get it how to implement this. 

But try to design schematic using OPA2333 for output and reference pin logic after reading multiple thread on TI.

Please review below schematic and let me know where i can make improvement for uV signal.

Thanks,

Gaurav

Hey Guarav,

The circuit would look something like the below. I'm not certain why you have a voltage divider on the output of the INA333, this seems unnecessary. You may use less gain in the second amplifier and simply not attenuate the input. You also must be careful to not saturate the output of the OPA333 secondary gain stage.This can be solved by referring the output to REF. Please note that the Rg has been changed to 100, as the maximum gain for INA333 is 1000

It may be helpful to review the Instrumentation Amplifier Overview: https://www.ti.com/video/series/precision-labs/ti-precision-labs-instrumentation-amplifiers.html?videoId=6118886700001

There are also many resources in this FAQ about Instrumentation amplifiers, and this may help in starting your design:

https://e2e.ti.com/support/amplifiers-group/amplifiers/f/amplifiers-forum/818894/faq-how-do-instrumentation-amplifiers-inas-fit-into-my-design

Best,
Jerry

Hey Jerry,

Thanks a lot for your suggestions and support.

I am going to connect output of OPA333 (VF2) with comparator. see below:

Please share your feedback on this.

Gerasimos Madalvanos said:

The circuit would look something like the below.

Can you please share simulation waveform for my better understanding? It will help me a lot.

Thanks,

GAURAV

Hello Guarav,

Please see below for the TINA simulation file. The OPA2333 should have no problem driving the comparator input. If R63 is 0 ohm, the comparator will be permanently latched. I would recommend populating a different value there. For the exact value, I will have to move this thread to the comparators team.

Guarav_INA333.TSC

Best,
Jerry

Hi Jerry,

Thanks a lot for your support.

Gerasimos Madalvanos said:

Please see below for the TINA simulation file.

Simulation file is not working for me.

Can you please suggest other way to see waveform?

Thanks,

Gaurav

Gerasimos Madalvanos said:

f R63 is 0 ohm, the comparator will be permanently latched. I would recommend populating a different value there

How can i use default hysteresis value of comparator? 

Thanks,

Gaurav

Hello Guarav,

I am able to run this circuit in TINA-TI. Do you have this downloaded? If not, you may download it here: https://www.ti.com/tool/TINA-TI

Best,
Jerry

Hi Jerry,

Thanks a lot for your support.

Gerasimos Madalvanos said:

For the exact value, I will have to move this thread to the comparators team.

Can you please provide feedback on hysteresis ?

Thanks,

Gaurav

Hi SIU,

Thanks a lot for your response.

HO SIU said:

What is the signal swing that you expect at the output of the OPAx333 (VF2 in TINA-TI simulation)?

between 0 to 3VDC

HO SIU said:

I believe this question was addressed in your other post: TLV3603: Need help to calculate resistor value for Hysteresis - Amplifiers forum - Amplifiers - TI E2E support forums

i will follow this.

HO SIU said:

R63 being 0Ω in schematic serves as a placeholder for your physical layout having pads for the hysteresis resistor.

correct.

Thanks,

Gaurav

Hi Jerry,

Review team put some questions on my design. see below:

1. Why i am using R6 and R5 in this circuit? it will consume current.

2. Replace INA333 with programmable gain INA --> can you please part number of programmable gain INA?

They also mentioned that 

"This device isn’t going to produce a very wide magnetic field. The coil design is a part of that factor. More importantly the circuit pumps very little current into your coil, so it won’t sense permeability changes far enough away.

One solution is to add some support circuitry to pump more current in. --> Can you please help me how can i do this?

Another would be to design a system based off how a true lcr meter works – feed in a sine wave and measure the peaks and phase shift when the current passes through an LR divider. "

Thanks,

Gaurav

Hi Jerry,

I have one thought about input for PGA. Let me know if i am doing something wrong or not.

If  i create a tank circuit and output of tank circuit will go to input of PGA. 

Can you please tell me what will happen at input and output of PGA? is it improve the strength of input signal for PGA?

Thanks,

Gaurav

Gaurav,

Because of the holidays, TI E2E design support forum responses will be delayed from Dec. 25 through Jan. 2. Thank you for your patience.

Hi Ron,

Thanks for your response.

Thanks,

Gaurav

Hi Jerry,

Can i use below reference circuit using LM111 ?

Please let me know if i need to modify something.

Thanks,

Gaurav