Signal conditioning
4. The data sheet quotes 1.6GHz GBP, but for TIA compensation, the noise gain will be crossing
over the amplifier’s Aol curve at a relatively high noise gain. Hence, for TIA design, the single-pole
projection to unity gain for the Aol curve is needed in the region where the Aol phase is approximately 90° to estimate the correct GBP number of
compensation solutions. That extracts to 1.3GHz for the online simulation model.
5. The example design of Figure 1 shows a 0.42pF
feedback capacitor as Cf. A few simple steps will show how to derive this estimate using the loop gain (LG) curve of Figure 2. This is the typical LG curve for most TIA designs, where the amplifier’s open-loop gain curve has the feedback noise gain (NG) response superimposed on it to show the key frequencies in the design.
The key frequencies are shown on the LG graph in Figure 2.
Fo will be the characteristic frequency of the second-order closed-loop Vout/Idiode frequency response and is the projection of the rising noise gain from a zero frequency
(Z1) intersecting the device Aol response. Mathematically, it is the geometric mean of
Z1 and the GBP of the amplifier. A single-pole op amp Aol model is usually adequate for this simplified design flow.
The noise gain will start rising at Z1 given by 1/(2π × Rf × (Cs+Cf)). A very useful approximation is to recognise that Cf is normally less than Cs, and perhaps Cf can be dropped from the Z1 expression for an approximate solution. Since that approximation on Z1 goes through a sqrt in the Fo expression, that error will usually be very small.
The compensation problem here is setting the
noise gain pole: P1= 1/(2πRfCf) or simply Cf in this case if Rf is already chosen. A detailed analysis of the second-order Laplace transfer function for Figure 1 will reveal the closed-loop second-order response
Q ≈ (P1/Fo). This very useful result gets even simpler if targeting Q = 0.707 (set P1 = 0.707 × Fo), the resulting closed-loop response will approximate a
maximally flat Butterworth with F-3dB = Fo. This yields a simple 4-step solution for Cf to give a TIA closed-loop Butterworth response. Using the example design above.
1. Find the approximate noise gain zero (1/ (2π × 20kΩ × 14.3pF)) = 556kHz. (This is neglecting the Cf that is being solved for in the exact Z1 equation.)
2. Use this noise gain zero (Z1) and the amplifier’s GBP, to estimate the Fo = √(556kHz × 1.3GHz) = 26.9MHz = F-3dB for this Q = 0.707 design target.
Figure 3. Simulated small signal response for the example of Figure 1.
3. Set the feedback pole at 0.707 × Fo, or 0.707 × 26.9MHz = 19MHz = P1. Or Cf = 1/(2π × 19MHz × 20kΩ) = 0.42pF.
4. Check that the high frequency noise gain is greater than the amplifier’s minimum stable gain, or 1 + (14.3pF/0.42pF) = 35V/V is greater than the specified minimum stable gain
of 10V/V. This gives an Fc in the LG plot of Figure 2 at 1.3GHz/35 = 37MHz.
A Butterworth second-order target design will give a 65.5° phase margin (for an ideal
single-pole Aol), which will be very stable if the higher-order Aol poles are far beyond the Fc frequency - which will be the case here. From
this simplified Butterworth target design, any other desired Q may be delivered by simply
scaling the feedback Cf value by that Q ratio. Many legacy TIA design flows have targeted a
Q = 1 by putting the feedback pole at Fo. To get that result, simply scale the Butterworth Cf down by 0.707/1 to produce that 1.2dB peaking
with 16% step overshoot result that comes with a Q = 1 second-order response.
Continued on page 60... Instrumentation Monthly March 2026 59
Or, solving this for max F-3dB given some Rf and GBP, assuming the P1 is being set at 0.707 of
And then manipulating the equations for max
Rf or max F-3dB given a device GBP and total source Cs gives
Testing the small signal AC response for the LTspice circuit of Figure 1 gives the reasonably flat response of Figure 3. This is not a second-
order shape as the Aol curve in the LT6200-10 model shows a higher frequency zero/pole pair,
but, at 33MHz, F–3dB is reasonably close to the simplified design flow predicting 27MHz F-3dB for this idealised Butterworth design.
Collapsing this design flow into a few simple
equations will give this solution for P1, the feedback pole.
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