Estimating differential error

AN4304

12/20

DocID024731 Rev 1

5

Estimating differential error

To estimate differential error (Vdiff), the example below considers that the input signal is a

triangular one.

Vin = triangular AC at 100 kHz, 5 Vpp

Rs = 10 Ω

Ccm = 2.2 µF with a tolerance of 20 %

Ccm1 = 2.2 µF +20 % = 2.64 µF

Ccm2 = 2.2 µF -20 % = 1.76 µF

Note that the fundamental amplitude of a triangular signal is lower than the actual signal by

a factor of

8/π

follows:

So, Vdiff = 122 mVpp.

Due to a tolerance of 20 % on the filtering capacitance, a theoretical ripple of 122 mVpp

appears on the differential input of the current sensing. Going back to the time domain,

Figure 13: "Differential voltage with Ccm1 and Ccm2 matched at 20 %" shows the impact

on the ripple due to the mismatching of the capacitances.

Figure 13: Differential voltage with Ccm1 and Ccm2 matched at 20 %

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