Transimpedance Amplifier Notes

Op-amp voltage noise, no Cf | Op-amp voltage noise, with Cf | Does noise current roll-off at 1/(2*pi*Rf*Cf)?

en: op-amp voltage noise (no Cf)

Simple model of TIA with no feedback capacitor.


At low frequency, inverting terminal follows non-inverting terminal, so voltage across Cin is eN. This generates current i=eN/ZCin through Cin, where ZCin=sCin (and s=j*2*pi*f). That current has nowhere to go but through Rf, and so the output voltage is eN + i*Rf. This is shown graphically below (red dashed line). The full calculation for e_out as a function of frequency (given by the non-inverting closed-loop gain) is shown as a black solid line. At low frequency, the two agree well, but deviations become apparent above fz = 1/(2*pi*Rf*Cin).
At low frequency, although i grows linearly with frequency, the gain was already unity at DC and so the perturbation to the noise gain at low frequency is small (perturbation << unity). In fact, in the linear approximation, the voltage gain Avcl is given by Avcl=1+f/fz. Eventually, f approaches fz and the noise gain deviates substantially from unity. The "eN*Cin" noise current approximation continues to rise

Here's a plot of the non-inverting closed-loop gain (Avcl), also known as the "noise gain". Also shown is the "linear approximation" (i.e. the gain assuming a current through Cin proportional to frequency). The dashed black line is the op-amp open loop gain, with unity-gain frequency fc (4MHz in this case).



To better see that the "en*Cin" current is a good approximation to the noise, we can push out the op-amp unity gain freqeuency (fc) to a huge value (e.g. 1e11 Hz) and see how well the linear approximation tracks Avcl.



You can see in the above examples that the noise gain peaks where the noise gain matches the opamp open loop gain (in this case, just above 0.2MHz). Adding a feedback capacitor damps this peak (see discussion/graphs below).

In summary: for frequencies below the crossover frequency (where the red and black dashed lines intersect), the "eN*Cin" approximation works very well.

en: op-amp voltage noise (with Cf)

Now, we look at the effects of adding a feedback capacitor.


When the feedback capacitor is chosen 'optimally' (according to Horowitz and Hill Art of Electronics 3rd edition, Section 8.11, explained below) then we have the following:

Here, the "optimal" choice of Cf is given by Cf=1/(2*pi*Rf*fp) where fp is the circuit's usable bandwidth (called fc in AoE, and equivalent to the intersection frequency of the red and black dashed curves -- 230kHz in this example).

If you push out the op-amp unity gain frequency (but don't modify Cf), then you find that the noise gain (Avcl) flattens out at fp = 1/(2*pi*Rf*Cf), and the linear approximation breaks down (because current then passes through Cf rather than Rf, which the levels off the output voltage even though the current through Cin grows with frequency). At these high frequencies, Avcl is given by 1+Ci/Cf (in this case 1+12/0.7 = 18). i.e. the feedback network is Rf || Cf and at these high frequencies, the impedance of Rf is huge compared to Cf, so ZRf || ZCf ~ ZCf.

Does the "en*C_in" noise current roll off at f=1/(2*pi*Rf*Cf)? No.

In the AoE discussion of the e_N*Cin current, it says: "It would continue to rise forever, except for the effect of the parallel capacitance Cf, which causes the noise current to flatten off at a frequency fc=1/(2*pi*Rf*Cf)"

This implies that the current through Cin levels off above 1/(2*pi*Rf*Cf), but I find that the current through Cin continues to grow with frequency well above that frequency (until the op-amp cannot ensure that the inverting voltage follows the non-inverting voltage). Instead, the voltage gain levels off above 1/(2*pi*Rf*Cf) because the noise current gain levels off at high frequency. In other words nearly all of the current goes through Cf rather than Rf because the impedance of Rf is huge compared to that of Cf. We can see this analytically by computing the current through the feedback network (which must be equivalent to the current through Cin). We define beta as the feedback fraction, to find that the current i is inversely proportional to Zi, and therefore directly proportional to frequency (as long as you are at frequencies less than the op-amp's open-loop unity-gain frequency):



in this derivation, we've used:
    e_- = beta * e_out
    beta = Zi / (Zi + Zf)
    e_out = e_N * Avcl
    Avcl = Avol / (1+beta*Avol) ~ 1/beta for f << open-loop unity gain freq