SBOA597 November 2024 OPA928
Figure 4-23 shows an example of the measurement at 25°C. Although most of the procedure is the same as for 85°C, the example shows the output voltage did not cross zero volts. In this case, one additional step is needed. To remove the effects of the leakage in the parasitic resistance of the capacitor, the parasitic resistance must be calculated.
In the example, the output voltage (buffer with a gain of ten) moved from 0.05V to 1.3V. As the curve does not cross zero volts, we need to take a leakage current into account. Leakage across the capacitor is proportional to the voltage across the capacitor multiplied by the resistance of the capacitor.
Plot current over the output voltage. The curve indicates dielectric relaxation settles at around output voltage 0.3V. The fitting curve shows -0.93aA/V. The resistance of the integration capacitor is calculated as 1 / (-0.93aA/V) / 10.1 (gain) = 106PΩ at 25°C. The intercept of the fitting curve indicates IB is 79aA. In this example, the resistance of the capacitor is used as the output voltage does not cross zero.
Go back to the original curve of output voltage over time. Zoom in the area where dielectric relaxation was settled. For example, take a derivative between 0.263V to 0.273V, dV/dt, to obtain a slope that is 6.73µV/sec. Apply the parasitic resistance of the capacitor of 106PΩ, on 0.268V / 10.1(gain) to obtain leakage of 0.25aA. IB is calculated as 6.73µV/sec / 10.1 x 117.4pF x (1-0.0002 x (25-20)) + 0.25 = 78.4aA. The calculation includes the temperature coefficient of the capacitor of -200ppm/°C.
Comparing the number between the intercept of the fitting curve (79aA) and (78.4aA) gives a delta of 0.6aA. With that, IB is most likely between 79aA to 78.4aA at 25°C.