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Passive Averager and Current

Discussion in 'General Electronics Discussion' started by Renato_Ferreira, Jan 4, 2013.

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  1. Renato_Ferreira

    Renato_Ferreira

    2
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    Jan 4, 2013
    Hi,

    I am trying to design a circuit that calculates the average of two sensor outputs and applies this voltage to the input of a voltage buffer, which I want to implement using a common collector transistor.

    The problem is that I don't know whether the output from the passive averager will remain the same once some current flows to the base of the transistor. To me, it seems that the resistors will cause a voltage drop that I can't have, as my voltage output would be lower, too.

    Is my reasoning correct, or is there a better way to accomplish what I am trying to do?

    Thanks!
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Google "summing amplifier" for *lots* of information on this.

    In general, you can use those 2 resistors, and the output of them is (V1 + V2) / 2 where V1 and V2 are much larger than the impedance of the input signals. This is a voltage output, so you need a high input impedance amplifier to buffer this.

    In many circuits you'll see, the amplifier has a gain of 2 so that the output is V1 + V2, but you want unity gain so you get the average, not the sum.
     
  3. Renato_Ferreira

    Renato_Ferreira

    2
    0
    Jan 4, 2013
    Thank you for your reply.

    So, supposing I get the average by using two resistors, can a common collector amplifier, which has unity gain, be considered a high input impedance amplifier? From my understanding, this scheme would provide me with an output with almost the same voltage as my average input, but with a higher current provided at the transistor collector, is that right?
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,496
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    Jan 21, 2010
    Common collector has voltage gain slightly less than 1. Its input impedance will load the output of the resistor network, reducing the effective gain still further.

    The approx 0.7V difference between input and output voltage may be an issue, especially if the signals are small.

    I would recommend an op-amp configured as a unity gain buffer.
     
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