3
\$\begingroup\$

I am trying to understand and calculate the expected noise behind a simple NTC-bridge circuit where the signal is amplified by a instrumentation amplifier.

Typical NTC- Bridge with Instrumentation Amplifier

My goal is to measure 1 mK temperature change which gives me around 23 µV of signal change out of the voltage divider.

Operating Parameters:
Op-amp Input Voltage Noise Density: \$ e_{n,in} = 50 \frac{nV}{\sqrt{Hz}}\$
Gain (also Noise Gain): \$ NG = 100 \$
Op-amp Bandwidth (also noise BW) at given Gain: \$ NBW = 2 MHz\$

My Requirements:
Input signal what I want to measure \$ NBW = 23 µV\$
Needed signal BW: \$ NBW = 1 kHz\$

For simplicity I take just the input voltage noise density into consideration and ignore the resistor thermal noise and input current noise.

The input RMS noise calculates as follows with the parameters above:
$$ E_{n,in} = \sqrt{NBW} * e_{n,in} = \sqrt{2 MHz}* 50 \frac{nV}{\sqrt{Hz}} = 70.71 µV RMS $$

So clearly the op-amp noise dominates my signal.

If I do the calculations with my needed BW I get a usable result and could amplify my signal:
$$ E_{n,in} = \sqrt{NBW} * e_{n,in} = \sqrt{1 kHz}* 50 \frac{nV}{\sqrt{Hz}} = 1.58 µV RMS $$

Questions:
A: Is there a way to limit the BW of the instrument amplifier to reduce its noise?
B: Should I consider using a standard op-amp where I can control its effective BW with an external RC circuit?

Note: I was planning to use the AD8421 without filter pins.


Thank you for the details.

Adding a RC filter in front of the instrument amplifier makes sense.
I assume this would look like so:

enter image description here

But how is a filter after the op amp affecting the input voltage noise of the op amp?

When calculating the output stage voltage noise I take the Gain into consideration and the input noise is then also amplified. But as far as my understanding goes; the input voltage noise of an instrument amplifier is not affected by an RC filter on its output!?

Thank you very much.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ 1kHz bandwidth seems insanely large for a lethargic NTC thermistor. \$\endgroup\$ Aug 6 at 17:56
  • \$\begingroup\$ Why does your question keep mentioning an op-amp? \$\endgroup\$
    – Andy aka
    Aug 6 at 19:14
  • \$\begingroup\$ 50nV/rtHz is an atrocious noise figure for all but absolute micropower opamps. if noise is at least a slight concern take something with ~5 nV/rtHz \$\endgroup\$
    – tobalt
    Aug 7 at 7:25
  • \$\begingroup\$ @tobalt I think the voltage noise is only a concern for higher BW. With my requested 1kHz bandwidth the 1/f noise is dominant. \$\endgroup\$
    – huababua
    Aug 8 at 6:08
  • \$\begingroup\$ Yes that is relevant, but then you cannot ignore the 1/f noise of the 4 bridge resistors. Or in other words, use metal film resistors, not thick film. \$\endgroup\$
    – tobalt
    Aug 8 at 7:14

3 Answers 3

3
\$\begingroup\$

As jp314 said, place a filter after the amplifier. This will reduce the noise bandwidth of the system. If you are measuring temperature, I'm guessing you can reduce the bandwidth even further to reduce the noise bandwidth.

Be careful of your assumptions. A single-pole low-pass filter's noise bandwidth is 1.57 times the filter's -3 dB bandwidth. Thus, the 2 MHz bandwidth amplifier will have an equivalent noise bandwidth of 3.14 MHz.

You need to consider the thermal noise of your input resistor network. The thermal noise from a 1000 ohm resistor is about \$ 4 \; nV/\sqrt{Hz} \$ at room temperature. If the resistor values are low enough, you may be able to ignore their noise contribution.

The filter after the amplifier can be a large source of noise. State variable filters (Sallen & Key, et. al.) are typically very noisy, multiplying the noise of the basic parts by about the Q factor. The common two-op amp gyrator filter is quieter, or you can use passive R-C or R-L-C filters, making sure you keep tabs on the noise caused by the resistor(s) in the filter network.

\$\endgroup\$
2
  • \$\begingroup\$ thank you. I have added a question to my original post. Could you please explain how a RC filter on the output is affecting the input voltage noise? As far as I understand there is no feedback from the RC filter on the output to the input right? \$\endgroup\$
    – huababua
    Aug 6 at 19:01
  • \$\begingroup\$ You are after equivalent input noise voltage of the system which is (output noise density) divided by the (gain from input to output). \$\endgroup\$
    – qrk
    Aug 6 at 21:10
2
\$\begingroup\$

You need to add the noise from the bridge resistor also.

You can limit the BW by filtering the output of the opamp. This is practical because the filter itself will not contribute as much additional noise as if it was at the input.

You could place a capacitor across the inputs also, but yor haven't described the bridge resistance and it might be too low to be practical to filter with a aon-electrolytic cap.

\$\endgroup\$
1
\$\begingroup\$

Is there a way to limit the BW of the instrument amplifier to reduce its noise?

Yes. Just because the input is differential doesn't make simple RC filters not work. Put a capacitor across the input, and you got a 1st order lowpass.

Should I consider using a standard op-amp where I can control its effective BW with an external RC circuit?

An instrumentation amplifier lets you implement the 1st order filter just as easily.

Unless the NTC is grain-of-sand tiny, the bandwidth you have in mind is too big by a couple orders of magnitude. A 0.5s time constant would be plenty, and that's only if whatever the NTC is attached to has relatively low thermal inertia.

\$\endgroup\$
2
  • \$\begingroup\$ thank you - understood. I have added a schematic how to add a RC filter on the input to my original post. \$\endgroup\$
    – huababua
    Aug 6 at 18:59
  • \$\begingroup\$ It is better in this case to add the filter to the output. The 4 k added at the input will introduce some noise; and mismatch of the 4K and/or the 1 nF will convert any VCC noise from common mode to differential. \$\endgroup\$
    – jp314
    Aug 7 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.