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I want to amplify / buffer a transformer coupled voltage with a fully differential op amp. When using e.g. instrumentation amplifiers with a transformer coupled input voltage, one side has to be grounded somehow. I believe that the circuit shown below should work though. What is it that makes a fully differential op amp work with a "floating" input voltage?

enter image description here

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A differential op-amp provides a path for the input bias current to local ground via the two resistors (feedback and input resistors).

An instrumentation amplifier does not require resistors at each input but, does expect input bias currents to be "soaked up" by the voltage source that connects to those inputs. However, if that voltage source is floating (as in the case of a transformer secondary), the bias currents can only flow to each opposing input terminal and you get a problem in that one input cannot necessarily act as a current sink to the other (because they are near-identical).

This is why such people as Analog Devices make recommendations like this: -

enter image description here

Taken from data sheet for AD8221 for example.

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This should indeed be fine. It is the preamp circuit that needs a ground, the transformer can simply connect to the +/- inputs of any balanced input amp. Sometimes (in other circumstances) people use a single ended input and ground one side of the transformer secondary.

Note however that this circuit doesn't seem to be giving you isolation, exactly, as the same ground is used for the preamp as for the transformer primary, so I'm wondering what the point of the transformer actually is (because to step-up/down the signal)?

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  • \$\begingroup\$ It's not a common transformer. It's part of an inductively coupled measuring probe. \$\endgroup\$
    – TonyDublov
    Jul 12 at 13:16
  • \$\begingroup\$ you haven't really given enough information for us to assess what the significance might be. Do the probe manufacturers give any specific recommendations. \$\endgroup\$
    – danmcb
    Jul 12 at 13:20
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    \$\begingroup\$ It was not about the specific probe. I wanted to understand in general why you could connect floating voltages to fully differential op amps. The question was answered with the return paths of the bias current. Thanks though. \$\endgroup\$
    – TonyDublov
    Jul 12 at 14:43

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