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I am having a bit of a hard time understanding why instrumentation amps are used over a single differential amp?

For example, the differential signal that I have is very small and comes from a Wheatstone bridge configuration.

I understand some advantages but I am still unsure of what the main reason is for why we use instrumentation amps for small differential signals.

The main advantage I see is that we can control the gain easily by the one resistor Rg, when compared to if a standard differential amplifier was used (since you need to select the resistors etc).

However, why else are these 2 buffer amplifiers with gain needed. What is the reason we add them and why do we want a high impedance input? Since without these amplifiers, the second stage is just a normal differential amplifier Another question is when do we use just a single differential amplifier (why do we need instrumentation amplifiers for small differential signals) enter image description here

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    \$\begingroup\$ If you used a single amplifier, the upper bound of the input impedance is R2+R3. With the instrumentation amplifier, it's the impedance of the bare OP input. \$\endgroup\$ – Janka Aug 7 '19 at 4:57
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    \$\begingroup\$ @Janka but after the initial buffer isnt the input impedance to the differential amplifier R2+R3 so the input signal would be changed anyway? \$\endgroup\$ – Student Aug 7 '19 at 5:02
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    \$\begingroup\$ The drive strength of the first stage outputs is much higher than the drive strength of your external source, so at the same impedance of the second stage, the error due to the current limitation is much smaller. \$\endgroup\$ – Janka Aug 7 '19 at 5:04
  • \$\begingroup\$ Also here electronics.stackexchange.com/questions/343096/… some insight \$\endgroup\$ – carloc Aug 7 '19 at 7:12
  • \$\begingroup\$ thanks, however, once the signal is buffered wouldn't the reading be a little inaccurate since the input impedance of the differential amplifier is low. As in for a standard differential amplifier the input impedance is low and so this may cause differences for the input signals. However with the buffer this solves this stages side but for the differential amplifier side wouldnt it be the same as just one differential amplifier? \$\endgroup\$ – Student Aug 7 '19 at 7:19
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The main difference is the very high input impedance of the IA compared to OA. Both inputs of IA have the same impedance (symetric) while for the OA this is not the case.

When measuring signals from high impedance transducers or wheatsone bridges, the current to the apmlifier shall be very low, in ideal case zero, so it doesn't affect the source itself. Once this differential voltage is buffered, you use OA to subrtact it. The asymetric impedance of OA is compensated by buffers (IA first stage buffers).

Also filtering is much simpler now. Since the IA has almost infinite and symetrical input impedance, you can add filters, voltae clamps - limiters, ...whithout affecting the measured source.

This Wheatsone bridge won't suffer too much about asymetry and input resistance. R4 has an additinal 200k ohm parallel impedance, while R2 additional 100k ohm, so this is the unbalance.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ thanks, however, once the signal is buffered wouldn't the reading be a little inaccurate since the input impedance of the differential amplifier is low. As in for a standard differential amplifier the input impedance is low and so this may cause differences for the input signals. However with the buffer this solves this stages side but for the differential amplifier side wouldnt it be the same as just one differential amplifier? \$\endgroup\$ – Student Aug 7 '19 at 7:17
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    \$\begingroup\$ Basiclly the non-inverting input has high impedance, while the inverting input has low impedance. The buffers have low output impedance and they compensate the impedance mismatch of the difference amp inputs. \$\endgroup\$ – Marko Buršič Aug 7 '19 at 7:38
  • \$\begingroup\$ When you say they compensate the impedance mismatch what do you mean? How does it compensate for this? Since I still do not understand why a differential amplifier cant be connected straight to the Wheatstone bridge? Since once buffered wouldnt that voltage for the diff amp inputs be changed by this low impedance and so it will be a little inaccurate. If not why isnt it? \$\endgroup\$ – Student Aug 7 '19 at 7:41
  • \$\begingroup\$ The differential amp has impedance on inverting input equal to R2, while the non-inverting has R2+R3. This is the unbalance that your Wheatsone will see when connecting directly. The IA buffer first and then feeds the difference amp, thus the current fed to the inverting input is higher compared to the non-inverting, voltages are not affected, this is this compensation or impedance match with other words. The OA as non-inverting amp is also used as impedance matching - you can put an arbitrary imedance on input (parallel) and you put an arbitrary resistor on output (series). \$\endgroup\$ – Marko Buršič Aug 7 '19 at 7:49
  • \$\begingroup\$ @Student The wheatstone bridge can be connected to the difference amp directly as long the bridge impedance is very low compared to the input imedance of the diff. amp. and the load unbalance of the Wheatstone is acceptable. \$\endgroup\$ – Marko Buršič Aug 7 '19 at 7:52

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