Ir = 120V/5,000,000 Ω = 24.0uA is the current through the 5MΩ resistor.
Ic1 = 120V/8,631,898 Ω = j13.9uA is the current through the 307.3uF capacitor.
Ic2 = 120V/2,652,582 Ω = j45.2uA is the current through the 1nF capacitor.
The current flowing through the DUT is a combination of the current flowing through
Ic2 (the 1nF capacitor) and the resistor:
Idut =(24.0Angle(0°)+45.2Angle(-90°))uA = (24.0 – j45.2)uA = 51.2Angle(-62.0°)uA
The total current is then Itot=(24.0Angle(0°)) + 13.9Angle(-90°) + 45.2Angle(-90°))uA
Itot = (24.0 – j13.9 – j45.2)uA = (24.0 – j59.1)uA
Thus Itot = 63.8Angle(-67.9°)uA
Once again the Offset feature is implemented to offset the current flowing through the
capacitance of the 620L system:
Offset = 13.9uA
The test is then run. The 620L will then calculate the displayed current value:
Displayed Leakage = 62.3uA
This is a result of the vector sum calculation: 62.3uA = √((63.8uA)^2 – (13.9uA)^2)
However, the calculated value of the current flowing through the DUT is 51.2uA. In the
first circuit the current through the DUT has a phase angle of 0° because it consists of
a pure resistance. The introduction of the 1nF capacitor to the DUT in the second
circuit causes the phase shift between the 620L system and the DUT to change from
90° to 62° (Idut = 51.2Angle(-62.0°)uA). As a result, the calculated value for displayed
current does not match the theoretical value calculated above.
The above example illustrates that the phase relationship between the offset current
associated with the 620L system and the current flowing through the DUT is critical.
This phase difference must be 90° in order for the vector subtraction to calculate
displayed current to represent the true value of the leakage current flowing through
the DUT.
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