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Common Mode and Differential Input Impedances

Discussion in 'Electronic Basics' started by Kingcosmos, Apr 4, 2007.

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  1. Kingcosmos

    Kingcosmos Guest

    I am trying to determine how the common-mode and differential input
    impedances of difference amplifiers are calculated. I am having a
    worse time finding out how these specifications are defined.

    For example, if I look at the INA133, the common-mode and differential
    input impedance are both 50k. All 4 internal resistors are 25k. The
    INA132 seems to follow suit where common-mode and differential input
    impedances are 80k, and all 4 resistors are 40k. Seems rather
    straight forward although I am not sure how these were really
    calculated.

    What throws me is the INA106 and the INA143. Both difference
    amplifiers have the same set of resistors. The input resistors on
    both devices are 10k and the feedback resistors are 100k. Yet the
    INA143 has a differential input impedance of 20k, and common-mode
    input impedance of 55k, and the INA106 has a differential input
    impedance of 10k and differential input impedance of 110k.

    What is the difference (no pun intended) and how are these values
    calculated/defined? Thanks in advance.
     
  2. Kingcosmos wrote:
    (snip)
    These specs make no sense to me. Assuming ideal opamps, to
    simplify the math, the basic assumption is that the inputs
    to the opamp have infinite impedance and the voltages of the
    two inputs match perfectly.

    So the non inverting input just has a pair of resistors in
    series to the reference voltage (let's say that is 0 V). So
    the impedance of that input must be 10 k plus 100k = 110 k.
    This impedance is independent of whether the applied
    signal is in common to both inputs or part of a differential
    voltage.

    The impedance of the inverting input is 10 k to a voltage
    source that has a value of 100 k / (10 k + 110 k) = 90.9% of
    the voltage applied to the inverting input (since the
    feedback forces the - input of the opamp to match the
    voltage applied to the + input of the opamp.

    So the problem is, how do you replace these two actual
    equivalent circuits with a common mode and differential
    impedance.

    If you tie the two inputs together, they look like 110k in
    parallel with 10 k / (1 - 100/110) = 110k. So the common
    mode impedance is 110 k /2 = 55 k. Perhaps the INA106 sheet
    shows the common mode impedance of each input, individually,
    while the INA143 sheet shows their parallel combination.

    Differentially, things are messier, since the individual
    impedances do not match.

    If we apply a voltage to the non inverting input and the
    inverse of that voltage to the inverting input, this would
    be a purely differential signal with an amplitude of the
    difference of the two voltages, but the two currents are
    very different. So how do you define a single differential
    impedance that involves two different currents, each driven
    by half of the voltage? One manufacturer might spec one
    half and one might spec the other.

    One way to define the differential impedance would be to
    apply a completely floating voltage source between the two
    inputs. With the uneven differential impedances, that
    voltage will be converted to some combination of common mode
    voltage and differential voltage. The common mode voltage
    should see the 55 k common mode impedance calculated, above,
    and the differential component would be loaded with the
    effective differential impedance.

    So lets say we apply a 1 volt floating source to the two
    inputs, with the positive side on the non inverting input.
    By virtue of the float, the current into one input must also
    be the current from the other. I'll call the voltage
    applied to the non inverting input V+ and the current into
    that input I+ and the voltage applied to the inverting input
    V- and the current into I-.

    I+ = - I- and V+ - V- = 1

    But I+ = V+ / 110k

    so I- = - (V+ / 110k)
    (the same current in the other direction)

    but from the equivalent circuit of the inverting input,

    I- = (V- - (V+ * 100/110))/10k

    so we can combine these two equations to find out how the
    floating input voltage divides between V+ and V-.

    V+ / 110k = ((V+ * 100/110) - V-) / 10k

    so V+ = V- * 11/9

    or equivalently, V- = V+ * 9/11

    Note that both V+ and V- have the same sign, rather than
    splitting across zero.

    But their difference must be 1 volt, so

    1 = V+ - (V+ * 9/11)

    So V+ = 11/2 and V- = 9/2

    Checking, the differential voltage is 1 volt or 2/2.
    The common mode voltage is 10/2.

    So each input should produce a common mode current of
    (10/2)/110k and a differential current of 1/Rdif.

    so I+ = (10/2)/110k + 1/Rdif

    but also, I+ = V+ / 110k

    therefore, (11/2)/110k = (10/2)/110k + 1/Rdif

    So Rdif = 220k

    Which matches neither data sheet, so I am satisfied. ;-)

    A more useful way to define the input impedances might be to
    hold one input at zero and apply voltage to the other, and
    define the input impedance under those conditions.

    As always, the non inverting input impedance is 110k.

    The inverting impedance under this condition is 10k.
     
  3. John Popelish wrote:
    (snip)
    (snip)

    This has to be wrong for several reasons.
    Can anybody spot them?
     
  4. Kingcosmos

    Kingcosmos Guest

    This makes perfect sense. I tried, with no much success, Thevenin and
    Kirchhoff for the common mode and differential impedances. I am going
    to pick the brain of TI, and see what they say.

    I was going to go through your analysis, but since you said it is
    wrong I guess I will pass :^). I will take a look at it later tonight
    to see what I can spot.

    Thanks for the reply.
     
  5. Guest

    (corrected)
    And since the signal is purely differential and floating, there can be
    no common mode current. Any input current is, by definition,
    differential.
    Correcton: all current (I+ or I- is differential current.
    So skip to bottom.
    Rdif = 1/I+ = 110k / V+ = 110k/(11/2) = 20k

    So if you drive the difference amplifier with a transformer winding or
    other floating source, it will see a 20k differential load impedance.
     
  6. The Phantom

    The Phantom Guest

    How do you get this? The second page of the INA133 data sheet says the
    differential impedance = 50k and the common mode impedance = 25k.
    Yes, that's what the data sheet says. Seems inconsistent with what the
    INA133 sheet says.
    It's too bad they don't say what circuit connections were used to determine
    those impedances. I assume that figure 1 of each data sheet would be the
    configuration to use.

    I would also assume that the common mode impedance is measured with the two
    inputs tied together and driven together.

    Let Ri be the two input resistors, and Ro be the two output resistors. My
    nodal analysis indicates that if the amplifier gain is very high, the
    common mode input impedance is (Ri + Ro)/2 and the differential input
    impedance is 2 * Ri.

    This is for the circuit of Figure 1 on each data sheet.

    The INA133 and INA143 data sheets are consistent with this analysis, but
    the other two aren't. I wonder why.

    For the INA143, they don't say whether the impedances are for a gain of 10
    or .1; my analysis indicates that the common mode impedance would be the
    same, but the differential impedance for the .1 gain configuration would be
    200k.
     
  7. Kingcosmos

    Kingcosmos Guest


    Ah, you are correct. I read far too many datasheets last night trying
    to correlate. My mistake. I am waiting for a response from TI, but I
    will look at your explanation later as well. Thanks for the reply.
     
  8. Kingcosmos

    Kingcosmos Guest

    A quick update. The information I received was not defintive;
    however, it seems that the impedances were derived from Spice analysis
    during testing. The other explanation given is what you guys have
    already explained. It depends on how the inputs are driven. So I
    guess in that manner, everyone is correct. :^). It is a shame that
    there are no figures in the datasheet showing the test circuits.
    Thanks for the help.
     
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