6
\$\begingroup\$

I have an IEPE accelerometer (this one) that I'd like to connect to an audio recorder. IEPE accelerometers have two cables (signal and ground) and require 18-30V DC applied to the signal cable to power the internal electronics. The signal (±5 V or so) hence has a DC bias that has to be removed before passing it on to a recorder. Think of the accelerometer as a contact microphone providing signals between 0 and 20 kHz.

When I first received the accelerometer, someone with experience using them provided a circuit diagram to use; here's my recreation:

my current circuit diagram

As I understand it, capacitor C1 and resistor R2 act as a high-pass filter and remove the DC bias; R1 limits the current to the accelerometer, connected at the BNC inputs. (The accelerometer manual specifies a constant-current diode instead, between 2-20 mA, but I don't yet have one.)

Initially I powered this with three 9V batteries in series and it worked very well. But I wanted to make a portable device that only used one 9V battery, so I got an adjustable boost-buck converter to boost it to 20 V, assuming that the switching noise would be above the audible range. Unfortunately I was wrong, and there's still some quiet-but-noticeable high-frequency noise at around 7 kHz.

My question: How do I filter the noise? I could use a low-pass filter after the power supply, by putting a capacitor between signal and ground, but would that also distort the accelerometer input in some way? Or would something like a T filter remove the noise without affecting the input from the accelerometer?

(If it's relevant, the accelerometer output impedance is specified to be ≤300 ohms; full specs are available here.)

\$\endgroup\$
11
  • 2
    \$\begingroup\$ You might find that taking a half decent load current from the output of the booster puts the switching artefacts above 20 kHz. \$\endgroup\$
    – Andy aka
    Aug 10, 2021 at 15:14
  • \$\begingroup\$ As in, putting a resistor from its output to ground just to draw a few milliwatts from it? I didn't show this in the diagram, but I do actually have a power LED, although presumably it draws a tiny amount of power. \$\endgroup\$
    – Alex Reinhart
    Aug 10, 2021 at 15:20
  • 2
    \$\begingroup\$ There are a couple of suggestions in Switching Supply Noise Filter for Audio Circuit. \$\endgroup\$
    – Andrew Morton
    Aug 10, 2021 at 15:22
  • \$\begingroup\$ How long are your wire connections from the output power to the rest of the circuit? The cable inductance could be coupling with the output capacitance of the voltage regulator module and creating a resonant rank circuit at 7 kHz. I would try twisting the power and ground wires and maybe adding a high ESR input capacitor to your circuit to damp the LC effect. \$\endgroup\$
    – Ryan
    Aug 10, 2021 at 15:41
  • 1
    \$\begingroup\$ Pulse skipping is being used because your converter is designed for multiple amps but you're only drawing a few milliamps, so essentially the converter is unloaded and it's trying to conserve power. You could pick a converter that is more appropriately sized for the load, or design better filtering to isolate the converter noise from your instrument. I would do both, and if noise is critical, consider adding a high PSRR linear regulator to further isolate your instrument from audio frequency noise. \$\endgroup\$
    – user1850479
    Aug 10, 2021 at 17:34

2 Answers 2

Reset to default
1
\$\begingroup\$

To add to the other comments, a quick and dirty solution would be to add a linear regulator to the your switched converter output. This will do very little for the MHz ripple frequency of the boost converter, but will substantially attenuate audio frequency noise. Combined with a low pass filter, this can be used to construct a very low noise power supply from a noisy switching converter.

To pick a completely random part from a parametric search, the Torex XC6701DJ02PR-G will take any voltage above ~ 18.5v and give almost exactly 18v output while rejecting audio frequency noise by 30-50 dB:

enter image description here

It currently sells for $1 on Mouser and would need a single 1uF MLCC capacitor (few cents) on the output. With slightly more effort you can probably find even better parts for this application.

\$\endgroup\$
0
\$\begingroup\$

First, R1 is the source resistance, that serves as "load" to the sensor, along with C1 & R2. "2 - 20mA" is the driving capability of the sensor. The sensor takes power through it anyway.

As @Andyaka pointed out, the switching frequency will go up when the power is loaded.

The total possible loading will be in "2 - 20mA", since the sensor only can operate in that range. So, if you'd like to filter out the power supply switching frequency, test the power supply with 2mA and 20mA loading, and measure the switching frequency.

Meantime, since the switching frequency is over many kHz, I would put a simple R&C, may be two stages, right at the power supply, then place the same circuitry as you would have otherwise.

Edited
Any loads to the power supply may go before the filters and R&C (filter). It still loads the power supply, and affects the switching frequency.

\$\endgroup\$
5
  • \$\begingroup\$ I'm not sure the sensor needs load; the manufacturer's suggested circuit (figure 7, page 9) does not include R1 or R2, just the current regulating diode and C1. I'll try this once I get the diode, but I expect to still need filtering on the power supply \$\endgroup\$
    – Alex Reinhart
    Aug 10, 2021 at 15:48
  • \$\begingroup\$ @AlexReinhart Thanks for your candor, sincerely. Your question makes me smile. :-) I am not sure if I can explain that in this limited space, but will try: The sensor needs to "signal out" while getting power through the only connection, and that is a single pair (BNC) of wire. So, the internal sensor circuitry has a similar structure of circuitry as the external circuitry, takes power from the portion of the current through BNC, which is 2mA. Meantime, the sensor circuitry has to send/drive the signal on the same BNC, which is the "load", that is "AC coupled or through galvanic isolation". \$\endgroup\$
    – jay
    Aug 10, 2021 at 16:13
  • \$\begingroup\$ @AlexReinhart The suggested current limit (diode) serves the same purpose as the resistor, R1. \$\endgroup\$
    – jay
    Aug 10, 2021 at 16:24
  • \$\begingroup\$ Thanks; you can tell my only experience was a freshman electricity & magnetism class. So I gather I should try putting a load on the supply so I'm drawing more than a few mA and see if the noise becomes inaudible. Or maybe add a passive filter between the power supply and the sensor. The filter won't affect the sensor output meaningfully? \$\endgroup\$
    – Alex Reinhart
    Aug 10, 2021 at 17:03
  • \$\begingroup\$ It all depends on what you need @AlexReinhart . In electronics, "S/N" is the Signal over Noise, or signal to noise ratio, that is a way to tell if my signal is good (strong/large) enough to stand out from the noise. That is one of the important criteria to choose a filter. What I would suggest is to try it on a "bread board" first. \$\endgroup\$
    – jay
    Aug 10, 2021 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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