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Improving Resolution of Sensor Output

by akhurash
Tags: improving, output, resolution, sensor
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akhurash
#1
Nov14-12, 10:37 AM
P: 25
I'm working on a sensor signal conditioning circuit. The sensor output has a 2.5V DC bias with a peak-to-peak voltage swing of about 50mV. The signal output is put through a demodulator and then filter (LPF) to obtain a DC voltage proportional to the out of the sensor. The problem is I need to be able to sense very small changes in the AC voltage output from the sensor (~ 2mv-5mV peak-to-peak changes). I'm new to signal conditioning and processing sensor outputs so I was wondering if anyone here knows how I can do this. I want ot learn about the basics and theory behind how to amplify the signal.

Currently my DC output goes from about 2.467V to 2.561V (with the midpoint being about 2.514V, the DV bias). I need to translate that into a DC voltage swing of 0V to 5V, or something better than 2.467V to 2.561V.

The way I'm thinking about doing this gaining the AC voltage BEFORE the LPF. I'm currently trying this approach but it isn't working because my output signal is all distorted after the AC gain.
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Floid
#2
Nov14-12, 12:19 PM
P: 235
You can use a series capacitor to block the DC bias on the signal. If you didn't mind a positive/negative signal you could then just use an inverting amplifier with a gain of say 100 to get you to a +/- 5V signal.

You should be able to just google series capacitor and inverting amplifier to learn more about them.
AlephZero
#3
Nov14-12, 02:39 PM
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P: 7,172
You didn't say what frequency range you are interested in, except for mentioning the LPF.

A simple "blocking capacitor" circuit to get rid of the DC is really acting as a HIGH pass filter, so you would need to be sure it wasn't affecting the low-frequency content of the signal that you want to keep.

You mentioned a demodulator as the first step in the signal processing chain, but didn't say anything about what it was for. Maybe you would be better to shift the signal level by about 2.5V (e.g. with a DC coupled unity gain amplifier) before the demodulator, and do the 100x amplification last, after the demodulation and LPF? But that's guessing, without knowing more about the demodulation step.

akhurash
#4
Nov15-12, 10:49 AM
P: 25
Improving Resolution of Sensor Output

Thanks for the replies. I kind of gave the wrong impression when posting this. I actually want to amplify/scale the DC range of 2.467V to 2.561V to make that range go from 0V to 5V with 2.5V being the mid voltage. Since the voltage is almost pure DC after the LPF (I have three, one active 2nd order and two passive) I can't block the DC, that is my information.

The demodulator is used to process the output signal of the sensor (which is modulated with a carieer frequency of 100 kHZ). My LPF's have a cutoff frequency of 10kHZ because my sensor output information is contained within the DC voltages after the demodulator.
skeptic2
#5
Nov15-12, 05:53 PM
P: 1,814
Here's a simple circuit that could be a starting point. Note that the final opamp must have rail to rail capability and a high enough slew rate to handle your signal.
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File Type: pdf Sensor.pdf (12.3 KB, 6 views)
vk6kro
#6
Nov15-12, 07:52 PM
Sci Advisor
P: 4,032
Here is a fairly standard opamp circuit:



It uses a precision opamp but almost any type would be OK. If you wanted the output to go to zero volts with a single supply, then you would pick an opamp which could do this.

The voltage source V1 would be removed and your input would be brought in here.

Capacitor C2 is optional. It wasn't necessary for the simulation, but may be advisable in practice if the wire lengths were significant.

One of the resistors shown as R2 and R3 would probably be made variable. Or you could replace both of them with a multi-turn pot.

You could increase the gain by increasing the resistance of R1. As shown, the gain is about 19.


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