The transimpedance amplifier is a common example of a circuit that has stability problems due to capacitance on the inverting node. While the stability concerns with a transimpedance amplifier can be solved using the methods described in previous sections, looking at this configuration separately is worthwhile as the configuration is very common and some additional detail is helpful. The transimpedance amplifier converts a current input to an output voltage. Transimpedance amplifiers are frequently connected to photodiodes because photodiodes output a current proportional to light. A reverse biased photodiode is generally connected directly to the inverting input with respect to ground. Reverse biased photodiodes can have significant capacitance because the cathode and anode act as a parallel plate capacitor with the depletion region acting as the dielectric. Note that the photodiode model contains a current source to emulate the diode output current versus light. The photodiode model also contains the junction capacitance (C_J) and a parasitic leakage resistance R_S. The op amp transimpedance gain is set by R_F, so often large values of R_F are used to amplify the small photodiode currents. The combination of the large feedback resistor with the photodiode capacitance often leads to instability. Figure 5-17 illustrates the transimpedance configuration connected to a photodiode as well as the photodiode equivalent circuit.

Figure 5-18 shows the uncompensated closed-loop transimpedance amplifier that is used for this example. Figure 5-19 shows the closed-loop AC response for Figure 5-18. Note that the gain shown in this example is the transimpedance gain (V_OUT/I_D ). The DC transimpedance gain is approximately equal to R_F,, or in decibels 20 × log(R_F). The large AC peaking and large rapid phase shift is an indication that the circuit is not stable. The reason this circuit is unstable is the large capacitance on the inverting node in conjunction with the large feedback resistor (C_J = 70pF, and R_F = 100kΩ).

The circuit for testing the open-loop response of the transimpedance amplifier is shown in Figure 5-20, and the open-loop AC response is shown in Figure 5-22. The open-loop response shows that a zero in 1/β causes the rate-of-closure to be 40dB/dec where A_OL intersects 1/β. The circuit stability challenges are confirmed by the 1.5° phase margin. As we saw earlier in this section, the correction is to cancel the pole with a zero by adding a feedback capacitor (C_F).

Equation 82 can be used to select the minimum value of C_F for stability (C_F(MIN) = 1.67pF in this example). The location of the zero and pole in 1/β can be found in Equation 83 and Equation 85, respectively. Figure 5-22 graphs the open-loop response with the minimum C_F. Notice that the phase margin for the circuit with minimum C_F is 51.4°, whereas phase margin was 1.5° without C_F.

Increasing the value of C_F beyond the minimum value of C_F generally improves the stability further. However, for larger values of C_F, the series combination of C_F and C_J can potentially create a capacitive load from the output to ground that can cause stability problems. The value of C_F also determines the closed-loop transimpedance bandwidth (see Equation 86) as well as the total output noise. Increasing the value of C_F reduces the bandwidth and noise. Figure 5-24 shows the closed-loop frequency response for the circuit in Figure 5-23 with different feedback capacitors. You can see that the circuit with no feedback capacitor (0F) has a large 34dB gain peak, and with the minimum (1.7pF) capacitor the gain peak is eliminated and the maximum transimpedance bandwidth is achieved. Using larger feedback capacitors further reduce the bandwidth but can introduce a stability issue at some point due to capacitive loading (C_L = C_F + C_J).