Most pro-audio microphones provide a balanced output -- this is intended to provide improved noise rejection, and provided the output and line impedances are appropriately balanced, it does. However, with conventional differential-to-single-ended conversion stages, source and line impedance imbalances degrade the noise rejection ability of the system. Transformers don't suffer from this, but have problems of their own (magnetic coupling, size/weight, low frequency response), so the Whitlock bootstrapped topology has come into use for differential-to-single-ended conversion, at least at line levels:

Figure 8. THAT 1200-series equivalent circuit diagram

Unfortunately, though, no microphone amp yet has adopted the Whitlock bootstrap for an internal differential-to-single-ended conversion function. However, at least some pro-audio microphone amplifiers are designed to be fully differential -- i.e. they have a differential input and a differential output. The intuitive thinking is that since a fully differential amplifier doesn't have a CMRR, it would not impact the ability of a downstream stage to reject the common mode signal, so cascading a fully differential microphone preamp with a line receiver is used to provide a high-CMRR balanced microphone input.

However, assuming a Whitlock bootstrapped line receiver (such as a THAT120x) is used for the differential-to-single-ended conversion function, does putting a fully differential preamp in front of it negate the ability of the Whitlock bootstrapped topology to maintain a high CMRR in the face of inbalanced input and line impedances?

  • \$\begingroup\$ If you're wondering why there's no schematic drawn for this question, it's because I can't find a "gain block" symbol in CircuitLab :/ (if someone has any other ideas than abusing opamp symbology, please comment here!) \$\endgroup\$ Commented Nov 12, 2016 at 16:38
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    \$\begingroup\$ since a fully differential amplifier doesn't have a CMRR I do not think that that is universally true. It depends on the amplifier design. Some have no CMM rejection, others do. I think there is no reason you could not cascade more than one CMM rejection amplifiers, so using a differential amplifier should not influence the CMRR ability upstream. \$\endgroup\$ Commented Nov 12, 2016 at 17:08
  • \$\begingroup\$ @Bimpelrekkie Whether they have CM rejection or not, they still always have a CMRR, no? \$\endgroup\$
    – endolith
    Commented Aug 8, 2019 at 16:39
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    \$\begingroup\$ Why do you say differential in-out amps have no CMR? \$\endgroup\$
    – MAM
    Commented Aug 8, 2019 at 17:05
  • \$\begingroup\$ @endolith Whether they have CM rejection or not, they still always have a CMRR, no? Your sentence makes no sense it is like: it can be yes or no, but it is always yes isn't it? If it is always yes, how can it be no? As I stated above, the CMRR of an amplifier depends on its design. I could design an amplifier with a very high CMRR or one with very poor CMRR, even CMM gain (that's amplifying the CMM). Most amplifiers suppress the common mode signal. That does not mean that that is universally true. \$\endgroup\$ Commented Aug 8, 2019 at 19:53

1 Answer 1


No, this would not work. The Whitlock amplifier works by bootstrapping the common-mode input impedance so that it is very high, so that the voltage divider effect from the source impedances becomes negligible, and therefore so do the effects of any imbalance in those source impedances.

Your fully-differential amplifier is essentially acting as a buffer amplifier, hiding the source and load impedances from each other, so all the Whitlock amplifier can do is make impedance imbalances of the diff amp negligible.

The source imbalance will only see the diff amp's input impedance, and if that is not extremely high, it will cause a voltage imbalance coming out of this stage.


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