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DC offset

  1. Mar 4, 2009 #1
    Hello

    I have a question regarding DC offset addition. I have a sine wave of Peak amplitude 2V. Now i have a DC signal of 0.4v. How do i add this DC signal to my sine wave?
    Similarly i'm having another sine wave of 2Vpeak but phase shifted by 180 from the first one. Now again i have 0.4v DC signal which needs to be subtracted from this sine wave.

    Can anyone provide a schematic or a link to any circuit which can do the above job? Please
     
  2. jcsd
  3. Mar 4, 2009 #2

    berkeman

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    Staff: Mentor

    Depending on frequencies and a bunch of other considerations, the easiest way to do it is via capacitive coupling. The AC signal will go through a capacitor, but the cap blocks DC. So say you have your signal comiong in the left side of the schematic, going through a horizontal capacitor, and to one side of a vertical resistor. The other side of the resistor goes to your + or - 0.4Vdc supply. The junction between the cap and the resistor is your AC+DC signal.

    Now, you have to think about the high-pass characteristic of that circuit, and ensure that you are getting your AC signal through, and also not loading it too much with the resistor, right?

    Quiz Question -- Can you show us how you would go about calculating the R and C values for this circuit, based on your signal source characteristics and the AC frequencies involved?

    You can also do this with opamps or other transistor buffer circuits, if the simple RC isolation/combiner is too limited for your application.
     
  4. Mar 5, 2009 #3
    Here is the opamp way. Get a good opamp. ground the + input to common (or use a 330 ohm resistor to reduce input bias current offsets). Attach two 1 k resistors to the - input. Tie the ac signal to one resistor and the dc signal to the other. Tie another 1k resistor from opamp output back to - input (feedback resistor). Output = sum of ac plus dc, times -1. The - input is called a summing junction. No ac coupling required. If you want an uninverted output, follow it with a gain of -1 inverter.
     
    Last edited: Mar 5, 2009
  5. Mar 5, 2009 #4
    Thanks for the replies

    @Berekman's Quiz question

    Is it like f=(1/2*pi*R*C), where f is the value of the frequency that the circuit has to allow.
    In my case source is 0f 50 Hz, so i should substitute 50 in place of f and calculate values for R and C.
    now that i have only one equation, i should have any of the values as a fixed one and calculate the other.

    Am i right? Correct me if i'm not

    @ Bob
    Thanks a lot. I have a small doubt though. The - input is from the source so i should ground my other terminal of the source and also the + input to this ground,right?
     
    Last edited: Mar 5, 2009
  6. Mar 5, 2009 #5

    berkeman

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    Staff: Mentor

    Yes, correct. 50Hz is a very low frequency, however, so it will take large capacitors to pass that frequency adequately. You are not working with AC Mains 220Vrms, 50Hz, are you? That can be quite dangerous for someone like you with only a very basic knowledge of electricity.
     
  7. Mar 6, 2009 #6
    No am not. My Sine wave is from an IC only. just 2V peak. Which of the method you suggest would be more accurate? The one with capacitors or with the Op-amp because this is one of the crucial steps in my project and i dont want it be less approximate.
     
  8. Mar 6, 2009 #7

    berkeman

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    Staff: Mentor

    You need to figure this one out on your own, anbullet. It's for school -- you need to do the work. What are the advantages and disadvantages of each approach? It's your assignment, do some work, and tell us your thoughts.
     
  9. Mar 6, 2009 #8
    Yeah i will. Just got overboard. sorry
     
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