ADA4897

Dear analog engineers,

I was looking in the datasheet of ADA4897.

There is an example of a low noise gain selectable amplifier with an extra switch in the datasheet.

In this example you placed an balance resistor, to get a more constant offset voltage.

I was looking at the circuit and I was wondering, Is it correct that the balance resistor is on the S2A line?

My feeling says it should be on the S2B line.

Can anyone of you give me a better explanation of how this compensation resistor lowers the offset voltage.

With kind regards,

Andre

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  • 0
    •  Analog Employees 
    on Jun 20, 2013 1:36 AM

    Hi, Andre.

    Here’s the explanation of the circuit shown below based on page 21 of the ADA4897 datasheet.

    Ideally, the voltages at the non-inverting and inverting input pins of an amplifier are equal which will produce zero volts at the output. However, in real world there are differences in the voltages. Also, ideally there is no current entering the input pins due to high input impedance but in real world there exists a small amount of current in order to properly bias the internal circuits of the amplifier. With these ideal concepts, the amplifier generates an output voltage, often called offset, in order to keep the input voltages equal thus cancelling the input bias currents at the inputs. So from the schematic above, the offset voltages of the first amplifier, U1, are V1=Rf1*Ibias and V2=Rf2*Ibias.

    Since the input bias currents of U2 exist, the on-resistance of S3B in the feedback path can be used as a route for this. In addition to this, U2 will still generate an output offset voltage due to the switching of the gain resistors. So to compensate this offset voltage, a series resistor connected at the non-inverting input pin of U2 can be used.

    To compute for the Rbalance at S2A, equate V1 and V2 (assuming that the output voltage of U1 is constant regardless of gain). So,

    V1 = V2

    Rf1*Ibias = (Rf2*Ibias) – (Rbalance*Ibias)

    Rf1*Ibias = (Rf2 – Rbalance)*Ibias

    Rf1 = Rf2 – Rbalance

    Rbalance = Rf2 – Rf1

    Rbalance = 225 ohms – 75 ohms = 150 ohms

    If the Rbalance will be connected at S2B, the resistance value will be negative which does not exist. Computation is shown below. If you want this configuration, you may interchange Rf1 and Rf2 in order to have a positive resistance value.

    V1 = V2

    (Rf1*Ibias) – (Rbalance*Ibias) = Rf2*Ibias

    (Rf1– Rbalance)*Ibias = Rf2*Ibias

    Rf1– Rbalance = Rf2

    Rbalance = Rf1 – Rf2

    Rbalance = 75 ohms – 225 ohms = -150 ohms

    The second amplifier, U2, is added at the output in order to transfer the constant voltage from high input impedance (Ron of S3B and Rbalance) to low output impedance so that the circuit can drive relatively low loads.

    I had attached a more accurate computation of Rbalance for your reference. Let me know if you have other questions.

    Regards,

    Anna

Reply
  • 0
    •  Analog Employees 
    on Jun 20, 2013 1:36 AM

    Hi, Andre.

    Here’s the explanation of the circuit shown below based on page 21 of the ADA4897 datasheet.

    Ideally, the voltages at the non-inverting and inverting input pins of an amplifier are equal which will produce zero volts at the output. However, in real world there are differences in the voltages. Also, ideally there is no current entering the input pins due to high input impedance but in real world there exists a small amount of current in order to properly bias the internal circuits of the amplifier. With these ideal concepts, the amplifier generates an output voltage, often called offset, in order to keep the input voltages equal thus cancelling the input bias currents at the inputs. So from the schematic above, the offset voltages of the first amplifier, U1, are V1=Rf1*Ibias and V2=Rf2*Ibias.

    Since the input bias currents of U2 exist, the on-resistance of S3B in the feedback path can be used as a route for this. In addition to this, U2 will still generate an output offset voltage due to the switching of the gain resistors. So to compensate this offset voltage, a series resistor connected at the non-inverting input pin of U2 can be used.

    To compute for the Rbalance at S2A, equate V1 and V2 (assuming that the output voltage of U1 is constant regardless of gain). So,

    V1 = V2

    Rf1*Ibias = (Rf2*Ibias) – (Rbalance*Ibias)

    Rf1*Ibias = (Rf2 – Rbalance)*Ibias

    Rf1 = Rf2 – Rbalance

    Rbalance = Rf2 – Rf1

    Rbalance = 225 ohms – 75 ohms = 150 ohms

    If the Rbalance will be connected at S2B, the resistance value will be negative which does not exist. Computation is shown below. If you want this configuration, you may interchange Rf1 and Rf2 in order to have a positive resistance value.

    V1 = V2

    (Rf1*Ibias) – (Rbalance*Ibias) = Rf2*Ibias

    (Rf1– Rbalance)*Ibias = Rf2*Ibias

    Rf1– Rbalance = Rf2

    Rbalance = Rf1 – Rf2

    Rbalance = 75 ohms – 225 ohms = -150 ohms

    The second amplifier, U2, is added at the output in order to transfer the constant voltage from high input impedance (Ron of S3B and Rbalance) to low output impedance so that the circuit can drive relatively low loads.

    I had attached a more accurate computation of Rbalance for your reference. Let me know if you have other questions.

    Regards,

    Anna

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