AC signal amplification in a presence of DC part

Discussion in 'General Electronics Chat' started by Crowbar, Dec 25, 2006.

  1. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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    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.
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    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.
     
  3. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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    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 :) ?
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    If the DC component is constant, you can subtract it. What kind of sensor do you have?
     
  5. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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    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.
     
  6. beenthere

    Retired Moderator

    Apr 20, 2004
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    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?
     
  7. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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    Here is a Sensor. DC component is a 'Constant' Earth magnetic field.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    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.
     
  9. thingmaker3

    Retired Moderator

    May 16, 2005
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    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.
     
  10. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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    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).
     
  11. thingmaker3

    Retired Moderator

    May 16, 2005
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    Perhaps it would help if you told us what you want to measure with the completed unit?
     
  12. beenthere

    Retired Moderator

    Apr 20, 2004
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    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).
     
  13. Distort10n

    Active Member

    Dec 25, 2006
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    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.
     
  14. beenthere

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  15. Crowbar

    Thread Starter Active Member

    Dec 19, 2006
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