Why does AC coupled OP amp circuit require a direct dc path to ground?

Discussion in 'General Electronics Chat' started by Romsha, Feb 20, 2017.

  1. Romsha

    Thread Starter New Member

    Nov 29, 2013
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    3
    To quote Microelectronic sedra smith[6th edition]
    One must always provide a continuous dc path between each of the input terminals of op amp and ground in an ac- coupled amplifier. This is the case no matter how small bias input current is.

    Can someone please explain the above statement. The circuit image is inserted.(click the image to see the AC Coupled circuit) The books goes on to say that the said circuit will not work without R3 connected to ground. [​IMG]
     
  2. bertus

    Administrator

    Apr 5, 2008
    18,608
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    Hello,

    The image stored on google does not show.
    Please upload the image to the forum using the "Upload a File" button.

    Bertus
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,953
    1,825
    Here's the first AC coupled op amp circuit I found on the web:

    [​IMG]

    We have an AC coupled input via the capacitor C.

    Assume for the moment that Vi is zero and hence the input is grounded. What is the voltage at the + input?

    For a real world op amp there exists an input bias current from the + input, either in or out. This current is only zero for an ideal op amp, real world op amps may have a very small current, but it is never zero. Same for the - input too.

    Without R1 the only path for this current is thru C. This is a DC current that will charge C.

    Eventually the + input will be stuck near either the positive or negative supply (depending on if the bias current is + or -).

    Now add R1. R1 will be the input impedance for the amp so the value must be chosen so it will not load the driving source, but 10K ohms may be a typical value. For a real world op amp the LM741 has a maximum bias current of 1.5 uA. Since steady state this current all flows thru R1 we can calculate the DC voltage at the + input as

    Ib * R1 = 1.5 uA * 10 K ohms = 0.015 Volts
     
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