AC signal amplification in a presence of DC part

Thread Starter

Crowbar

Joined Dec 19, 2006
32
Hi a little question- I have a signal from a sensor (DC part is around 1V and AC part is near 10mV)- some times AC part must be amplified 10 times- can it be done without using of HPF (what will be with such configuration in a presence of DC signal, active part is powered from +/-5V)? Hope, the question is clear :)

Best Regards, Konstantin.
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

One method is to place a capacitor in the signal path to block the DC offset. Just about every AC amplifier does this. Select the capacitor such that its reactance is small with respect to the input impedance of the amplifier so the AC component of the s1gnal is not attenuated.
 

Thread Starter

Crowbar

Joined Dec 19, 2006
32
Let me explain: can I amplify AC signal by a factor of 10 in a presence of DC signal without any filtering(attenuation) or gain OpAmp will set to supply rail in a whole spectrum (DC to 1KHZ) cause 1x10>5 :) ?
 

Thread Starter

Crowbar

Joined Dec 19, 2006
32
Yes I have also think about it: if sensor is fixed- this method works, but if we will rotate it, DC part will change in a range of -2V to 2V.
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

Amplifina signal in the presence of a common-mode voltage is not a problem. Your last post seems to indicate that the DC signal component is significant. What is this sensor and could you post up the circuit?
 

Ron H

Joined Apr 14, 2005
7,014
To measure small, low-frequency variations, you probably will need to use a DSP. I don't think this can easily be done (if at all) with a strictly analog approach.
 

thingmaker3

Joined May 16, 2005
5,084
Just use a blocking cap, as Beenthere suggested earlier. No worry about polarity of the DC component. Make the cap a high enough value to be of low impedence at the frequency of the signal you are measuring.
 

Thread Starter

Crowbar

Joined Dec 19, 2006
32
You misunderstood me: whole device must measure DC and AC components and It' is not a problem for me to use Butterworth HPF when AC signal is needed- I asked such a question because i whant to minimize circuit and as it possible to remove switching parts(like HPF switching on/off).
 

beenthere

Joined Apr 20, 2004
15,819
Hi,

What you might do is to use a low pass filter to eliminate the AC component from the DC level. Apply it to one input of an instrumentation amp, with the composite on the other input. This will make the DC signal level common-mode, and thus become invisible to the IA.

The filtered signal will give you the DC information, and the IA will give you the AC ( amplified to whatever level is convenient).
 

Distort10n

Joined Dec 25, 2006
429
After several responses I think I finally ge the gist of this. You are saying that you need to measure an AC output while eliminating the DC output from this sensor or measure a DC output while eliminating the AC output. Do I understand correctly?

The datasheet itself is a little confusing. It seems that the output of this sensor is DC referenced to a voltage determined by half the voltage between the supplies:

Reference output OUT- (V+ - V-)/2 +/-1%
Output voltages X, Y , Z ref. to OUT- +/-1 V/35 uT, max. +/-(V+ - V-)/2


It does go on to mention a ripple component:

Ripple @ excitation freq. = 17 kHz typ. 1.6 mVPP

Which I do not see as being part of a 'useful' signal. Sadly, there is no application section to the datasheet.

The description mentions intergration with a microprocessor; however, the output voltages are large in magnitude. If one uses +12V supply, then the REF would be 6V and output voltages are in reference to this. -1V would imply a 5V output assuming 35 uT. Much to large for a microcontroller or even DSP since MAX input voltages on their I/O's are usually 3.6V. Use a lower supply to get around this apparently.

That is the way I interpret the datasheet.
 
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