operational-amplifierpower supplytransimpedance
Was going through this webpage and noticed that a trans-impedance amplifier using single supply. However, I am unable to understand how it would work?
How would the opamp see inverting input(photodiode – on non inverting) current and reference it(non inverting in ground)? Is there any advantage of using single supply in this case?
Note that the OPA355 input common mode range extends to 0.1 Volt below ground BUT only to within 1.5V of Vdd. We'd need to see your circuit to know if this was significant. Also Vout is "not quite" rail to rail. OPA355 datasheet
I'll not claim to be a master in this context, but the following may be useful:
Similar specs to OPA355in many areas.
Apparently much superior CMRR.
2.75 vs 5.8 nV/Hz^2
Similar GBW
Superior output voltage swing (especially +ve)
280 vs 360 V/uS slew (ie slower)
LTC6252 Datasheet
720 MHz GBW
400MHz -3dB point as buffer
Input common mode rail to rail.
2.5V to 5.25V operation and fully specified at 2.7V and 5V.
CMRR 105 dB ! (!!!!!)
Noise 2.7 nV.Hz^-2
280 V/uS slew :-) :-) :-) :-) :-)
$4.59/1 in stock Digikey (Cheap considering spec).
Single, dual and quad versions available.
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
No, it's not using a single supply as per the question linked in your comment, the Maxim circuit is generating a mid-rail - there is an important distinction: -
R1 and R2 bias the non-inverting input at Vcc/2. This sets the inverting input at Vcc/2 and the output of the amplifier also at Vcc/2 when zero photodiode current is flowing.
Basically you can make a single supply TIA with inputs referenced to the most negative rail providing the input current to the TIA is negative i.e. as from a photodiode with the cathode at the input. The photodiode current flows into the cathode and out of the anode and this means the op-amp output will rise above 0V (most negative rail) to re-balance the situation. The re-balance is the output.
Clearly, rail-to-rail op-amps are required for this to work effectively. Take a look at this picture below. It doesn't show power rails but think about the signals. Light generates a photodiode current that flows away from the inverting input. This has to make the output rise higher than the non-inverting input and here lies the basis for a true single-supply, ground referenced TIA. The only caveat is that there will be a small (a few mV) dead-band until sufficient current flows in the photodiode.