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The characteristic of the above converter is relatively linear because :

  • A.\(\mathrm{V_{01}}\) Vs. T graphs are linear
  • B.Uses same capacitor for charge and discharge
  • C.High performance linear comparator
  • D.High performance op-amp

Answer: B

The key is uses same capacitor for charge and discharge.

In the first phase the integrator charges through \(R\) for a fixed time \(T_1\). In the second phase the same \(R\) and the same \(C\) discharge it to zero in time \(T_2\). The two slopes are \(v'_A/RC\) and \(V_{\mathrm{REF}}/RC\), so the factor \(RC\) cancels in \(T_2 = T_1\,v'_A/V_{\mathrm{REF}}\). Any error or drift in \(R\) or \(C\) therefore has no effect on the result, and the conversion stays linear and accurate.

That the ramps are straight lines (A) is a property of an integrator, not the reason the whole conversion is linear. A good comparator (C) and a good op-amp (D) help accuracy but do not explain it.

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