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I have a Wheatstone bridge I want to use to measure a resistive sensor. At the moment, all four arms are fixed resistors, while I get the circuit set up. The noise in the circuit is much larger than I expected, and I can't work out why.

schematic

simulate this circuit – Schematic created using CircuitLab

The AFG is configured to source two sine waves in antiphase on the two outputs. The Stanford is making a DC-coupled differential measurement with an effective noise bandwidth of 0.5 seconds. The two instruments are phase locked using the TTL ref. I'm then looking at the variation of the measurements to work out the noise. I find a noise component which scales with excitation voltage, which I can't identify.

  • The unidentified noise is white from 10 Hz to 1 kHz
  • The amplitude of the noise is about 1 µV/√Hz per volt applied across the bridge
  • When the excitation voltage is zero, I measure a noise consistent with Johnson noise in the bridge.

I have lots of experience of measuring R less than 1 Ω, but 200 kΩ is new to me, so I might be missing the obvious.

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  • \$\begingroup\$ How long are the cables? \$\endgroup\$ – Scott Seidman Nov 20 '15 at 12:31
  • \$\begingroup\$ Cables are a 10cm or so. The bridge is in a diecast box, excitation cables are BNC, measurement wires are a shielded twisted pair. \$\endgroup\$ – Jack B Nov 20 '15 at 13:08
  • \$\begingroup\$ Yep, microvolts per root hertz. The Johnson noise is about 50nV/√Hz though, and with the Stanford I should be able to push it down to about that. \$\endgroup\$ – Jack B Nov 20 '15 at 13:16
  • \$\begingroup\$ If the noise scales with the excitation voltage, then it is probably being generated by the function generator. \$\endgroup\$ – Barry Nov 20 '15 at 13:20
  • \$\begingroup\$ If you're using a Lock-in amp, do you really care? \$\endgroup\$ – Scott Seidman Nov 20 '15 at 13:23
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I'm betting a component is quantization noise or other distortion on the AFG3000, which has 14bit output. The harmonic distortion is listed as -60dB in your freq range.

You might try using a cleaner sine wave and seeing what you get, or condition the blazes out of the excitation wave you have. In any case, after asking about long cables that may be passing on noise, that would be my first debugging step. Next would be using shielded cables with twisted pairs to drive your bridge, and the next would be using a real bridge amp with sense circuitry.

My money is on the quality of the signal generator, though. I have yet to be pleasantly surprised by the signal quality of an arbitrary waveform generator.

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  • \$\begingroup\$ If the sig gen itself is a noise source, why isn't this common mode noise, effectively cancelled out by the bridge? (OK I see the bridge is 0.1% off balance, but that should mean of the order of 60dB attenuation of quantization noise.) \$\endgroup\$ – Brian Drummond Nov 20 '15 at 14:24
  • \$\begingroup\$ I think this might be it. The signal from the bridge is about 1/1000 of the excitation, so 1uV/√Hz noise on the output would be 1mV/√Hz on the input. That sounds like a lot for a very expensive AFG (16 times the expected quantization noise, for example), but it could be. \$\endgroup\$ – Jack B Nov 20 '15 at 15:01
  • \$\begingroup\$ If I set the AFG to give a DC output, the noise drops by a factor of 10-15 (and of course my signal disappears). Noise still scales linearly with excitation amplitude. So it looks like the AFG is kicking out noise in a narrow band near the output frequency, close enough for the lockin to detect. I guess I need to find a better sine source. \$\endgroup\$ – Jack B Nov 20 '15 at 16:33
  • \$\begingroup\$ @JackB -- so far as bench equipment goes, I find that function generators are where you get the absolute least bang for your buck! The price of the AFG3000's I'm seeing online don't look high enough to bring this piece of gear into the Cadillac category. It's tough to say where the quantization noise would be, because you don't know where Vref is set, and it may be set inappropriately for your output size. Also, it only has a 14 bit output. If Vref is 5V, for whatever reason, you're talking 100's of uV of quantization noise. \$\endgroup\$ – Scott Seidman Nov 20 '15 at 16:55
  • \$\begingroup\$ I noted 3 hours ago that the AWG was the source of the excessive noise. I'm glad you guys now seem to agree. \$\endgroup\$ – Barry Nov 20 '15 at 16:59
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Having recommended reading about partition noise I'm finding nothing terribly useful online myself!

So I'll try running a few numbers. In the absence of a better reference model, I'll model it simply as 2 shot noise sources, one in each leg of the bridge. (The sum will be simply the shot noise inherent in the same total current from the sig gen).

Applying 1V of excitation, the bridge current is 5 uA, or 2.5 uA each leg. Noise is simply sqrt(N) where N is the number of electrons/second.

I = 2.5 uA = 2.5e-6 * 6.25e18 = 15.6e12 electrons/second.

So I(noise) = sqrt(15.6e12) = 3.95e6 electrons/sec = 0.63pA/rtHz in each leg.

Now taking the bottom of the bridge as an 0V reference, this current develops a voltage across 200kilohms of 0.126uV. Now because it's partition noise I'll assume the opposite happens in the other leg (the noise sources are NOT independent) I'll double my estimate of voltage noise to 0.25 uV/rtHz.

If this is correct, partition noise is in the right ballpark, but I can't see how it accounts for all of your noise (1uV/rtHz).

Furthermore, the least useless link I've found (assuming it works elsewhere) suggests (page 94) that partition noise has a 1/f distribution, which contradicts your observation. (I don't know why they suggest 1/f).

One line of experiment : if you replace the bridge with 2k resistors (2 orders of magnitude lower) shot noise (ditto partition noise?) should diminish in importance by 1 order of magnitude. What happens in practice?

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  • \$\begingroup\$ Maybe I've missed something here, but isn't the noise you have calculated the noise relative to the bottom of the bridge? In which case it's common mode noise, and should be rejected by the Stanford at 100dB? If I calculate the voltage dropped over 2kΩ by the noise current (to repesent the difference between the two arms of the bridge) I get 4nV/√Hz - less than Johnson. \$\endgroup\$ – Jack B Nov 20 '15 at 14:39
  • \$\begingroup\$ What you've missed is that by its nature, it's coherent and opposite in sign in each leg of the bridge giving a differential output. \$\endgroup\$ – Brian Drummond Nov 20 '15 at 14:42
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    \$\begingroup\$ I think I'm with you now. From the description above, I think this will scale as √I = √V, not V as I observe? Combined with the lower calculated amplitude I think this may be lurking in the background, but I don't think it's the dominant noise source. \$\endgroup\$ – Jack B Nov 20 '15 at 14:58
  • \$\begingroup\$ Another experiment : if you balance the bridge by hand (with a trimpot) does the excess noise go away? If so, Scott's answer seems all the more likely. Partition noise would remain. \$\endgroup\$ – Brian Drummond Nov 20 '15 at 15:31

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