buffergainoperational-amplifier
When I searched for op amp buffer circuits on internet, I found this picture.
I just want to use it as a buffer circuit. So, obviously the gain of this configuration will be 1. I would like to put 0 Ω resistor as Rf in that location.
Is it ok or do I need to connect those two lines without resistor in the middle. If I put a 0 Ω resistor in that place what will happen? Will there be any instability problem?
If you have a reference voltage of (say) 3V, the range your input will see is: -
If you used a 50uA current excitation and a grounded sensor, the range your input will see is: -
With a current excitation the percentage of your 3V reference range used is 95%. With voltage and resistor excitation you only get 63% of the range.
If your reference voltage is lower or higher, the above "range" statements are still true.
Here is an example. An ADC input is directly connected to the sensor. The sensor is fed with 50uA via the PNP transistor. The 50uA is measured across "R" and compared with "V" by the op-amp. The op-amp keeps the current through R at a value that generates a voltage "V" across it. Values could be R=10k and V=0.5V or R=20k and V=1V. Op-amp should be chosen that has close-to-either-rail IO performance like an AD8605 (used many times by me in this same configuration for strain gauge excitation).
Here's a quick DC simulation for 60k and 3k loads: -
Transistor used is BC547C or BC847C for those with good eye-sight.
Note that due to small base currents in the transistor, the current isn't 50uA but 49.846uA with 60kohm load and 49.849uA when loaded at 3kohms. Note also the voltages across the sensor - 2.991V on 60k load and 149.5mV on 3k load.
It doesn't work.. at least it needs a power supply on the op-amp.
The TL081 has a common-mode input range that goes down about 3V above the negative rail, so it cannot deal with voltages at the negative rail (in this case, ground).
The reason to have the 10uF there is to have a DC gain of 1 and an AC gain that is higher (in this case the AC gain is 11 and it rolls off at the 16Hz -3dB cutoff frequency).
With the capacitor there, the DC gain is still one, so you would need to bias the input to +4.5V or so and capacitively couple the input to get it to work (say two resistors across the supply, with a capacitor to the input- but that has no power supply noise rejection).
Best Answer
If the op-amp has significant input bias currents (rather than offset currents) then making \$R_F = R_1||R_2\$ has some DC accuracy benefit.
If the op-amp has an unstable unity gain bandwidth, then using \$R_F\$ might just make the difference between the circuit oscillating or not because gain will naturally rise at higher frequencies due to the parasitic input capacitance at the inverting input.
Adding \$R_F\$ to op-amps that don't suffer from either of the above will degrade noise performance because the inverting input's parasitic capacitance will make the AC gain at higher frequencies significantly higher than unity. This may or may not be regarded as a problem.
Personally, since 1980 when I first started designing with op-amps, I bet I haven't used a feedback resistor in any unity-gain op-amp circuit since the mid 1990s (because of the availability of quality op-amps with really low bias currents).