Adding Offset to an Analog signal

Discussion in 'The Projects Forum' started by loganversele, Apr 11, 2008.

  1. loganversele

    Thread Starter New Member

    Apr 1, 2008
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    I need to take an analog signal with a Vmin ~-90mv and a Vmax ~1.5v and add about 2 volts to the entire signal. The signal is going to be converted by a microprocessor that requires an input between 0 and 5volts I've tried a couple things and have had no luck. This is for an ECG circuit, powered with two 9v batteries. thanks
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    A simple voltage divider and capacitor would do it.
    [​IMG]

    If you need to maintain DC characteristics then an op amp adder.

    [​IMG]
     
  3. SgtWookie

    Expert

    Jul 17, 2007
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    See the attached schematic.

    Note that the cap should be a poly type. Electrolytic won't work well.

    With R1 set to midpoint, you should have about a 2.25v offset.

    If you need to go much lower than that, disconnect R1 from the ground, and add another 100k resistor from Vee to the lower side of R1.

    Make sense?
     
  4. SgtWookie

    Expert

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    Bill beat me to it ;) "Variations on the same theme..." :)
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    The mouse is quicker than the keyboard!

    Besides, I draw prettier pictures. :p
     
  6. RaoulDuke

    New Member

    Apr 11, 2008
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    I was going to answer this thread......but Bill and the Marine beat me to it!

    They are both right......use an opamp to add an offset.
     
  7. RaoulDuke

    New Member

    Apr 11, 2008
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    SgtWookie........let me guess...you learned electronics in the military.

    Right?;)
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    I've been accused of having done that before, yes. :D
     
  9. loganversele

    Thread Starter New Member

    Apr 1, 2008
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    thanks guys...i'll try your suggestion bill, I'm trying to avoid using op-amps to reduce the amount of battery pwr being used so i'll try your first circuit
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    Oops - something neither of us thought about before is the battery voltage dropping over time. This will cause the "float" voltage to constantly drop as the battery discharges.

    A couple of adds, and a couple of changes:
    1) A 3.9v Zener diode was added to keep the voltage across the pot constant. Instead of a 100k pot, you could use a couple of high-value resistors.
    2) A 1nF cap was added across the pot to quiet things down a bit. Ceramic would be a fine choice here.
    3) R2 reduced from 100K to 4.7k to get enough current flowing through the Zener for regulation.
    4) C1 was increased from 1uF to 27uF. Increasing the value of the pot or replacing with high-value resistors may make that unnecessary. In any event, I suggest adding a 0.1uF cap in parallel with C1 so that higher-frequency signals are passed along as well. The larger a cap is, the more parasitics come into play (inductance, resistance).

    I threw some datapoints into a simulated signal generator, using your +1.5v and -90mV, along with some overshoot tossed in to give a rough idea on how it might perform.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Sarge is right, battery power will change over time. Here's another way...

    [​IMG]

    We're almost doing the same way, I'm just eliminating the extra parts. Sarges design lets you tweak it while in use. Do you need help calculating values?

    If there are low frequency components these circuits will cause some distortion, but it will be slight. The op amp version will have a lot higher fidelity.
     
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